Physics 215 Final ExamSolutions Winter2018

[Pages:11]Physics 215

Final Exam Solutions

Winter 2018

1. Consider a particle of mass m subject to a one-dimensional potential of the following form,

V (x) =

1 2

m2

x2

,

,

for x > 0, for x 0.

(a) Determine the possible the bound state energy values of the particle.

The Schro?dinger equation for this problem in the interval 0 < x < is,

2

- 2m

d2 dx2

+

1 2

m

2x2

(x) = E(x) .

(1)

This is the Schro?dinger equation for the one-dimensional harmonic oscillator, whose energy

eigenvalues and eigenfunctions are well known. However, in this problem, there is an infinite

barrier at x = 0, so we must impose an additional boundary condition, (0) = 0.

The energy eigenvalues of the one-dimensional harmonic oscillator without the infinite bar-

rier at the origin are En =

(n

+

1 2

),

where

n

is

any

non-negative

integer.

The

corresponding

energy eigenstates, n(x), are also eigenstates of parity and satisfy n(-x) = (-1)nn(x).

That is, the energy eigenstates are parity even for even n and parity odd for odd n. In

particular, if n is odd then (0) = 0. In contrast n(x) = 0 for even values of n. We can see this behavior explicitly from the expressions for the harmonic oscillator wave

functions. From eq. (B.4.3) of Sakurai and Napolitano,

n(x) = (2nn!)-1/2

m

1/4

exp

-

m 2

x2

Hn(x

m/ ) ,

(2)

where H0(z) = 1, H1(z) = 2z, . . . are the Hermite polynomials. The recursion relation satisfied by the Hermite polynomials, given in eq. (2.5.25) of Sakurai and Napolitano,

Hn+1(z) = 2zHn(z) - 2nHn-1 ,

can be used to compute Hn(z) for all n = 2, 3, . . ., given H0(z) = 1, H1(z) = 2z. It immediately

follows that Hn(0) = 0 for odd n and Hn+2(0) = -2(n + 1)Hn(0) for even n, which yields H2n(0) = (-2)n(2n - 1)!! = 0 for all non-negative n, after noting that H0(0) = 1.

We conclude that only the odd parity harmonic oscillator wave functions vanish at the

origin. These are the allowed square integrable solutions to eq. (1) subject to the boundary

condition (0) = 0. The corresponding energy eigenvalues are En =

(n+

1 2

)

for

odd

positive

integers n. Writing n = 2N + 1, we conclude that the possible bound state energies are

EN =

(2N

+

3 2

)

,

for N = 0, 1, 2, . . .

1

(b) What is the normalized ground state wave function in the coordinate representation?

The ground state wave function for this problem is proportional to the n = 1 parity odd energy eigenstate of the one-dimensional harmonic oscillator. Using eq. (2) with H1(z) = 2z,

Ax exp N=0(x) =

-

m 2

x2

,

0,

for x > 0, for x < 0.

where A is determined by the normalization condition,

|N=0(x)|2 dx = 1 .

(3)

-

For our problem, due to the fact that N=0(x) = 0 for x < 0, the square of the normalization constant A must be twice that of the corresponding n = 1 harmonic oscillator wave function in order to satisfy eq. (3), i.e., |A|2 = 2|AHO|2. Since eq. (2) implies that (for n = 1),

AHO =

4

1/4

m

3/4

,

it follows that normalized ground state wave function for our problem is,

N=0(x) =

16

1/4

m

3/4

x exp

-

mx2 2

,

0,

for x > 0, for x < 0.

after setting the overall phase (by convention) to unity. One can verify explicitly that this result satisfies eq. (3).

2. Consider a one-dimensional problem with a Hamiltonian,

H

=

P2 2m

+

V

(X) ,

(4)

where X is the position operator and P is the momentum operator. Assume that the spectrum of the Hamiltonian is discrete. Denote the normalized energy eigenstates by |n , and the corresponding energy eigenvalues by En.

