Post on 16-Jan-2016
3.1 Schrödinger Equation
Postulate 6: Schrödinger EquationThe time evolution of a quantum system is determined by the Hamiltonian or total energy operator H(t) through the Schrödinger equation
)()()( ttHtdt
di
= Operator corresponding to total energyDerived from classical Hamiltonian•What kind of terms do we expect it could contain?•How is H different from E?
Let’s see what this equation tells us about our states if H is time-independent
is a functional: traditionally a map from a vector space of functions usually to real numbers. In other words, it is a function that takes functions as its argument and returns a real number (or it can be written in matrix form as an operator)
H is an observable Hermitian operator Eigenvectors form a complete basis.
i
iii
ii EtctEc )()(
Let’s write the eigenvalue equation for this operator (assume H is time-independent for now)
iii EEEH Ei : energy eigenvalue
•As an operator corresponding to an observable, what properties do we expect it to have?•If we diagnonalize it, what will the eigenvalues correspond to?
Combine the above equation with the Schrodinger equation to determine what we can say about the eigenvectors (eigenstates)
3.1 Schrödinger Equation
Postulate 6: Schrödinger EquationThe time evolution of a quantum system is determined by the Hamiltonian or total energy operator H(t) through the Schrödinger equation
)()()( ttHtdt
di
= Operator representing total energy
Argued Friday: If H is time-independent, and we write this in the continuous functional form: ih(dψ(t)/dt) = Hψ(t)
The time derivative returns itself, with an i out frontSo we can write ψ(t) = eiαtn, where n is time-independent
So in Dirac notation we may expect: |ψ(t)> = |eiαtn> is a functional: a map from a vector space of functions to numbers. In other words, it is a function that takes functions as its argument and returns a number – often used in the calculus of variations to find the minimizing function – here that would be finding the state that minimizes energy
i
iii
ii EtctEc )()(
Let’s write the eigenvalue equation for this operator (assume H is time-independent for now)
iii EEEH Ei : energy eigenvalue
As an operator corresponding to an observable, we know it’s Hermitian, so it must contain a complete basis of eigenkets and eigenvalues which correspond to energies of our system
Here |Ei> is our |n>, we’ve taken out the time-dependence**NOTE H is an operator, Ei are real, scalar,
eigenvalues – the Hamiltonian is NOT the energy!Let’s put time back into the states – if the full eigenstates |ψ(t)> had no time-dependence then d/dt would yield zero, but the time dependence doesn’t “live” in the eigenstate of the hamiltonian since H has no time dependence. AND, the eigenvectors must satisfy the completeness relationship, so we must be able to write:
Shove into the Schrödinger equation
Use orthonormality to simplify
Verify what our understanding of continuous functions told us
Simplify
Get general form of |ψ(t)> for all time-independent Hamiltonians:
|ψ(t)> = ΣiCoe-iwit|Ei>
that’s an wi in the exponential! (Yucky microsoft… )
i
iii
ii EtctEc )()(
What will happen if our system starts in one particular energy eigenstate at t=0, say |E2>, then we watch it for some time t – how do we write the state at time t? •How will the probability of finding the state with energy E2 (or any energy) after time t differ from at time t=0?•How will the probabilities associated with any observations of this state after time t differ from at t=0?
What name might you give such an initial “pure” state to explain how it behaves with time?
If our initial state at t=0 is a|E1> + b|E2>, what will be our state after some time t? •How will the probability of finding the state with energy E2 (or any energy) after time t differ from at time t=0?•How will the probabilities associated with any observations of this state after time t differ from at t=0?
Reminders: found that the energy eigenstates are “stationary states”, and that if H is time independent, we can expand any state as: |ψ(t)> = ΣiCie-iE
it/ħ|Ei>
Refresher calculation: our initial state at t=0 is a|E1> + b|E2>, what will be our state after some time t, and what is the probability of finding the state with energy E2 (or any energy) at any time t?
Your turn: limit to 2-level system for simplicity•How will the probabilities associated with any observations of this state after time t differ from at t=0?
Energy eigenstates• Regardless of our initial energy eigenstate, the
probability of observing the energy of the state does not change with time (that’s because our Hamiltonian, which represents total energy, is time independent!)
• (This is true for any observable that commutes with H as well – since commuting observables share eigenstates)
• However, if we are considering a general observable, our probability will oscillate with time!
• angular frequency of the “time evolution” = Bohr Frequency ω12 = E2-E1/ħ
Measurement of some other observable A – say position, momentum, or spin!!
