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"; Quantum Mechanics & Linear Hermitian Operators Exam Practice - Chem Homework Help
Quantum Mechanics & Linear Hermitian Operators Exam Practice

4—37. Show that the average energy of a particle described by Equation 4.78 is a constant.

The particle is described by

IV (x, t) — word image 2499 1 1/2 Elt/h Sin — TX + 1 1/2 —i E2t/h sin word image 2500

a a a a


word image 2501 —iE1t /h + word image 2502 -i E2t/h

The average energy is


word image 2503



1 a

word image 2504iE lt / h + i E2t/h word image 2505 —iElt/h + word image 2506 —iE2t/h word image 2507


1 a

word image 2508



word image 2509 El word image 2510 dx + dx


a a

word image 2511 dx


word image 2512 (El + E2)



word image 2513


The average energy is a constant. The orthonormality of the wave functions is used in deriving this expression.

4—39. Derive an expression for the average position of a particle in a box in a state described by

1/2 2m x 1/237TX W(x, t) = -iE2t/h sinsin

With what frequency does the particle oscillate about the midpoint of the box?

This problem is similar to Problem 4—34. The average position of x is

word image 2514 word image 2515 word image 2516 1/2 1/2 37tx iE2t/1j sini E3t/h stn

( x )

word image 2517 word image 2518 1/2

277 x word image 2519 iE3t/h sin 37TX

—i E2t/h sin word image 2520

word image 2521

word image 2522 x sin dx + e word image 2523x sin word image 2524

word image 2525 2 21TX277 x 377 x

word image 2526 27TX 3rrx 2 3rrx

word image 2527 word image 2528 x sin sin dx + x sin dx

2 27TX word image 2529

word image 2530 x sin — dx+—

x sin2 37TX

— dx

a o

word image 2531 word image 2532 word image 2533 word image 2534 word image 2535 E3—E2)t/h x sin 2nx sin

+ ei (

2 Yr x 2 3Jtx

word image 2536 x sin — dx + — x sin dx

word image 2537 37TX

x sin sin word image 2538

where the relation cos 0 (e l + e -10 )/2 and the definition of E2)/h have been applied. The first two integrals have been evaluated in Problem 3—14′

word image 2539

x sin dx word image 2540

The third integral can be evaluated by using the trigonometric identity (Problem 3—2 1):

Sin sin f} cos(a H) COS(CY F)



2nx Yr x 1


rx 5nx

x sin sin dx — word image 2541x cos — x cos
0 a a 20 a a
I2 2 a IT X ax aa

57tx ax 57TX

word image 2542— cos — + — sin word image 2543 cos— — sin
2word image 2544 word image 2545 2257T2 aaa a

word image 2546



word image 2547

Combining all the integrals,

l a2 l a2 2 cos 023t 24a 2 word image 2548 a 25T2

a 48a

word image 2549 cos 0)23t

2 25Tt 2

The particle oscillates about the midpoint of the box with an angular frequency of (or a frequency of 0)23/2n).


The Harmonic Oscillator and Vibrational Spectroscopy

5—13. In the infrared spectrum ofH1271, there is an intense line at 2309 cm¯l. Calculate the force constant of H1271 and the period of vibration ofH1271

Equation 5.39 can be rearranged to give

k = (2rrcöobs) IL

The reduced mass for H1271 is

(1.008 amu) (126.9 amu)

word image 2550 = 1.000 amu

1.008 amu + 126.9 amu


word image 2551 2

(2.99792458 x 10 10 cm.s- 2309 cm-I (1.000 amu) (1.660 5402 x 10¯27 kg•amu¯ word image 2552


The period of vibration is r = 2r/0) = 27T (U/ k) 1/2 (Problem 5-4).

word image 2553 r = 27T [(1.000 amu) (1.660 5402 x 10¯27 word image 2554 = 1.445 x 10— 14 s

5—14. The force constant of 35C135Cl is 319 word image 2555 Calculate the ftlndamental vibrational frequency and the zero-point energy of 35C135Cl.

From Problem 5—8, the reduced mass for 35C135Cl is = 34.97 amu/2. This is used in Equation 5.39,

1 k 1/2

bobs = word image 2556


word image 2557 1/2

= 556.5 cm ¯l

The zero-point energy is


E hcöobs


word image 2558 (6.626 0755 x 10 34 2.997 924 58 x 10 10 cm.s- 556.5 cm¯ word image 2559

= 5.527 x 10 21 J

Chapter 5

5—15. The fundamental line in the infrared spectrum of 12c160 occurs at 2143.0 cm and the first overtone occurs at 4260.0 cm¯l. Calculate the values oföe and for 12c160 word image 2560

Equation 5.43 gives (bobs = bev — ieöev (v + 1).

Thus, for the two lines in this problem,

Fundamental: – word image 2561 = 2143.0 cm 1

First overtone: 2öe — word image 2562 = 4260.0 cm—1

Multiply the fundamental frequency by 3 and subtract the overtone to get

word image 2563 = 3(2143.0 cm -I ) — 4260.0 cm -I = 2169.0 cm—1

Multiply the fundamental frequency by 2 and subtract the overtone to get

– – = 26.0 cm ¯l



ieöe — 13.0 cm—1

5—21. Verify that YITl(x) and 1/12(x) given in Table 5.4 satisfy the Schrödinger equation for a

Ivarmonic oscillator.

