Some Basic Notions of Quantum Mechanicsdkeren/qm-01.pdf · Some Basic Notions of Quantum ......
276
Transcript of Some Basic Notions of Quantum Mechanicsdkeren/qm-01.pdf · Some Basic Notions of Quantum ......
Sources:1. Feynman’s Lecture Notes
2. Shankar, “Principles of Quantum Mechanics”
3. Eisberg & Resnick, “Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles”
4. http://www.upscale.utoronto.ca/GeneralInterest/Harrison/SternGerlach/SternGerlach.html
5. http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polar.html
6. http://www.symonds.net/~deep/stuff/qp/chap03.php
7. http://www.nobel.se/physics/articles/ekspong/
8. http://physics.berea.edu/~king/Teaching/ModPhys/QM/Duality.htm
9. http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/mulslid.html
10. http://physics.ucsd.edu/was-sdphul/labs/2dl/exp6/exp6-BACK.html
11. http://www.microscopyu.com/articles/formulas/formulasresolution.html
12. http://newton.phy.bme.hu/education/schrd/2d_tun/2d_tun.html#animation
13. Brandt & Dahmen , The Picture Book of Quantum Mechanics.
14. www.falstad.com
Some Classical Mechanics
• Lagrangian formulation: particle wants to minimize the total of the Lagrangianwhere:
),,,( txxVT &ℑ=−=ℑ
time)( velocity
location energy potential
energy kinetic
==
===
tdtdxx
xVT
&
• Why?
• Given initial and final location/time
the path classically taken minimizes (extremizes) the action
),,(),,( ffii txtx
)(tx
∫ℑ=f
i
t
t
VdtxxtxS t)independen timeis (here ),()]([ &
????
Isaac Newton
• Via calculus of variations, this leads to the following equation, which holds in the classical path for every t:
∂∂ℑ
=∂∂ℑ
)()( txdtd
tx ii &
• Often, the following notations are used:
(momentum) (force) iqip
iqiFiqix&∂∂ℑ≡
∂∂ℑ≡≡
• And so
====
dtdp
dtmvd
dtdvmma
dtdpF i
i)(
Force
• In the continuous case, often and one minimizes
),,,( tfff ′ℑ=ℑ &
∫∫ ⇒′ℑ dxdttfff ),,,( &
.0=′∂
∂ℑ−
∂∂ℑ
−∂∂ℑ
fdxd
fdtd
f &Stands for the coupling of
“infinitely close particles”(e.g. points of a string).
f ′
q
p&∂
∂ℑ=• We have
• Suppose we want to find an expression, call it such that
,Η
:So .qp
&=∂Η∂
The Hamiltonian
∫
∫∫ℑ−=
∂∂ℑ
−
=−=⋅=Η
qpdpdp
qdq
qp
dpdp
qdpqpdpq
&&
&&
&&&
1
• What is in physical terms? Look at a particle moving under a simple potential:
Η
mgxmvmgxmvxmvqp +=
−−=ℑ−=Η
22)(
22
&&
• So is the TOTAL ENERGY (kinetic + potential).Η
• What are the Hamiltonian equations? As we saw
qp
&=∂Η∂
• And also
pqq
qpq
&&
−=∂∂ℑ
−=∂
ℑ−∂=
∂Η∂ )(
• So
qp
&=∂Η∂
pq
&−=∂Η∂
Example: Free Fall
mgxm
pmgxmv+=+=Η
22
22
• Note that we have to write the energy as a function of the momentum and the coordinates, as opposed to the coordinate and its time derivatives (the latter is appropriate for the Lagrangian formulation).
• The equations are (remember:
new) (nothing vmpq
p=⇒=
∂Η∂
&
gamavmmgpq
−=⇒−=−=⇒−=∂Η∂ )( &&
Which is indeed the free fall equation.
) , vqxq == &
Example: Harmonic Oscillator
• The point at which the spring is at zero potential energy will be taken to be The Hamiltonian is then
.0=x
+
=
⇒−=⇒−=−=
⇒−=∂Η∂
+=Η
tmkBt
mkAx
xmkxxmvmkx
pq
kxm
p
sin cos
,22
22
&&&&&
&
• A,B are determined by the initial conditions.
Example: Coupled Harmonic Oscillator
[ ] 2 , 2
)(222
212211
221
22
21
22
21
xmkx
mkxx
mkx
mkx
xxxxkm
pm
p
−=+−
=
⇒−++++=Η
&&&&
• Let us discuss differential equations of the form
xx &&=ΩWhere:
• is a vector of functions.
• is Hermitian with negative eigenvaluesand eigenvectors Ωx
22
21 , λλ −−
21,ee
matrix) 22 a is ( 2
1 ×
≡ E
ee
EDenote
m.equilibriu from ntsdisplaceme theare , 21 xx
unitary is ,0
0 , ,
22
21 ED
ExyyExDEEyyEExx
tt
t
−−
=
===Ω=Ω⇒=Ω
λλ
&&&&
• After diagonalizing, the system is easy to solve. If we assume that the initial velocity is zero,
)cos(),cos( 21 tty λλ=• Initial conditions:
0)0(,0)0(,)0(,)0( 2121 ==== xxbxax &&
• For y, the initial conditions are then
′′
=
ba
ba
Eso the solution is
′′
≡
′′
=
ba
Cba
tt
y)cos(0
0)cos(
2
1
λλ
• To return to x,
[ ]IIIItcIItctUxtUtx
ba
eetceetc
ba
Etc
tcE
ba
CEEx
tt
tt
)()()( ,)0()()(
)()(
)(000
000)(
2211
22221111
22
11
+==
⇒
+
=
+
=
=
Where we have taken the opportunity to introduce a common QM notation, Dirac’sfamous bra and ket:
• stand for column and row vectors.
• stand for eigenvectors.
• are solutions to the differential equations with one variable, where are eigenvalues.
• is called the propagator and is of fundamental importance in QM.
?,?III , sΩ′
)(),( 2211 tctcffff 21 , ωω == &&&&
sΩ′21,ωω
)(tU
Example: Vibrating String
• Once we realize that the propagator formulation carries over to the infinite dimensional case, we can write down in a straightforward manner the solution to more general problems, such as the vibrating string:
L• First, we have to write down the corresponding differential equation. Think of the string as being composed of many small springs, that is, the limiting case of many masses coupled by springs:
iq
Ni 1−
Ni
Ni 1+
• The Lagrangian is the limit of
[ ]...)()(...21...
21... 2
122
122 +−++−++−++ +− iiiii qqhqqhqm &
≡ h
NL denote
hence the equations are
Assume the spring constant kto be equal to 1.
iiii q
mqqq
&&=+− +− 11 2
but in the limit, is just the well-known approximation to the second derivative, so –ignoring constants – the spring equation is
11 2 +− +− iii qqq
2
2
2
2 ),(),(t
txfx
txf∂
∂=
∂∂
,),(),(2
2
2
2
ttxf
xtxf
∂∂
=∂
∂• To solve proceed as before:
• First, find the (normalize) eigenvectors/eigenvalues
of the operator Since they’re restricted to be
zero at 0 and L, they are
.2
2
xf
∂∂
2
2221
eigenvalue ,sin2L
mxL
mL
Imππ
−=
=
• What are the equivalents of As before, if we restrict the initial velocity to be 0,
the solutions are and – also as
before – the general solution can be written with a propagator:
?)(),( 2211 tctc
,cos
t
Lmπ
∑
∑
∑
∞
=
∞
=
∞
=
=
=
⇒=
=
1
1
1
)0,(cos
)0,( cos),(
)0,()(),( ,cos)(
mmm
mmm
tmm
m
IxItL
m
xIItL
mtx
xftUtxfIItL
mtU
ψπ
ψπψ
π
But in the limit the internal product is just an integral, so
∑ ∫∞
=
=
1 0
)0,( cos),(m
m
L
m IdxIxftL
mtxf π
• Later, we will see that in QM, every single particle has a function – the probability amplitude – associated with it, and the main goal is to determine how it changes over time. It turns out that this function satisfies a first order (in time) differential equation (as opposed to our examples which had a second time derivative), called Schrödinger’s equation, and one seeks the propagator which – given the probability amplitude at – computes it for every t:0=t
)0,()(),( xtUtx ψψ =
Why all this?....
Poisson Brackets and Canonical Transformations
Let be some function of the state variables p,q, where we assume no explicit dependence on t.Then its change over time is given by
),( qpω
∑
∑
Η≡
∂Η∂
∂∂
−∂Η∂
∂∂
=
∂∂
+∂∂
=
i iiii
ii
ii
i
qppq
pp
qqdt
d
,ωωω
ωωω&&
• For any two define the Poisson bracket by
),( ),,( qpqp λω
∑
∂∂
∂∂
−∂∂
∂∂
=i iiii qppq
λωλωλω ,
• Note that
ijjijiji pqppqq δ=== ,0,0,
• We may want to define new variables
),( ),,( qpppqpqq →→
A straightforward calculation yields that if we want Hamilton’s equations to hold in the new coordinates, we must have
Such a transformation is called canonical.
ijjijiji pqppqq δ=== ,0,0,
[ ]Hidt
d
dtd
,
mechanics quantumin And
,
mechanicsclassicalin :similarity theNote
Ω−
=Ω
Η=
h
ωω
Examples:• Rotation is a canonical transformation (check it).
• Important: the two body problem. Assume we have two bodies in space, with locations and masses given by Assume also that there’s a potential which depends only on their relative distance, We can then define a canonical transformation that reduces the analysis to that of two independent bodies:
• Thus the center of mass is a free particle, and the reduced mass moves under the potential V. One can then solve for them separately. One such system, for example, is the hydrogen atom.
• We will later show how one can guess such a transformation; it follows from the requirement that the Hamiltonian be diagonal in the new coordinates.
).,(),,( 2211 rmrm
).( 21 rrV −
mass) reduced (the )( , ,
mass) ofcenter (the 0 , ,
21
2121
2121
2211
rVVmm
mmmrrr
Vmmmmm
rmrmr CMCMCM
=≡+
=−=
=+=++
=
µ
The Electromagnetic Lagrangian
• The force acting on a charge q, moving at speed v, due to an electric field E and a magnetic field B, is
( ) ( ) ( )
( )( )
( ) ( )
( ) ( )
potential. areally not isA v
at however th Note .Bv E
BvAv
AvAv
A 1 AvA
vAv Av
yieldsit , check that tohave We
Avvv21 Lagrangian a define Next,
.A 1 E B,A satisfying
A, potentials with the,Bv EF
AvA
momentum" canonical"
⋅−
×+=
×=×∇×
∇⋅−⋅∇
+
∂−+∇−=
⋅∇+−+∇−
=⇒⋅∇+∇−=
+
∂∂ℑ
=
∂∂ℑ
⋅+−⋅=ℑ
∂∂
−∇−==×∇
×+=
∇⋅+∂∂
cqq
cqcq
dtcqqdtd
cqq
mdtd
cqqc
qmdtd
xvdtd
cqqm
tc
cq
t
ii
φ
φφ
φ
φ
φ
φ
44 344 21
43421
The Electromagnetic Hamiltonian
.2
A pH
so A,vp of in terms expressed
be tohasit correctly,it writeorder toin But
vv21vpH ,before As
2
φ
φφ
qmcq
cqm
qTqm
+−
=
+=
+=+⋅=ℑ−⋅=
Max BornErwinSchrödinger
Werner Heisenberg
Paul Dirac Richard Feynman
Max Planck Albert Einstein Niels Bohr
QM was developed in order to explain physical phenomena which were not
consistent with classical physics.
Louis de Broglie
Part I: Physical Intuition and the Finite-
Dimensional Case
Prelude: The Nature of LightLight was thought to be made of particles, which move in a straight line. Later, light interference was discovered (Young, 1801), which led to the conclusion that light is a wave:
The mathematics behind this: a wave can be described by the equation Such a wave moves through space at a speed of is the frequency, the wave number. If another wave is behind, their sum is
which peaks/is zero when Denoting (the wavelength), there is a maximum if which is hardly surprising – it means that if the lag between the two waves is a multiple of the wavelength, they contribute to each other (they are in phase).
.)( tkxie ω−
.kω ω k∆
∆∆−−∆+− +=+=+ ikiktkxitxkitkxi eeeee 1)1()())(()( ωωω
.)12(/2 ππ +=∆=∆ nknkkπλ 2=
,λn=∆
Two slit experiment
Wave interference: as opposed to the drawing, assume Then the lag between the two waves is
so, the maxima correspond to which implies that the distance between adjacent maxima is (this can actually be used to measure the light’s wavelength).
• Mathematically, if the (complex) wave functions are then the energy arriving when slit i is open is and the energy arriving when both are
open is NOT but
., xad >>.)( 2222 ∆=≈−+−+
daxaxdxd
,λndax =
adλ
,2ih
,, 21 hh
,22
21 hh +
4434421 termceinterferen
212
22
12
21 )cos(2 δhhhhhh ++=+ δ 2h1h
So far so good – but it turns out that the wave model fails to explain certain phenomena.
When you shine light upon certain metals, electrons are emitted (the photoelectric effect).
It turns out that
• The emitted electrons move with greater speed if the light has a higher frequency.
• No electron is emitted until the light passes a threshold frequency – no matter how strong the light is.
• The electrons are emitted too quickly.
All this contradicts the hypothesis about the wavelike nature of light.
• Einstein solved this problem (and was awarded the Nobel prize for that), by suggesting that light is composed of little “quanta”, which impact like “baseballs”, and not like waves. These were later termed photons, and it turned out that a photon with frequency has an energy and momentum given by
ω
kpE
h
h
== ω note: .pcE =h is a very
small constant.
It also turns out that not only don’t the lightwaves behave like waves – also, particles don’t behave like particles.Let’s go back to the two slit experiment. How will it look with particles (electrons)? We may assume it will look like
i.e., the number of electrons hitting any specific point on the wall when both slits are open, will be the sum of the numbers when only the first/second slit is open.
But that doesn’t happen; the electrons also interfere. That is, at some points, LESS
electrons hit when BOTH slits are open!! This happens even if the electrons are fired
very slowly (one at a time).
There is no way to explain this – it just happens.
• Note: we can immediately realize that each electron cannot go through one slit. However, they don’t “break”, because when they arrive at the wall, the energy corresponds to a “whole” electron.
• Fortunately, the simple model that describes wave interference ( ) also explains particle interference. The problem is, of course, to compute the so-called “probability amplitudes”
221 hh +
., 21 hh
electron gun
There are such
that and 21,hh
2ii hP =
.22112 hhP +=
What happens if we try to watch the electrons? If we see them all, there’s no interference!
To be explained later. Intuitively, the measurement forces the particle to assume a definite position.
Some QM axioms:• Probability is the squared absolute value of the amplitude.
• If there are several possibilities for an event to happen, the amplitudes for these possibilities are summed. So, to get the total probability, you first sum and then take absolute value squared – thus there is interference.
• When it is possible to decide what alternative was taken, you sum the probabilities (as usual – no interference).
• There is uncertainty; we cannot predict what will happen, even if we know everything in advance. This uncertainty is reflected, for example, in the inability to exactly describe both a particle’s position and momentum – the famous Heisenberg uncertainty principle. All we can do is provide probability distributions. But, as opposed to our standard understanding of what uncertainty is, here it’s much deeper; it’s not “I know the particle is somewhere, I just don’t know where” – the particle isn’t anywhere, so to say. And if we try to nail it down, we lose its momentum. There’s no way out of this.
It’s strange – but it works.
To quote Feynman…
“That’s how it is. If you don’t like it, go to a
different universe, where the laws are simpler” – R.P
Feynman, NZ, 1979.
The Heisenberg Uncertainty Principle
• Take the limiting case of a wave with amplitude
It has exact momentum ( ) and energy ( ) but there’s no information on its location, because for every x the probability is 1.
• Real particles have amplitudes that are not uniform in space. They may look like a Gaussian, meaning that there’s a far higher probability to find the particle near the Gaussian’s center. But in order to get a Gaussian, you have to add many “pure waves”. Each “pure wave” has definite momentum – but when you mix them, the uncertainty in the momentum increases. The narrower the “position Gaussian”, the broader the “momentum Gaussian” (more later).
.)( tkxie ω−
ωhkh
Momentum certain, but zero knowledge on position
Position certain, but zero knowledge on momentum
x
Momentum uncertainty
Position uncertainty
Trying to measure both location and momentum
The smallest distance between two points in an object that will produce separated images is If a photon has energy it possesses momentum To be collected by the lens, the photon must be scattered through any angle between to So, the x-component of the momentum may have any value between to Thus the uncertainty in the electron's momentum is So, the product of the uncertainties is
).sin(θλ≈∆ x,ωh .λh
θ− .θ
λθ )sin(h− .)sin( λθh.)sin( λθh≈
.h≈
Intuitively, we need a “powerful” photon (high frequency) to get good position measurements; but the higher the energy, the more it knocks off the electron, hence the momentum uncertainty becomes larger.
Interlude – the Resolution of a Microscope
The resolution of an optical microscope is defined as the shortest distance between two points on a specimen that can still be distinguished by the observer or camera system as separate entities. When becomes close to , it’s impossible to distinguish. That’s why electron microscopes are used (much smaller wavelength).
3
4
10 be should peaks between Distance
0.01 ,1.0 ,10
=
===
ad
ad
λ
λ
4
4
102 be should peaks between Distance
0.01 ,005.0 ,10
⋅=
===
ad
ad
λ
λ
4
4
10 be should peaks between Distance
0.01 ,01.0 ,10
=
===
ad
ad
λ
λ
λa
a
θadλa
d
)sin(θd≈
In order to distinguish the objects, the two maxima should fall in the view area, so we must have
)sin()sin(
θλλθ ≥⇒≥ a
add
Another experimental demonstration of the uncertainty principle – shooting a particle through a single slit (Fraunhofer diffraction):
slit width = W
θ
wavelength = λ
One can prove that .Wλθ ≈
y
Before the particle (wave) hits the barrier, we can assume that its y-momentum is zero. After it passes the slit, we know its position with an accuracy of ,but there’s an uncertainty in the y-momentum, (where p is the original momentum). So, but QM tells us that
• The narrower we make the slit, the more confident we are about the location; but then the diffraction pattern becomes wider, and we lose confidence in the momentum.
Wy ≈∆Wppy λ≈∆,λppy y ≈∆∆
).2( hπλ == hp
De Broglie Waves: The Wavy Nature of Particles
In the previous slide, we adapted to particles a result connecting the wavelength and momentum of a photon. This is based on De Broglie’s ingenious “hunch” in 1924: “if waves (light) behave also like particles, then particles should also behave like waves”. It turns out that a moving body behaves in certain ways as though it has a wavy nature. Its wavelength is defined, just like
for a photon, by
If so, electrons for example must diffract. This was proved experimentally by Davisson and Germer in 1927: they shot electrons at a crystal made of layers. An electron is scattered from different layers, hence there’s a phase difference, hence there’s diffraction.
λ.
mvh
ph==λ
Note: larger objects (baseballs) also show interference, but is so small that the interference pattern is practically impossible to create and observe.
λ
Application: the Radius of an Atom
• Let the “average radius” of the hydrogen atom be denoted by a.
• The momentum p should be, on the average, this implies that the average kinetic energy is
• The potential energy is where e is the electron charge. Hence the total energy is
• It’s easy now to find the a minimizing this expression, and it’s indeed pretty close to the true radius.
• Why then can’t we squash an atom, and why doesn’t the electron collapse onto the nucleus? Because then the uncertainty in the position will go to zero, and the uncertainty – and hence the average – of the momentum (and the energy) will go to infinity.
.ah
2
22
22 mah
mp
=
,2 ae−
ae
mahE
2
2
2
2−=
More on Probability Amplitudes
Two slit experiment again:
Denote the amplitude that a particle from s will arrive at x by Dirac’s “bra and ket” notation:
We will later see that it’s just an inner product.
We know that the probability is the square of the amplitude. We also know that amplitudes add:
open 2slit open 1slit open slits bothsxsxsx +=
sxsx ≡ leaves particleat arrives particle
Another law about amplitudes: the amplitude for a path is the product of the amplitudes to go part of the way and the rest of the way.
siixsxcba
iαα
α
∑==
=,,
2,1
• We will later address amplitudes with a time variable (e.g. what is the amplitude for a particle to be somewhere at a given time).
• Another amplitude rule – if two particles don’t interact, the amplitude for them to do two things is the product of the individual amplitudes.
Applying the amplitude rules to the two-slit experiment with a light source (trying to find
what slit the electron went through)
sx
If there’s no light, sxsxsx 221121 +=+= φφ
If there’s light, there are certain amplitudes for a photon to arrive at if the electron went through 1 – call these amplitudes a(b). Applying the principles and some symmetry considerations:
)( 21 DD
4847648476 21 at D photon
212
at D photon
221)at electronPr( φφφφ babax +++=
If we can detect the slit via which the electron went with totalcertainty, then b=0 and the probability is hence no interference at all. If we try to decrease the photon energy, wewill have to increase its wavelength, hence b will grow and we’ll have more and more interference.
,22
21 φφ +
Question: how can a light source affect the interference of baseballs?
Another Example: Scattering from a Crystal
There are some crystals for which the scattering has both “classical” and interference components:
neutron source
crystal θ
θ
number of neutrons
Why does this happen?
neutron counter
• Sometimes the crystal nuclei have a property called “spin” (they may be “spin up” or “spin down” – more later).
• Usually, the neutron is scattered without “leaving a mark” on the specific atom which scattered it. That case is like the two slit experiment without light, and we get interference.
• But sometimes the neutron can “leave a mark” on the atom that it scattered from – by changing the latter’s spin (and also its own spin). That is equivalent to the two-slit experiment with light; in that case, we can distinguish what path the neutron took, and therefore we should add probabilities and not amplitudes (so no interference).
• In this crystal the neutron sometimes changes the nuclei spin and sometimes it doesn’t – hence we have a mixture of both distributions: interference and no interference. Note: it makes no difference whether we try or don’t try to find the atom from which the neutron scattered!
= +
spin changedresult Spin not changed
Bosons and FermionsLet’s look at a scattering of two particles:
• If the amplitude for the left event is then the probability of any particle arriving at the detector is
),(θf
1D 1D
2D 2D
1D2
22
22Pr
2 :)()(Pr
=⇒=−+=ππθθπθ fff
• But what if the particles are identical? Then we cannot differentiate between the left and right events, and we have to add amplitudes before taking the squared absolute value to find the probability.
•. Exchanging the particles does not change the physics (probabilities), but it may change the phase. Since exchanging again brings us back to the original system, the amplitude must either remain the same or be multiplied by It turns out that there are two types of particles: for Bosons the amplitudes are the same, for Fermions they are inverse.