(a) Show that the matrix elements of X and P , with respect to a basis consisting of orthonormal energy eigenstates, satisfy the following relation,

k| P |n = c(k, n) k| X |n ,

(5)

where c(k, n) depends on Ek and En. Determine an explicit expression for c(k, n).

2

We begin by considering the commutator [H, X], where H is given by eq. (4). Using the canonical commutation relation, [X, P ] = i I, and noting that X commutes with any function of X, it follows that

[H, X] =

P2 2m

,

X

=

P 2m

[P,

X]

+

[P,

X

]

P 2m

=

-

i m

P

,

(6)

after employing the commutator identity, [AB, C] = A[B, C] + [A, C]B. Hence, it follows that

P = im[H, X] .

Inserting this result into k| P |n yields, k| P |n = im k| [H, X] |n = im k| (HX - XH) |n = im (Ek - En) k| X |n ,

where we have used the fact that |k and |n are energy eigenstates, e.g., H |n = En |n . Comparing with eq. (5), we conclude that

c(k, n) = im (Ek - En) .

(b) Evaluate the infinite sum,

S

m

2

(Ek - En) k| X |n 2 ,

(7)

k

where the sum is taken over the complete set of energy eigenstates, |k . Show that the sum is equal to a dimensionless constant and determine its value.

Note that the equation H |n = En |n implies that the energy eigenstates |n are dimensionless. Moreover, X has the same dimensional units as /P (cf. the canonical commutation relations). Hence, it follows that S is dimensionless since P 2/m has the dimensions of energy.

It is convenient to introduce the following shorthand notation,

kn Ek - En ,

Xkn k| X |n .

Then, the result of part (a) can be rewritten as,

k| P |n = imknXkn .

(8)

Since X is a self-adjoint operator, it follows that Xkn = Xnk. Hence, eq. (7) can be written

as,

S=m

knXknXnk .

(9)

k

Noting that kn = -nk, we can rewrite eq. (9) as,

S

=

m 2

(kn - nk)XknXnk .

(10)

k

3

We now can use eqs. (8) and (10) to obtain,

S

=

-

i 2

Xnk k| P |n - n| P |k Xkn .

(11)

k

k

Inserting the definitions of Xnk and Xkn in eq. (11) and employing the completeness relation, k |k k| = I (where I is the identity operator), it follows that

S

=

-

i 2

=

-

i 2

n| X |k k| P |n -

k

k

n| (XP - P X) |n

=

1 2

,

n| P |k

k| X |n

after using [X, P ] = i I and n|n = 1 (since the energy eigenstates are normalized to unity).

REMARK:

The identity,

2m

2

(Ek - En) k| X |n 2 = 1 ,

(12)

k

is called the Thomas-Reiche-Kuhn sum rule, and it plays an important role in quantum

radiation theory. It is mentioned on p. 368 of Sakurai and Napolitano, where another method

of derivation is proposed. Sakurai and Napolitano suggest considering,

n|

X, [X, H]

|n

=

i m

n| [X, P ] |n

2

= -m

n|n

2

= -m ,

(13)

after making use of eq. (6) and [X, P ] = i I. Then after inserting a complete set of states in appropriate places,

n| X, [X, H] |n = n| (X2H - 2XHX + HX2 |n

= n| X |k k| XH |n - 2 n| XH |k

k

= 2 (En - Ek)| k| X |n |2 ,

k

k| X |n + n| HX |k

k| X |n (14)

after using H |k = Ek |k and H |n = Ek |n . Comparing eqs. (13) and (14), we end up with eq. (12).

There are numerous related sum rules that are similar to the Thomas-Reiche-Kuhn sum rule, which can be derived with similar techniques. For more details, see pp. 211?217 of Hans A. Bethe and Roman Jackiw, Intermediate Quantum Mechanics, 3rd edition (Westview Press, Boulder, CO, 1997).

