Lecture 6: Operators and QuantumMechanics The material in this lecture covers the following in Atkins. 11.5 The informtion of a wavefunction (c) Operators
Lecture on-line Operators in quantum mechanics (PDF) Operators in quantum mechanics (HTML) Operators in Quantum mechanics (PowerPoint) Handout (PDF) Assigned Questions
Tutorials on-line Reminder of the postulates of quantum mechanics The postulates of quantum mechanics (This is the writeup for Dry-lab-II)( This lecture has covered postulate 3) Basic concepts of importance for the understanding of the postulates Observables are Operators - Postulates of Quantum Mechanics Expectation Values - More Postulates Forming Operators Hermitian Operators Dirac Notation Use of Matricies Basic math background Differential Equations Operator Algebra Eigenvalue Equations Extensive account of Operators Historic development of quantum mechanics from classical mechanics The Development of Classical Mechanics Experimental Background for Quantum mecahnics Early Development of Quantum mechanics
Audio-visuals on-line Early Development of Quantum mechanics Audio-visuals on-line Postulates of Quantum mechanics (PDF) (simplified version from Wilson) Postulates of Quantum mechanics (HTML) (simplified version from Wilson) Postulates of Quantum mechanics (PowerPoint ****)(simplified version from Wilson) Slides from the text book (From the CD included in Atkins ,**)
(Ia) A Quantum mechanical system is specifiedby the statefunction Ψ(x)
(Ib) The state function Ψ(x) contains allinformation about the system we can know
We now have Operators and Quantum Mechanics
(Ic) A system described by the state function HΨ(x)=EΨ(x)has exactly the energy E
Review
O
X
We have seen that a ' free' particle movingin one dimension in a constant (zero) potentialhas the Hamiltonian
−
h2
2mδ2ψ(x)
δx 2 = Eψ(x)
ˆ H =−
h2
2mδ2
δx 2
The Schrodinger equation is
with the general solution :
ψ(x) = Aexpikx + Bexp−ikx
and energies E =
h2k 2
2m
Operators and Quantum MechanicsReview
How does the state function Ψ(x,t) give us information about an observable other than the energy such as the position or the momentum ?
Any observable 'Ω' can be expressed in classical physicsin terms of x,y,z and px,py,pz.
Examples:Ω=x, px, vx, px
2, T, V(x), E
Operators and Quantum MechanicsGood question
Classical Mechanics Quantum Mechanics
x px ˆ x −> x ; ̂ px−>hiδδx
y py ˆ y−> y ; ̂ py−>hiδδy
z pz ˆ z−> z ; ̂ pz−>hiδδz
We can construct the corresponding operatorfrom the substitution:
as ˆ Ω (x,y,z,
hi
ddx
,hi
ddy
,hi
ddz
)
Such as:ˆ Ω =ˆ x , ˆ p x, ˆ v x, ˆ p x
2, ˆ T , ˆ V (x), E
Operators and Quantum MechanicsReview
Ω ψn = ϖnψn
For an observable Ω with the correspondingoperator ̂ Ω we have the eigenvalue equation :
Operators and Quantum Mechanics
(IIIa). The meassurement of the quantity represented by Ωhas as the o n l y outcome one of the values
ϖn n = 1,2,3 ....