(1) If [A, H ] = 0, then observations won’t change with time – “stationary states” since A and H must share eigenstates
(2) If [A, H ] 0, then observations will oscillate with time
Look at example of time evolution of square well states using the PhET simulation (http://phet.colorado.edu)
Explain what you observe in terms of our new understanding of “time evolution” of quantum states
Recipe for solving a standard time-dependent quantum mechanics problem with a time-independent Hamiltonian
Given Hamiltonian H and an initial state |y(0), what is the probability that an is measured at time t?
• Diagonalize H (find eigenvalues Ei and eigenvectors |Ei).
• Write |y(0) in terms of energy eigenstates |Ei.
ii
tii
ii
tE
i
ii
ii EecEectEc i
i )()0(
tE
i i
e
• Multiply each eigenstate coefficient by to get |y(t).
2)()( taaP nn • Calculate probability
3.2 Spin Precession
Scm
e
Scm
qg
BScm
eBUKH
e
e
e
2where
Hamiltonian of a spin-1/2 system in a uniform magnetic field (The electron g-factor is a bit more than two, and has been measured to twelve decimal places: 2.0023193043617)
B
SN
Find our measurements depend on energy differences, so our Hamiltonian needs to only include terms that will involve energy differences in the two possible spin states
•Only the dipole potential energy does this: recall that U = -μ·B•Choice for zero point of potential energy is arbitrary if we only care about energy differences•No kinetic term is needed here, K=0 below
3.2.1 Magnetic Field in z-direction
cm
eBSS
cm
eBzBS
cm
eBH
sozBB
ezz
ee
000
00
0
,
,
The uniform magnetic field is directed along the z-axis.
0, zSH H and Sz share common eigenstates. Therefore, what would we expect about measurement of Sz?
This is a time independent hamiltonian!!
10
01
2: oHor
Eigenvalue equations for the Hamiltonian
ESH
ESH
z
z
2
2
00
00
Eigenvalues andEigenvectors
EE
EE
,2
,2
0
0
We can just “write down” the eigenstates since they must be the same as for H – since Sz and H commute
The probability for measuring the spin to be up along the z-axis
)0(
20
)(t
itE
ieet
The initial state is an energy eigenstate.
The time evolved state simply has a phase factor in front.
therefore there is no physical change of the state with time
Example (1) The initial state is spin up along z-axis:
1)()(
2
22 0
ti
etP
time independent
|+ and |- are stationary states, if we start purely in one of them, we stay there!.
Example (2) The most general initial state:
2
sin2
cos)0( i
ne
In matrix formalism
2sin
2cos
2sin
2cos
)(
2sin
2cos
)0(
2
2
0
0
it
i
ti
itE
i
tE
i
i
ee
e
ee
et
e
2sin
2cos
0
0
2
ti
ti
ee
2sin
2cos)(
itE
itE
ieeet
This overall phase out front does not effect measurements
Only this term changes – it is rotated to a new angle phi, but theta has stayed the same!
White board activity
• Find the probability finding the general state in spin up in z, in x, or in y (row 1, 2 and 3):
|ψ(t)>
2sin
2cos
0
0
2
ti
ti
ee
2
cos2
cos
2sin
2cos
01)()( 2
2
2
2
)(
22 0
0
0
t
i
ti
ti
ee
etP
Probability for measuring the spin projection along the z-axis:
• Time independent since the Sz eigenstates are also the energy eigenstates, and are therefore stationary states.• Consistent with the fact we found the polar angle q to be constant.
2
)(
2
)(
22
2sin
2cos
2
1
2sin
2cos
112
1)()( 0
0
0
ti
ti
ti
xx ee
etP
Probability for measuring the spin projection along the x-axis:
• Time dependent since the Sx eigenstates are not stationary states ([Sz,Sx] is not zero).
2sin
2sin
2cos
2cos
2
1 2)()(2 00 titi ee
)cos(sin12
10 t cossin2)2sin(: use
White board activity
• Find <Sx>
• <Sy>
• And <Sz>
• For the general state |ψ(t)>
2sin
2cos
0
0
2
ti
ti
ee
Expectation Value of Spin Angular Momentum:
)()()(2
)(2
tStPPS zz
2sin
2cos
10
01
22sin
2cos
)(
2)(2
0
0
0
0
ti
titi
ti
eeee
cos
22sin
2cos
222
2sin
2cos
2sin
2cos
2 )(
)(
0
0
ti
ti
ee
)cos(sin22
sin2
cos2
2cos
2sin
2sin
2cos
2
2sin
2cos
01
10
22sin
2cos
)()(
0)()(
)(
)(
)(
2)(2
00
0
0
0
0
0
0
tee
ee
eeee
tStS
titi
ti
ti
ti
titi
ti
xx
)sin(sin22
sin2
cos2
2cos
2sin
2sin
2cos
2
2sin
2cos
0
0
22sin
2cos
)()(
0)()(
)(
)(
)(
2)(2
00
0
0
0
0
0
0
tieie
i
iee
ee
i
iee
tStS
titi
ti
ti
ti
titi
ti
yy
),(ˆ2
cosˆ)sin(sinˆ)cos(sinˆ2
ˆˆˆ)(
000 tnztytx
zSySxStS zyx
zB ˆ0
x
yf +w0 t
q
f
Spin Precession
),(ˆ2
)0(
n
S
q
w0 t
),(ˆ2
)( 0tntS Larmor frequency:
frequency of precession
cm
eB
e
00
),(ˆ2
cosˆ)sin(sinˆ)cos(sinˆ2
ˆˆˆ)(
000 tnztytx
zSySxStS zyx
What is <S(t)>:
• if we start in a |+>x state?