• So, for a scattering of two bosons, the probability of one of them arriving at is
• And for two Fermions it is
• So, two Fermions cannot scatter at an angle of
.1−
22
24Pr
2 :)()(Pr
=⇒=−+=ππθθπθ fff
1D
!2π
0Pr2
:)()(Pr 2 =⇒=−−=πθθπθ ff
a
1
b
2
• Let’s look at a double scattering. We know that the
probability for this is
• Assume that the particles are distinct. The probability for
.21 22 ba
22
21
24 2121 baabba
→→+
)12( )21( →∧→∨→∧→ baba
• But if they are Bosons, we cannot differentiate which went where, so we have to add amplitudes and then square:
• This yields a probability two times bigger than for distinct particles. So: Bosons like to go to the same place, Fermions like to stay afar from each other. This is the Pauli exclusion principle, and without it atoms would have looked very different (electrons would stick together).
is
• If 1 approaches 2, we can denote
.21 ,21 bbbaaa ====
so the probability for both particles to go to 1 approaches .2 22 ba
.2121 2222 abba +
Light PolarizationWaves can be characterized by being either:
• Transverse: the thing that is "waving" is perpendicular to the direction of motion of the wave (light waves).
• Longitudinal: the thing that is "waving" is in the direction of motion of the wave (sound waves).
Two possible polarizations of light waves. The arrows depict the oscillations of the
electromagnetic wave. These will be called x and ypolarizations.
ZZ
Polarizers:
Polarization by reflection.
There are apparatuses that allow only light as a certain polarization to go through (LCD, polaroidsunglasses).
• From the QM point of view, every single photonhas a mixture of the two polarizations associated with it, and we can only compute the probability that it goes through a certain polarizer.
• and resp. refer to the state of a single photon in an x resp. y beam which is polarized in the classical sense.
• Filtering with a polarizer in angle relative to x-y yields the state
x y
θ.)sin()cos(' yxx θθ +=
If we place a second polarizer at angle 0, then the probability that goes through is , which is in perfect accordance with classical physics – but here, it’s not that the wave energy is decreased by this factor, but that only of the photons go through – and those which do, don’t lose any energy.
'x )(cos2 θ
)(cos2 θ
Thus – a classical, macro, continuous phenomena is explained by a (rather
different) micro, discrete, QM phenomena.
Light is (classically) said to be RHC polarized if its x and y components are equal but out of phase (in this case, the field is not linear as before, but oscillates in a circle which lies in the plane perpendicular to the light’s direction). The QM interpretation for a single photon is
Where does the i come from? We know that a phase shift, when repeated twice on yields So, we can see why it is represented by a multiplication with i.
o90
( )yixRHC +=2
1
o90
,y .y−
Interlude: Wiesner’s Quantum Money
• Bank has a list of bills; each has a serial number and 20 photons embedded in the bill, each with one of four possible polarizations:
• A would-be counterfeiter cannot copy the bill, since any attempt to measure the polarization will not only probably destroy the original polarization, but will also leave too many possibilities. For example, if a photon measured with a green polarizer gives zero (the photon doesn’t pass), it can be either a black, red, or blue photon. As with the electron, the measurement operation changes the state; this is why the method works.• Note that the bank may say “yes” even if some photons are not correctly polarized – but the probability of the counterfeiter to obtain this can be made arbitrarily small.
The total amount of deflection is a function of
• The total amount and distribution of electric charge on the ball.
• The orientation and rate of spin. As the rate of spin increases, so does the deflection. As the axis of the spin becomes more vertical, that amount of deflection also increases.
If the beam from the electron gun is directed to the magnets, as shown to the right, the beam is split into two parts. One half of the electrons in the beam are deflected up, the other half were deflected down. The amount of deflection up or down is exactly the same magnitude. Whether an individual electron is deflected up or down appears to be random. Stern and Gerlach did a version of this experiment in 1922.
This is very mysterious. It seems that the "spin" of electrons comes in only two states. If we assume, correctly, that the rate of spin, total charge, and charge distribution of all electrons is the same, then evidently the magnitude of the angle the spin axis makes with the horizontal is the same for all electrons. For some electrons, the spin axis is what we are calling "spin up", for others "spin down".
You should beware of the term "spin." If one uses the "classical radius of the electron" and the known total angular momentum of the electron, it is easy to calculate that a point on the equator of the electron is moving at about 137 times the speed of light! Thus, although we will continue to use the word "spin" it is really a shorthand for "intrinsic angular momentum.“ It has no classical counterpart.
Introduction to base states: spin and the Stern-Gerlachexperiment
Building a Spin Filter
The blocking mechanism is really a “filter” – allows only one spin type to go through.
Shorthand notation for “spin up” filter. On the average, it allows half the electrons through. We can rotate the filter. Still, on the average, it
allows half the electrons through.
All those which passed the first will pass the second.
None of those which passed the first will pass
the second (nothing passes both of them).
Half of those which passed the first will pass the
second: in general, it’s
2cos2 α
One quarter will pass.
To realize that something strange is going on, think about a filter with both paths unblocked being here:
Correlation measurements in a radioactive substance that emits a pair of electrons in each decay:
If the right electron passes, then its left hand companion does not pass its filter, and vice-versa. We say that each radioactive decay has a total spin of zero: if one electron is spin up its companion is
spin down.
Again, one-half of the right hand electrons pass through their filter and one-half of the left hand electrons pass through their filter. But this time if a particular right hand electron passes its filter, then its companion left hand electron always passes its filter. Similarly, if the right hand electron does not pass its filter, its companion electron doesn't pass through its filter either.
1. One-half of the right hand electrons emerge from their filter.
2. One-half of the left hand electrons emerge from their filter.
3. If a particular right hand electron passes its filter, one-half of the time its companion left hand electron will emerge from its filter, one-half of the time it will not.
It turns out that in the “world of Stern-Gerlachapparatuses” (SGA), the states U and D (denoting spin up and down) form something that resembles a basis in linear algebra. What does this mean? For example:
• If an electron is in one of these base states (B), we can predict the probability it goes through any SGA, regardless of its previous history.
• Every state and every transition amplitude can be describe as a mixture of these states:
• Since the base states are complete (i.e. they span all states), we must have that the sum of probabilities of a state to be in any of the base states is 1, or
But the probability of a state to go to itself must be 1, so according to , we must have
φχφχ iiBi∑∈
=iiBi∑∈
= χχ
12
== ∗
∈∈∑∑ χχχ iii
BiBi
∗
∈
=∀⇒==∑ χφφχχχχχ φχ ,1iiBi
Just like in linear algebra, we have many sets of base states; e.g. the outputs of SGAs which are rotated in various angles, each time leaving only one beam unblocked. And just as in linear algebra, we can switch from description in to a description in if we know for all i,j. For example, for spin ½, and for rotation around the longitudinal axis, the transition matrix is
1 iB =,2 jB = ij
−
2cos
2sin
2sin
2cos
θθ
θθ
Operators and base statesSuppose we do something to a particle – for example, make it go through a mess of SGAs. Let’s call this mess A an operator.
• If we have such an operator, we may ask questions such as: given two states , what is the amplitude for which will be denoted We get another familiar result from linear algebra (remember that it makes more sense if you read from left to right):
Where i,j range over base states.
• Other “linear algebra laws” follow – the composition of operators is the matrix product of their individual matrices, we have the usual laws for transforming between bases, etc.
• The matrix product representation encapsulates a rather strange result – composition of operators (like position and momentum) is not necessarily commutative, as opposed to classical physics. This has profound consequences.
φχ ,,χφ →→ A
.φχ A
φχφχ iiAjjAji∑=
,
Dependence of Amplitude on Time: The Hamiltonian
• How does the amplitude change with time?
• For a particle in a state of definite energy the amplitude is independent of position and is given by (intuition: it’s like a wave, and it has zero uncertainty in momentum, hence infinite uncertainty in position).
• Note that any probability question for such a particle is independent of time; hence such a particle is said to be in a stationary state.
• In order to have probabilities which change in time, the particle has to have more than one energy state: note that
depends on the time t.
• The amplitude of a particle in uniform motion
is where E is the particle’s total energy.
,0E
tEiae )( 0 h−
2)(2
)(1
21 tEitEi eaea hh −− +
,)()( xpEtiae ⋅−− h
Interlude: Jumping Over a Potential Barrier
Suppose that a particle with a certain energy somehow goes over a potential barrier which is larger than its energy (this is, of course, impossible in classical physics). Its energy has to be negative, and since its momentum has to be imaginary. Since the amplitude behaves like
,22 mpE =
,)()( xpEtiae ⋅−− h we may expect it to look like .xe α−
Incredibly enough, this is true. It explains, for example, particles getting into places which are forbidden by classical physics. More on this later.
A Little More on Bras and Kets• Recall that if ranges over base states, then
• Inspired by this, we write
• And further:
• is a bra and is a ket (these notations are valid not only for base states).
• If and then
• So it’s just like an inner product. These notions extend naturally to an infinite number pf base states, with the structure (plus some functions which are not in the classical ).
• For an operator A we have
• It’s common in QM to denote the state vector by anything which identifies it. For example, a state with a definite momentum p will be denoted by etc.
iφχφχ ii
i∑=
φφ iii∑=
iii∑= |
ii
ii
Ci∑=φ ,ii
Di∑=χ
∑∑ ∗∗ ==i
iii
i CDiD φχχ and
2L2L
jjAiiAji∑=
,ψ
,p
∫ ∗= gfgf ),(
How does the amplitude change under an apparatus, or operator? Passage of time is also an operator; denote by the change which takes place between time and By that we mean that if a system is at state at time then its amplitude to be at state in time is
),( 12 ttU
φχ ),( 12 ttU
1t .2tφ
χ,1t
2t
., ji
),,(),(),( 122313 ttUttUttU =
So, it’s enough to know
for base states
Since it’s enough to know Denote as before ).,( tttU ∆+ ,)()( titCi ψ=and All this yields .|, jUiU ji =
∑ ∆+=∆+j
jjii tCtttUttC ).(),()( ,
Letting guess that ,0→∆t tKU jijiji ∆+= ,,, δ
For reasons to be explained, this is written as
ttHitttU jijiji ∆−=∆+ )(),( ,,,h
δ
so ∑ ∆−=∆+j
jjijii tCtHittC )(][)( ,,h
δ
jttUi ),( 12
∑−=∆
−∆+
jjji
ii tCHit
tCttC )()()(,
h
and
∑=j
jjii tCH
dttdCi )()(
,h
H is called the Hamiltonian. It could also have been called the energy matrix: look at the equation we saw for the amplitude for a particle at rest, with energy Then we have
,)( )(1
0 tEietC h−=
.0E
01,111,1101 )()()( EHtCHtCE
dttdCi =⇒==h
hi−This is why H represents energy, and why the factor was inserted before. In classical mechanics, the function giving the total energy of the system is called the Hamiltonian, hence the name here.
Interlude: the Hamiltonian is Hermitian
Recall that the adjoint of a matrix A is defined by∗+ = ijji AA ,,][ and that A is Hermitian if .AA =+
Since we can write the solution as),()( tHCdt
tdCi =h
)0()( CetC iHt−= (forget for the moment, it’s only a constant).
h
so tiHeCtC+
= )0()(
multiplying yields
)0()0()()(1 CeeCtCtC iHttHi −+
==
=+−++= +−+
...)...)(( iHtItiHIee iHttHi
.HH =+So, we must have that
)()( 2toHHitI +−+ +
Some useful facts: A Hermitian matrix has only real eigenvalues, and the eigenvectors corresponding to different eigenvalues are orthogonal. Also, if H is Hermitian, )( e thereforand iHtiH ee −−
is unitary. So, the evolution of a state is actually a rotation.
A Two-State Example: the Ammonia Molecule
The molecule can flip, and every state is a combination of the “up”and “down” states:
=+= ψψψ 2211
21 21 CC +
The question – what is the Hamiltonian? Based on symmetry considerations:
tAEitAEi ebeatC
ACCEdt
dCiACCEdt
dCi
))(())((1
1202
2101
00
22)(
,
+−−− +=
⇒−=−=
hh
hh
tAEitAEi ebeatC ))(())((2
00
22)( +−−− −= hh
• What do these solutions mean? It’s always useful to look for states with one frequency (or a definite energy). These occur in two cases: if the energy is and the amplitudes to be in and are the same. If ,the energy is and the amplitudes to be in and have opposite signs (but the probabilities are equal!).
• Suppose we know that when the state is This implies
,0 AE −,0 AE +
12
1 2
.1
=
−=
=
+=
⇒==⇒==
−−
−
−−
−
h
h
hhh
h
hhh
h
AtieeeetC
AteeeetC
baCC
tEiAtiAti
tEi
tEiAtiAti
tEi
sin2
)(
cos2
)(
10)0( ,1)0(
00
00
)()()(
)(2
)()()(
)(1
21
So Hence, if we know that at the state is “up”, we also know that at
).(sin)( 222 hAttC =
At
2hπ
= the state is “down” (!!).
0=t
0=t
0=a
0=b
Interlude: Symmetry and Conservation
A hydrogen ion can be in one of the two states
The operator P is defined by a reflection with the property
Assume that the physics of the ion is symmetric with respect to P. What does this mean? If we reflect by P, let the system evolve
.2 and 1
.1 2 ,2 1 == PP
for some time, and reflect again, it will be just like letting the system evolve and then reflecting. If we denote, as before, the operation of “waiting” by U, it means that P and U commute: In general, an operator Q will be said to have a symmetry if it commutes with U.
.UPPU =
Recall that So Q commutes with U iff Q commutes with H (remember that U is a sum of powers of H).
Suppose there’s a state is physically the same state as ; so, it differs from only by a phase factor. For example, look at the two states
.h
iHt
eU−
=
ψψψψ Q=′ that such
ψ
, :2
21,
221
IIIIPIIPIII −==−
=+
=
so, the phase change is 0 for and for
• Another example: look at an RHC polarized photon, and rotate it by an angle about the z-axis. Then its state is multiplied by
• It’s straightforward to see that if the symmetry operation of Q on a state multiplies by a phase factor , then this is true forever:
I .IIπ
δ.δie
0ψ 0ψδie
==⇒= ))(0,())0,(( 0000 ψψψψ δ QtUtUQeQ i
But this is conservation!t
it
ii eQtUeetU ψψψψ δδδ =⇒= ))(0,())(0,( 00
Parity
Look at the operator
Since it is immediate that if
).,,(),,( zyxzyxP −−−= ,ψψ δieP =,2 IP =
.1or 1 then −== δδ ii ee
• P is called the parity operator and the two cases above correspond to states having even or odd parity.For the hydrogen ion, has even parity, has odd parity, and has no definite parity.
• It turns out that parity is not always conserved; it is violated by • If is not involved, it turns out that any state with definite energy which is not degenerate (meaning that there are no other definite energy states that have the same energy) must have either even or odd parity. As a matter of fact, let Q be any symmetry operator, and let be a non-degenerate state with energy E. So and
I1
II
decay. βdecay β
) ( )( ψψψψ QEQEQHQH ===
ψ ,ψψ EH =
So, is a state with definite energy E, but we assumed that is the only such state. So we must have but as we saw before, if then
ψQ ψ,ψψ δieQ =
,PQ = .1or 1 −=δie
The conservation laws in QM•Various conservation laws can be expressed in QM as multiplication by a certain phase factor which depends on time, location etc. with a constant which equals the conserved quantity. Note: since there are states without a definite momentum for example, we cannot always talk about “conservation of momentum” as in classical physics (the most we can do for such states is hope that the average momentum is conserved). Some examples:
)(aDx0ψ .)( 00 ψψ ika
x eaD =Then corresponds to the classical momentum. Similarly, there may be special states such that after the passage of time , the state turns into Then corresponds to the classical energy.
If you find this perplexing, check what happens to the wave equation
hk0ψ
τ .0ψωτie− hω
τω +→+→− ttaxxe tkxi or when)(
at the equations for a photon’s momentum and energy).
• Let denote translation by a, and assume that there’s a special state such that
• If denotes rotation by around the z-axis, and )(φzR φ,)( 00 ψψφ φim
z eR = is the angular momentum forhm .0ψIt turns out that for an RHC/LHC photon this is equivalent to the classical notion of angular momentum, with values For linearly polarized light, there’s equal amplitude to be in RHC or LHC, so a single photon doesn’t have a definite angular momentum, and that’s why (classically) a beam of linearly polarized light has zero angular momentum.
.h±
(and look again
Infinitesimal Transformations
As in Lie group theory, it is customary in QM to study transformations with an infinitesimal displacement. For example, let ).,()( ttUDt ττ +=
As we saw before, so for a small ,)0()( ψψ h
iHt
et−
= t∆
⇒
∆−−≈∆=∆
∆−
)0(1)0()0,()( ψψψh
htiHetUt
tiH
.)(1)( HtitDt ∆−=∆h
But we know that for a state of definite energy E the passage of time multiplies the state by So, it makes sense to refer to H as the energy operator (more on this later).
energy. sphoton' 1 ==⇒∆−≈∆− hωωω Htitie
• Similarly, for a small displacement we may assume that
the state changes linearly in
),( xDx ∆
ψψ
∆+=∆∆ xpixDx xx
h1)( :
kpx h=but
for a definite momentum state; so, call the momentum operator.xp( ) ⇒∆+≈=∆ ∆ ψψψ xikexD xik
x 1)(
Note that behaves like should, so we can guess
That That turns out to be true (more later). xpi )( h dxdψ
.dx
dipx
ψh−=
• Similarly, we can define the angular momentum operator by
ψψψφψψφ φ hh
mJJieR zzim
z =⇒
∆+==∆ ∆ 1)(
zJ
And it immediately follows that seems like zJ .xy ypxp −
Part II: the Continuous (Infinite-Dimensional)
Case: the Wave Function
The Wave Function and the Axiomatic Foundations of QM
• So far, we have dealt with a finite number of base states. That is, a system could be described by a finite linear combination of probability amplitudes.
• This does not always suffice. Think for example about the location of a particle along the x-axis. There is a continuum of possible locations, and they cannot be expressed as a finite linear combination of anything.
• Same holds for momentum.
• This led to the definition of the wave function.For a particle in a one-dimensional world, it is a continuous function which defines the amplitude to be, for example, at any location x.
• Next, we present the axioms of QM for what a wave function is and how it behaves over time. They are given side-by-side with the classical physics axioms, to emphasize both the similarity and difference between the two.
changes according to Schrödinger’s equation:
where H is the quantum Hamiltonian, created by substitutingIn the classical Hamiltonian.
The system changes with time according to Hamilton’s equations, where is the classical Hamiltonian (total energy):
If a particle is in the state a measurement of the variable
yields an eigenvalue of the operator with a probabilityand the state changes from to ( collapses to
A measurement of a dynamical variable yields a uniquely defined result, and the state is unchanged by the measurements.
The x of classical mechanics is replaced by the operator X, defined by and p is replaced by the operator P defined by
Every dynamical variable is a function of x and p.
X and P are Hermitian.
The state of a particle at time t is given by a vector (function) in Hilbert space,
The state of a particle at time t is given by its position and momentum
QMClassical
).(tp)(tx
.)(tψ
,)()( xxfxfX =
.)(dxdfixfP h−=
),( pxΩΩ
,ψ
),( PXΩω
,)Pr(2
ψωω ∝ψ
ω
xp
px
∂∂
−=∂∂
=ΗΗ , &&
)()( tHtdtdi ψψ =h
)(tψ
PpXx →→ ,
Η
ψ ). ω
Points for Discussion:
• What is the wave function? Inner product:
• What are the X and P operators, and what are their eigenvectors and eigenvalues? Relation to Fourier transform!
• The meaning of the collapse of the state vector for the X and P operators. What happens if an eigenstate is measured? Reminder: two-slit experiment with measurement. Question: how can a photon cause the wave function of a battleship to collapse? Reminder: Stren-Gerlach experiment.
• The Schrödinger equation for a free particle as the limit of a finite dimensional state: where does the derivative come from? (Later).
Complications:1. What does the classical xp go to: XP or PX? Answer: average.
2. What if is degenerate? Answer: where is the projection on the subspace of vectors with eigenvalue
3. The eigenvalue spectrum of is continuous. Answer: replace all sums by integrals, and “probability” by “probability density”.
Ω ,|)Pr( ψψω ωP∝ωP
.ωΩ
∫=≤≤=Ωb
a
dxxbaX 2)() position Pr( : ψ
=−′= ppPxxxxXipx
e ,21 : , ),( : h
hπδ
∫∞
∞−
∗= φψφψ
Example:So we should have (normalization).1)( 2 =∫
∞
∞−
dxxψ
Expectation and Uncertainty
• We saw that the measurement process in QM is probabilistic. Yet, we can ask what the average result of measuring in the state is: following the usual definition of expectation, it equals
ψΩ
ψωωωψωψψωωψωωωωω
==≡Ω ∑∑∑
2
where range over the eigenvalues/eigenvectors of Since is Hermitian, its (normalized) eigenvectors are orthonormal. It’s easy to see that can be resolved by
ωω , .ΩΩ
ωωωω∑=Ω
Ω
(hint: verify that both sides are equal when operating on every , and use the fact that span the entire space).
ω ω
So, the expectation is This is nice, since we don’t need to find eigenvalues and eigenvectors in order to compute the expectation.
• The average has a meaning only when computed over an ensemble of identical particles. It cannot be computed for a single particle, because the first measurement changes it.
• For an eigenstate, the expectation is equal to the corresponding eigenvalue. Thus, the Energy/Position /Momentum measurement of a state with definite EPM yields the “correct” and unique EPM result.
.ψψ Ω=ΩsΩ′
The uncertainty for in a state is also defined justlike the classical variance:
Ω ψ
( ) ( )[ ] 21222 )Pr()( ψψωω
ω
Ω−Ω=Ω−=∆Ω ∑
Example – the Gaussian centered at a with width
2222 )(
212
2)(
412 )(
1)Pr()(
1)( ∆−−∆−−
∆=⇒
∆= axax edxxex
ππψ
:∆
The average of the position is
adxxxxXX === ∫∞
∞−
∗ )()(| ψψψψ
which is hardly surprising. As for the uncertainty:
( ) ( ) =+−=−=∆ ψψψψ |2|)( 2222 XXXXXXX
2 is this
22
const. theis const.
a of nexpectatio
2
just is
2 ||2|aX
XXXXXX −=+− ψψψψψψ43421
So we have to compute 2
22)(22)(
212
2
2)(
1|2222
adxexeX axax +∆
=∆
= ∆−−∞
∞−
∆−−∫π
ψψ
and finally, (also hardly surprising). 2∆
=∆X
What about the momentum operator P?