(c) Verify your calculation of S in part (b) in the case of the one-dimensional harmonic

oscillator.

That

is,

assuming

V (X)

=

1 2

m2X

2,

compute

explicitly

k| X |n

and then use the

energy eigenvalues of the one-dimensional harmonic oscillator to explicitly evaluate S.

4

Using eq. (2.3.25a) of Sakurai and Napolitano,

k| X |n = 2m n k,n-1 + n + 1 k,n+1 .

Thus, for fixed n, the only two non-zero values of k| X |n are,

n - 1| X |n =

n 2m

,

n + 1| X |n =

(n + 1) 2m

.

Note that if n = 0, only the second matrix element above is relevant.

The energy eigenvalues of the one-dimensional harmonic oscillator are En = Hence, it follows that

(15)

(n

+

1 2

).

En-1 - En = - ,

En+1 - En = .

(16)

Once again, if n = 0, only the second energy difference above is relevant.

Employing eqs. (15) and (16) in evaluating the sum S [eq. (7)], only two terms in the sum

survive if n = 0,

S

=

m

2

2m

-n + (n + 1)

=

1 2

.

If n = 0, we replace -n + (n + 1) with 1 since only one term contributes to the sum. The end

result

is

the

same,

and

we

again

confirm

that

S

=

1 2

.

3. One would perhaps conclude from the lack of angular momentum in the ground state of the hydrogen atom that the electron is stationary.

(a) To show that this is not so, calculate the probability that the electron's momentum, if measured, would be found to lie in a momentum element d3p centered at momentum p.

The wave function for the ground state of hydrogen is

(r, , ) = 1 a-0 3/2e-r/a0 ,

where a0 2/(?e2) is the Bohr radius and ? is the reduced mass. To compute the probability that the electron has momentum between p and p + d3p, we need to compute the momentum

space wave function,

(p)

=

1 (2 )3/2

d3x e-ip?x/ (x) .

To perform the above integral, we shall make use of spherical coordinates and choose the

z-axis to lie along p. Then, p?x = pr cos , where r |x|. Hence,

(p)

=

1 (2 )3/2

1 (a30)1/2

r2 dr d cos d e-ipr cos / e-r/a0

=

1 (2 )3/2

1 (a30)1/2

2

r2 e-r/a0 dr

0

1

d cos e-ipr cos /

-1

=

(22

1 3a30)1/2

2 p

r e-r/a0 sin

pr

dr .

0

(17)

5

To evaluate the remaining integral, consider the following function,

F (a) = e-ay sin(my)dy ,

0

where a > 0 (the latter is needed for the integral to be convergent). Then,

F (a) = Im

e-ay eimy dy = Im

e-(a-im)y dy = Im

0

0

1 a - im

= Im

a + im a2 + m2

Taking the derivative of the above equation and multiplying by -1 yields,

-

dF da

=

0

y e-ay

sin(my)dy

=

2am (a2 + m2)2

,

Applying this result to eq. (17), we end up with

for a > 0.

(p) = 8

a0 2

3/2

1+

1 a0p

2

2.

=

a2

m + m2

.

(18)

Note that (p) depends only on the magnitude p |p|. That is, the momentum space ground state wave function is spherically symmetric (which is also true for its coordinate space wave function).

The probability that the electron lies between p and p + d3p is given by |(p)|2 d3p.

REMARK 1: A useful check of our calculation is to verify that the momentum space wave function is normalized to unity (this is a consequence of Parseval's theorem of Fourier analysis).

d3p |(p)|2 = 4

p2 dp |(p)|2 = 2562

0

a0 2

3

0

p2 dp 1 + a0p

2

4.

The integral is recognized as a Beta function,1

B(p,

q)

=

(p)(q) (p + q)

=

0

up-1 du (1 + u)p+q

.

After a change of variables, u = x2, it then follows that

0

xp dx (1 + x2)r

=

1 2

(p

+

1)

r-

1 2

(p

+

1)

2(r)

.

