Lecture 6: Operators and Quantum
Mechanics
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 ,**)
Operators and Quantum Mechanics
We now have
Re view
(Ia) A Quantum mechanical system is specified
by the statefunction Ψ( x )
( Ib) The state function Ψ( x) contains all
information about the system we can know
(Ic) A system described by the
state function HΨ( x) = EΨ( x)
has exactly the energy E
Operators and Quantum Mechanics
Re view
We have seen that a ' free' particle moving
in one dimension in a constant (zero) potential
has the Hamiltonian
2
2
δ
2m δx 2
X Ĥ = − h
O
The Schrodinger equation is
h δ ψ ( x)
−
= Eψ ( x)
2
2m δx
2
2
with the general solution :
ψ ( x ) = A exp
ikx
h 2k 2
and energies E =
2m
+ B exp
− ikx
Operators and Quantum Mechanics
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 ?
Good question
Any observable ' Ω' can be expressed in classical physics
in terms of x, y, z and px , p y , pz .
Examples :
Ω = x, p x , v x , p2x , T, V(x), E
Operators and Quantum Mechanics
We can construct the corresponding operator
from the substitution :
Classical Mechanics Quantum Mechanics
x
px
y
py
z
pz
h δ
i δx
h δ
yˆ − > y ; pˆ y − >
i δy
xˆ − > x ; pˆ x − >
hδ
zˆ − > z ; pˆ z − >
i δz
h d h d hd
ˆ
as Ω(x,y,z,
,
,
)
i dx i dy i dz
Such as :
ˆ V(x),
ˆ
ˆ = x,
ˆ pˆ x , vˆx , pˆ 2x , T,
E
Ω
Re view
Operators and Quantum Mechanics
Im por tan t news
For an observable Ω with the corresponding
operator Ω̂ we have the eigenvalue equation :
Ω ψn = ϖn ψn
(IIIa). The meassurement of the quantity represented by Ω
has 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 of Ω will result in the
value ϖn
Quantum mechanical principle.. Operators
For any such operator Ω̂
Im por tan t news
we can solve the eigenvalue problem
Ω̂ ψn = ϖn ψn
We obtain
eigenfunctions and eigenvalues
The only possible values that can arise from measurements
of the physical observable Ω are the eigenvalues ϖn
Postulate 3
Operators and Quantum Mechanics
Im por tan t news
The x - component ' p x ' of the linear momentum
r
r
r
r
p = px ex + p y e y + pz ez
Is represented by the operator p̂x = h δ
iδx
With the eigenfunctions
Exp[ikx ]
and eigenvalue hk
h δ Exp[ikx ]
= hkExp[ikx ]
i
δx
We note that k can take any value
−∞ > k > ∞
Operators and Quantum Mechanics
ψ ( x ) = A exp
For A = 0
ikx
+ B exp
− ikx
New insight
h 2k 2
and energies E =
2m
−ikx
ψ − ( x ) = B exp
this wavefunction is also an eigenfunction to pˆ x
With eigenvalue for pˆ x of - hk
h 2k 2
Thus ψ- ( x ) describes a particle of energy E =
2m
2
Px
and momentum px = − hk ; note E =
as it must be.
2m
This system corresponds to a particle
moving with constant velocity
px
We know nothing about its position
= -hk/m
vx =
m
since | ψ (x) |2 = B
Operators and Quantum Mechanics
ψ ( x ) = A exp
For B = 0
ikx
+
New insight
− ikx and energies E = h2k 2
B exp
2m
ψ + ( x ) = A expikx
this wavefunction is also an eigenfunction to pˆ x
With eigenvalue for pˆ x of hk
h 2k 2
Thus ψ ( x ) describes a particle of energy E =
2m
Px2
and momentum px = hk ; note E =
as it must be.
2m
This system corresponds to a particle
moving with constant velocity
p
We know nothing about its position
v x = x = hk/m
m
since | ψ (x) |2 = B
Operators and Quantum Mechanics
New insight
What about : ψ ( x ) = A expikx + B exp −ikx ?
It is not an eigenfunction to pˆ x since :
h d
h d
ikx
− ikx
ˆp x ψ (x) = A
exp + B
exp
i dx
i dx
= Ahk expikx − Bhk exp− ikx
How can we find
p x in this case ?
Quantum mechanical principles..Eigenfunctions
ˆ will have a set of
A linear operator A
eigenfunctions fn ( x ) {n = 1,2,3..etc}
and associated eigenvalues kn such that :
ˆ n ( x ) = k n fn ( x )
Af
The set of eigenfunction {fn ( x ), n = 1..}
is orthonormal :
* f ( x )dx = δ
f
(
x
)
∫ i
j
ij
all space
= o if i ≠ j
= 1 if i = j
Quantum mechanical principles..Eigenfunctions
An example of an orthonormal set is the Cartesian unit vectors
ei
e i • e j = δ ij
ei
ei
An example of an orthonormal function set is
ψ n (x) =
1
nπx 

sin
L  L 
n = 1, 2, 3, 4, 5....
L
*ψ (x) = ∂
ψ
(
x
)
∫ n
m
nm
o
Quantum mechanical principles..Eigenfunctions
The set of eigenfunction {fn ( x ), n = 1..}
forms a complete set.
That is, any function g(x) that
depends on the same variables
as the eigenfunctions canr be written
ei ; i = 1, 2, 3 form a complete set
all
g(x) = ∑ anfn ( x )
i=1
ei
ei
where
* g(x)dx
(
x
)
an =
f
∫ n
all space
ei
r
For any vector v
r r r
r r r
r r r
v = (v • e1 )e1 + (v • e2 )e2 + (v • e 3 )e 3
Quantum mechanical principles..Eigenfunctions
all
In the expansion :
g(x) = ∑ aifi ( x )
(1)
i=1
we can show that : an = ∫ fn ( x )* g(x)dx
V
*
from the orthonormality : ∫ fi ( x ) fj ( x )dx = δij
V
A multiplication by fn (x) on both sides followed by
integration affords
all
*
*
g(x) = ∑ aifi ( x ) ⇒ ∫ fn (x) g(x)dx = ∑ ai ∫ fn ( x ) fi ( x )dx
i=1 V
i=1
V
all
or
or : : aann==
* dx
(
x
g(x)f
(
x
)
dx
∫
n
n
all space
space
all
δij
Operators and Quantum Mechanics
ψ( x ) = A expikx + B exp− ikx is a linear combination
of two eigenfunctions to pˆ x
p x = hk
How can we find
p x in this case ?
p x = −hk
What you should learn from this lecture
1. Postulate 3
ˆ
For an observable Ω with the corresponding operator Ω
we have the eigenvalue equation : Ω ψ n = ϖn ψ n
(i) The meassurement of the quantity represented by Ω
has 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 ψ n
a meassurement of Ω will result in the value ϖn
Illustrations :
ψ + ( x ) = A expikx is an eigenfunction to pˆ x with eigenvalue hk
ψ − ( x ) = A exp− ikx is an eigenfunction to pˆ x with eigenvalue - hk
Both are eigenfunctions to the Hamiltonian for a free particle
h 2 (pˆ x )2
h2k2
H=
with eigenvalues E =
2m
2m
ψ + ( x ) represents a free particle of momentum hk
ψ − ( x ) represents a free particle of momentum - hk
What you should learn from this lecture
2. Postulate 4.
The set of eigenfunction {fn ( x ), n = 1..}
forms a complete set.
That is, any function g(x) that depends on the same
variables as the eigenfunctions can be written :
all
g(x) = ∑ anfn ( x ) where
i=1
an =
∫ g(x)fn ( x )dx
all space