∫∞
∞−
∗ =′−=⇒−= 0)( dxiPdxdfixfP ψψhh
( )∆
=∆⇒∆
=′′−=∆ ∫∞
∞−
∗
22 2
222 hhh PdxP ψψ
22 2)(
412 )(
1)( ∆−−
∆= axex
πψ
as can be directly verified by substituting and integrating.
• It is instructive to see how looks in momentum space. Since the orthonormal basis corresponding to momentum space is
given by then is represented in momentum space by
)(xψ
,)2( 21 hh ipxep −= π )(xψ
222 241
2
2
hh
h∆−−
∞
∞−
∗
∆== ∫ pipa eedxpp
πψψ
So, the narrower the position distribution (small ), the broader the momentum distribution. This not only agrees with the physics discussed before, it is also no surprise to anyone who studied the Fourier transform! Note:
∆
44 344 21h
principley uncertaint
. 2=∆⋅∆ XP
• This happens because P and X don’t commute, and don’t have common eigenvectors (“eigenkets”). Since only the eigenkets of an operator are unaffected by a measurement of the operator, we cannot have a system for which we know both the position and momentum.
• If two operators commute, there is a basis of common eigenkets, and the order of the measurements doesn’t make a difference.
• The commutator of two operators is defined as
• It is straightforward to see that
[ ] ΛΩ−ΩΛ≡ΛΩ,
[ ] hiPX =,
Interlude: Proof of Heisenberg’s Uncertainty Theorem
We want to obtain a lower bound on
( ) ( ) ψψψψ 2222 )()( PPXXPX −−=∆∆
The trick is to reduce this to an expression depending on
[ ] hiPXXPPX =−=,
define PXXPBAABPPBXXA −=−⇒−=−= ,So we should bound
ψψψψψψψψ BBAABA =22
Since A,B are Hermitian
From the Cauchy-Schwartz inequality, this is larger than22
ψψψψ ABBA =
Now comes the trick. Denote
DCABBAABDBAABC +=⇒+
=−
=2
,2
So we should bound2
ψψ DC +
Now:
ψψψψψψψψ
ψψψψ
ψψψψψψ
DCDCDC
DCDC
∗∗+
++
=+=+22
22
But and since is normalized, ,2hiC = ψ .
4|
22 h=ψψ C
To get the inequality, we will throw away
which is of course positive, and prove that
,|2
ψψ D
!0 thisCall
=+ ∗∗
4444444 34444444 21G
DCDC ψψψψψψψψ
ψψψψψψψψ
BAABBAABBAABBAABG
+−
++−≡∗
∗
Ignoring the constant factor of 2, this last expression equals
Now we have to remember that ψψψψψ+∗
Ω Ω=Ω∀ :,
And that and that ,, BBAA == ++ +++ΩΛ ΩΛ=ΩΛ∀ )( :,
And it immediately follows that G is indeed 0. Thus the proof is complete, and after taking roots: 2))(( h≥∆∆ PX
• Is the minimum attained? We can assume The Cauchy-Schwartz inequality is not strict only if the two vectors are linearly dependent, so we must have
.0=X
( ) ( )
+
=
⇒+=′⇒=−
xPcxi
ex
cxPicXPP
2
2
)0()( h
h
ψψ
ψψψψ
But to minimize, we must also have
( ) ( ) 0=−+− ψψ PPXXPP
but
( ) ( ) ⇒=−=− ∗ XcPPcXPP ψψψψ ,( ) ( ) ( ) 02 =+=−+− ∗ ψψψψ XccPPXXPP
So c must be pure imaginary, and so (after correcting by shifting with we get
icc =)X
( )( )
hh 2
2
)(Xx
cXx
Pi
eeXx−
−
−
=ψψThat is, a Gaussian with a phase shift proportional to the momentum saturates the uncertainty.
Interlude: Finite-Dimensional Example: the effect of measurement (Shankar Ex. 4.2.1)
Consider the following operators on :3C
−=
=
100000001
, 010101010
21
zx LL
• What are the possible values obtained if is measured?
Answer: the eigenvalues/normalized eigenvectors are
zL
zL
−
100
,1,010
,0,001
,1So the possible results when is measured are 0,1,-1.
zL
• In the state in which what are
Answer: if the state must be According to what we proved before:
,1=zL ?,, 2xxx LLL ∆
,1=zL .)0,0,1( t
21
21)0,0,1()0,0,1( 0)0,0,1()0,0,1(
22
22
=−=∆
====
xxx
txx
txx
LLL
LLLL Note: we should take
conjugate transpose, but everything here’s real.
• Find the eigenvalues and normalized eigenkets of in the basis.
Answer: the basis is the standard one. An immediate computation yields that the eigenvalues/normalized eigenvectors are
zL
xLzL
−−
−
12
1
21,1,
12
1
21,1,
101
21,0
• If a particle is in the state with , and is measured, what are the possible outcomes and their probabilities?
xL
Answer: if the state must be According to the axioms, the possible outcomes are the eigenvalues, with probabilities equal to the square of the inner product of with the corresponding eigenket. A straightforward calculation yields that we get 0 with probability 1 with probability and with probability
,1−=zL .)1,0,0( t=ψxL
ψ
,21 ,41 1−. 41
• Consider the state in the basis. If is measured in this state and the result is 1, what is the state after the measurement? If is then measured, what are the possible outcomes and respective probabilities?
( ) t 21,21,21=ψ zL 2zL
zL
Answer:
=
100000001
2zL
So, there’s a degenerate eigenvalue, 1, with an eigenket subspace and a 0 eigenvalue with an eigenket (0,1,0). If 1 was measured, the result is the (normalized) projection of on V, which equals
),,0,( baV =
ψ ( )( ) . 2,0,1 31 t
1−=zL xL
Next, is measured. The possible results, and their probabilities, are: 0 is measured with probability 0, 1 is measured with probability and –1 is measured with probability
NOTE: the measurement of does not determine the subsequent measurements of
zL
,31 .322zL
!zL
• A particle is in a state for which the probabilities are
note that the most general state with this property is.41)1Pr(,21)0Pr(,41)1Pr( =−===== zzz LLL
12
02
12
321
−=+=+== z
i
z
i
z
i
LeLeLe δδδ
ψ
does this mean that one can choose arbitrary values of without affecting the physical state?
321 ,, δδδ
Answer: no. The physical state is immune only to global phase shifts: are physically equivalent in the sense that they generate the same probability distribution for any observable. However, if for example then
and if then
ψψ δie and
,0321 === δδδ ,1== xLψ
, 0 231 and πδδδ === .1−== xLψ
A Note on the Position, Momentum, and Energy Operators
• We saw that for a state of definite position (delta function) the operator X indeed returns the position, and that for a state of definite momentum (a pure wave, ) the operator P indeed returns the momentum.
• What about the other states? The most we can ask for is that the average of X (P) will be what we consider intuitively to be the average position (momentum). That is indeed the case.
• Let be a state. The probability density to be at the position x is so the average position
is defined by but the average of the X
operator is defined by
And they are indeed equal.
)( tkxie ω−
ψ),()()( 2 xxx ψψψ ∗=
,)()( dxxxx ψψ ∗∫
dxxxxxXXX )()( ψψψψψψψψ ∗∫====
Note: if the integration limits are unspecified, they are assumed
to be . to ∞∞−
What about the momentum? Remember that the momentum operator is defined by
dxdiP ψψ h−= h
h
ipx
epπ2
1 =and that its eigenkets, with eigenvalues p, are defined by
So the probability for the state to be with momentum p isψppp ψψψ =
2
.444 3444 21
I
dpppp ψψ∫So the average momentum is
We want to prove that this equals the average of the operator P, which is
4434421h
II
dxiPP ψψψψ ′−== ∫ ∗)(
To prove that I=II, note that dxxpipx
e )(21 ψπ
ψ ∫−
= h
hWhich is just Fourier transform (up to the factor). Now, remember that (up to constants which we won’t bother with here), the product of the Fourier transform with p is the transform of the derivative; this reduces in I to the transform of the derivative. Lastly, remember that the Fourier transform is unitary, so the inner product of the transforms equals the inner product of the respective functions. This immediately reduces I to II.
sψ ′ h
ψpp
In very much the same way, the Hamiltonian H is the energy operator, and is the system’s average energy.ψψ |HH =
• Example: Gaussian wave packet ( ) 412
2)()( 220
∆=
∆+−+
πψ
axaxik ee
A straightforward calculation yields
• Average position =
• Average momentum =
• Average energy =
∫ −== ∗ adxxX ψψψψ
( )∫ =′−= ∗ hh 0 kdxiP ψψψψ
∫ =′′−= ∗ dxm
H ψψψψ2
2h
2
2220
42 ∆+
mmk hh
• Question: what happens to the average energy when , and why? It looks odd that the mass m is in the denominator –can you explain it?
0→∆
(for a free particle)
Schrödinger’s Equation
4444 34444 21
hhh
)22(
2
22
particle free aFor
2)()()(
mPH
xmt
dtditHt
dtdi
=
∂∂
−=⇒=ψψψψ
• Tells us how a state changes over time, hence is very important. Where does it come from? There are various heuristics to justify the equation; for example, suppose a free electron is moving through a crystal, in which the atoms are close to each other:
δ1C 2C
. ), 20 CC
0 1 2
Let be the amplitude to be at atom at time t. Now look again at the equations for the ammonia molecule. Here, we have three states (if we assume that, as a very short time passes – remember, we’re looking at the time derivative – the change in depends only on it and on So we can guess the following equation (up to constants):
iC i
1C
0C
20101 ACACCE
dtdCi −−=h
Now, use the well-known approximation for the second derivative at x:
2)()(2)()(
δδδ ++−−
≈′′ xfxfxfxf
And thus the right-hand side equals, up to constants, the second derivative of the wave function by x.
• The simplest (and very common) case for a non-free particle has the Hamiltonian
)(2 2
22
xVdxd
m+−
h
Where V is some potential.
• It is impossible to prove Schrödinger’s equation. It does not follow from the axioms.
ψψψψ HitHtdtdi == &hh ,)()(
Solving Schrödinger’s Equation
• This is a differential equation, which resembles equations in classical physics.
• The goal is to find the propagator which satisfies
• As often happens, a good direction to proceed is to first find the (normalized) eigenkets and eigenvalues of the Hermitian operator H.
• Since H is the energy operator, we will often denote the eigenkets/eigenvalues with the letter E:
• Suppose we have these eigenkets/eigenvalues. Since they are orthonormal and span the space, then for every state
)(tU.)0()()( ψψ tUt =
EEEH =
)()()( taEtEEt EEE∑∑ ≡= ψψ
• Plugging this last form into the equation yields
• So,
• And
h&h iEtEEEE eatatEatai −=⇒= )0()()()(hiEt
EE eatatE −== )0()()(ψ
hh
hh
h
iEt
E
iEt
E
iEt
EE
iEtE
iEtE
EE
eEEtUeEE
eEEEeE
EeaEtat
−−
−−
−
∑∑
∑∑∑∑
=⇒
==
===
)()0(
)0()0(
)0()()(
ψ
ψψ
ψ
• So we have a closed-form expression for the propagator. In the infinite dimensional case, the sum is replaced by an integral.
• The so-called normal modes, are also called stationary modes, because they change only by a phase factor, hence have the same probabilities to be in a certain state regardless of the time. They correspond to systems which start off in an eigenket of H (definite energy state).
,)( hiEteEtE −=
• Another expression for the propagator is given by
But it is not always manageable. Note that it implies that is a unitary operator.
hiHtetU −=)()(tU
• Schrödinger’s equation on the line is usually solved in the X-basis.
• Note that is an operator. By)(tU )0()()( ψψ tUt =
xdxxtxUtx ′′′= ∫∞
∞−32143421
conditions initialpropagator
)0,()0,;,(),( ψψwe mean that
where hiHt
E
exEExxtxU −′=′ ∑)0,;,(
Or, in the infinite dimensional case,
∫ −′=′E
iEt dEexEExxtxU h)0,;,(
• How does one derive such equations? It’s straightforward
from multiply by on : )( hiEt
E
eEEtU −∑=
the left and by on the right.
x
x′
xdxxtxUtx ′′′= ∫∞
∞−32143421
conditions initialpropagator
)0,()0,;,(),( ψψ
• The propagator as the evolution of simple states: let’s look at the equation
And take to be a state with definite position, that is, a delta function at a. Then
)0,(x′ψ
)0,;,()()0,;,(),( atxUxdaxxtxUtx =′−′′= ∫∞
∞−
δψ
taking yields bx =
=)0,;,( atbU The amplitude to be in b, at time t, when at 0=t the state is a delta function at a.
• And in general, is the factor weighing the contribution of the amplitude to be in at time 0 to the amplitude to be in x at time t.
)0,;,( xtxU ′x′
Solving Schrödinger’s Equation for a Free Particle
• If there’s no potential, the equation is relatively simple:
ψψψm
PHi2
2
==&h
• Is it trivial? Not at all; even for relatively simple initial states, the development of the wave function over time may be surprisingly complicated.
• To solve the equation, first find the eigenkets of H:
• The kets , which are eigenstates of P, suggest themselves. Remember that
EEEm
PEH ==2
2
,p
pppPpipx
e == ,21 h
hπ
mEppEpm
ppm
P 222
22
±=⇒==
and so
Hence for every E there are two corresponding eigenstates:
mEpEmEpE 2,,2, −==−==+
• Physically, this means that a particle with energy E can be moving to the left or right. But in QM, the following state –which has no meaning in classical mechanics – also exists:
mEpmEpE 22 −=+== βα
Note that this is a single particle which can be caught moving either left or right!
• The next step is to compute the propagator. In the position eigenkets it equals
dpeee mtipxipipxxtxU
hhh
h
22
1)0,;,(
2−′−∞
∞−
=′ ∫π
∫ −′=′E
iEt dEexEExxtxU h)0,;,(
The integral evaluates totxxime
itm h
h
22)(21
2′−
π
and so for an initial state the state at time t is),0,(x′ψ
xdxtxxim
itmtx e ′′
∞
∞−
′−
= ∫ )0,(
2)(2
),(22
1
ψπ
ψh
h
• What happens if We get the integral of a function multiplied by things like where A is very large:
?0→t),cos( 2Ax
Plot of multiplied by If we integrate, only the value at 0 remains.
).cos( 2Ax2xe−
Interlude: The Free Particle Propagator in the Momentum (Fourier) Domain
• For a free particle, Schrödinger’s equation is
2
2
2
22
22
dxd
mi
dtd
dxd
mdtdi ψψψψ hh
h =⇒−=
• Switching to the Fourier domain yields
),(2
),( 2 tuum
idt
tudΨ=
Ψ h
• Which is solved by
)0,(),( 22
uttu mihtu
e Ψ∝Ψ
• This observation highlights the “convolution nature” of the free particle propagator, and the manner in which it blurs spatially, and sharpens in the frequency (momentum) domain, as time goes on. The convolution kernels are like those in the previous slide.
Interlude: The Density Matrix• Expresses the expectation as a linear function of the operator being measured.
• Allows to include “usual” uncertainty about the state.
• Closely related to Boltzmann distribution and quantum statistical mechanics.
[ ]. )(,1)(
y thatimmediatel followsit equation,er Schroeding theFrom
.propertiesrelevant physically thecapturesit hence factor, phasearbitrary an havenot does )( :Note
.Hermitian) is ( and , Also,
.1)()())(Tr( and operator, positive a is )( :Note
.matrixdensity theis )()()( where
),)((Tr)()()()(
then , and operator,an is
If .)()( where,)( is state in the (t) finding ofy probabilit The basis. lorthonormaan be Let
2
,
2
tHi
t
t
tctctt
ttt
AttctcAtAtA
uAuAA
utcttcuu
nnn
pnpn
npt
nn
npp
nn
nnn
n
ρρ
ρ
ρρρρρ
ρρ
ψψρ
ρψψ
ψψ
h& =
==
==
=
===
=
=
+
∗
∗
∑
∑∑
∑
• Reminder – what is the probability to obtain a value when measuring A (at time t)? na
( ).Tr)( )( is obtain y toprobabilit that the
immediate isIt . eigenvalue with rseigenvecto of subspace on the projection thebe Let :Answer
nn
n
n
n
PtPta
aP
ρψψ =
• Incorporating “classical” uncertainty: suppose the starting conditions of a physical system are not exactly known, and that it has probability to be in state Then, the density matrix can be naturally extended by defining
And linearity immediately proves that the expected value of an operator A is
kp ∑ =≥k
kkk pp ).1 ,0( ψ
kkk
kp ψψρ ∑=
).Tr( Aρ
• Some further properties:
• is Hermitian.
•
•
• The meaning of is the probability to obtain in a measurement. It is termed “population”, while are termed “coherence”.
• If we’re working in the basis of eigenstatesof the Hamiltonian H, then from the equation for the development of over time, it follows immediately that
ρ.1)Tr( =ρ
).state" pure" a (i.e. identical are s' theof
all ifonly holdingequality with ,1
)Tr( basis, lorthonormaan is since and
,
,
,
2
,
,,
2
2,
22
ψ
ψψψψψψ
ψψψψρρ
ψψψψψψρ
∑∑∑
∑∑
∑∑
=
≤=
===
=
=
lklk
lklklk
lkkllklk
nlknlkn
lklknn
n
n
lklk
lklkk
kkk
pp
pppp
uuppuuu
ppp
nnρnu
npρ
npρ
( ) ).0(exp)( npnpnp tEEit ρρ
−=h
• Relation to quantum statistical mechanics: in thermodynamic equilibrium, it turns out that
, where k is Boltzmann’s
constant and T the temperature. Z is chosen so to normalize the trace to 1 (it is the partition function). If is the basis of H’s eigenstates, then
−=
kTH
Z exp1ρ
nE
on).distributi s'(Boltzmann 0 and
, exp1 exp1
=⇒≠
−=
−=
np
nnnnn
pnkTE
Zu
kTHu
Zρ
ρ
Path Integrals: Feynman’s Formulation of QM
∑∝′ h)]([)0,;,( txiSextxUWhere the summation ranges over all paths connecting to and S is the classical action (the integral over of the kinetic minus potential energy).
• What does this mean?
• The classical path is the one with the smallest action. In Feynman’s formulation, the propagator is the sum of allpaths, each weighted by the exponent of times the action over What happens for objects with a large mass, for example, is that due to the in the denominator, the contributions around any path but the classical one are in very different phases and therefore cancel out.
• This formulation is equivalent to the one we had shown, but highlights the connection to classical physics. Also, in some cases (but not usually) it is easier to use it to calculate the propagator.
)0,( x ′),,( tx )( tx
i.h
h
• Proof of the path integral formulation: look at the propagator with a fixed potential V. Proceeding much like as in the free particle case, this time is an eigenstate with energy so the propagator is
p,V22 +mp
txximtiit
m
dptmpi
ppe
dptmpi
pp
dptpiEpptU
ee
e
e
e
ti
hh
h
h
h
h
h
22)(V2
22
V22
)( )(
21
V
′−−
=−
=+−
=−=
∞
∞−
−
∞
∞−
∞
∞−
∫
∫
∫
π
equals this,and between interval in the speedfor stands )( and
,potential) fixed a assumingfor ion justificat theis (this small, very is timeelapsed theif Now, .propagator
particle free for the one eexactly th is integral thesince
xxxv
t
′
∆
( ) ( )
following. see -factor scale a toup *
s.r'Schrodinge toequivalent isn formulatio thisout that It turns .*proof thecompleteswhich
, )]([)(V2
)( to
add exponents in the powers theso .successionin events of amplitude about the postulate
thefollowing above, theofproduct get the we and zero, togo paths-sub limit the In the paths.
-sub itsover go toamplitudes ofproduct theis amplitude thepath,every For slits). more and
more with screens more and more of(think pathsdifferent all of amplitudes theof sum
theas ),( to)0,( from getting of amplitude thecompute weI,Part in postulates theUsing
2))((2)()V()(2
0
2
21
txSidtxxmvi
txx
tximvtxiti
m
t
ee
hh
hh
h
=
−
′
∆∆−
∆
∫
π
• Derivation of the free particle propagator via path integrals:
4444 34444 2144 344 21
h
&h
434214434421
&&h
&h
&h
h
96 slide toCompare
2)(
0 to0 from Propagator
0)(0)0(
)(2
0)(0)0(
)(2)(2
to from paths the to0at ending and starting loops the
maps )()(ation transformthefact that theusing
0)(0)0(
)(2
)()0(
)(2
)()0(
)]([
)0,(
)0,(
)0,(
)0,()0,(
)0,()0,;,(),(
2
0
2
0
22
0
2
0
2
xdxee
xdxe
xdxe
xdxexdxe
xdxxTxUTx
Txxmi
Txx
dttxmi
Txx
dtT
xxtxT
xxtxmi
xx
tTxx
xtxtx
Txx
dtT
xxtxmi
xTxxx
dttxmi
xTxxx
txSi
T
T
T
T
′′
∫
=′′
∫
′′
∫
=′′
∫=′′
=′′′=
∫∑
∫ ∑
∫ ∑
∫ ∑∫ ∑
∫
∞
∞−
′−
==
∞
∞−==
′−
+
′−
+
′
′−+′+→
∞
∞−==
′−
+
∞
∞−=′=
∞
∞−=′=
∞
∞−
ψ
ψ
ψ
ψψ
ψψ
)()0( since 0, to
integrates
Txx =
( )T
xx
t
2
tointegrateshence , of
tindependen
′−
Path Integral Normalization Factor
• The question arises as to how to normalize the summation over all paths.
• Approximate each path by a piecewise linear function:
t
x
ax =0
bxN =
1+ixix
at it 1+it bt
. .
.
. .
...