(19)

Recall

that

(n)

=

(n - 1)!

for

positive

integer

n,

(

1 2

)

=

and

(x + 1)

=

x(x).

It

follows

that

0

x2 dx (1 + x2)4

=

32

.

One then readily verifies that

d3p |(p)|2 = 1 .

1See, e.g., N.N. Lebedev, Special Functions and Their Applications (Dover Publications, Inc., New York, 1972).

6

REMARK 2: The following alternative derivation of eq. (17) is especially noteworthy, since this method is quite useful in a number of applications. We begin with

(p)

=

1 (2 )3/2

1 (a30)1/2

r2 e-r/a0 dr

0

d e-ik?x ,

(20)

where k p/ . Using the expansion of the plane wave in terms of spherical waves,2

eik?x = 4

i Ym(r^)Ym(k^)j(kr) ,

=0 m=-

where r^ x/r (with r |x). Integrating over the solid angle = (, ), where and are the polar and azimuthal angles of r^ with respect to a fixed z axis,

d eik?x = 4

i Ym(k^)j(kr)

=0 m=-

d Ym(r^) ,

(21)

Recall that the spherical harmonics satisfy the orthonormality relation,

d Ym(r^)Ym (r^) = mm .

Using the fact that Y00(r^) = (4)-1/2, it follows that

d Ym(r^) = 4 0m0 .

Using this result in eq. (21), it follows that

d eik?x

=

4j0(kr)

=

4 sin(kr) kr

.

(22)

Taking the complex conjugate of eq. (22) and inserting this result into eq. (20) (with k = p/ )

yields

(p) =

(22

1 3a30)1/2

2 p

r e-r/a0 sin

pr

dr ,

0

which confirms the result of eq. (17).

(b) What are the electron's mean kinetic and potential energy?

To compute the mean kinetic energy, it is convenient to perform the calculation in momentum space. Thus,

p2 2?

=

p2 2?

|(p)|2

d3p

=

2 ?

0

p4

dp |(p)|2

=

1282 ?

a0 2

3

0

p4 dp 1 + a0p

2

4.

Employing the result of eq. (19), we end up with,

p2

2

2? = 2?a20 .

(23)

2See the class handout entitled, Expansion of plane waves in spherical harmonics.

7

Next, we turn to the computation of the mean potential energy. The potential energy of

the hydrogen atom is

V

(r)

=

-

e2 r

.

Hence, the mean potential energy is

V (r) = -e2 1 . r

It is straightforward to compute

1 r

=

|(x)|2

1 r

d3x

=

4

1 a30

r e-2r/a0

0

dr

=

1 a0

,

after making use of the well known integral

0

xn-1e-ax

dx

=

(n) an

,

(24)

under the assumption that a > 0. For positive integer n, (n) = (n - 1)!.

Hence, it follows that

V (r)

=

-

e2 a0

=

2

- ?a20

,

where we have replaced e2 = 2/(?a0) using the definition of the Bohr radius, a0. Comparing with eq. (23), we obtain

p2 2?

=

-

1 2

V (r)

.

(25)

REMARK 3: An alternative calculation of the mean kinetic energy

One can also compute the mean kinetic energy by evaluating the expectation value in

position space,

p2 2? =

(x)

-

2

2?

2

(x)

d3x .

Note that in an = 0 state, (x) depends only on the radial coordinate r |x|. Hence,

2(x) =

2 r2

+

2 r

r

-

L2 r2

(r) =

2 r2

+

2 r r

(r)

since L 2 is a purely angular differential operator. Hence,

p2 2?

2

= 4 -2?

1 a30 0

d2 + 2 d e-r/a0 r2 e-r/a0 dr dr2 r dr

2

= 4 -2?

1 a30 0

1 a20

-

2 ra0

2

r2 e-r/a0 dr = 2?a20 ,

after employing eq. (24).

8

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