Transcript Chapter04
Commutators and the Correspondence Principle
Formal Connection
Q.M.
Classical Mechanics
Correspondence between
Classical Poisson bracket of
functions f ( x , p) and g( x , p)
And
Q.M. Commutator of
operators f and g .
Copyright – Michael D. Fayer, 2007
Commutator of Linear Operators
A, B A B B A
(This implies operating on an arbitrary ket.)
If A and B numbers, = 0
Operators don’t necessarily commute.
A B C A B C
AQ
Z
B A C B A C
B S
T
In General
Z T
A and B do not commute.
Copyright – Michael D. Fayer, 2007
Classical Poisson Bracket
f , g
f g g f
x p x p
f f ( x , p)
g g( x , p)
These are functions representing classical
dynamical variables
not operators.
Consider position and momentum, classical.
x and p
Poisson Bracket
x p x p
x , p
x p p x
x, p 1
zero
Copyright – Michael D. Fayer, 2007
Dirac’s Quantum Condition
"The quantum-mechanical operators f
and g, which in quantum theory replace
the classically defined functions f and g,
must always be such that the commutator
of f and g corresponds to the Poisson
bracket of f and g according to
i
f , g f , g ."
Dirac
Copyright – Michael D. Fayer, 2007
This means
i
f ( x , p), g( x , p)
f , g
Poisson bracket of
classical functions
(commutator operates
on a ket)
Commutator of
quantum operators
Q.M. commutator of x and p.
x , p i
commutator
Therefore,
x , p
Poisson bracket
x, p 1
x , p i
Remember, the relation implies operating on
an arbitrary ket.
This means that if you select operators for x and p such that they
obey this relation, they are acceptable operators.
The particular choice
a representation of Q.M.
Copyright – Michael D. Fayer, 2007
Schrödinger Representation
p P i
x
momentum operator, –i times derivative with
respect to x
xx x
position operator, simply x
Operate commutator on arbitrary ket
x, P S
( xP P x ) S
x i
x S
S i
x
x
Using the product rule
i x
S S x
S
x
x
i
s .
Therefore,
x, P S
i
S
and
x, P i
because the two sides have the
same result when operating on
an arbitrary ket.
S
Copyright – Michael D. Fayer, 2007
Another set of operators – Momentum Representation
xxi
p
position operator, i times derivative with
respect to p
p p
momentum operator, simply p
A different set of operators, a different representation.
In Momentum Representation, solve position eigenvalue problem. Get
x , states of definite position.
They are waves in p space. All values of momentum.
Copyright – Michael D. Fayer, 2007
Commutators and Simultaneous Eigenvectors
A S S
B S S
S are simultaneous Eigenvectors of operators A and B
with egenvalues and .
Eigenvalues of linear operators
observables.
A and B are different operators that represent different observables, e. g.,
energy and angular momentum.
If S are simultaneous eigenvectors of two or more linear operators
representing observables, then these observables can be
simultaneously measured.
Copyright – Michael D. Fayer, 2007
A S S
B S S
B A S B S
AB S A S
B S
A S
S
S
Therefore,
Rearranging
AB S B A S
A B B A
A B B A is the commutator of
S 0
A and B, and since in general S 0,
A, B 0
The operators A and B commute.
Operators having simultaneous eigenvectors commute.
The eigenvectors of commuting operators can always be constructed in such
a way that they are simultaneous eigenvectors.
Copyright – Michael D. Fayer, 2007
There are always enough Commuting Operators (observables) to
completely define a system.
Example:
Energy operator, H, may give degenerate states.
H atom 2s and 2p states have same energy.
J
2
square of angular momentum operator
j 1 for p orbital
j 0 for s orbital
But px, py, pz
J z angular momentum projection operator
2
H , J , J z all commute.