(IIIb). If the system is in a state described by ψn a meassurement ofΩ will result in the
valueϖn
Important news
Quantum mechanical principle.. Operators
we can solve the eigenvalue problem
ˆ Ω ψn =ϖnψn
For any such operator ˆ Ω
We obtain eigenfunctionsand eigenvalues
The only possible values that can arise from measurements of the physical observable Ω are the eigenvalues ϖn
Postulate 3
Important news
The x - component 'px ' of the linear momentum
r p =px
r ex + py r ey + pz
r ez
Is represented by the operator ˆ p x =
hiδδ x
With the eigenfunctions Exp[ikx] and eigenvalue hk
hiδ Exp[ikx]
δ x =hkExp[ikx]
We note that k can take any value −∞ > k > ∞
Operators and Quantum MechanicsImportant news
ψ(x) = Aexpikx + Bexp−ikx and energies E =
h2k 2
2m
For A = 0 ψ−(x ) = B exp−ikx
this wavefunction is also an eigenfunction to ˆ p x
With eigenvalue for ˆ p x of - hk
Thus ψ- (x ) describes a particle of energy E=h2k 2
2m
and momentum px =−hk ; note E= Px2
2m .as it must be
This system corresponds to a particle moving with constant velocity
vx = pxm
=-h /k m We know nothing about its positionsince |ψ( )x |2=B
Operators and Quantum MechanicsNew insight
ψ(x) = Aexpikx + Bexp−ikx and energies E =
h2k 2
2m
For B = 0 ψ+(x ) = A expikx
this wavefunction is also an eigenfunction to ˆ p x
With eigenvalue for ˆ p x of hk
Thus ψ (x ) describes a particle of energy E=h2k 2
2m
and momentum px =hk ; note E= Px2
2m .as it must be
This system corresponds to a particle moving with constant velocity
vx = pxm
=h /k m We know nothing about its positionsince |ψ( )x |2=B
Operators and Quantum MechanicsNew insight
What about : ψ(x ) =A expikx+B exp−ikx ?It is not an eigenfunction to ˆ p x since :
ˆ p xψ(x)=Ahi
ddx
expikx+Bhi
ddx
exp−ikx
=Ahkexpikx−Bhkexp−ikx
How can we find px in this case ?
Operators and Quantum MechanicsNew insight
A linear operator ˆ A will have a set of eigenfunctions fn(x ) {n = 1,2,3..etc}and associated eigenvalues kn such that :
The set of eigenfunction {fn(x),n=1..} is orthonormal :
fi(x)*
all space∫ fj(x)dx=δij
ˆ A fn(x ) =knfn(x )
=o if i ≠ j
= 1 if i = j
Quantum mechanical principles..Eigenfunctions
Quantum mechanical principles..Eigenfunctions
ei •ej =δij
ei
ei
ei
An example of an orthonormal set is the Cartesian unit vectors
An example of an orthonormal function set is
ψn(x)= 1L
sinnπxL
⎛ ⎝
⎞ ⎠
n=1,2,3,4,5....
ψn(x)*
o
L∫ ψm(x)=∂nm
That is, any function g(x) thatdepends on the same variables as the eigenfunctions can be written
g(x) = anfn (x )i=1
all∑
where
an = fn(x)*g(x)dxall space
∫
Quantum mechanical principles..Eigenfunctions
The set of eigenfunction {fn(x ),n =1..} .forms a complete set
ei
ei
ei
r e i ; i=1,2,3 form a complete set
For any vector r v
v=(r v •
r e 1)
r e 1+(
r v •
r e 2)
r e 2+(
r v •
r e 3)
r e 3
we can show that : an = fn(x)*g(x)dxV∫
In the expansion : g(x) = aifi (x )i=1
all∑ (1)
from the orthonormality: fi(x)*
V∫ fj(x)dx=δij
Quantum mechanical principles..Eigenfunctions
g(x)= aifi(x)i=1
all∑ ⇒
V∫ fn(x)*g(x)dx= ai
V∫ fn(x)*fi(x)
i=1
all∑ dx
A multiplication by fn (x) on both sides followed byintegration affords
or : an = ( )g x fn(x )dxall space
∫ δij
A multiplication by fn (x) on both sides followed byintegration affords
or : an = g(x)fn(x)*dxall space
∫
ψ(x)=Aexpikx+Bexp−ikx is a linear combination of two eigenfunctions to ˆ p x
How can we find px in this case ?
Operators and Quantum Mechanics
px =hk px =−hk
1. Postulate 3For an observable Ω with the corresponding operator ˆ Ω we have the eigenvalue equation : Ω ψn =ϖnψn(i) The meassurement of the quantity represented by Ωhas as the o n l y outcome one of the values ϖn n=1,2,3 ....(ii) If the system is in a state described by ψn a meassurement of Ω will result in the value ϖn
What you should learn from this lecture
Illustrations:
ψ+(x)=Aexpikx is an eigenfunction to ˆ p x with eigenvalue hk
ψ−(x)=Aexp−ikx is an eigenfunction to ˆ p x with eigenvalue -hkBoth are eigenfunctions to the Hamiltonian for a free particle
H=h2(ˆ p x)2
2m with eigenvalues E=
h2k2
2m ψ+(x) represents a free particle of momentum hk
ψ−(x) represents a free particle of momentum -hk
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