The Schrödinger equation for a harmonic oscillator is given by Equation 5.31 :

word image 2564 1

—kx 2 = 0


h word image 25651/2
where E — word image 2566 word image 2567


t,+ – . From Table 5.4,

word image 2568 3 1/4


word image 2569 1/4

1/12 (x) (2ax2 word image 2570 —ax /2

The Harmonic Oscillator and Vibrational Spectroscopy 239

where a = (kg) 1/2/h. Substituting into the Schrödinger equation with v = I gives

1 word image 2571

dX2 word image 2572 ti2 E 2kx2 VI

—ax2/2 2 —ax2/2—ax2/2

word image 2573


word image 2574



4a_3 1/4

word image 2575


word image 2576 —ax — 2ax + a 2 x3 e-ax2/2 xe—ax2/2

word image 2577 1/4

—3ax + a 2 3x + 3ax —a x word image 2578

Substituting into the Schrödinger equation with v = 2 gives


word image 2579






— ax




I la x2 + 2a x4

Both and 1/12 are solutions to the Schrödinger equation,

5—22. Show explicitly for a harmonic oscillator that is orthogonal to word image 2580 and t//3(gE) and that VI(C) is orthogonal to and (see Table 5.4).

From Table 5.4,

240 Chapter 5


word image 2581 —ax2/2 l/fo(x) =

3 1/4

—ax 2 /2

y/’l(.v) = xe

1/4 2

Mf2(x) = (2ax2 l)e—ax /2


word image 2582 3 1/4 —ax 2/2

1/f3(x) (2ax — 3x)e

Of the five integrals that must be evaluated to show orthogonality, three have integrands that are odd functions of x, and so are zero.

word image 2583 dx = 0

This leaves the integrals with even integrands to be evaluated explicitly.

word image 2584 dx = 2

word image 2585


word image 2586 1 2

— 2


1/4 3 1/4 00

word image 2587


4 2

(2ax — 3x d.v

word image 2588 3 1/4 3 1/4 3 1/2

word image 2589


802) C) 3

— o

5—24. Prove that the product of two even functions is even, that the product of two odd functions is even, and that the product of an even and an odd function is odd.

Recall that an even function is one for which f (x) = f(—x) and an odd function is one for which f (x) —f(—x). Let P (x) be the product of two functions f (x) and g(x). For two even functions,

P (x) = f(x)g(x) = f(—x)g(—x) = P ( —x)

so the product of two even functions is even. For two odd functions,

P (x) = f(x)g(x) word image 2590

so the product of two odd functions is also even. For one odd and one even function,

P (x) = f(x)g(x) — word image 2591 ¯f(—x)g(—x) — word image 2592

so the product of one odd and one even function is odd.

5—25. Prove that the derivative of an even (odd) function is odd (even).


The Harmonic Oscillator and Vibrational Spectroscopy

If f (x) is even, it can be represented by a power series of the form

        1. (X) CO -+- C2X C4X4 O (X 6)

where the only allowed values of n in xn are even. The derivative of this function is

f'(x) = 2c2x + 4c4x 3 -F O(x 5 )

which is an odd function expressed in a power series.

Similarly, ifg(x) is odd, it can be represented by

        1. (X) = C IX C3X CSX5 O (X7)

where the only allowed values of n in x’l are odd, and its derivative is

g/ (X) = Cl + 3C3X 2 + 5C5X4 + O (X 6)

which is an even fttnction.

5—26. Show that

word image 2593

2) 1/2 is the square root of the mean of the square of the for a harmonic oscillator. Note that (x displacement (the root-mean-square displacement) of the oscillator.

word image 2594 =—ax2/2 SO

From Table 5.4, V2(x)

word image 2595 dx


word image 2596




word image 2597



5-27. Show that (p) = 0 and that

word image 2598 = —1101k) 5 l/2


Chapter 5

for a harmonic oscillator.

Recall that = —ili— and P word image 2599 Then from Table 5.4,


Y/J2(x) — word image 2600 —ax 2/2

Note that I/f2(x) is an even function of x. Problem 5—25 showed that the derivative of an even ftmction is odd, and Problem 5—24 showed that the product of an even function and an odd function is odd, so in evaluating

word image 2601 word image 2602 word image 2603 word image 2604 (p) word image 2605 dx = —ih 1/f2(x) — 1/0 (x) dx

the integrand is seen to be overall an odd function. Consequently, (p) = 0.

For (p word image 2606

word image 2607 1/J2(x) P 2 V2(x) dx

1/2 word image 2608

word image 2609 -2h2 word image 2610 —ax2/2 d Saxe—ax 2/2 — 20 2 x 3 e—ax-/2 word image 2611

1/2 00

word image 2612 —2112 word image 2613 50 — I la 2x 2 -h 2cy v word image 2614

1/2 word image 2615

word image 2616 -2112 — 240 3x 4 -l- 2 la 2x 2 word image 2617 —cy.v-

word image 2618 1/2


—h2a = —h (11k) 1/2

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