( ) .2 , ......1lim
as integralpath the writetocustomary isIt
.2
So, .2)(
22
equalmust it ,propagator particle free with the thiscomapring
2
toevaluates integralGaussian The
. with replacingby y immediatel followswhich
...)(2
exp...lim
...2
exp...lim
factor) gnormalizin a is ( then is propagator particle free The
. so , Denote
21
121],[
0
2021
)(22
121
21
1
111
210
111
2
0
1
20
1
=
=
−
−
∑ −
≅
∑ ∫
≡−=−=
−∞
∞−
∞
∞−
∞
∞−→
−−
−−
−
−
−
∞
∞−
∞
∞− =−
∞
∞−→
∞
∞−−
∞
∞− =
∞
∞−→
+
∫ ∫ ∫
∫ ∫∫
∫ ∫∫−
miB
Bdx
Bdx
Bdxe
B
imA
xxT
mi
Tim
ei
mAN
xxx
dxdxxxmiA
dxdxdtxmiA
A
TttNtt
NabSi
NN
xxT
miNN
jj
N
N
jjj
N
N
j
x
x
abii
cl
N
j
j
e
επ
εππ
πε
ε
ε
εε
ε
ε
ε
h
hh
h
h
&
h
&h
h
h
Equivalence to Schrödinger’s Equation
.in order second work toshould we,in order first to workingre we'if that means This
.2region on the econcentrat ll we'so rapidly,very
oscillatesexponent the,2
When . where
)0,(0,2
exp2
exp2
be should ),( So, over. integrate omidpoint t no is theresince factor, gnormalizin external no is e that therNote .
inorder first the tostandard are ionsapproximat thewhere
0,22
)(exp2
ely approximat is )0,;,( propagator integralpath the, interval timemalinfinitesian For
21
2
221
221
ηε
πεη
πεηη
ηηψηεεη
επ
εψε
εεεπ
εε
≤
≥−′=
+
+−
′+
−′−
′
∫∞
∞−
m
imxx
dxxViimi
m
x
xxVxxmii
m
xxU
h
h
hhh
hh
( )
( )
.predictionequation sr'Schrodinge with agrees indeedwhich
)0,()0,(2
)0,(),(
get weEventually .in integralGaussian a is This
dropped). are termsetc. (again, 2
)0,(0,)0,(2
exp2
is
integral The ). )( re they'since ,order or terms
neglect (we 0,10,2
exp
2)0,()0,(
in order first toExpanding
)0,(0,2
exp2
exp
is ),(for compute tointegral The
2
22
2
22
221
23
2
22
2
xxVxm
ixx
dxx
xxViximi
m
O
xVixVixx
xx
dxxViim
x
ψεψεψ
η
ηεηψηψη
ψεψεη
επ
εηε
εηε
ψηψηψηψ
ε
ηηψηεεη
εψ
+
∂∂−−
=−
∂∂
+∂∂
+−
−=
+−
∂∂
+∂∂
+=+
+
+−
∫
∫
∞
∞−
∞
∞−
h
h
hhh
hh
hh
Calculating the Path Integral
[ ]
factor. scale a toupfunction theknow wegiven afor since extracted, be stillcan function
waveabout then informatio ticprobabilis that theNote
on only Depends
)]([D)()()()(21exp
exp equals integral
theSo, out. factors exp Also, extremum.an is
)( since vanish,mslinear ter the, expandingWhen
)]([D)()(exp)0,;,(
So .0)()0( endpoints, at the agree )( and )( Since ).()( as )(path theWrite
path. classical thearound expandingby computed becan propagator the,
form theof potentialsFor .impossible sit' Usually,
0
0 0
22
0
0
2
T
T
tydttyteytcytymi
Si
SitxS
tytytxSixtxU
Tyytxtxtytxtx
xxexdcxbxaV
T
cl
cl
cl
cl
cl
cl
444444444 3444444444 21
&&h
h
h
h
&&
∫ ∫
∫
−−
×
+=′
==+
++++=
[ ]
[ ]
).propagator particle free thes(it' 0 aslimit the
know wesince possible ision Normalizat .)sin()(
1)(1
...),, g(forgettinor
,22
exp integralsGaussian theofproduct
infinite theis integral theconstants,other andJacobian about themoment for the forgetting and tion,representa this
toGoing .22
)()( that see
easy to isIt .sin as )(Represent ).0
)()0( where( )]([D)()(2
exp
equalsit oscillator harmonic for the example,For
?)]([D)()()()(21exp
compute toHow
21
21
122
2221
122
2222
21
1
222
2222
222
1
2
0
222
11
0
0 0
222
0
0 0
22
→
=
−=
−
=
−
−
−=−
≡
=
=
−
−−
−
−∞
=
−∞
=
−∞
=
∞
∞−
∞
=
∞
=
∞
=
∏∏
∏
∫
∑∫
∑∑
∫ ∫
∫ ∫
ωωω
πω
πωπ
ωππ
ωπ
ωπω
π
ω
TTTK
nTTK
nT
Tn
TTni
daTTnai
TTnadttyty
yaT
tnaty
Tyytydttytymi
tydttyteytcytymi
nn
n
nn
nn
T
nnn
nn
T
T
h
h
&
&h
&&h
Perturbation Theory and Feynman Diagrams
[ ][ ]
[ ]
equals integral theSo, .it denote me,at that tilocation the and time the:over integrate to variables twohave wesummands,
all toextended ision considerat When this. to from and to 0 from spropagator particle free ofproduct theleft with re We'1. to
goes which )()(2
exp with termsthe
dropcan weand well,as 0 zero, togoes successive any twobetween time thesince then , variationbounded assume
weif limit, in the that note ,)( with summand thecompute To... )(...)(
)(...)(2
exp...
:etc.) (ignoringion approximat discrete at thelook summand, second thecalculate To .propagator particle
free just the is summandfirst thepaths, allover gIntegratin
...)(2
exp2
exp
as integrand the writeoexponent t theexpand
][D)]([)(2
exp )0,;,(
21
21
1
110
21
201
00
2
0
2
,
0, 0
2
y
Ttt
xxxxmixxx
xVdxdxxVxV
xxxxmi
dtxVdtxmiidtxmi
txdttxVtxmixTxU
jj
jjjj
iii
j
NN
NN
TTT
Tx
x
T
τ
ε
−+−
→−
++
×
−++−
+
−
−=′
+−
+
−
∞
∞−
∞
∞−−
′
∫ ∫
∫∫∫
∫ ∫
h
h
&hh
&h
&h
τττ dydxyUyVyTxUxTxUT
∫ ∫∞
∞−
′≡′0
001 )0,;,()(),;,()0,;,(
.propagator particle free theis where 0U
( ) ( )
( )
diagram:FeynmanUU
Tx
yx
following by the drepresente is )on"perturbatiorder (second)first " (a ) ( etc. integral), quadruple aby ed(represent potential by the
twicescattered particle a toscorrespond summand third themanner, same In the ., freely to
propagates then and potential, by the )(scatteredupon acted then is ,, to0, fromfreely
propagates particle the:is meaning physical The
21
τ′
time
)0,(x′
),( Tx
),( τy
V
)0,(x′
),( 22 τy
),( 11 τy
),( Tx
V
V
equalswhich
]),([]),([2
exp!2
1
equalson perturbatiorder second The
000
22
′′′
− ∫∫∫
TTT
sdssxVdsssxVdtxmii&
hh
τττ
τττ
τττττττ
τττ
ττττ
dydxyUyVyTxUi
xTxUxTxU
dydxyUyVyTxUxTxU
TU
nddyddyxyU
yVyyUyVyTxUi
T
k
T
k
T
)0,;,()(),;,(
)0,;,()0,;,( :equation integral theexponent theofexpansion thefrom deduce easy to isit and
)0,;,()(),;,()0,;,(
GeneralIn orders.
higher for same theand ,by above integral quadruple in the replace toallows which ,0)?,;(?,
that assume tocustomary isIt .set over the carried is integral thebecause vanished,factor) !1 thegeneral,in (and,factor 2!1 that theNote
)0,;,(
)(),;,()(),;,(
00
0
10
0
212
1221
2211110
1112200 0
2220
2 2
′
−′=′
′=′
=⇒<≤
′
−
∫ ∫
∫ ∫
∫ ∫ ∫ ∫
∞
∞−
−
∞
∞−
∞
∞−
∞
∞−
h
h
A Path Integral Formulation for the Density Matrix
( )
.)0( ,)(such that paths"" allover isn integratio thewhere
)())(()(2
1exp),(
:matrixdensity for the expression integralpath aevaluate toused becan propagator for the integral
path a evaluatingfor procedure that thesees One
exp),(
)(exp),;,(
matrixdensity theand propagatormechanics quantumebetween th similarity theNoting
12
0
212
1212
12121122
xxxx
uDxduuxVuxmxx
xEExExx
xEExttEitxtxU
kkk
k
kkk
k
==
+−=
−=
−−
=
∫ ∫
∑∑
h
&h
h
h
β
ρ
βρ
β
If the initial state is a Gaussian wave function
( ) 412
2 220
0,(∆
=′∆−′
πψ
xxip eexh
then a direct integration yields
( )[ ]
−
∆+∆−−
∆+∆= m
tpxipmtimtpx
mti eetx 2)1(2
)(
2100
22
20
21
,( hhhπψ
with the probability density
( )[ ]
1),Pr(22222
20((
2122222∆+∆
−−
∆+∆= mt
mtpx
emt
tx h
hπ
• The mean position at time t is (like the classical case).
• The uncertainty in position grows from at time 0
mtp0
2∆21
42
22
12
∆
+∆
mth
to at time t.
Evolution of a Gaussian Packet
tmt
mt largefor
21
2
21
42
22
∆≈
∆
+∆ hh
Position uncertainty =
This can be viewed as t times the initial uncertainty in the velocity.
• Note also that the uncertainty in position grows very slowly – especially for a heavy particle.
• Note that there is a non-zero probability for the particle to “go backwards”, that is, be captured at the opposite direction in which it is “generally moving”.
• The Gaussian packet has an “average” velocity, associated with the motion of its center. This is the so-called “group velocity”.
• Let a (stationary) packet be defined by
where is the phase of Assume that has a strong peak at then the integral can be approximated by
This wave packet has a peak at the location in which the phases of the exponentials change the slowest, since then they interfere the least. This occurs for the x at which the derivative
of the phase by k is zero, that is,
∫+≡∫=
∞
∞−
∞
∞−dkkxkiekgdkikxekgx ))(()()()0,( αψ
)(kα ).(kg,0k
)(kg
∫+′−+∆+
∆−
kk
kkdkkxkkkkiekg
0
0
))()()(()( 000 αα
).( 00 kx α′−=• If the packet is moving, we have
So the packet’s center is at
and its “group velocity” is
∫−+=
∞
∞−dktkkxkiekgtx ))()(()(),( ωαψ
tkktx )()( )( 000 ωα ′+′−=
).( 0kω′
Operator Power Series for the Propagator
• We saw that the propagator for a free particle can be expressed as
n
nn
nn
nn
n
hiHt
dxxd
miht
ntx
dxd
miht
ne 2
2
02
2
0
0,(2!
1,(2!
1 ψψ
=⇒
= ∑∑
∞
=
∞
=
−
• Sometimes, this can be used to compute . ,( txψ
An apparent counterexample
≤
=otherwise 0
2 sin0,(
LxLx
xπ
ψ
It appears as if the spread of the wave function is not growing over time, because all the derivative are nonzero only at the interval How is this possible? . 2Lx ≤
The operator power series doesn’t converge (at the boundaries).
n
nn
n dxxd
miht
ntx 2
2
0
0,(2!
1,(ψ
ψ
= ∑
∞
=
Look more carefully at
You can try and sum a few of the first terms, and obtain a reasonable approximation, if the higher derivative don’t diverge and if t is small enough.
• But one has to be cautious: look at a particle moving under the influence of a force f. The Hamiltonian is then (assuming we fix all constants to obtain a simple form):
ψψ4434421
nfx
dxdHfx
dxdH n
+=⇒+= 2
2
2
2
The calculation of this operator is tricky because the operators of second derivative and multiplication by x do not commute!
Solving Schrödinger’s Equation for a Particle in a Box
• For a particle under the influence of a potential the states of definite energy satisfy
),(xV
ψψψψ 2
2 ))((2)(2 h
xVEmExVm
P −−=′′⇒=
+
• Next, we solve it for a particle bounded by an infinite potential (“walls”) on both sides:
∞
≤=otherwise
2 0)(
LxxV
2L− 2L
I IIIII
∞→≈′′⇒−
−=′′ 002 ,))((2 VVxVEm ψψψψh
• In regions I and III,
• So the solutions are
xVxV BeAe 00 −+=ψ
• In region III, for example, the part is not physically allowed, as it diverges to infinity. Only the part is allowed, but as the solution is zero. Same holds for region I.
• So the solution can be non-zero only in region II. The equation there is
xVAe 0
xVBe 0−
,0 ∞→V
)sin()cos(
2 22
kxBkxA
kmE
+=
⇒−=′′≡−=′′
ψ
ψψψψh
• Boundary conditions: .0)2()2( =−= LL ψψ
• It immediately follows that the (normalized) solutions are
=
=
=
...6,4,2 sin2
...5,3,1 cos2
21
21
nL
xnL
nL
xnL
n
π
π
ψ
• There is no solution corresponding to since it equals 0.
• The energy (E) corresponding to is
• There is no state with definite energy 0!
,0=n
nψ . 2 2
222
mLnEn
πh=
First (green) and second (red) states: left=amplitude, right=probability.
• A state with energy 0 would violate the uncertainty principle:
• The particle is bound between and so the position uncertainty is bounded by
• We must have else the particle will drift away to infinity, but that’s impossible, as it is bound in the box. But
• What about the average energy? We know that
2L− ,2L.2L
( ) 2
22
2 ,
2 LP
LPPXLX hhh
≥∆⇒≥∆⇒≥∆∆≤∆
,0=P
( ) 2
2222
LPPPP h
≥=−=∆
2
2222
22 mLmP
HE h≥==
While the lowest energy level is obtained when and it equals (the ground state energy). The uncertainty principle can sometimes be applied to find a rough estimate of the ground state energy.
1=n222 2mLπh
Example Question (Shankar, 5.2.1):
• A particle is in the ground state in a box of length L. Suddenly the box expands to twice its size. What is the probability to find the particle in the ground state of the new box?
=
=
box big 2
cos1
box small cos2
21
2
21
1
Lx
Lg
Lx
Lg
π
π• Answer: the two ground state are
And the probability to find in is1g 2g
72.038 222
221 ≈
=
∫
−
∗
π
L
L
dxgg
SHOW MAPLE FILE HERE
( )raster) is ordering(
2 is state the0at amplitude,
y probabilitofin timet Developmen
21 ψψ +=t
Quantization of Energy
• We saw that the energy levels of the particle in a box are quantized.
• A particle which is bound by a potential from escaping to infinity is said to be in a bound state.
Formally,
• It turns out that the energy levels of a bound particle are always quantized.
• Why? Look again at Schrödinger’s Equation:
• Assume that is everywhere bounded. It follows that both are continuous. Assume also that there’s a bounded region (a “well”) in which the particle’s energy is larger than the (fixed) potential, but that outside the well the potential is greater than the energy:
ψψ 2))((2
h
xVEm −−=′′
.0)( lim =±∞→
xx
ψ
)(xV and ψψ ′
EV <EV >EV >
ψψ 2))((2
h
xVEm −−=′′
0, >−=′′ IIII cc ψψ
Solution is sum of sines and cosines
0, >=′′ IIIIII cc ψψ0, >=′′ II cc ψψ
Solution is sum of rising and decreasing exponentials
Solution is sum of rising and decreasing exponentials
III
III
But one of them has to be zero because the rising exponentials go to infinity as .∞→x
• So we’re left with one parameter (degree of freedom) for each of regions I and III, and two d.o.f for II. We have four constraints – the continuity of the function and its derivatives at each of the borders (just like splines). It appears that we can solve for every E; but note that there is a redundancy for multiplying by a scalar, hence we really have 3 d.o.f and four constraints – so there will be a solution only for “special” cases. This is why the energy levels for a bound state are quantized. This argument can be extended to any finite potential. If the potential is infinite (particle in a box) we have to enforce only the continuity of the function, not the derivative.
Finite Potential Well• Like particle in a box, but the walls are lowered to a finite height: we’ll assume the energy to be which is smaller than
ax −=
0V
ax =
I IIIII
,0E.0V
• The solution has the form
20
12h
mEk =
200
2)(2
h
EVmk −= 2
002
)(2h
EVmk −=
+−
−
xk
xikxik
xk
DeIIICeBeIIAeI
2
11
2
: region : region
: region
• As noted, enforcing continuity of the function and its derivative yields four equations in A,B,C,D, but these can be solved only up to a scale factor, hence the system is overdetermined and can be solved only for special values of energy. The resulting equations are transcendental and don’t have a closed-form solution.
The Probability Current
• For particles, or classical flux, the current is the amount of something crossing, say, a unit area during unit time.
• In QM, it makes sense to talk about the probability current. If many particles are present, it will be equivalent to the classical definition.
• The probability current is defined for a one-dimensional particle as the amount of probability crossing over: how did the probability to find the particle to the right of change between time and
0x 0t?0 tt ∆+
0x 0x
0t tt ∆+0
• The probability change is
[ ]∫ ∫
∫∫∞ ∞
∞∞
∂∂
∆≈−∆+
=−∆+
0 0
00
),Pr(),Pr(),Pr(
),Pr(),Pr(
000
00
x x
xx
dxt
txtdxtxttx
dxtxdxttx
• Assume that there’s a function J such that
xJ
t ∂∂
−=∂∂ Pr
Then the probability change is and J is therefore the probability current.
• A straightforward application of Schrödinger’s Equation yields
,tJ∆
−=
∗∗
dxd
dxd
miJ ψψψψ
2h
• Initial state: Gaussian packet centered at 0, moving to the right at unit speed.
State at 2=t
State at 3=t
Probability current at 3=t
• Why is the current negative at certain points?• Because the packet is moving to the right, but is also expanding! So there’s an area in which the probability to the left of x is increasing.
• For the right moving wave packet
( )Etpxitm
ppxi
AeAe−
−
= hh 2
2
• The probability current is
• Intuitively, this is true because is the speed of the packet, and the packet is uniform, so the current depends only on the speed.
• It can also be immediately verified that current for two such wave packets moving at the opposite directions is
mpAJ 2=
mp
( )mpBAJ 22 −=
Where A,B are the coefficients of the right and left moving packets.
Introduction to Scattering
When a classical particle hits a potential barrier, it either passes it (if its energy is higher than the barrier’s energy), or is reflected (if its energy is lower). The situation in QM is rather different.
• Start by considering a step barrier
0=x
0VIψ 0=t
0>>tRψ
Tψ
An incoming packet hits the barrier and disintegrates into a left and right going packets.
incoming
transmitted
reflected
Straightforward Solution• Find the basis of the normalized eigenstates of the step potential Hamiltonian.
• Project the initial wave function on that basis.
• Propagate each of the basis functions in time (with the corresponding coefficient).
• As the time tends to infinity, identify the right and left moving components.
It turns out that, under certain assumptions, a much simpler method can
be used – which is rooted in classical optics, and in the wave-particle duality.
• The problem can be solved in principle for any packet, but it is impossible to obtain a closed-form expression for the solution.
• Often, one is interested in cases in which the incoming wave is very wide, and therefore has a sharp peak in momentum space.
• In that case, it is possible to derive the ratio of the reflected and transmitted parts with relative ease by essentially reducing it to a stationary problem, dependent only on the energy.
• The main trick is to look at all the components –incoming, reflected, and transmitted – as coexisting, and to measure the ratio of their probability currents. The justification is: if the incoming wave is very wide, it is impossible to tell when it hits the barrier.
• We’ll assume that is the wave number of the incoming packet, hence its energy and momentum are
0k
0
200
momentum2 energy
kmkE
h
h
=== And its stationary state
behaves like .0xike
• The reflected part has the same wave number (and frequency, of course) as the incoming wave. The transmitted part has a wave number which must satisfy 1k
202
012100
22h
hmVkkmkVE −=⇒=−
• So, we have the following situation:
0=x
xikAe 0
xikBe 0−
xikCe 1
• We can find A,B,C up to a scale factor because the function and its derivatives are continuous at 0:
• The currents associated with incoming, reflected, and transmitted are . , , 1
20
20
2 mkCmkBmkA hhh
TAB
R −== 12
2So, the probabilities are: 0
002
2
02
12
EVE
AC
kAkC
T −==
for reflectedfor transmitted
10
0
10
10 2 ,kk
kAC
kkkk
AB
+=
+−
=
Tunneling
• What happens if ?00 EV >
0=x
0V “the forbidden zone”ψ
energy = 0E
• The solution to Schrödinger’s Equation in the forbidden zone is
200
22
200
)(2 ,
0)(2
h
hEVmk
VEm
xke −=∝
⇒=−
+′′
−ψ
ψψ
III
,2 201 hmEk =If
the solutions are +
−
−
xk
xikxik
CeIIBeAeI
2
11
: region : region
where
+=⇒
−=
=⇒−+
=
22
21
21
2
21
1
2
21
21
42
1
kkk
AC
kikik
AC
AB
kikkik
AB
• So, a particle can break into the forbidden zone (although it cannot go too far into it). If the potential drops again to zero, the particle can escape through the barrier and then it is free (like some particles escaping from a nucleus).
• It is immediate to see that the probability current going into the right half is zero; yet –some particles manage to escape into it.
• How can this happen? It seems to violate energy conservation!
• From the uncertainty principle, conservation of energy is violated only if we can measure this violation. The position uncertainty for the particle in the forbidden zone is very small (a decreasing exponential), hence the uncertainty in momentum is large enough – and it doesn’t allow the measuring process to determine that energy conservation was violated.
A 1D particle hitting a step with a potential higher than the particle’s energy. Note short penetration
into the forbidden zone.
Tunneling: a 2D particle hitting a barrier. The probability amplitude is depicted.
ax −=
0V
Incoming (energy =
ax =
)0E
I IIIII
00 EV >
• In I,
• In II,
• In III,
• As before, enforcing the function and its derivative to be continuous at yields four equations, which can be solved up to a scale factor, and then
200 2 ,00 hmEkBeAe xikxik =+= −ψ
2001 )(2 ,11 hEVmkDeCe xkxk −=+= −ψ
xikEe 0=ψ
ax ±=
)2(sinh)2(cosh1
122
12
2
akakAET
α+==
)(42
000
00
EVEEV−
−=α
Potential Barrier
Scattering off a potential barrier. Horizontal axis is vertical axis is T. Scales are chosen so that
(following Eisberg & Resnick, p. 202). Lower is detail of upper; note that T is equal to 1 at certain energies (strange, because we can have moretransmission at a lower energy!).
