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Physics 324 Solution to Problem Set # 6

1. The Parity operator in one dimension, P1, reverses the sign of the position coordinate, x: P1(x) =(x). Note that P2

1(x) = P1P1(x) = ( x) = (x)

(a) Find the eigenvalues of the Parity operator, P1. That is, find those values, , such that theequation P1| = | has a solution.

Solution:

P1(x) = (x) = (x) => P12(x) = 2(x) = (x) => 2 = 1

Therefore, the eigenvalues of the parity operator are = 1.

(b) Find the eigenfunctions ofP1.

Solution: For = 1, the eigenfunction is a solution to:

P1(x) = (x) => (x) = (x)All functions, e(x), that are even in x (e(x) = e(x)) are eigenfunctions ofP1 with an eigenvalue

of +1.

For = 1, the eigenfunction is a solution to:P1(x) = (x) => (x) = (x)

All functions, o(x), that are odd in x (e(x) = e(x)) are eigenfunctions ofP1 with an eigenvalueof 1.

(c) Is the Parity operator, P1, Hermitian? Why, or why not. Remember, A is the Hermitian Adjoint

of an operator A if and only if A

f|g = f|Ag for all vectors |f and |g.Solution: Hermitian requires that P1f|g = f|P1g for all vectors |f and |g. In the x basis,

P1f|g =

P1f(x)

g(x) dx =

f(x)g(x) dx

Letting x = x,

P1f|g =

f(x)g(x) dx =

f(x)P1g(x

)

dx = f|P1g

Therefore, P1 is a Hermitian operator.

(d) Derive an expression for [P1, x], the commutator of the parity operator with the position operator.(Hint, make use of a trial vector.)

Solution :P1, x

f(x) = P1

xf(x)

xP1f(x) = (x)f(x) xf(x) = 2xf(x) = 2xP1f(x)Therefore,

P1, x

= 2xP1.

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2. Problem 3.40 in your textbook.

(a) Given that

A, B

= 0, and A = SAS1, B = SB S1:

A, B

= AB BA = SAS1 SB S1 SBS1 SAS1 = SABS1 SBAS1

= S

ABS1 BAS1 = SAB BAS1 = SA, BS1 = 0because S S1 = I.

(b) For diagonal matrices, Aij = Aiiij and Bjk = Bkkjk .

For C = AB, Cik =j

AijBjk =j

AiiijBkkjk = AiiBiiik

Similarly, for D = BA, Dik =j

BijAjk =j

BiiijAkkjk = BiiAiiik

Therefore Cik

= Dik

for all i, k, ie AB = BA, giving us A,B = 0.

(c)A,B

= 0, and A| = a|:

A,B

| = 0 = ABBA| = AB| aB| => AB| = aB|Therefore, B| is also an eigenvector ofA with eigenvalue a. But the spectrum ofA is nondegenerate

(for this problem), which means that if | and | are two eigenvectors ofA with the same eigenvalue, then| = c| where c is a constant.

Therefore, B| = c|, ie | is also an eigenvector ofB.

3. For operators, Q, that do not depend expliciltly upon time, we have:

d

dtQ = i

hH, Q

Using the results from Problem 3.41 in your text,

(a) For H = p2/2m + V(x), compute dx/dt.

Solution :

H, x

= p2

2m, x

+

V(x), x

But

V(x), x

= 0, and from problem 3.41 in the text,

p2, x

= p

p, x

+

p, x

p = 2ihp. Therefore:d

dt x

=i

hH, x

=i

2mh2ihp

=

pm

(b) For H = p2/2m + V(x), compute dp/dt.

Solution :

H, p

= p2

2m, p

+

V(x), p

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But

p2, p

= 0, and from problem 3.41 in the text,

V(x), p

= ih dV(x)/dx. Therefore:

d

dtp = i

hH, p = i

hih dV(x)

dx = dV(x)

dx

(c) For the harmonic oscillator Hamiltonian (Eqn. 2.39), compute dP1/dt, where P1 is the parityoperator from Problem 1.

Solution :

H, P1

= p2

2m, P1

+

12

m2x2, P1

but

x2,P1

f(x) = x2P1f(x) P1

x2f(x)

= x2f(x) (x)2f(x) = 0

and

p2,P1

f(x) =h22m

d2

dx2P1f(x) P1

h22m

d2

dx2f(x)

=

h22m

d2

dx2f(x) h

2

2m

d2

d(x)2 f(x) = 0

Therefore,dP1

dt= 0

4. Problem 3.48 in your textbook.

Solution: At time t = 0, the particle starts in the ground state of an infinite square well potential, wherewe know the wave function can be written as: (x, 0) =

2/a sin(x/a), for 0 x a and (x, 0) = 0 for

a x 2a.The energy eigenstates of the infinite square well potential with width 2a can be written as: n(x) =

2/2a sin(nx/2a).

We can write (x, 0) =n=1

cnn(x) where cn =

2a

0

n(x)(x, 0) dx

=> cn =

2

a

a0

sin(nx/2a)sin(x/a) dx =2

2

/20

sin(ny) sin(2y) dy

where we substituted y = x/(2a).

=> cn =

2

/20

cos(n 2)y cos(n + 2)ydy =

2

sin(n 2)yn 2

/20

sin(n + 2)yn + 2

/20

where for n = 2 we interpret sin

(n 2)y/(n 2) = y.(a) The probability of measuring energy En = n

2

2

h2

/(8ma2

) is given by |cn|2

. Looking at our resultfor cn, we see that the most probable result will be to measure E2, with a probability of|c2|2 = 1/2.

(b) We need to evaluate the other values of cn:

c1 =4

2

3, c3 =

4

2

5...

The next most probable result will be to measure E1 with a probability of 32/(92) 0.36.

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(c) Because we moved the wall suddenly without disturbing the particle momentarily, we did no workon the particle and did not change its energy. The expectation value of the energy after the wall was movedwill then be the same as its expectation value before the wall was moved: E = 2h2/(2ma2), the groundstate energy of the original well.

5. Problem 3.57, parts (a) and (c), in your textbook.

(a) For any vector, |:

P2| = PP| = |P| = ||| = || = P|because | = 1. Therefore, P2 = P.

P| = | => | | = | => | = c|where c is a constant (ie the vectors | and | must be parallel). Then:

|c

= c = c => = 1

Therefore, the eigenvalue of P is 1, and the eigenvectors are c|.

(c) Q|ej = j |ej . Any vector | can be written as | =n

j=1 aj|ej =>

Q| =n

j=1

ajQ|ej > =n

j=1

ajj |ej > =n

j=1

ajj I|ej >

Using the results from part (b), we can substitute for I:

Q| =n

j=1

ajj

nk=1

|ekek||ej =

nj=1

aj

nk=1

|ekek|ejj

Butn

k=1

|ekek|ejj =n

k=1

|ekek|ejk because ek|ej = kj

=> Q| =n

j=1

aj |ej nk=1

k|ekek|

= nk=1

k|ekek| nj=1

aj |ej = nk=1

k |ekek||

=> Q =n

k=1

k|ekek|