• If we start in a |+>y state?
• If we start in a |+> state?
What does this mean?
),(ˆ2
cosˆ)sin(sinˆ)cos(sinˆ2
ˆˆˆ)(
000 tnztytx
zSySxStS zyx
Classical expectations:
• Assume the magnetic moment is aligned with its angular momentum
• γ is the gyromagnetic ratio, proportional to q/2m
• Torque will be perpendicular to the moment and to B! It will change the direction of the angular momentum – just like precession of a top
• If γ>0, precession is clockwise. In the spin-1/2 case for electron, it is negative (spin and magnetic moment are anti-parallel) so the precession is counterclockwise
• This precession shows our system has angular momentum!!
Bdt
d
Bdt
d
dt
dL
L
B
)/(
3.2.2 Magnetic Field in general direction
Larmor frequencies:cm
eB
cm
eB
ee
11
00 ,
xzzxee
SSBSBScm
eBS
cm
eBH 1001 )(
Hamiltonian:
01
1010
201
10
210
01
2
H
In matrix representation
xBzBB ˆˆ 10 B
0B
1B
q0
1tanB
B
This will not have the same eigenvalues and eigenvectors as the last case!We must diagonalize it!!
21
20
2
1
2
02
2
100
01
10
20
22
0222
0
22
22
21
20
0
21
20
1
21
20
1
21
20
0
21
20
01
10
22
H
NOW DO A TRICK:Rewrite the Hamiltonian 0
1
0
1tan
B
B qw0
w1
21
20
1
21
20
0
sin
cos
Characteristic equation: 0 HI
What does this become if B is along z, so Bx = 0?
n
n
zxn
SH
S
SSnSS
xzn
H
21
20
21
20
cossin
sincos
2cos
10
01
2sin
01
10
2
cossin
sincos
cossin
sincos
2
Now we simply get:
Let n be the unit vector in the direction of the magnetic field:
qw0
w1
21
20
1
21
20
0
sin
cos
Now we’re using our field direction to define the coordinate system:This makes sense – we’re shifting our perspective to a new axis z’ which has been rotated from z by an angle of theta
2cos
2sin
2sin
2cos
n
nEigenstates:21
202
E
21
202
E
Initial state
nn
nnnn
nnnn
nnnn
2sin
2cos
)0(
n
tE
i
n
tE
ieet
2sin
2cos)(
Time evolved state
zB
We already know how to deal with this system, because we know the eigenstates corresponding to Sn:
We want to see if it is possible for a system that starts in the state |+> to end up in the state |-> (or vice versa) in this magnetic field that is aligned along z and x Let’s start in |+>:
0B B
1B
q
x
z
zB
)(t
2cos
2sin
2sin
2cos
n
n
2)()( tP
Probability of a spin flip
)0(
tE
tEE
tEE
e
ee
ee
ee
tEE
i
tE
itE
i
n
tE
i
n
tE
i
n
tE
i
n
tE
i
2sinsin
2
)(sinsin
)(cos1sin
2
11
2cos
2sin
2cos
2sin
2sin
2cos
2sin
2cos
2sin
2cos
2222
2
2)(22
2
2
2
Probability oscillates with frequency dependent on delta(E)
21
20
1sin
21
202
ESince and, we can rewrite the probability:
tP2
sin)(21
202
21
20
21
Rabi Formula
t
2sin
)( P
E2
tE
P2
sinsin)( 22
zB
tE
P2
sinsin)( 22 Note if B = Bz, θ is zero, and no spin flip occurs – which is what we would expect. That system just precesses around the z-axis!However, if B = Bx, sin(θ)=1, and we get maximum probability for the flip
The probability is always less than one if the angle is between 0 and 90, so the flip is never 100% certain to happen
This is generalizable to any 2-state system with non-diagonal H – the probability of being in a non-eigenstate will oscillate between two values
Understanding “spin space”
• You know 3-d space isn’t the same as our spin space– Evidence: the “length” of the spin vector is longer
than we can ever measure – more or less meaning we can’t align real space perfectly parallel to spin space
– We must be careful when interpreting any physical picture of spin! (they are all “unsatisfactory” and can be misleading if this fact is forgotten)
In words of QM theorist at Los Alamos:
• Quantum objects are like what classical things look to someone with bad eyesight: a quantum spin half object is like a cylinder with one red end and one green end looked at by someone with such bad eyesight that it looks more like a fuzzy sphere. So, the red side basically points out a hemisphere towards which it points. The abstract vector space is a space in which every classical hemisphere (or smaller solid angle for larger spin value) is a point. You can naively associate the central direction as the classical vector, as long as you remember the irreducible fuzzyness.