,0VE
220 92 h=amV
assume(a,positive); assume(k1,real); assume(k2,real);psi1:=A*exp(I*k1*x)+B*exp(-I*k1*x); # (-infinity,-a/2) k1=sqrt(2*E*m/h^2)psi2:=C*exp(k2*x)+DD*exp(-k2*x); # (-a/2,a/2) k2=sqrt(2*(V0-E)*m/h^2)psi3:=F*exp(I*k1*x); # (a/2,infinity)
# now force continuity and first derivative continuity
eq1:=subs(x=-a/2,psi1-psi2);eq2:=subs(x=-a/2,diff(psi1,x)-diff(psi2,x));eq3:=subs(x=a/2,psi2-psi3);eq4:=subs(x=a/2,diff(psi2,x)-diff(psi3,x));sol:=solve(eq1,eq2,eq3,eq4,F,C,B,DD);Fsol:=subs(sol,F)/A; numF:=numer(Fsol);help1:=expand(simplify(numF*conjugate(numF)));denomF:=denom(Fsol);help2:=expand(simplify(denomF*conjugate(denomF)));
T:=help1/help2;
T1:=subs(k1=sqrt(2*m*E/h^2),k2=sqrt(2*m*(V0-E)/h^2),T);# Following Eisberg & Resnick p. 202T2:=simplify(subs(h=sqrt(2*m*V0*a^2/9),T1));T3:=simplify(subs(E=V0*r,T2));
plot1:=plot(T3,r=0..1,thickness=3,axes=boxed,axesfont=[TIMES,BOLD,16],numpoints=1000):
Xmaple code for potential barrier, tunneling
EV >0
#################################### end V0>E, start E>V0
assume(a,positive); assume(k1,real); assume(k2,real);psi1:=A*exp(I*k1*x)+B*exp(-I*k1*x); # (-infinity,-a/2) k1=sqrt(2*E*m/h^2)psi2:=C*exp(I*k2*x)+DD*exp(-I*k2*x); # (-a/2,a/2) k2=sqrt(2*(E-V0)*m/h^2)psi3:=F*exp(I*k1*x); # (a/2,infinity)
# now force continuity and first derivative continuity
eq1:=subs(x=-a/2,psi1-psi2);eq2:=subs(x=-a/2,diff(psi1,x)-diff(psi2,x));eq3:=subs(x=a/2,psi2-psi3);eq4:=subs(x=a/2,diff(psi2,x)-diff(psi3,x));sol:=solve(eq1,eq2,eq3,eq4,F,C,B,DD);Fsol:=subs(sol,F)/A; numF:=numer(Fsol);denomF:=denom(Fsol);
T:=simplify(abs(numF/denomF)^2);
T1:=subs(k1=sqrt(2*E*m/h^2),k2=sqrt(2*(E-V0)*m/h^2),T);#following Eisenberg&Resnick p. 202T2:=simplify(subs(h=sqrt(2*m*V0*a^2/9),T1)); T3:=simplify(subs(E=V0*r,T2));
plot2:=plot(T3,r=1..10,thickness=3,axes=boxed,axesfont=[TIMES,BOLD,16],numpoints=1000):
with(plots):display(plot1,plot2);plot(T3,r=2..10,thickness=3,axes=boxed,axesfont=[TIMES,BOLD,16],numpoints=200);
EV <0
A Delta Function Potential
• Assume the same situation as before, but instead of a step potential we have a “Delta function”potential – that is, a very narrow infinite wall.
0=x
xikAe 0
xikBe 0− xikCe 0
• Note that the wave number on the right of the barrier is the same as on the left, because there is no region in which is loses energy.
• As before, there are two equations: the function has to be continuous at 0 (hence A+B=C). The first derivative equation is a bit more tricky, because Schrödinger’s Equation is
Potential = )(xaδ
ψδψψψ 220
200 )(22)(2
hhh
xmamEVEm+−=
−−=′′
• What happens to at Since it is the integral of which contains a product of the Delta function, it jumps by when crossing the barrier from left to right. So, the equation for is
?0=xψ ′,ψ ′′
22 hmaC
ψ ′
22)(h
maCBACik =+−
• Solving this together with A+B=C yields
RTkam
amABR −=
+== 1 ,42
022
222
h
• As the particle is heavier, and its wave number smaller, it will tend more to behave classically (that is, reflect).
Arbitrary Barrier
• Express as the limit of rectangular barriers of infinitesimal width, and calculate the limit of the product of the corresponding transmission coefficients.
Some results on Degeneracy
• Recall that an eigenvalue is degenerate if there are two linearly independent states with eigenvalue
Theorem: there is no degeneracy in one dimensional bounded states.
Proof: assume
ω.ω
222
2
111
2
2 ,
2λψψψλψψψ =+′′−=+′′− V
mV
mhh
Multiply by and and subtract to get1ψ2ψ
( )
2.
1212
2
1
1
2121
21212121
.)log()log( So,
0.get to take.,00
ψψψψψψ
ψψ
ψψψψψψψψψψψψ
consteconst
constxconst
=⇒+=⇒′
=′
=∞→=′−′⇒=′′−′⇒=′′−′′
But a difference in scale factor means that the states represent the same physics.
Theorem: If parity is conserved (that is, the Hamiltonian commutes with the parity operator), then every non-degenerate state of definite energy has definite parity.
Proof: Let Q be the parity operator.
( ) ( )( ) ( )
parity. definite s there'so ,11but 2degeneracy-non
±=⇒=
=⇒===⇒=
δ
δ ψψψψψψψψ
i
i
eQ
eQQEEQHQQHEH
There can be degeneracy in two dimensional systems – such as a particle in a 2D box:
• The states of definite energy are products of the 1D states for x and y, and their energy is the sum of energies of the corresponding 1D states.
• So, if we know for example that the energy is
then the state is not uniquely determined,
but instead it can be any normalized linear combination of
,2
102
22
mLπh
=
=
Ly
Lx
L
Ly
Lx
Lππψ
ππψ
cos3cos2
or ,3coscos2
1,3
3,1
That is, the state has to be of the form
1 , 221,33,1 =++ βαψβψα
The Harmonic Oscillator
• Very important – models the behavior of a system around an equilibrium point, even for many particles (as follows by diagonalizing the Hamiltonian).
• The Hamiltonian is the same one as in classical mechanics, with the usual substitution :, XxPp →→
22
222 Xmm
PH ω+=
• Since X and P are Hermitian, it follows that Hcannot have negative eigenvalues.
• To find the definite energy states, we must solve
02
2 22
22
2
=
−+ ψωψ xmEm
dxd
h
• Define constants and a variable change
byxEmw
b ==
= , ,
21
ωε
h
h
• To obtain
• To solve it, look at the asymptotic behaviors:
• But is ruled out from physical considerations.
• At the other end
0)2( 2 =−+′′ ψεψ y
222
0y
meAyyy±
≈⇒≈−′′⇒∞→ ψψψ
2
2ymeAy
εεεψ
εψψ
2 ),()2sin()2cos(
020
2 BcyOcyAyByA
y
=++=+≈
⇒≈+′′⇒→
• Next, guess a solution
Where
2
2
)()(y
eyuy−
=ψ
0)12(2powers)(lower )(
powers)(higher )(0
=−+′−′′
+→
++→
∞→
→
uuyuyyu
cyAyum
y
y
ε
• Guess further
• And obtain
• This in itself would not be very bad, but it’s trivial to verify that behaves like when which implies that
∑∞
=
=0
)(n
nn yCyu
kkkk Ck
Ckk
kC 2)1)(2(
2122 ∞→+ →
++−+
=ε
)( yu 2ye,∞→y
,)( 2
2y
yey
∞→≈ψ but that is physically untenable.
• Fortunately, there’s a way out: look at the recursion
kk Ckk
kC)1)(2(
2122 ++
−+=+
ε
This has to stop somewhere in order to prevent the solution from diverging, so we must have
...2,1,0 ,2
12=
+= kkε
• So, energy quantization somehow came (again) into the picture. The solution for energy E is
22
21
41
22 )!(2)(
xm
nnE exmHn
mx h
hh
ωωπ
ωψ−
=
Where ωh
+=
21nE Hermite, not
Hamiltonian!
• The H’s are the Hermite polynomials. The first are
)21(2)(2)(1)( 2210 yyHyyHyH −−===
+−=
−−= 42
43
3 344112)(
3212)( yyyHyyyH
A few examples. The upright arrows mark the classical turning point for an oscillator with the given energy; note that, much like as with the potential barrier, the oscillator enters the classically forbidden zone with non-zero probability. Note also that as the energy increases, the behavior starts to resemble the classical behavior.
Interlude: Finding the Oscillator’s Ground State from the Uncertainty Principle
22
222 Xmm
PH ω+=
• Let’s search for the minimal energy state, that is, the normalize which minimizes :ψψ Hψ
ψψωψψψψ 222
21
21 XmPm
H +=
Recall that (just as with ordinary average and variance), and the same for P. Since we want a minimum, take and then
( ) 222 XXX +∆=ψψ
,0== PX
( ) ( )222
21
21 XmPm
H ∆+∆= ωψψ
( ) ( ) ( )( )2
22
222
44 XPXP
∆≥∆⇒≥∆∆
hhWe know that
( )( )22
2
2
21
8 Xm
XmH ∆+
∆≥ ωψψ h
And therefore
The minimum is obtained when
( )2
for which ,2
2 ωω
hHmhX ==∆
• We proved that the minimum is obtained for a Gaussian; since the Gaussian is centered at 0 and has no phase (momentum) shift, which after normalizing equals
,0=∆=∆ PX
h
h2
2 4
1
min )(xm
emxω
πωψ
−
=
• Due to the non-degeneracy of one dimensional bounded states, must indeed be the ground state (there’s only one state with the minimal energy equal to its eigenvalue). Thus, to quote Shankar, we have found the ground state by “systematically hunting in Hilbert space”.
minψ
• Example question: a potential barrier is added, so that the potential for What is the ground state energy?
• Answer: no even solutions are possible, since they are non-zero at So the ground state energy is
. is 0 ∞≥x
.0=x( ) .23 ωh
Analysis of the Oscillator with the “Raising” and “Lowering” Operators
• An important and particularly attractive technique due to Dirac for solving the oscillator in the energy basis.
• In order to solve
• Define the two operators
EEEXmm
PEH =
+=
22
222 ω
Pm
iXma
Pm
iXma
21
21
21
21
21
2
21
2
−
=
+
=
+
hh
hh
ωω
ωω
• It readily follows that their commutator is
• And they “nearly factor” H (“nearly” because X,P do not commute):
[ ] 1 , =−= +++ aaaaaa
( )21ˆ define so ,21 +==+= ++ aaHHaaH
ωω
hh
• Let’s try and solve
• The action of on the eigenvectors of is very simple, because
εεε =H aa and + H
[ ] [ ][ ]( )
( )))(1(ˆ Similarly,
).)(1(ˆ ˆ,ˆˆ ,ˆ
ˆ, ,ˆ,
εεεεεε
εεεεε
++
++
+=−=−
=−==−==
aaHaaHa
HaHaaHHaHaaHa
• So, operating with lowers and raises the eigenvaluesof But since H (and therefore are positive definite, the operation by a has to stop somewhere:
• Due to the non-degeneracy of one dimensional bounded states, every such “decreasing sequence” has to end with the same denote it
• Label the successive operations of on by
+aa,.ˆ H )ˆ H
( ) 021
00
00
ˆ021ˆ00
0
εεεεεε
=⇒=−⇒=⇒=∃ +
HHaaa
,0ε+a 0ε .n
( )( )
( ) ( ) ( )( )( )
( ) .212121But .
get omultiply t and ,1obtain toadjoints Take .1
such that constant a s there'1211ˆ and degenerate-non
is ˆ since and 21ˆˆ Now, .21ˆ have we1,by raises every and 0210ˆ Since .normalized
0 is :because ,1 1 ,1 that followsIt
2 nnnHnnaanCnaan
CnannCna
CnnnH
HnannaHanaHnnnH
aH
annnnannna n
=−+=−==
−=−=
−−=−
−=−=+=
=
++=−=
++
∗+
+
++
( )( )
0!
So: 21nan
n+
=
• So, the representations of in the basis are very simple, and so are those of X and P:
Haa ,, + n
( ) ( )aamiPaam
X −
=+
= ++ 2
121
2 ,
2hh ω
ω
.0
Example question:
• What are in the state
• Up to a constant Ignoring c for the moment,
Since lower and raise and since
Now, use the identities
To obtain (after multiplying by
2, XX ?n
.++= aaX
000 =+=+=+= ++ nannannaanX+aa, n nmmn ,δ=
( ) ( )naannaannaannaan
nannannaanX++++
++
+++=+++=+=
00
2222
1 1 ,1 ++=−= + nnnannna
( ) ,2 21
ωmc h=
)2c
ωmnX h
+=
212
Passage from the Energy Basis to the Coordinate Basis
( )
2 ,2 so
denote 22
21
2
21
21
2121
21
21
21
−=
+=
=
+
⇔
+
=
−− +
dydya
dydya
xmydxd
mxm
Pm
iXma
hh
h
hh
ωω
ω
ωω
• To find the ground state in the coordinate basis:
ion)normalizat(after
is state ground the
241
200
00
22
00
h
h
xmy
emeA
dydya
ω
πωψ
ψψ
−−
==
⇒=
+⇒=
And the other eigenstates can be derived (up to a normalizing factor) by
( )( ) ( )
!
2! 021
2
021 ψψψn
dydy
nna
nn
n
−==
−+
Coherent States
• Seek the “classical limit” of the oscillator, i.e. states with a relatively small uncertainty in energy etc.
• Time development of classical oscillator:
( )( )
( )
( )[ ]
( )[ ]( )[ ]
20
*00
*00
2
) all(for energy The2)(ˆ
21)(ˆ)0()()()(
)(ˆ)(ˆ21)( Define ).(ˆ)(ˆ ),(ˆ)(ˆ
)(1)(ˆ ),()(ˆ Define
)()( ),(1)(
αωαα
ααααωαα
αωω
ωβββ
ω
ωω
ωω
ω
h
&
&&h
h&&
=−−=
+==⇒−=
+=−==
===
−==
−
−
−
teeitp
eetxettit
tpitxttxtptptx
mtptptxtx
txmtptpmtx
titi
titi
ti
• There is a lot of similarity between these relations and the average behavior of the quantum oscillator:
( )( )( ) ( )( )
( )2121ˆ
21ˆ
+=−−==
+==
+
+
+
aaHaaiPP
aaXX
ωβ
β
h
h
( )( )
[ ].21that
then Assume . 21
but -- )0( have should wesimilarly, behave
energy to For the .(0)(0)ask that therefore We. of role the
playing with equations, oscillatorclassical the toidenticalalmost is This
)0()0(2)(ˆ)0()0(21)(ˆ
20
20
20
0
0
0
*
*
][][
>>
+=
==
=
−−=
+=
+
−
−
ααω
αωω
αψψα
α
αα
ααωω
ωω
h
hh
H
aaH
a
eeitP
eetXtiti
titi
• This analysis suggests to require
.!
:ionnormalizatafter And
)(!
)(
)()()(
1)()( So
.)( denote ,Let like?look of rseigenvecto thedo How
.)0()0( require tois thisenforce ay tosimplest w The
)0()0( ,)0()0(
2
0
1
0
200
2
nn
e
cn
c
nccnc
nncnaca
ncaa
a
aaa
n
n
n
n
nn
nn
nn
nn
nn
∑
∑
∑∑∑
−
−
+
=
=
⇒=⇒
=−==
==
=
==
αα
ααα
ααααα
ααα
ααααα
ψαψ
αψψαψψ
α
• It follows that:
( )
forth. andback moveshich Gaussian w
small"" a isfunction waveThe change.t don' ,that means also This . ofr eigenvectoan remains state theSo,
02022
02
242222
!
!
:evolution Time22
, 2
)(Im 2 ),(Re2:derive torwardstraightfo is
it ,, of in terms eexpressibl are , Since
.1 so And
412 ,
21
20
20
αα
ωωω
αω
α
αααα
αα
α
αα
αα
αα
α
ωω
αωαω
αα
αω
ααωαω
PXa
tititiE
n
ntiti
tiE
n
n
eenen
eee
nen
e
PXmPm
X
mPm
X
PXHHH
HH
n
n
∆∆
−−−−
−−
−−
+
=
=
=∆∆⇒=∆=∆
==
<<∆
⇒=∆
++=
+=
∑
∑h
h
hhh
hh
h
hh
The Operator and the Wave Function)(αD )(xαψ
[ ]
+
∆
−−
=
==
==
==
=
−==
∑
∑−
+−
−+−
−+
+
hh
xPiXXxmex
Dxxx
nn
eD
nn
ee
eeeD
aaDeD
i
n
naa
aa
aa
αα
αθα
α
πωψ
ααψ
ααα
α
α
ααααα
α
α
αα
ααα
αα
241
2*
2exp)(
isresult The.0)()(function wave thecalculate
toallows which ,!
0)(
so ,!
0 and ,00
.)( that commutatortheir
withcommute which operators for two formulasGaluber' from followsit ,, Since
unitary. is )( , )( Define
2
*
*2
*
Field Quantization• For two coupled oscillators, new variables can be defined which decouple them. This can also be done for an infinite chain of oscillators, a continuous string, and the electromagnetic field.
• The Hamiltonian for a pair of coupled oscillators equals
where controls the strength of the coupling.
• Defining new variables and new masses
decouples the resulting differential equations, and the Hamiltonian can be written as
( ) ( ) ( )22122
222
12
22
21
21
21
22xxmaxmaxm
mp
mp
−+++−++ ωλωω
λ
2,2,
2)()()(,
2)()()( 2121 mmtxtxtxtxtxtx RGRG ==
−=
+= µµ
λωω
ωω
λλω
λωµωµ
µµ
41
,,2
)()()(),()()( where
414
412
21
21
2221
21
222
22222
+=
=−
=+=
++
+−+++
R
GRG
RRRGGGR
R
G
G
tptptptptptp
amaxxpp
• The quantum mechanics case proceeds similarly: new observables are defined by
• From the commutation relations of the original observables
It follows that the same relations hold for the new observables. It also follows that the Hamiltonian
equals
Note that this is true only because of the commutation relations between the new observables.
2,,,
221
212121 PPPXXXPPPXXX RRGG
−=−=+=
+=
[ ] [ ] 0) scommutatorother the(and ,, 2211 === hiPXPX
( ) ( ) ( )22122
222
12
22
21
21
21
22XXmaXmaXm
mP
mP
−+++−++ ωλωω
λλω
λωµ
µωµ
µ 414
412
21
221
222
22
222
2
++
+−+++ amaXPXP
RG H
RRRR
R
H
GGGG
G
4444 34444 2144 344 21
A basis for the system’s wave functions is given by the tensor product of the corresponding eigenvectors of which are given by the respective raising and lowering operators:
the eigenvectors of are given by
The eigenvectors of the system are given by
, and RG HH
λωµωµ
ωµωµ
412,
21
21
+−=′
−′=
−=
+
+
aXXPiXa
PiXa
RRRRR
RRR
R
GGG
GGG
G
hh
hh
( )
.for similarly and ,21
of eigenvaluean with , !
10
RnG
GnG
Gn
n
an
ψω
ψψ
h
+
= +
( ) ( )
λλωωω
ψψ
414
21
21
ofenergy an with , !!
1
22
0,0,
++
++
+
= ++
ampn
aapn
RG
pR
nGpn
hh
GH
• In the (0,0) state, both the average and the distance between the two masses are Gaussians around zero.
• In the (1,0) state, the average’s distribution has two peaks away from zero, and the distance is a Gaussian around zero.
• In the (0,1) state, the average’s distribution is a Gaussian around zero, and the distance’s distribution has two peaks away from zero.
• In the (1,1) state, the distributions of both the average and the distance have two peaks away from zero.
)0,0( )0,1(
)1,0( )1,1(
Continuous String
. at time string theof state the
determine )(),()()(),(
:sin2)(with expansion Fourier
),(2
energy Kinetic
),(2
energy Potential
,0),( 1
1
0
20
2
2
2
2
2
2
t
tqtqxftqtxu
Lxk
Lxf
dxt
txu
dxx
txuF
Fvtxuxtv
kkk
kk
k
L
L
&⇒=
=
∂∂
=
∂∂
=
==
∂∂
−∂∂
∑
∫
∫
∞
=
π
µ
µ
• In Fourier domain, the equations are
decoupled:
• It’s straightforward to see that the
Lagrangian equals
• The momentum conjugate to is
• The Hamiltonian is
Lvktqtq kkkkπωω ≡=+ ,0)()( 2&&
[ ]∑=
−=1
222
2 kkkk qqL ωµ
&
kq
kk
k qqLp &&
µ=∂∂
=
∑=
+=
1
222
222 kkk
k qpH ωµµ
µ
[ ]22 ˆˆ21 yields
1ˆ ,ˆ Denoting
kkk
kkk
kkkkk
pqH
ppqq
+=
≡=≡
∑ ω
µωββ
β
h
hh
A schematic proof that are canonical (most constants ignored,
kkk qpq &=,
( ) ( )[ ]
( )
.)()(
)()()()(,
.,for same ,0)()()()(
, :relationsn Commutatio
.)(),(,)(),(
so ,)()(),()()(),(
:sin)(with expansion Fourier .L is )( toconjugate momentum theand
,),(),(LLagrangian
2
0
2
000
2
000
2
0
2
0
11
2
0
22
kllk
lklklk
lklklk
lk
kkkk
kkk
kkk
k
dxxfxf
dxxu
pxu
qxu
pxu
qpq
ppdxxu
qxu
qxu
qxu
q
dxxftxupdxxftxuq
xftqtxuxftqtxu
kxxfuuxu
dxxtxuttxu
δπ
π
π
ππ
π
=
=
∂∂
∂∂
−∂∂
∂∂
=
=
∂∂
∂∂
−∂∂
∂∂
=
≈≈
=⇒=
=≈∂∂
∂∂−∂∂==
∫
∫
∫
∫∫
∑∑
∫
∞
=
∞
=
321321&&
321&
321&
&
&&
&&
:)2π=L
• Now, quantize by promoting to
Operators(note: as for one particle, the classic variables are time-dependent, but the operators are not).
• Denote It has
eigenstates and eigenvalues
• Since the commute, a general state can be defined by with energy
kk pq ˆ,ˆ
.ˆ,ˆ satisfying ˆ,ˆ2121
][ kkkkkk iPQPQ δ=
( ). ˆˆ21 22
kkkk PQH += ωh
.21
kkkk nnH ωh
+=
kH,...21 knnn
( )( )( )( ). ˆˆ21
,ˆˆ21 :operatorson annihilati andcreation define before, As
state. ground theabove
kkk
kkk
kkk
PiQaPiQa
n
−=+=
+
∑ ωh
kk qq ,&
( ) ( )
( ) ( )
equation. classical thesatisfies )()( that show tohardnot isit And
)(12
1
)()(operator an topromoted is ),( observable The
0 ... !
... !
......
...1...1.........1.........
1
11
11
11
1
txU
kaaxf
QxfxUtxu
na
nann
nnnnnannnnna
kkkk
kk
kkk
k
nk
n
k
kkkk
kkkk
k
≈+
==
=
++=−=
+
++
+
∑
∑
ββ
• What is the ground state Its k’th Fourier coefficient is an oscillator with frequency proportional to k, at its ground state – i.e. a Gaussian with variance So a sample of the ground state is a function with slowly decreasing Fourier coefficients. This decrease reflects the coupling in the spatial domain, and the samples are not random noise, but have some measure of smoothness associated with them.
