Physics 505 Homework No. 1 Solutions S1-1 1. Some...

Physics 505 Homework No. 1 Solutions S1-1

1. Some Preliminaries. Assume A and B are Hermitian operators.

(a) Show that (AB)† = B†A†.

Solution∫

dxφ∗ABψ =∫

dx (A†φ)∗Bψ =∫

dx (B†(A†φ))∗ψ =∫

dx (B†A†φ)∗ψ.

End Solution

(b) Show that AB = [A,B]/2+{A,B}/2 where the anticommutator {A,B} = AB+BA.Further, show that the anticommutator is Hermitian and the commutator is anti-Hermitian (that is, [A,B]† = −[A,B]). We know that expectation values of Hermitianoperators are real. What can you say about the expectation value of an anti-Hermitianoperator?

Solution

[A,B]/2 + {A,B}/2 = (AB −BA)/2 + (AB +BA)/2 = AB.

{A,B}† = (AB)† + (BA)† = B†A† + A†B† = {A,B}, so the anticommutator isHermitian.

[A,B]† = (AB)† − (BA)† = B†A† −A†B† = −(AB −BA) = −[A,B], so the commu-tator is anti-Hermitian.

Now let A be an anti-Hermitian operator. Then 〈A〉 =∫

dxψ∗(Aψ). Also 〈A†〉 =∫

dx (Aψ)∗ψ = −∫

dxψ∗(Aψ) which means that 〈A〉 is pure imaginary.

End Solution

Copyright c© 2012, Edward J. Groth


2. Generalized Uncertainty Principle. Let A and B be Hermitian operators. Definethe “uncertainty” in A by the square root of the mean square deviation from the mean:∆A =

√

〈(A− 〈A〉)2〉 =√

〈A2〉 − 〈A〉2. Show that

∆A∆B ≥ 1

2|〈[A,B]〉| .

Hint: Use the Schwarz Inequality and the results from problem 1. Comment: The com-mutator of x and px is ih̄, so ∆x∆px ≥ h̄/2.

Solution

Define new Hermitian operators A′ = A−〈A〉 and B′ = B−〈B〉. Then by the Schwarzinequality,

〈A′2〉〈B′2〉 ≥ |〈A′B′〉|2 ,

or∆A∆B ≥ |〈A′B′〉| = |〈[A′, B′]〉/2 + 〈{A′, B′}〉/2| ≥ |〈[A′, B′]〉|/2 .

Since the expectation value of the commutator is imaginary and the anticommutator isreal, each makes a positive contribution to the absolute value, and the anticommutatorcan be dropped without changing the inequality in the last step. So,

∆A∆B ≥ |〈[A′, B′]〉|/2 = |〈[A,B]− [A, 〈B〉]− [〈A〉, B] + [〈A〉, 〈B〉]〉|/2 = |〈[A,B]〉|/2 .

〈A〉 and 〈B〉 are just numbers, so they commute with the operators and the commutatorsinvolving them are 0.

End Solution



3. We saw in lecture that the eigenfunction of the momentum operator with eigenvaluep is fp(x) = (1/

√2πh̄) exp(ipx/h̄). We are working in one dimension here and we are

assuming (almost) the δ-function normalization described in lecture. (The extra√h̄ has

been inserted for convenience below.) We also stated that the eigenfunctions of a Hermitianoperator form a complete set. This means that for any wave function, we should be ableto write

ψ(x) =∑

p

cpfp(x) ,

where cp are the expansion coefficients. Since p is a continuous variable, the coefficientsbecome a function and we should write:

ψ(x) =

∫ +∞

−∞

ϕ(p)eipx/h̄dp√2πh̄

,

where ϕ(p) plays the role of the expansion coefficients. Either ψ(x) or ϕ(p) is suitable fordescribing the state of the system. If we are using ψ(x), we are using a position space orconfiguration space description. If we are using ϕ(p), it’s a momentum space description.

(a) Show how to determine ϕ(p) from ψ(x). (The above already gave an expression forψ(x) in terms of ϕ(p).)

Solution

We take the dot product of fp(x) with ψ(x).

〈fp|ψ〉 =

∫

dx1√2πh̄

e−ipx/h̄ψ(x)

=

∫

dx1√2πh̄

e−ipx/h̄∫

dp′ ϕ(p′)1√2πh̄

eip′x/h̄

=

∫

dp′ ϕ(p′)

∫

dx1

2πh̄ei(p

′ − p)x/h̄

=

∫

dp′ ϕ(p′)δ(p′ − p)

= ϕ(p) .

The configuration and momentum space wavefunctions are just Fourier transforms of eachother!

End Solution



(b) What is the meaning of∫ p2

p1

|ϕ(p)|2 dp ?