Copyright – Michael D. Fayer, 2007
Commutator Rules
A , B B , A
A , BC A , B C B A , C
AB , C A , C B A B , C
A , B , C B , C , A C , A , B 0
A, B C A, B A, C
Copyright – Michael D. Fayer, 2007
Expectation Value and Averages
A a a
eigenvector
normalized
eigenvalue
If make measurement of observable A on state a will observe .
What if measure observable A on state not an eigenvector of operator A.
Ab ?
normalized
Expand b in complete set of eigenkets a Superposition principle.
Copyright – Michael D. Fayer, 2007
b c1 a1 c2 a2 c3 a3
(If continuous range
integral)
b ci a i
i
Consider only two states (normalized and orthogonal).
b c1 a1 c2 a2
A b A c1 a1 c2 a2
c1 A a1 c2 A a2
1c1 a1 2c2 a2
Left multiply by
b .
b A b c1* a1 c2* a2
c
1 1
a1 2c2 a2
1c1*c1 2c2*c2
1 c1 2 c2
2
2
Copyright – Michael D. Fayer, 2007
The absolute square of the coefficient ci, | ci|2, in the expansion
of b in terms of the eigenvectors ai of the operator (observable)
A is the probability that a measurement of A on the state b
will yield the eigenvalue i.
If there are more than two states in the expansion
b ci a i
i
b A b i ci
2
i
eigenvalue
probability of eigenvalue
Copyright – Michael D. Fayer, 2007
Definition: The average is the value of a
particular outcome times its
probability, summed over all
possible outcomes.
Then
b A b ci i
2
i
is the average value of the observable when many measurements are made.
Assume: One measurement on a large number
of identically prepared non- interacting systems
is the same as the average of many repeated
measurements on one such system prepared
each time in an identical manner.
Copyright – Michael D. Fayer, 2007
b Ab
Expectation value of the operator A.
In terms of particular wavefunctions
b Ab
b A bd
Copyright – Michael D. Fayer, 2007
The Uncertainty Principle - derivation
Have shown -
x , P 0
and that
x p
Want to prove:
Given A and B,
Hermitian with
A , B i C
C another Hermitian operator (could be number –
special case of operator, identity operator).
Then
1
AB
C
2
with
C SC S
short hand for expectation value
S and S
arbitrary but normalized.
Copyright – Michael D. Fayer, 2007
Consider operator
D A B i B
arbitrary real numbers
D S Q
Q Q S DD S 0
Since Q Q is the scalar product of vector
with itself.
Q Q S DD S A2 2 2 B 2 C ' C 0
(derive this in home work)
C ' AB BA
is the anticommutator of
A and B .
AB BA A , B
2
A
S A S
anticommutator
2
Copyright – Michael D. Fayer, 2007
B S 0
for arbitrary ket S .
Can rearrange to give
1 C'
2
2
A B
2 B2
2
2
2
B 1 C
2 B2
1 C'
4 B2
2
2
1 C
0
2
4 B
Holds for any value of and .
Pick and so terms in parentheses are zero.
Then A2
B
2
1
C
4
2
C'
2
1
C
4
2
(Have multiplied through by
2
B and transposed.)
Positive numbers because square of real numbers.
The sum of two positive
numbers is one of them.
Thus,
A
2
B
2
1
C
4
2
Copyright – Michael D. Fayer, 2007
A
2
B
Define
2
1
C
4
A
B
2
2
A A
2
B
2
2
2
B
Second moment of distribution
- for Gaussian
standard deviation squared.
2
For special case
A B 0
1
AB
C
2
Average value of the observable is zero.
C
1
2 2
C
square root of the
square of a number
Copyright – Michael D. Fayer, 2007
Have proven that for A , B i C
1
AB
C
2
Example
x, P i
Number, special case of an operator.
Number is implicitly multiplied by the
identity operator.
x P 0
Therefore
x p / 2.
Uncertainty comes from superposition principle.
The more general case is discussed in the book.
Copyright – Michael D. Fayer, 2007