• The ground state is a fractal of dimension 2.
• An excited state is the superposition of sinusoids with the ground state.
?0
. 1 k≈
Quantization of the EM Field
tE1J4 B
tB1 E
J0 B4 E
: Equations sMaxwell'
∂∂
+=×∇∂∂
−=×∇
∂∂
−=⋅∇=⋅∇=⋅∇
ccc
tπ
ρπρ
⇒∂∂−=⋅∇
====⋅∇∂Λ∂+=′Λ∇−=′
∂∂−−∇==×∇
tc
tc
tc
φρφ
φφ
φ
1 A :gauge Lorentz
0).J 0, if 0( 0,A :gauge Coulomb
1 ,AA : transformGauge
A1 EBA : Potentials ,
2 πρφ 4−= 2
cA J4 π
−=
22
2
22 1
tc ∂∂
−∇≡
• Experimental motivation for quantizing the EM field: spontaneous decay of hydrogen atoms when no field is present.
[ ] r81 : field EM theofEnergy
: particle charged aon Force
322 dBE
BcvEq
∫∫∫ +
×+
π
relations.commtationsimilar with operators tothempromote and ,, ,
such that momenta and scoordinatecanonicalfind:onquantizatifor Recipe
ijjiAiALi
jiA
δ=Π∂∂=Π
Π•
&
• The EM Lagrangian is
and indeed it yields Maxwell’s equations when varied with respect to the potentials.
• But since does not appear in L, has no conjugate momentum.
• Since we’re working in free space (no charges/currents), this problem can be alleviated by working in Coulomb gauge with
• The Lagrangian is then simpler, but it doesn’t yield the equation any more. This condition now has to be enforced.
• This will result in one less degree of freedom for the vector potential A.
[ ]
rA A181
r81
322
322
dtc
dBEL
∫∫∫
∫∫∫
×∇−∇−
∂∂
−
=−=
φπ
π
φ& φ
.0=φ
0E=⋅∇
• The coordinates are , and the
conjugate momenta are
• Alas, they are not canonical. We’d like to have
but taking the divergence with respect to ryields zero on the left-hand side, but not the right-hand side.
• The way out of this is to parameterize A by independent variables, so that will hold. As in some cases in classical mechanics, this is easier to do in the Fourier domain, in which the condition is algebraic and not differential, and in which the coordinates can be decoupled.
)rr()r(),r(A ′−=′Π δδijji
. 4
)r(E)(
ciri π
−=Π
)r(Ai
0A=⋅∇
• Since and are real, there is a complex vector function satisfying:
• The existence of is guaranteed by the redundancy in the Fourier transforms of the real functions A and : for a real f,
• The choice of is guided by the following identities for the oscillator:
)(A r )(rΠa(k)
[ ] k(k)a*a(k))r(A 3rkrk dee ii∫ ⋅−⋅ +=
[ ] k(k)a*a(k)4
1)r( 3rkrk deekic
ii∫ ⋅−⋅ −=Ππ
Π
a(k)
k).(k)( *FF =−
a(k)
( ) ( ) 2
, 2
2121
aamiPaam
X −
=+
= ++ hh ω
ω
• The conditions are expressed in the Fourier domain by
which suggests to expand a(k) in a basis orthogonal to k.
• Define for every k an orthonormal basis
• For every k expand a(k) as follows:
This yields
0A =Π⋅∇=⋅∇
[ ] [ ] 0a(k)k0(-k)aa(k)k(-k)aa(k)k =⋅⇒=∗−⋅=∗+⋅
( ) ( ) ( ) ( ) .kk3such that ,k3,k2,k1 εεεε
( )∑=
==3
1
2122 )( )(k)k(4a(k)λ
ωλελωπ kcac
[ ] k)k()k()k()k(4
A(r) 3rkrk21
2
2
deaeac ii∑∫ ⋅−⋅ ∗+
=
λ
λελλελωπ
[ ] k)k()k()k()k(64
1(r) 3rkrk21
4 deaeai
ii∑∫ ⋅−⋅ ∗−
=Π
λ
λελλελπω
• Before imposing transversality
And it follows that
)r-r()r(r),(A
,0)r(r),()r(Ar),(A
′=
′Π
=
′ΠΠ=
′
δδijji
jiji
[ ][ ]
etc. , )k(by given )(r),(rA
by sderivative with theusual, as continue Now
)k(r(r)rA(r))k(
)k(r(r)rA(r))k(
constants, all ignoring :proof of Outline)k-k( )k(),k(
,0)k(),k()k(),k(
0rk00
3rk3rk
3rk3rk
3
λε
λελ
λελ
δδλλ
λλλλ
λλ
ji
jj
ii
ii
e
dedeia
dedeia
iaaaaaa
⋅−
⋅⋅
⋅−⋅−
Π
⋅Π+−=∗
⋅Π+=
′−=′′∗
=′′∗∗=′′
∫ ∫∫ ∫
′
• Next, discard which must be zero due to the transversality.
• The EM Hamiltonian is
and it equals
)3k(a
[ ] r 81 322 dBEH ∫∫∫ +=π
k)k()k( 32
1daa∑∫
=
∗λ
λλω
• Guided by the oscillator equations, define
[ ]
[ ]
relations.bracket Poisson canonicalesatisfy th that they verifiedbecan it
)(ka)a(k2
)k(
)(ka)a(k)(2
1)k(
21
21
λλωλ
λλω
λ
∗−
−=
∗+=
ip
q
• It immediately follows that
so the field is a sum of a continuous family of oscillators, two at each frequency.
• Next, promote to operators
satisfying the canonical commutation relations
and define raising and lowering operators
[ ] k)k()k(21 3222 dqpH ∑∫ +=
λ
λωλ
)k(,)k( λλ qp
)k(,)k( λλ QP
[ ] ( )k-k)k(,)k( 3 ′=′′ ′δδλλ λλhiQP
[ ] )k-k()k(),k(satisfywhich
)k(2
1)k(2
)k(
)k(2
1)k(2
)k(
3
2121
2121
′=′′
−
=
+
=
′+
+
δδλλ
λω
λωλ
λω
λωλ
λλaa
PiQa
PiQa
hh
hh
A and are also promoted to operators:
To find the Hamiltonian, symmetrize
by to obtain
[ ] k)k()k()k()k(4
A(r) 3rkrk21
2
2
deaeac ii∑∫ ⋅−+⋅ +
=
λ
λελλελωπ
h
[ ] k)k()k()k()k(64
1r)( 3rkrk21
4 deaeai
ii∑∫ ⋅−+⋅ −
=Π
λ
λελλελπωh
Π
aa∗
( )( )aaaa ∗+∗21
k21)k()k( 3daaH ωλλ
λ
h∑∫
+= +
• In the ground state, ALL the oscillators are in their ground state, so for all
• Formally, the energy of the ground state
(“vacuum”) is
.00)k( =λaλk,
( ) ??k21 30 =≡∑∫ dE ω
λ
h
• It can be verified from the definition of Hand from the commutation relations that
(this is computed “above” ).
• The momentum operator of the field can be computed by promoting the classical EM
momentum to
And it can be verified that the momentum of is Since
it follows that is a massless particle – the photon. Its wave function is
( ) ( ) 0)k(0)k(0)k( λλωλ +++ == akcaaH hh
0E
( ) rBE41 3d
c ∫ ×π
[ ] kk )k()k( 3daaP h∑∫ +=λ
λλ
λλ k0)k( ≡+a k.h42222 cmcpE += λk
( ) 23
rk
2)k(k
πλελ
⋅
→ie
• The state of the field is given by the number of photons at each that is, by successive application of creation operators (Fock state):
• Expected value and variance of vacuum electric field in location r:
,k,λ
( ) ( )
states.such ofnscombinatiolinear and ,on polarizati
and number with wavephotons
0 .... !
! 2
22
1
1121
i
ii
nn
knn
kan
ka
λ
λλ=
++
∞== 0r)(0but ,00r)(0 2EE
• Why? It’s enough to prove for :r)(Π
• For simplicity’s sake, write
[ ]
[ ]
( )[( )]
[ ]
. cutofffrequency afor 0k0
0kk
)kk()(k)(k0
0kk)(k)(k0
0kk)(k)(k
)(k)(k0
: variance the toNow0.(k)0 ,00(k) :k allfor but
0k(k)(k)0
0r)(0 So
k(k)(k)r)(
0401
31
23
13
213
12r)kk(21
221
1
23
13
21r)kk(21
221
1
23
13r2k
2r2k
2
rk1
rk1
212
211
3rkrk21
3rkrk21
21
21
11
kkOdk
dd
aaekk
ddaaekk
ddeaea
eaeakk
aa
deaeak
deaeak
i
i
ii
ii
ii
ii
+⋅−
+⋅−
⋅−+⋅
⋅−+⋅
+
⋅−+⋅
⋅−+⋅
=
=
−+
=
=−
−
==
−
=Π
−≈Π
∫
∫∫∫∫
∫∫
∫
∫
δ
The Classical Limit
• We should hope that, as mass and energy scales increase to “everyday scale”, QM laws should converge to Newtonian mechanics laws. Look for example at the harmonic oscillator when the energy increases – it resembles the behavior of a “Newtonian” oscillator.
• It makes sense, in this regard, to look at the average, or expectation, of position and momentum. Therefore let us look at how the average changes over time: let be a function of X and P. Then Ω
ψψψψψψ
ψψ
&&& Ω+Ω+Ω
=Ω=Ωdtd
dtd
.0=Ω&• Assuming has no explicit time dependence, We then obtain, using Schrödinger’s Equation
Ω
[ ]
−=Ω Η=Ω , , ωω
dtd
Hi
dtd
hRemember!Ehrenfest’s theorem
• For example
• and
• which look just like Hamilton’s equations. Alas, we cannot argue that the latter hold for average quantities, since in general
• However, it’s still a good approximation.
[ ]
)(2
assuming
2
,
+=
∂∂
==
−=
XVm
PH
PH
mPiX
dtd
HXh
XH
dxxdVP
dtd
∂∂
−=−=)(
etc. ,PH
PH
∂∂
≠∂∂
Rotational Invariance and Angular Momentum
• Translations in 2D are given by
and linear combinations of these. Much like in the 1D case, translation by a vector a is given by
yiP
xiP yx ∂
∂−→
∂∂
−→ hh ,
jiP ,a)( Payx
i PPeT +== ⋅− h
2D TranslationsMotivation: Central Potential (Atoms)
h
h
hh
iaPdxda
UUdxdU
dad
Udxd
dxaxdaxax
UUda
dU
iaPdxda
dxda
dxdaI
xaxax
axxU
Pidxd
dxdiP
ee
ee
aaa
a
aaa
a
−
=−
=⇒−=
−=
−−=
−−−−
=−
=
−
=−
=
−+−
=−′′+′−
=−=
=−=
→
+
→
so,
)()()(lim
)()(lim
:ons translatimalinfinitesi via thissee alsocan One .
...!2
...)(2
)()(
)())((
, , :1DIn
0
0
2
22
2
ψψε
ψεψε
ψψψ
ψ
ψψψ
ψψ
ψψ
ε
ε
ε
• Let a 2D rotation be described by
−→
yx
yx
)cos()sin()sin()cos(
00
00
φφφφ
• The operator that rotates the 2D vectors is denoted by and the corresponding
Hilbert space operator is denoted
For an infinitesimal rotation,
),k( 0φR[ ]. )k( 0φRU
[ ] [ ]
φ
φεε φ
∂∂
−=−=
=−= −
h
hh
iYPXPL
eRULiIRU
xyz
Lizzz
z
coordinatepolarin
00 )k( , )k(
• Physical interpretation: if H is rotationally invariant, and experiments will yield the same results in the coordinates are rotated. Also, there is a common basis for H and
[ ] ,0, =HLz
. zL
sEigenvalue s'zL
postulate. a sIt' Hermicity? impose
can we Why .number quantum magnetic thecalled is ...2,1,0 ,1)2,()0,(
implies , y,Hermiticit Imposing
) w.r.t lenormalizab be tohas ( )(),(
),(),(
2
1221
0
mmmle
LL
dReR
lilllL
zil
zz
ill
lzl
zzzz
z
z
z
z
z
±±==⇒=⇒=
=
=
⇒=∂
∂−⇒=
∗
∞
∫
h
h
h
h
π
φ
πρψρψ
ψψψψ
ρρρφρψ
φρψφ
φρψ
• There are still an infinite number of degrees of freedom for choosing This is solved by demanding that the eigenvectors of are also the eigenvectors of, say, a rotationally invariant Hamiltonian. Physically, this means that the state is determined not only by its angular momentum, but also by its energy.
).(ρRzL
=ΦΦ=Φ ∫ ′′
−π
φ δφφφπφ2
0
21 )()( )2()( Define mmmmim
m de
Examples (Shankar p. 315):
• A particle’s wave function is
Prove that
).(cos),( 22 22
φφρψρ ∆−
= Ae
.61)2()2(,
32)0( =−===== hh zzz lPlPlP
±==
=Φ==
=
∫ ∗
otherwise024
0)(),()(
ion,normalizat toup So, factor. gnormalizin aonly ispart the),( calculate to:Solution
2
222
0
mm
dmlP
mlP
mz
z
ππ
φφφρψ
ρ
π
h
h
• A particle’s wave function is
.)sin()cos(),(22
2
+∆
= ∆−
φφρφρψρ
Ae
Prove that .21)()( =−=== hh zz lPlP
Solution: the same as above – note that
±=
=Φ+∫ otherwise01.
)())sin()cos((22
0
mconstdA m φφφφ
π
Rotationally Invariant Problems
),(),()(112
:mass) thedenotes ( is for equation eigenvalue Then the ).(),( that Assume
2
2
22
22
φρψφρψρφρρρρµ
µρφρ
EE EV
HVV
=
+
∂∂
+∂∂
+∂∂−
=
h
• Note that in this case H commutes with , since is represented by and we must have
zLzL ,φ∂∂
( ) ψφφ
ψ VV∂∂
=
∂∂
which will hold if V is a function of only.
• We seek common eigenvectors of H and of the form (separation of variables):
ρ
zL
)()()(12
:equation radial The )()(
2
2
2
22
ρρρρρρρµ
φρψ
EmEm
mEmEm
ERRVmdd
dd
R
=
+
−+
−
⇒Φ=
h
Angular Momentum in 3D
[ ] [ ] [ ]
[ ] tensor.ricantisymmet , ,
or or
, ,, ,,
:satisfy which ,,
, :generators )(Hermitian Three
3
1∑=
==
=×
===
−=−=
−=
kijkkijkji
xxzxzyzyx
xyzzxy
yzx
LiLL
LiLL
LiLLLiLLLiLL
YPXPLXPZPL
ZPYPL
εεh
h
hhh
• The total angular momentum operator (squared) is defined by
[ ] [ ] [ ] 0,,, 2222222 ===⇒++= zyxzyx LLLLLLLLLL
The square root of is regarded as the total angular momentum of the system. The goal is to characterize a “pure” state by its (definite) total angular momentum, energy, and magnetic quantum number (the last being the angular momentum around the z-axis).
2L
Finite Rotations in 3D
• Rotations are represented by
or)(unit vect axisrotation ˆ angle,rotation ,ˆθ
=
==
θ
θθθ
• Turns out that [ ] hh LθLˆ)θ(
⋅−⋅−==
ii eeRU θθ
Proof: when a vector r is rotated by an infinitesimal angle it moves to (because the addition to the vector is perpendicular both to the vector and the rotation axis). So, theinfinitesimal change induced by is
θ,δrθr ×+δ
[ ])θ(δRU
( ) ψδψδψψ ∇⋅×+=×+→ rθ(r)r)θ(r(r)Putting aside for the moment, we should show that this change in is also induced by the vector product term is
i,h(r)ψ ).ˆ)(( θδθ ⋅+ LI
zyx
zyx rrr
ˆˆˆ
ψψψ
θθθδθ
zyxand the change induced by is
)ˆ)(( θδθ ⋅+ LIψθθθδθ )ˆˆˆ)(( zzyyxx LLL ++
To prove that the two changes are equal, compare e.g. the coefficients
of (and remember that zθ ).xyz YPXPL −=
. ,, ediagonalizusly simultaneo toiesusually tr one so
,0],[But commute.t don' thesince ,,,,esdiagnolaiz which base a findcannot one However,
conserved. aremomentum,angular total theand , of components threethe
all Hence, .0],[ so and ,0],[ rotations,
malinfinitesiby drepresente are ,, since then,]),[][(rotation under invariant is If
2
2
2
z
iizyx
i
zyx
LLH
LLLLLLH
LLHLH
LLLRHUHRUH
=
==
=
Common Eigenvalues for and 2zLL
[ ] [ ]( )
bounded. be tohas )( around momentumangular the),(
momentumangular lgiven tota afor since somewhere, end tohas This
, createsit on ofaction the,given So,
,),(similarly And
,),( ).()(
).)(()(
.0, Note . ,:operators raising and lowering are there too,Here,
, rseigenvectocommon tooscillator for the used paradigm theFollowing
2
2
2
βα
βααβ
βαβααβ
βαβααβαβααβ
αββαβαβ
αββαβαβααβ
z
kL
CL
CLLLL
LLLLLL
LLLLLiLLL
LL
zz
zyx
z
h
h
h
hh
h
±
−=
+==
+=+=
=±=⇒±=
==
±
−−
++++
++++
±±±±
momentum.angular totalan thesmaller th sit' mechanics, classicalin Unlike
state. theof momentumangular thecalled is 2
.122
, ...2,1,0 ,2
havemust we, size of steps in to
fromgot weSince .maxmin that followsIt
)min(minsimilarly , )max(max
get we, using and sidesboth
to Applying .0 :raised becannot
thatan s there'So . havemust we
so ,Hermitian) are , (since definite positive is
but ,
max
2max
maxmin
22
max
max2
222222
h
hh
h
hh
h
β
αβ
αβαβ
ββ
ββαββα
αβ
αββα
αβαβαβαβ
=
+
===
−=
+=+=−−=
=
≥
++=−
+−
−+
k
kkkk
k
LLLLL
LL
LL
LLLLLL
zz
yx
yxyxz
::::......3/2
1
1/2
000
Angular
Momentum
2k
αmaxβ αβ
0,0
2
23
21
h
h
2h
2,
43
2,
43
2
2
hh
hh
−
( )( ) 221 hhh
h
hh
−,2
0,2
,2
2
2
2
But we proved that can only have integer multiplies of ??h
zl
• Answer: here we only used the commutation rules, not the operators’ specific shapes.
44 844 76 hLLL i=×
• There are particles whose wave function is more complicated (non-scalar). For these particles, there are rotation-like operators J which satisfy the commutation rules
JJJ hi=ו For example:
k),,(j),,(i),,(),,ψ( zyxzyxzyxzyx zyx ψψψ ++=
• Rotations of such a function involve rotating both the components and the coordinates: under an infinitesimal rotation k,zε
.,.r.t operator widentity ,components vector .t thematrix w.ridentity 22
spin. e.g. .
00
00
1001
),(),(
),(),(
)1(
)2(
)2()1()2()1(
yxII
SSLSIILJ
iii
LLi
xyyxxyyx
xyyxxyyx
zzzzzz
y
xz
z
zz
y
x
yzzyzzxzy
xzzyzzzxx
=
×=
=+=⊗+⊗=
−−
−
=
′′
′=−++−+→
′=−+−−+→
ψψεε
ψψ
ψεεψεεψεψ
ψεεψεεεψψ
h
h
hh
• Just as before:
( )
( )( )[ ]
:feasiblemoreofn calculatio themakes
This blocks. fixed in the diagonalblock are matricesThese well.as basis in the matrices their writeto
rwardstraightfo isit ,2
,2
Since
1,1 so
)1(
),(
1,),(
1,),(
1,),( before, As
,...0,...1, ,
...23,1,
21,0 ,)1(
Jθ
21
2222
222
adjoint
22
h
mh
hhh
h
h
h
⋅−
−+−+
±
+−+
∗+−
++
±
±±
−=
+=
±+±=
−−+
=−−==
⇒+=
⇒+=
±≡
±=
−−==
=+=
iej
jmiJJJJJJ
mjmjmjjmJ
mmjj
jmJJJjmjmJJjmmjC
mjmjCJjm
mjmjCjmJ
iJJJmjmjCjmJ
jjjmjmmjmJ
jjmjjjmJ
yx
zz
yx
z
Finite Rotations• In order to compute the transformation induced by finite (as opposed to infinitesimal) rotations, one has to calculate
[ ] .)θ(JθJˆ
hh
⋅−⋅−
==ii
eeRUθθ
• The calculation is facilitated by the fact that the matrices for are all block diagonal in the constant j blocks.
• Also, since the eigenvalues of (in the constant-j block) are
zyx JJJ ,,
Jˆ ⋅θ
( ) ( )0). is any vectoron action itsobviously (since
0ˆˆ then ,,...1, =−⋅⋅⋅⋅+⋅+−− jIθJjIθJjjj
• This allows to express as a linear
combination of , so the powers in the exponential usually have a simple periodic behavior which allows to write it down as simple functions of (see also discussion here on Stern-Gerlach apparatuses).