• the superposition principle that lies behind the construction of the state vector space is the most difficult bit in QM. The hemisphere picture works somewhat but is not very clean… I know that almost everybody involved in understanding why quantum computation is faster than classical computation for some problems would like to understand this. It is also at the heart of most controversies about interpretation of quantum mechanics. So, if textbooks don't have a good picture, its because the author doesn't.
Some possibly interesting papers
• “Delirium Quantum: Or where I will take quantum mechanics if it will let me” by Christopher Fuchs at Bell Labs, 2009
• http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1968v1.pdf
• “Nonlinear Quantum Mechanics, the Superposition Principle, and the Quantum Measurement Problem” by Kinjalk Lochan and T.P. Singh atTata Institute, 2009
• http://arxiv.org/PS_cache/arxiv/pdf/0912/0912.2845v2.pdf
Extra practice to help with lab
• Write down the projection operator needed for combining the two ports from the S-G along X
• Calculate the probabilities you’d get out of the |+> and out of the |-> ports on the last S-G device
• Consider the spin-1 case, and write down the projection operator for combining multiple ports as needed for lab 3
Neutrino Oscillations• What are neutrinos?• http://www.particleadventure.org/index.html• Tau neutrino discovered at Fermilab in 2000 – strong
evidence for 3 “generations” of standard model• 1998 was first report of evidence for neutrino
oscillation (at Super-K in Japan)• “missing solar neutrinos” in 2001 – realized to be
due to oscillations– Proton mass is ~1GeV– Electron mass ~1MeV– Neutrino mass <1eV!
Neutrino Oscillations:Produced via the weak interaction:• |Ve> and |Vμ> are eigenstates of the hamiltonian for
the weak interaction – in weak processes they will be stationary states
However, in “free space” (i.e., when they are produced in the sun and travel to the earth) their energy is described by the Hamiltonian for their relativistic energy
• Eigenstates for this hamiltonian are NOT the same – IF these two neutrinos don’t have the same mass - these two hamiltonians won’t commute!
• Write the “free space” eigenstates as |V1> and |V2>
“mixing”• 2-level system (if we consider only electron and
muon neutrino, tau is less common) – expand it using our spin ½ knowledge:– |Ve> = cos(θ/2)|V1> + sin(θ/2)|V2> – |Vμ> = sin(θ/2)|V1> - cos(θ/2)|V2>
• Call θ/2 the “mixing angle”• If we start with an electron neutrino |Ve> write
how our state would look with time. (think carefully: What eigenvalues are you using in your exponent – the ones from the weak hamiltonian or the relativistic hamiltonian?)
Simplify the relativistic energy eigenvalues
• Ei = √[(pc)2 + (mic2)2]• (note the energies would be the same if they
had the same mass)• Simplify using the fact that mc2 << pc• Write the probability for finding this neutrino
that started as an electron neutrino to later be a muon neutrino
• Simplify E1-E2, and use t = L/c, where L is the distance from the sun to the earth
“Super-K” (Kamiokande)• http://en.wikipedia.org/wiki/Super-Kamiokande
Why are neutrinos massive?• In Standard Model, fermions have mass because of
interactions with Higgs field (Higgs boson), but this can’t explain neutrino mass
• Mass of neutrino is at least 500,000 times smaller than the mass of an electron, so any correction to standard model wouldn’t explain why the mass is SO small
• This remains unexplained
• http://en.wikipedia.org/wiki/Neutrino_oscillations
Extending our knowledge• Cabibbo–Kobayashi–Maskawa matrix• specifies the mismatch of quantum states of quarks
when they propagate freely and when they take part in the weak interactions
• IF this matrix were diagonal, there would be no “mixing”
• Necessary for understanding Charge Parity violation (important research topic)
• (Nobel prize for 2008)• http://en.wikipedia.org/wiki/Cabibbo
%E2%80%93Kobayashi%E2%80%93Maskawa_matrix