Solution

It’s the probability that a measurement of the momentum produces a result betweenp1 and p2. (Remember, the square of an expansion coefficient is the probability of beingin the corresponding state.)

End Solution

(c) What are the position and momentum operators in momentum space? (You canprobably make a good guess, but justify your answer!)

Solution

We know the operators in configuration space, so let’s operate in configuration spaceand transform to momentum space.

〈fp|xψ〉 =

∫

dx1√2πh̄

e−ipx/h̄xψ(x)

=

∫

dx1√2πh̄

e−ipx/h̄ x∫

dp′ ϕ(p′)1√2πh̄

eip′x/h̄

=

∫

dp′ ϕ(p′)

∫

dx x1

2πh̄ei(p

′ − p)x/h̄

=

∫

dp′ ϕ(p′)

∫

dxh̄

i

∂

∂p′1

2πh̄ei(p

′ − p)x/h̄

=

∫

dp′(

− h̄i

∂

∂p′ϕ(p′)

)∫

dx1

2πh̄ei(p

′ − p)x/h̄ integration by parts

=

∫

dp′(

− h̄i

∂

∂p′ϕ(p′)

)

δ(p′ − p)

= − h̄i

∂

∂pϕ(p) .

The position operator is (−h̄/i)(∂/∂p) as you might have guessed. By doing essentially thesame thing but operating with (h̄/i)(∂/∂x) we would find that the momentum operatoris multiplication by p, also as you might have guessed. Note that it’s still the case that[x, p] = ih̄.

End Solution



4. Projection operators. It’s often the case that we want to find the “component” of afunction “parallel” to another function. We just take the dot product with the secondfunction, but then we also need to multiply by the second function. A handy notation is

|ψ〉〈ψ| .

This projects onto ψ. Operating on |ϕ〉, we get

|ψ〉〈ψ|ϕ〉 ,

which is what we want! Remember 〈ψ|ϕ〉 is just a number and |ψ〉 is the vector. Similarly,operating on 〈ϕ| we get

〈ϕ|ψ〉〈ψ| ,

which is the desired expression for the adjoint vector. Suppose you have a complete setof orthonormal basis vectors |ψn〉. What is a compact expression for transforming anarbitrary vector |ϕ〉 into this basis set? (This is much easier to write down than to ask!)

Solution

We want to project |ϕ〉 onto each basis vector (this gives the expansion coefficients)and then sum the coefficients times the basis vectors:

|ϕ〉 =

(

∑

n

|ψn〉〈ψn|)

|ϕ〉 .

End Solution



5. We often need to exponentiate a matrix! As an example, let A be the 2 × 2 matrix

A =

(

0 11 0

)

.

Compute eitA using two different methods:

(a) Use the Taylor expansion for the exponential:

eitA =∞∑

n=0

(itA)n

n!.

Solution

Note that

A2 =

(

0 11 0

) (

0 11 0

)

=

(

1 00 1

)

= 1 .

So the even terms in the Taylor expansion involve the identity matrix and the odd termsinvolve A.

eitA = 1 − t2/2! + t4/4! − · · · + iA(t− t3/3! + t5/5! − · · ·)

= cos t

(

1 00 1

)

+ i sin t

(

0 11 0

)

=

(

cos t i sin ti sin t cos t

)

.

End Solution

(b) Use the spectral decomposition of A: write A =∑2

k=1 λk|ψk〉〈ψk| with 〈ψk|ψl〉 = δkl

and use

eitA =

2∑

k=1

eitλk |ψk〉〈ψk| .

Hint: |ψk〉 are just two element column vectors that are eigenvectors of the matrix A.

Solution

The first thing we need to do is find the eigenvalues and eigenfunctions. The eigenvalueequation for this matrix is

(

0 11 0

)(

ab

)

= λ

(

ab

)

.

or(

−λ 11 −λ

)(

ab

)

= 0 .

This has a solution (that’s not all 0s) only if the determinant vanishes. Taking the deter-minant and setting it to zero we get the characteristic equation

λ2 − 1 = 0 ,



with solutions λ = ±1 and eigenfunctions (eigenvectors) belonging to λ = ±1,

(

1/√

2±1/

√2

)

.

So

eitA = eit(

1/2 1/21/2 1/2

)

+ e−it(

1/2 −1/2−1/2 1/2

)

=

(

(eit + e−it)/2 (eit − e−it)/2(eit − e−it)/2 (eit + e−it)/2

)

=

(

cos t i sin ti sin t cos t

)

.

End Solution


Physics 505 Homework No. 1 Solutions S1-1 1. Some...

Documents

Transcript of Physics 505 Homework No. 1 Solutions S1-1 1. Some...