( ) 12ˆ +⋅
jJθ
( ) ( ) ( ) jJJJI
22 ˆ ...ˆ,ˆ, ⋅⋅⋅ θθθ
θ
The Matrices of Angular Momentum Operators
( ) ( ) ( )
( )
( )
( )( )( ) 2
2
2
2
2
2000001,10200000,10020001,1
000430021,
21
000043021,
21
000000)0,0(
1,10,11,121,
21
21,
21)0,0(
h
h
h
h
h
−
−
−
−
jm
mj ′′
2J
( ) ( ) ( )
( )
( )( )( )( ) h
h
h
h
−−
−
−
−
−
000001,10000000,1000001,1
000210021,
21
000021021,
21
000000)0,0(
1,10,11,121,
21
21,
21)0,0(
jmmj ′′
zJ
Spin half Spin one
( ) ( ) ( )
( )( )( ) 0200001,1
2020000,10200001,1
00002021,
21
00020021,
21
000000)0,0(
1,10,11,121,
21
21,
21)0,0(
h
hh
h
h
h
−
−
−
−
jm
mj ′′
xJ
( ) ( ) ( )
( )( )( ) 0200001,1
2020000,10200001,1
00002021,
21
00020021,
21
000000)0,0(
1,10,11,121,
21
21,
21)0,0(
h
hh
h
h
h
iii
i
i
i
−−
−
−
−
−
−
jm
mj ′′
yJ
Spin half Spin one
Angular Momentum Functions in the Coordinate Basis
( )
( )
., )momentum(s with and radiusat particle thefind toamplitude theis )( is,That
)(),)1(Pr(
then,),,(),()(
where),,()(),,( If
.))(cos(
productinner under the lorthonorma areThey
).harmonics spherical( ),( momentum, andangular totaldefinite with ionseigenfunctfor search and part,
theseparate webefore, As . with operations successiveby found becan others theand , find toallows This
). )exp( with gmultiplyinby ofcare taken ispart theand out,left ispart that the(note
cot basis, coordinate In the
.0 satisfying, state topmost"" theStart with
0
2222
0
1
1
2
0
mlrrC
drrrCmLllL
drYrC
YrCr
ddd
Yzr
Lll
ilr
ieL
llLll
ml
mlz
ml
ml
ml
l
l
lm
ml
ml
i
∫
∫
∑ ∑
∫ ∫ ∫
∞
∗
∞
= −=
−
−
±±
+
==+=
Ω=
=
=Ω
−
∂∂
±∂∂
±=
=
hh
h
φθψφθ
φθφθψ
φθ
φθ
φφ
φθ
θ
π
φ
lml
mlm
lm
l
ml
immml
xdxdx
lxP
Peml
mllY
)1()1(!2
1)(
))(cos()!(4
)!)(12()1(),(
222
21
−−=
+−+
−=
+
+
θπ
φθ φ
• Example question (Shankar, 12.5.13): a particle’s state is proportional to Prove that
• Answer: It’s immediate to see that
.)2( arezyx −++.61)Pr()Pr(,32)0Pr( =−===== hh zzz lll
( ) ( )11
11
11
11
01 6
4 ,6
4 ,3
2 YYriyYYrxYrz +=−== −− πππ
So all that’s left is to compare the coefficients of .2in ,, 1
11
10
1 zyxYYY ++−
( )1)(cos3)165(
)cos()sin()815(
)(sin)3215(
)cos()43(
)sin()83(
)4(
22102
2112
222122
2101
2111
2100
−=
=
=
=
=
=
±±
±±
±±
−
θπ
θθπ
θπ
θπ
θπ
π
φ
φ
φ
Y
eY
eY
Y
eY
Y
i
i
i
m
m
Note that as m grows, there is more of the momentum in the z direction, and the probability to have large zvalues decreases.
Solution of Rotationally Invariant Problems
[ ]
ElEllEl
ElEl
El
ElEl
mlElmElmz
EE
EUUrDUr
llrVdrd
Ur
llrVEdrd
rUR
mERRrVrll
rr
rr
YrRrLLHLH
rErrVr
rrr
rr
rVrV
=≡
+++−
=
+−−+⇒=
=
+
+
−∂∂
∂∂−
==
=
+
∂∂
+
∂∂
∂∂
+∂∂
∂∂
−=
)(2
)1()(2
toequivalent iswhich
02
)1()(2
). (no )()1(12
:equation radial aget 2D,in As ).,()(),,(:,, of
eigenkets ussimultaneoseek ,0, case in this Since
),,(),,()()(sin
1
)sin()sin(
112
becomesequation dinger oSchr the),(),,( If
2
2
2
22
2
2
22
2
22
2
2
2
2
2
22
22
2
2
µµ
µµ
µ
φθφθψ
φθψφθψφθ
θθ
θθµ
φθ
hh
h
h
h
h
&&
Note: is mass (to avoid confusion with m). µ
( ) ( )[ ]( )( )
→ →
→r
c
r
URconstcrElU
dr
dUU
dr
dUU
dr
dUU
dr
dUU
ikrerElUrElU
drEl
UdrEl
Rr
ElRElRElU
dr
dUU
dr
dUU
drUUUU
drUlDUdrUlDUElU
lD
∝∝=→
=
∗
−∗
∞∗
−∗
∞→∞→
∞∞=
=
∞∗
−∗
⇒=∫∞ ″∗
−′′∗
⇒∫∞ ∗
=∫∗
∫∫
∞
zero,not is c If .0
if satisfied is This .0)0(12
21whether
on dependsy Hermiticit theand vanishes
0
12
21
in limit upper thecases,both In state). (unbound
or state) (bound 0either havemust we
0le,normalizab be toFor this .
0
2 must take one
integrate to:, ofion normalizat theaddress Next,
0
0
12
21
partsby nintegratio
00 2121
*
0 1221 ,
toregard with ofy Hermiticit Imposing
0
.0havemust weSo, r).(4)1( as origin,
at theequation sdinger'oSchrsatisfy not does
since impossible is which origin, at the Diverges
0
32
→−=∇
→rElUr
rc
πδ
&&
General Properties of ElU
• The goal is to further study the asymptotic behavior of as r tends to zero or infinity, depending on the potential.
ElU
limit). in the ntialinconseque
becomes ( )1( :barrier
lcentrifuga called-so by the dominatedisequation then the,han singular tless is )(that assuming and ,0 As
2
2
EUrllU
rrVr
ll+
≅′′
→−
≅
≅
−
+
analysis) previous followingout ruled ,(irregular (regular)
yields solution a Guessing1
l
l
l
l
rr
U
rU α
• Intuitively – as before, as the angular momentum increases, the particle avoids the origin more and more.
• Next, let us study the behavior at infinity. Assume that the potential does not dominate the behavior at infinity; a condition which will be justified later is
0)( → ∞→rrrV
infinity).at behavior theinfluencenot does (
2 becomesequation the, largeFor 22
2
l
UEdrUdr E
E
h
µ−=
. as decreaselly exponentia toexpect weforbidden, classicaly is region The :0
infinity.at oscillate toexpect weinfinity, toescapemay Particle :0
:cases then twoare There
∞→∞→<
>
rUrE
UE
E
E
( )
into form above theplug that,do To . as .whether check and ,)( writenow usLet
2,
are solutions The .0 Assume212
∞→→=
=+=
>
±
−
rconstferfU
EkBeAeU
E
ikrE
ikrikrE hµ
′′
=
=
=−′±′′
=
+−−+
∫r
r
El
rdrVkirfrf
drrVki
fdf
frVfikf
Ur
llrVEdrd
0
)(exp)()(
)(
:small is derivative second theAssuming
0)(2)2(
infinityat obtain To
02
)1()(2
20
2
2
2
2
22
2
hm
hm
h
h
h
µ
µ
µ
µµ
then, potential Coulomb in the as
,)( e.g. ifBut . asconstant a
totendsintegralthe,0)( If2
rerVr
rrV r
−=∞→
→ ∞→
( )
origin. thefrom isit far howmatter noparticle theinfluences potential theSo,
lnexp)(
lnexp)()(
2
2
02
2
0
+±∝
±=
rk
ekrirU
rr
keirfrf
Eh
h
µ
µ
( )
state. a determinefully momentumangular axis theand momentum,
angular total theenergy, that themeans This.,,different for orthogonal are ),()(
so and ,different for orthogonal are vhence degeracy),on discussion in the proof
(see degenerate-non is )(operator The unity. tolenormalizab are eigenketsenergy the,0For
22)ln(exp
before asexpect weatom)(Hydrogen case coulomb In the
.0)( ifonly trueis thisbefore, Asusual. as 0 and ,
,2 , change the
over withcarry results the,0 case For the
2
2
212
−
<
±=
±∝
→
=+ →
=→
<
−∞→
z
mlEYrR
EU
rDE
kre
krreU
rrVBBeAeU
Eik
E
mlEl
El
l
E
rrrE
ere
φθ
κµ
κµ
µκκκκ
mmh
h
h
The Free Particle in Spherical Coordinates
( )( )
).cos()(or )sin()( : start with to
need Weoperator. raising a is So)(
,
,1,1: operators raising and lowering define so
,oscillator harmonic theresembles This
)1(
2 ,0)1(get we0 since and before, as Define
),()( start with : before As
0
00
111
11
22
2
22
22
2
2
ρρρρ
ρρρρ
ρρρ
µ
φθψ
==
=⇒=⇒==
=+
+−=+
+=
=
++−⇒=
==
+−+
==
++
++++++
+++
++++
++
B
Al
llllllllll
llllllllll
llllll
ll
El
El
mlElElm
UUU
dUcUdUdUddd
ddddUdUddd
UUddldddl
ddd
UUllddkr
EkUrllk
drd
VUYrR
h
µφθφθψ
ρρ
ρρρ
ρρ
ρρρ
ρρρ
2 ),,()(),,(
by given are space entirein theregular are which solutions particle free So,
(Neumann) )cos(1)(
Bessel) (spherical )sin(1)(
solutions generates which , 1)(
thatprove easy tosit' , gintroducin-re and unity, be to Taking
22
0
kEYkrjr
jR
jR
RR
Rc
mllElm
ll
lBl
ll
lA
l
ll
l
ll
h==
−
∂∂
−=≡
∂∂
−=≡
∂∂
−=
• Example (Shankar 12.6.8): Find the energy levels of a particle in a sphere of radius , given l=0.
• Solution: for l=0, we must have at the boundary, hence
0r0)sin( =ρ
increases) radius theasfiner ison quantizati that thenote(
22
20
222
02 since
2
==⇒=
h
h Ekr
nEnkr nµ
µππ
:scoordinate sphericalin computed seigenstate theof sum a asit express topossible
bemust it so)cos()2(
1),,( , momentum
with axis theup going is particle theIf .22
,rp)2(
1),,( s,coordinateCartesian In
,23
222
23
θπ
φθψ
µµ
πψ
ikrrp
zkpE
izyx
e
e
E
E
h
h
h
h
=
−==
⋅=
Relation to Cartesian Coordinates Solution
))(cos()()12(
thatso ),12( shown that
becan It .4
12 )),(cos()(
then)),(cos(4
12)( Since .direction)that
in momentumangular no has axis theup going particle a y,(physicall on dependt doesn' side hand-left theas
relevant, is 0only But ).,()(
0
)cos(
210
0
)cos(
210
0
)cos(
θ
πθ
θπ
θ
φ
φφθ
θ
θ
θ
ll
llikr
ll
llll
llikr
ll
l
l
lm
mll
ml
ikr
Pkrjlie
liC
lCCPkrjCe
PlY
z
YkrjCe
∑
∑
∑∑
∞
=
∞
=
∞
= −=
+=
+=
+
==
+
=
−
==
The Isotropic Oscillator
( )
∑∞
=
+
−
−
=
==
+−−+′−′′
⇒=
=≈⇒∞→
=
+−−+
⇒=
+++++
=
0
1
2
2
21
2
222
22
2
2222222
)(
thatguess origin, near thebehavior about the analysis previous theingIncorporat
,0)1(122
)()( assume So
,
02
)1(212
),(
21
2
2
2
n
nn
l
y
y
El
ml
ElElm
zyx
yCyyv
Evyllvyv
eyvyU
ryeUr
Ur
llrEdrd
Yr
U
ZYXPPP
H
ωλλ
µω
µµωµ
φθψ
µωµ
h
h
h
momentum).angular theodirectly t state theiessolution t
But this .)21( is oscillator 1D afor GSE since and , variablesof separationby )23(isenergy state ground that theguess could (one
...2,1,0,232)1(2
are levelsenergy quantized the thereforeand , start with we becauseout ruled is 0 case The expansion.
in the powerseven only are therehencezero), is all (since zero be tohas which ,
is oft coefficien at the verify theasy to sit'
However, .21 that implieswhich
, and between relation recursion ayieldsequation theinto thisngSubstituti
2
1
2
ωω
ω
λ
hh
h =
++=⇒+=
=
−+=
−
klkEkn
Cn
Cy
ln
CC
l
nn
( )
n.Hamiltonia in the symmetries ofresult a isIt later. explained be willin degeneracy The
0,1,2,3 ;0,1 )0(3),1(1 3
0,1,2 ;0 )0,2()1(0 20,1 1 1
0 0 0are seigenstatefirst The
.0or 1...4,2,2 are valuesallowed the,each At .23
energy, thedeterminesit and ,2by defined isnumber quantum principle The
31
20
l
mkkln
mkklnmlnmln
nnnknllnnE
lkn
ll
ll
43421
43421
h
==
==
±±±±=====
±±=====±======
−−=−=+=
+=ω
The Hydrogen Atom
( )[ ]
( )
recursion successive a cases),other the toopposed (as
yields This . assume 0,at behavior the
gConsiderin .)(,2 Define.)( have we0 as and
electron), theliberate to takeit would(which energy binding theis where,)(
have einfinity wat factor, polynomial a toupthen potential, asuch under that before proved It was
),()(),()(),,(
usual as and ,02
)1(2
0
1
212
1
2
2
2
22
2
2
k
kk
lEl
ElEl
lEl
rmWEl
ml
ElmlElElm
El
Cv
verUrmWrrUr
EWerU
Yr
rUYrRr
Umrll
reEm
drd
ρρ
ρ
φθφθφθψ
ρ
∑∞
=
+
−
+
−
=
==
≈→
−=≈
==
=
+−++
h
h
h
• Assume in the meanwhile that the proton’s mass is infinitely larger than the electron’s, so only the electron’s wave function is sought. The potential is induced by the attraction between the proton and electron, and the equation has the form
( )
2...1)2(21)1(2 is degeneracy total
thelevel,energy an specifies which given afor So, values.momentumangular axis 12 eappropriat
theare there thoseofeach for and,0...2,1
are values allowed theeach At .2
are energies allowed the,1 number, quantum principle theof In terms
..2,1,0..,2,1,0,)1(2
by defined levelson quantizatienergy theget weand terminate tohas series theusual as
,)1()1)(2(
)1(2
:2 ,0)1(
2equation for therelation
22
4
22
4
21
2122
2
2
2
nnn
nzl
nn
lnn
meE
lkn
lklk
meWE
lllklklke
CC
Wmvlle
ddv
dvd
n
k
k
=++−++−
−+−−
−=
++=
==++
−=−=
+−+++++++−
=
==
+−
+−
+
h
h
h
λ
λρρ
λ
ρρ
( ) ( ) ( )
θ
ρ
θπ
ψ
θπ
ψ
πψ
πψ
ρρ
ρ
iar
ar
ar
ar
crnnl
kpkkkk
ppxppx
p
llnnl
eear
a
ear
a
ear
a
ea
mea
erRr
LdxdLxedxdeL
LeRn
lnk
±−±
−
−
−
−−
+−
+−−
−
+
=
=
−
=
=
=
≈∞→
−==
≈−
−−=
)sin(32
1
)cos(32
1
232
1
1
. Define
.,When
1),(
:spolynomial Laguerre associated thearelatter the where),2()( :1 degree of
spolynomial are solutions theSo outside.factor as there'and ,2at s terminateseries The
0
0
0
0
2
0
21
30
1,1,2
2
0
21
30
0,1,2
2
0
21
30
0,0,2
21
30
0,0,1
220
1
00
121
1l
h
[ ])1(32
general,In
state. ground in the nucleus thefromelectron theof distance average the
is hence , when maximum a is This. is thicknessand radius of shell
spherical ain be y toprobabilit theTherefore,),(,1For
20
002
22
11
,1,
0
0
+−=
=
∝
∝−=
−
−−−
−
llnar
aanrredrr
Yrenl
nnar
mn
nnarmnn φθψ
Multi-Electron Atoms
• Due to the Pauli exclusion principle, two electrons cannot occupy the same state if their parameters (energy + angular momentum + z-axis momentum + spin) are identical.
• Hence, first the lowest energy level, is filled: Hydrogen and Helium.There are two electrons there (two spins). Then, for the degeneracy is 4, multiplied by two spins given 8, and this accounts for the following 8 elements (Lithium to Neon). For first the are filled (two electrons: there’s a degeneracy, multiplied by two spins). Same for l=1, for which there are six electrons (3 “l levels”multiplied by two spins). By then we are at 2+8+8=18 (Argon). However, the next electron (in Potassium) does not choose n=3&l=2, but n=4&l=0. WHY?
,1=n
,2=n
,3=n
,1=n
112 =+l0=l
• The answer lies in the interaction of the electron’s charges with each other (which we have neglected). The n=4&l=0 is farther away from the nucleus then the n=3&l=2 electron (see equation for average distance). So, it is partially screened from the Coulomb potential by the charge of the other electrons.
Spin• Spin has no analogue in classical physics.
• An electron is prepared in a state of zero linear momentum (i.e. space-independent) state. Since all the angular momentum operators involve derivatives w.r.t x,y,z, we expect that its angular momentum(s) will all be zero. However, upon measurement along, say, the z-axis, the results of measuring are
• So, the electron has some kind of “intrinsic”angular momentum. Same holds for other particles.
• The wave function of a particle with spin has a few components, and the spin operators linearly mix these components – as opposed to the orbital momentum operators, which mix the coordinates.
• The number of components is determined by the number of values which the “intrinsic” angular momentum can attain. The wave function is then just a direct product unifying the “ordinary” wave function with these different “intrinsic” values.
zL .2h±
[ ]
.,,for satisfied also are relations thesecommute, and since and
,, relationsn commutatio thesatisfies J that assume also We2. isdimension theassume we2, is oft measuremenfor results possible ofnumber
theSince .,, operators theof form theas well as recovered, be tohas components ofnumber theNow
,
matrix. an also is where
.
.
.
...0...
.
.
.0...
.
.
.
toaccording changesfunction wave theaxis, thearoundrotation malinfinitesian Under operator.
spin theS and momentumangular orbital is L where S,LJ then isoperator rotation dgeneralize The
11
zyx
kkijkji
z
zyx
zzzz
z
n
z
n
SSSSL
JiJJ
S
SSS
SLJJiI
nnS
Si
i
i
iI
z
∑=
+=
−=′
×
−
∂∂−
∂∂−
−=
′
′
−
+=
ε
ψεψ
ψ
ψ
ε
θ
θ
ε
ψ
ψ
h
h
h
h
h
h
( )
.21 as defined isspin electron The particle. the tointrinsic isit and changed
becannot spin total themomentum,angular orbital total the Unlikestate. anyon 43 of valuea yieldsit so and
1001
43Sby given is spin total The
.,, oft independenare , that deducecan wezero, yields Poperator
momentum thesince momentum, zero of case For the.componentscomplex twohasIt .spinor a called is This
10
),,(01
),,(),,(),,(:function wave
nent two-compoaby described iselectron theSo10
012
,0
02
,0110
2
:momentumangular on discussion general in the analysis general in the did they asjust
operators, S thedetermine rulesn commutatio The
2
22222
h
h
hhh
=++=
+
=
=
−
=
−=
=
−+
−+−
+
zyx
zyx
SSS
zyx
zyxzyxzyxzyx
Si
iSS
ψψ
ψψψψ
ψ
( )
1 ,21,
21
21,
21
by state general a and10
21,
21,
01
21,
21,
bygiven is basisA axis. thealongspin the and case),our in 21(spin total theis where
,,,, are kets whosespace,Hilbert 2D a todiscussion limit thecan then We
spin. thefromtly independen evolvesfunction waveorbital t themoment tha for the Assume
.1
thereforeiscondition ion normalizat theand ,amplitudesposition andspin thecombine
will),,(notation theHereafter
22
basis
basis
basis
22
=+
→−+=
→−=
→=
−
≡=
=+∫∫∫ −+
±
βαβα
βαχ
ψψ
ψ
z
z
z
S
S
S
z
ms
ms
zms
msmsss
dxdydz
zyx
h
Remember that we’re assuming that
)(),,,(),,,,( ttzyxtzszyx χψψ =
)(tχwhere is a two component spinorindependent of the position.
( )
( )
( )( )
−=−≡
=+≡
±
−
=++=⋅
=
±=±±⋅±
++==
−
−
−
)2cos()2sin(
ˆdown ˆ
)2sin()2cos(
ˆup ˆ
:) 2 seigenvalue (and rseigenvecto
with,)cos()sin(
)sin()cos(2
Sˆ then,)cos(),sin()sin(),cos()sin(
ˆ,ˆ,ˆ Denote direction. ˆ theup/down spin have tosaid are states these
and n)computatiodirect by proved be(can ˆ2,ˆS,ˆ ,ˆr unit vecto
afor S,ˆ of seigenstate theare ,ˆ Ifkji,,S Define
2
2
2
2
2
2
θθ
θθ
θθθθ
θφθφθ
φ
φ
φ
φ
φ
φ
i
i
i
i
i
izzyyxx
zyx
zyxzyx
ee
nn
ee
nn
ee
SnSnSnn
nnnn
nnnnnn
SSSSSS
h
h
h
• Note that knowing determines the state. Actually, instead of specifying a state by we can specify a unit vector such that the state is an eigenstate of with eigenvalue
S
Sˆ ⋅nn, and βα
. 2h
direction.at that momentumangular themeasuringre we'because is This angle.polar ingcorrespond
at the axis its with appartus,-Gerlach Stern aby prepared becan stateany that indicates this:Note
. Sˆ of rseigenvecto theof form general theis this2) offactor aby
division the to(up and ),sin(),cos(can take we,1 Since .
, :opposite are angles that theirassume and s)meaninglesy (physicallfactor
phase aby multiply can We.1that such , numberscomplex twobe let there
:following theisthisseetodirect waymost The
21
22
212
1
22
⋅
===+=
=
=+
−
•
n
ee
i
i
θρθρρρρβ
ρα
βαβα
φ
φ
( )
[ ] [ ]
( )( )
( ) ( ) ( )
( ) ).(T21product inner under the basis lorthonormaan get we,identity theadd weIf
.σˆ4Sˆ so
02
Sˆ2
Sˆ
thatfollowsit ,2 seigenvalue has Sˆ matrixsuch each since :σˆgenerally
more and identity, theequal squaresTheir . tracelessareThey
2,,)2(, :derived yimmediatel becan which nspermutatio
cyclic and ,satisfy They other.each with commute-antiThey
1001
,0
0,
0110
:σ2S σ, matrices Pauli theof definition the toleads operatorsspin theof form The
222
2
ABrI
InIn
InIn
nIn
Ii
i
ii
ijjik
kijkji
zyx
zyx
•
•
•
•
•
=⋅⇒=⋅
=
−⋅
+⋅
±⋅=⋅
==
=
−
=
−=
=
=
+∑
h
hh
h
h
δσσσεσσ
σσσ
σσσ
The eigenvectors span the space, and if the operator is zero on them then it’s zero on all the space.
( )
[ ]
ts).coefficien equatingby y immediatel (follows
σˆ2
sin2
cosσˆ2
exp
2σθSθ)(
that recall latter, For the operator.rotation theand propagator thegcalculatin
in helps which S,ˆ of powers thecalculate easy to relatively sit' ,Sˆ Since 2
⋅
−
=
⋅
−
=⋅−=⋅−=
⋅=⋅
θθθθθ
θ
iIi
iiRU
nIn
ee h
Spin Dynamics (How Spin Behaves in a Magnetic Field)
Bµ
I
• The torque induced by the magnetic field B on the current carrying loop is to align its normal with the field. The torque is given by
Where is the magnetic moment,
B,µT ×=
⊥⋅
= eIµcA
A the loop area, and a unit vector perpendicular to the loop’s plane. See slides 27-8 in electrodynamics file.
• The interaction energy between the loop and Bis given by (so that the gradient of this expression is the force acting on the loop).
⊥e
B⋅− µ
B. aroundµ of precession ain resultsit instead, B;µ with align not to is here T ofeffect The
l.µ parallel, are l andµ Since momentum.angular
of magnitude the ,22
:area theandcurrent theofproduct theiswhich
magnitude a hasmoment magnetic theand ,2
thenis current"" The . radius of circle ain speedat moving , charge and mass of particle aby
replaced loop with theanalysis preceding Carry the
ratio icgyromagnet
2
⋅=
==
=⋅=
=
γγ
ππ
µ
π
mvrllmcq
cr
rqv
rqvI
rvqm
. Bωfrequency aat precessesit so
,)()sin(
of B) to(relative anglean by moves l vector theof tip
theso product),vector their toalproportion
s(it' B and l tolar perpendicu is lBut
)sin(B)l(lB)l(Bµ
lT ismotion
ofequation The
0 γ
γθ
φ
θγγγ
−=
∆−=
∆−=∆
∆=∆⇒
×=∆⇒×=×
==
tBl
l
∆tlBl∆t
dtd
B
ll ∆+
l
φ∆
θ
Explanations:
1. The particle rotates around l
2. T is torque (two slides earlier).
The “real” reason this happens is the colinearityof the magnetic and angular moments.
• What is the equivalent of these ideas in QM?
• The Hamiltonian of a particle of mass m and charge q in a magnetic field is
( ) ( )4444 34444 21
nHamiltonian interactio2
APAAP
22P
2AP
2
2222
mcq
mcq
mmcqH +⋅+⋅−=
−=
( )
L.2
µ take weSo
.expression same by thegiven isn Hamiltonian interactio hein which t
case classical thefrom adopted isanalogy the and operator,moment magnetic dconjecture
a isµ whereB,µBL2
be out to part turnsn interactio the,in term quadratic theDropping k.BA
j)i(2A Taking B.A satisfying potential vector theisA where
int
mcq
mcqH
HB
xyB
≡
⋅−≡⋅−=
==×∇⇒+−==×∇
• The spin magnetic moment is assumed to be of the same shape:
.Bωfrequency aat B around precessesS So, B.)θ( anglean by state therotate toiselectron on theeffect theso
8)-(slides207 S as properties same theshares which J,for proved welikejust θby rotates
operator Sθ but the ,(0)B)S((0) )(by usual asgiven
is (0) stateelectron theofevolution Thecase. orbital thetwice iswhich
)( B,S: theory)and experimentby both justified
(assatisfy out to urnsconstant t related TheBBµ
0
int
γγ
ψγψψ
ψ
γγ
σ
−=−=
−−
⋅−⋅=−=
−=⋅−=
⋅∝⋅−=
tt
iti
Htit
mceH
H
eee
hh
h
( )( )
( )( )
( ) ( ) ...22,02, :etc , alongspin tochange it will axis,
thealongspin with startsit if example,For 2sin2cos)(then
2sin2cos,ˆ)0(in starts
electronan If . 2002
)(
equalsexponent thediagonal is since and,
2)( So k.B axis,
thealongbetoBconsider example,For
of eigenstate of eigenstate
2)(
2)(
2
2
0
0
0
0
0
→==→==
−
=
→+=
−=
===−
−
−−
−
44 344 214434421
h
ySxS
ti
ti
i
i
yx
t
n
titi
tU
zti
BztSitUBz
ee
ee
ee
ee
z
πφπθφπθ
θθψ
θθψ
ωωσ
σω
γ
ωφ
ωφ
φ
φ
Which is precession around the z-axis.
Energy Levels of an Electron in a Magnetic Field
• As noted before, if the Hamiltonian separates into a spin and orbital parts, we have twice the number of orbital states – each one will now have a spin.
• Now assume there’s a magnetic field B=Bkpresent. For the hydrogen atom, the coupling of the proton to B can be ignored because its mass to charge ratio is much higher than the electron’s. For the electron, the combined Hamiltonian is
( )21±
sss
ss
zz
zz
nlmmmmmc
eBn
nlmmH
mnlmmSLLH
H
SmceBL
mceBHH
++
−=
−−
−−=
)2(2
R:different are seigenvalue theHowever,
spin). theis ( :before as states same theby eddiagonaliz isit , and ,,, with commutes Since large). as twiceisspin
theoft coefficien coupling that the(remember 2
2y
2Coulomb
Coulomb
h
( )
.effectZeeman thenamed is splitting This2
)21,1( 2 )1,0,21,0( 1
)21,1or 21,1( 0 )1,0,21,0( 1 )21,1( 2
R
:equal are 2in which statesin remains degeneracyonly the:levels five into splits
now ,degenerate eightfold which waslevel, 2
The ).degeneracy no (so 2
R now
are levelsenergy the thereforeand ,21but ,0 levels; twohas now 2, of leveldegenracy a had which ,)1( levelfirst theexample,For
degenracy. 2 theofreduction in the results This
)2(2
R
1
1y2
y1
2
2y
=
−=−=−=−==−
=−=−=====
==
⋅±−=
+
=
±−=
±==
=
++
−=
=
=
mceBc
mmlmm
mmmmlmm
mm
cE
mm
nmc
eBE
mm
nn
nlmmmmmc
eBn
nlmmH
s
s
ss
s
s
n
s
n
s
sss
h
h
h
The Stern-Gerlach Experiment Revisited
• As in classical electrodynamics, an inhomogenousmagnetic field creates a force on a dipole, which is
given by k (assuming the field is in the z-
direction). So particles with positive spin will go upwards, and those with negative spin downwards. By blocking one of the paths, we can ensure that the particle leaving the apparatus will have a definite and known spin in the z-direction (i.e. we know that it is an eigenstate of and we know the eigenvalue). This results in a space quantization of the emerging particles (they’re not continuously scattered, but emerge only at distinct points corresponding to the different spins; spin half will concentrate at two points etc). Note, especially, that not only the total spin is quantized, but the spin at every axis as well.
zBz
z ∂∂µ
,zS
• Suppose we have a sequence of SG apparatuses. We may ask what percentage of those leaving apparatus i will leave apparatus i+1 (see slides 48-51). We can now solve this problem rigorously, since the equations for the behavior of spin under rotation of the axis have been derived.
• The electron is in a mixed spin state .10
01
+
βα
Due to the heavy proton, the behavior is semi-classical, so the spin up part induces an upwards moving wave-packet, and similarly for the spin down part. So the wave packet has really two components – upwards and downwards. The blocking is like a measurement, and it forces the spin into a single state; if the lower beam is blocked, we get “pure” spin up.
Examples• Spin half particles are moving along the y-axis (see drawing) enter two collinear SGAs, the first with its field along the z-axis and the second’s field is along the x-axis (i.e. it is the first, rotated by What portion of those leaving the first will exit the second, if the lower beams in the two SGAs are blocked?
• Answer: we can think of the particles leaving the first being rotated by and then entering an SGA identical to the first. The equation for rotation of spin half particles was developed before (slide 210):
).2π
2π
[ ]
σˆ2
sin2
cosσˆ2
exp
2σθSθ)θ(
2 ⋅
−
=
⋅
−
=⋅−=⋅−=
θθθθθ iIi
iiRU ee h
In our case
( )
−=−
=⇒==
1111
21
21
θ)]([2),0,1,0(ˆ
2 yiI
RU
σ
πθθ Note: all calculations are carried out in the
basiszS
• Since the lower path of the first SGA is blocked,
those emerging from it are in a state and after
acts on them they become
The probability for them to exit the second SGA (which after the rotation also has its field pointing up the z-axis), equals their probability to yield 1 when their z-spin is measured, i.e. the square of the projection of the state vector on the eigenvector of
with eigenvalue 1, but this eigenvector is so the probability is clearly
• Now, assume that another SGA is placed after the second, which transmits only spin down along the z-axis. How many of those entering it will exit? We know that those leaving the second have a spin up along x, so their state vector is the eigenvector
of with eigenvalue 1, which equals
,01
. 11
21
)]([ θRU
zS ,01
. 21
xS . 11
21
The probability of this particle going through a z-spin down SGA is clearly and so of those leaving the first will exit the third.
• Note that if the middle SGA has both beams unblocked (i.e. all those entering it leave it), then no particle which leaves the first will exit the third (they leave the first with spin up and all are therefore blocked by the third). So, the blocking in the middle one increases the percentage of those leaving the third.
• Other way to compute: to see what happens with the particle after it leaves the first (and is in a state in the z-basis) upon entering the rotated SGA: simplest way to understand it is to rotate the entire system by since then we’re back with an SGA with its field along z, calculate there, and rotate back by
,21 41
t)0,1(
,2π−
.2π
SGA’s Acting on Spin One Particles
• The computation proceeds very much like for spin half. Suppose that the first SGA passes only z-spin up particles, and the second one is rotated around the y-axis by an angle What percentage of those exiting the first will leave the second?
• Answer: as before, we compute what such a rotation does to the particle’s state. Just like for spin half, the rotation operator is defined by
.θ
[ ] hSθ)θ( ⋅−= iRU eTo compute the exponents of spin one operators, note that, since the eigenvalues are then
The Pauli matrices are defined just like for spin half, but here we shall denote for example,
( ) ( ) S)ˆ(S)ˆ(0SˆS)ˆ(Sˆ 3 ⋅=⋅⇒=−⋅⋅+⋅ nnInnIn hh
h±,0
).2(not σS hh=
−
−=
000
00
21σ
iii
i
y
See slides 207-8
[ ]
[ ]
SGA.
second exit the llportion wi that and , 2
2)cos(1
is upspin in be y toprobabilit its and2
)cos(1,2
)sin(,2
)cos(1 tormsit transfo, up,
spin with particle a with started wesince So,2
)cos(12
)sin(2
)cos(12
)sin()cos(2
)sin(2
)cos(12
)sin(2
)cos(1
)sin(1)cos( : termsCollecting
) :(remember ...!4
1!3
1!2
1Sθ
)θ( and ,0),(0,θ case,our in So
001
23
3243
223
+
−
−+
−
+−
−
−−
+
=−−+
=+
+−−=−
=⋅−
==
θ
θθθ
θθθ
θθθ
θθθ
σθσθ
σσσθσθ
σθθσθσ
θ
z
z
iI
i
iIyii
RU
yy
yyyy
yyee h
Addition of Angular Momenta
• Question: how does a system with two spin half particles appear (in terms of spin) when viewed as a single object?
• Such systems are naturally described in QM by the direct (tensor) product. For example, the system with two spin half particles is the vector space of dimension 4 spanned by
−
−−+−−+++
+=
1000
0000
0000
0001 is basisproduct in this
ofmatrix Then the . Define
etc. component, second on theidentity the toequal is andcomponent first on the usual as acts
21
1
h
z
zzz
z
SSSS
S
−−+−−+++
++=+
⋅+=
2000
0110
0110
0002
is SS2SS )SS(
)SS(S ofmatrix theand
2
2122
2121
212
h
321
As can be verified by directly computing the tensor product.
, , , −−+−−+++
zS
zS
yS
yS
xS
xS
21
21
21
+
+
However, the following basis diagonalizes
( )
( )
1. is spin totaleclearly th here so ,)1( of eigenvalue
withseigenstate hasoperator momentum total theoperators, momentumangular generalfor
developed formalism the toaccordingBut .2by state themultipliesit it,on operates Swhen
because in 1 Thus, axis. in thespin totalandspin totalfor the stand , that Note
2 :0,0
:1,1
2 :0,1
:1,1
:S),SSS(S,S,S
2
2
2
21
21
22
21
21
21
21
2
h
h
+
++=−
+−−−+==
−−−==
+−+−+==
++==
++=
−
−
ll
szms
ms
ms
ms
mszyxz
.0,0,1,1,0,1,1,1:by spannedisstateshalfspin twoof system theSo==−====== msmsmsms
• This means that the tensor product of two spin-half particles behaves like the direct sum of a spin one and spin zero particle – since each state can be viewed as a linear combination of the three spin one states (l=1,m=1,0,-1) and the spin zero state (l=m=0) (careful – the spin one particles requires three, not one, scalars to describe its state!).
• And in general, for two particles of spin and
0121
21
⊕=⊗
1j :2j
( ) ( ) )(...1 21212121 jjjjjjjj −⊕⊕−+⊕+=⊗
As follows by counting arguments and from considering the upper and lower bounds for the total spin. It remains to calculate the coefficients of the basis transformation. All this carries over to systems with more than two particles, e.g.
21
21
23
21
21
21
⊕⊕=⊗⊗
• Formally, we write
The Hyperfine Interaction in Hydrogen
• In addition to the Coulomb interaction, there’s an interaction Hamiltonian of the form between the proton and electron, due to their magnetic moment. It splits the ground state to two
levels:
21SSAHhf =
Singlet
Triplets
01
10
,3
1000
1,
0001
1,
0110
1,
rseigenvecto and seigenvalue haswhich 1000012002100001
43421444444 3444444 21−+−−+−−++−++−+
−−+−−+++
−−
−−
=⊗+⊗+⊗ zzyyxx σσσσσσ
43 ,4 22 hh ARARE yy −−+−=+
isit constants toup So
SSSSSSSS ofmatrix thecompute :Proof
21212121 zzyyxx ⋅+⋅+⋅=⋅
A Very Short Introduction to Scattering
• Assume that a plane wave hits a potential centered around the origin, and which decreases faster than (so the potential has no effect at infinity).
• Asymptotically the scattered wave will behave like a free particle. The solutions are known to be of the form
for the wave to be purely outgoing, this must equal
and the total wave is therefore
)exp(ikz
r1
),()2/cos()2/sin(,
φθππ ml
ml
ll Ykr
lkrBkr
lkrA∑
−
+−
( )∞→r
),,()exp( φθfrikr
ions.considerat physicalfrom guessed becan expression This parts. reflected and
incident theof sum theis, that ),,()exp()exp( φθfrikrikz +
• The stationary states satisfy (up to a choice of constants) ( is proportional to the energy).
• The cross-section (amount of particles scattered per angle) is proportional to It also turns out that, asymptotically, the scattered current is only radial.
[ ] 0r)( r)(2 =−+∆ ψUk 2k
.),( 2φθf
The Integral Scattering Equation and Green’s Function
[ ] [ ] r)()r(r)( 0r)( r)( 22 ψψψ UkUk =+∆⇒=−+∆
• Try to solve with a Green function for the operator:[ ]
[ ] ( )
which, with rr replacecan r wedenominato in the r; of
direction in ther unit vecto a is where),r,(rrthen thatsee easy to isIt . integrand, theof range meaningful
in the that assumecan we,hen behavior w thein interested re we'and limited,spatially is potential theSince
r)r()r(rr
)rrexp()exp()r(
thereforeisequation Thehere. interested ree'in which w wave
outgoing theis )r( ),)iexp(- constant) a to(up )r(and )exp( are problem scattering for the
solutions Meaningful ).r( of powersin solution iterativean allowsequation integral The ).r()r()r()r((r)
)r()r()r()r()r()r()r()r(A:0)r( equation, homogenous theto
solution a is )r( where),r()r()r()r()r(satisfies alsosolution a then and (r),(r)
such that G(r) find ),r()r()r(equation andoperator an given :idea The .)r()r(
3
0
0
0
00
2
r
uurrr
r
dUik
ikz
GrkrGikz
UUU
UGAUGAA
UGAG
UAAGk
′−
′−≈′−
′>>∞→
′′′′−
′−−=
±=
=∗=∗=∗+=
=∗+=
==
=+∆
∫
+±
ψψ
ψ
ψψδψψψψ
ψψψψψ
δψψ
δ
( )
.,on only depends since
,,)exp()exp( form in the indeed iswhich
r)r()r()ruexp()exp()exp((r)
ion approximat asymptotic theyields
3
φθ
φθ
ψψ
r
frikrikz
dUikrikrikz
+
′′′′−−≈ ∫
• It is customary to use the following notations:
.kkK is ,direction in theor
The u.kby defined is ,direction in the theSimilarly, r).kexp()exp(
so beam, theof axis the thealong directed isk which modulus of vector theis ,k , The
id
d
i
i
ector ed) wave v transferr(scatteringkvectorwavescattered iikz
rwave vectoincident
−=
=⋅=
φθ
φθ
( )
( )
( )( )
etc. potential, thefromagain scattered s"scatteringsecondary " torelateexpansion
order Higher amplitude. scattering the topotential theof ansformFourier tr therelatesIt ion.approximat
Born theis which ,r)r()r Kexp(,
r)rkexp()r( rkexp)exp(
r)rkexp()r( ruexp)exp(r)rkexp()r()rr( using and
expansionBorn order first theinto ,)exp(r)kexp( Plugging small. is potential
theif useful is which ,expansionBorn theis This...rr)r()r()r()rr()rr(
r)rkexp()r()rr(r)kexp(
r)r()r()rr(r)kexp(
3
3
3
3
33
3
3
′′′⋅−−=
⇒′′⋅′′⋅−−
=′′⋅′′⋅−−
≈′′⋅′′−
+⋅=
+′′′′′′′′′′−′′−
+′′⋅′′−+
=′′′′−+=
∫∫
∫∫
∫∫∫∫
+
++
+
+
dUif
diUirikr
diUikrikr
diUG
frikr
i
ddUUGG
diUGi
dUGi
id
i
i
i
ii
i
φθ
φθ
ψ
ψ
ψψ
The Born Approximation
Some examples of the Born approximation (left – the potential, right – dependence of on . This is 2D, the incoming wave is at the x-direction).
)(θf θ
Partial Waves
( )( )
( )( )
( )( )
( )( )θδπδπψ
θπ
π
θπθ
θθ
φθπ
cos))2/(exp()2/(exp()(
denote so phase; in the becan particle free the torelative differenceonly heinfinity tat then present, is potential a If
state).steady in the flowy probabilit no be should there(sinceout cancel which currentsy probabilit opposite induce waves twoThe barrier. lcentrifuga the todue difference phase awith
amplitude, same theof wavesoutgoing and incoming are there
each for thus;cos))(exp()exp()12(21
)exp( potential) (no particle free
for theget we,)2sin()( Since .
l thecalled are )( .)(12
4cos
cos)()12(),( can write weSimilarly,
).dependence no( )()()12(4)exp(
0
0
021
0
0
0
lll
llrk
ll
r
rl
lll
lll
lll
l
Pr
lkrilkriAr
lPr
lkriikrlik
ikzkr
lkrkrjamplitudewave
partialthkaYl
P
Pkalkf
Ykrjliikz
+−−+−→
−−+
→
−→
+=
+=
+=
∑
∑
∑
∑
∞
=∞→
∞
=
∞→
∞→
∞
=
∞
=
• If the potential is rotationally symmetric (central), the scattered wave can be studied by separately analyzing its components which have fixed angular momentum.
[ ]
( )
origin). thefromdistance sparticle' theis where,roughly is momentum
angular the thereforeand is momentum the(sincescatter don't s'such potential theofextent n thelarger tha is this
if and , than moreorigin thecloser to comedoesn't particle thefor since ,considered be toneed )( ofnumber small aonly such that are potential theof range the
and ,, when is wavespartial use tomotivationmain Theorigin. theof vicinity the
at statesteady theofsolution theandinfinity at solution ebetween th continuity imposes one shift, phase thefind To
)(sin)12(4
: theoffunction a as expressedsimply becan section crossThe wave.outgoing the to)2exp(shift phase aattach to
thereforeis potential theofeffect The .2
1)2exp()(
defined, was),( how Recalling .)exp(
)(cos(2
1)2exp()12()exp())(cos(
)(exp()2exp()exp()12(2
1)r(
and )),2/(exp(2
12 so same, thebe
must waveincomingtheparticle,free with theComparing
00
max
max
0
22
0
0
rkrk
lkl
llka
kl
lk
iikika
frikr
PikilikzP
lkriiikrlikr
liiklA
l
ll
l
l
ll
ll
ll
llrk
ll
≈
>
+=
−=
×
−++=
×−−−+→
++
=
∑
∑
∑
∞
=
∞
=
∞
=∞→
δπσ
δδ
δ
φθ
θδθ
πδψ
δπ
Example: scattering of l=1 from a hard sphere
( ) ( ):at continuity enforce toremainsit Now ).tan(
so ),sin(sin)cos(cos)2sin(factor)constant a to(up equal should , as
which,)sin()cos()cos()sin()(
2 toreduces
equation radial theoutside, In the infinite). is potential the(since zero isfunction wave thesphere, theInside
.1 and sphere hard for theshift phase thecompute sLet'
shift. desired theis ).2sin()( :2 shift, particle free the torelative )(in shift
phase thecompute tohasonly one ),scatteringon slidesprevious (see by divison ealready th includes )( Since
).(2
)()(2
)1(2
isequation The ).Problems"Invariant lyRotational ofSolution " of slides (see)(in
equation radial theofsolution theplus part) homogenous(the )exp( form theof isIt . as waveoutgoingat thelook :follows as issolution of method general The
.for )( ,for 0)( is potential The
01
111
222
2
,
,
,
,
22
,2
2
2
22
,
00
rrakrkrkr
r
krkr
krakrkr
krCru
ukurdr
ud
l
lkrCrulru
rru
rukrurVr
lldrd
rRru
ikzr
rrrVrrrV
llrlk
lk
lk
lklk
lk
==+−=+−
∞→
++−=
⇒=
+−
=
+−≈
=
+
++−
=
∞→
<=∞>=
∞→
δδδδπ
δδππ
µµµhhh
( )
( ) .)( ,every for out that it turns222) slide (see functionsNeumann and Bessel spherical theof properties asymptotic Using).( oforder thealso is
this),tan( angles smallfor Since .order theof is
)0( energies lowin very which ,)sin()cos(
)cos()sin(
0)sin()cos()cos()sin(havemust weso there,zero be tohasfunction waveThe
120
1
30
00
0
00
0
00
00
0
0
+≈
≈
→+
−−=
⇒=
++−
ll krkl
kkr
kkr
krkr
krkr
kr
a
krkr
krakrkr
kr
δ
δθθ
![Eisberg R & Resnick R - Quantum Physics - Atoms, Molecules, Solids, Nuclei, And Particles - Solutions Supplement Accompany (2Ed Wiley)[1]](https://static.fdocuments.us/doc/165x107/5572071b497959fc0b8ba297/eisberg-r-resnick-r-quantum-physics-atoms-molecules-solids-nuclei-and-particles-solutions-supplement-accompany-2ed-wiley1.jpg)
