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) — 1 1/2 Elt/h Sin — TX + 1 1/2 —i E2t/h sin
a a a a
—iE1t /h + -i E2t/h
The average energy is
iE lt / h + i E2t/h —iElt/h + —iE2t/h
El dx + dx
(El + E2)
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
1/2 1/2 37tx iE2t/1j sini E3t/h stn
( x )
277 x iE3t/h sin 37TX
—i E2t/h sin
|x sin dx + e||x sin|
2 21TX277 x 377 x
27TX 3rrx 2 3rrx
x sin sin dx + x sin dx
|2 27TX |
x sin — dx+—
|x sin||2 37TX|
E3—E2)t/h x sin 2nx sin
+ ei (
2 Yr x 2 3Jtx
x sin — dx + — x sin dx
x sin sin
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′
x sin dx
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
|x sin sin dx —||x cos — x cos|
|0 a a 2||0 a a|
|I||2 2 a IT X ax a||a|
57tx ax 57TX
|— cos — + — sin cos||— — sin|
|2||2257T2 aa||a a|
Combining all the integrals,
l a2 l a2 2 cos 023t 24a 2 a 25T2
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)
= 1.000 amu
1.008 amu + 126.9 amu
(2.99792458 x 10 10 cm.s- 2309 cm-I (1.000 amu) (1.660 5402 x 10¯27 kg•amu¯
The period of vibration is r = 2r/0) = 27T (U/ k) 1/2 (Problem 5-4).
r = 27T [(1.000 amu) (1.660 5402 x 10¯27 = 1.445 x 10— 14 s
5—14. The force constant of 35C135Cl is 319 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
= 556.5 cm ¯l
The zero-point energy is
(6.626 0755 x 10 34 2.997 924 58 x 10 10 cm.s- 556.5 cm¯
= 5.527 x 10 21 J
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
Equation 5.43 gives (bobs = bev — ieöev (v + 1).
Thus, for the two lines in this problem,
Fundamental: – = 2143.0 cm 1
First overtone: 2öe — = 4260.0 cm—1
Multiply the fundamental frequency by 3 and subtract the overtone to get
= 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
The Schrödinger equation for a harmonic oscillator is given by Equation 5.31 :
—kx 2 = 0
|where E — |
|t,+ – . From Table 5.4,|
1/12 (x) (2ax2 —ax /2
The Harmonic Oscillator and Vibrational Spectroscopy 239
where a = (kg) 1/2/h. Substituting into the Schrödinger equation with v = I gives
dX2 ti2 E 2kx2 VI
—ax2/2 2 —ax2/2—ax2/2
—ax — 2ax + a 2 x3 e-ax2/2 xe—ax2/2
—3ax + a 2 3x + 3ax —a x
Substituting into the Schrödinger equation with v = 2 gives
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 and t//3(gE) and that VI(C) is orthogonal to and (see Table 5.4).
From Table 5.4,
240 Chapter 5
—ax2/2 l/fo(x) =
—ax 2 /2
y/’l(.v) = xe
Mf2(x) = (2ax2 l)e—ax /2
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.
dx = 0
This leaves the integrals with even integrands to be evaluated explicitly.
dx = 2
1/4 3 1/4 00
(2ax — 3x d.v
3 1/4 3 1/4 3 1/2
802) C) 3
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)
so the product of two odd functions is also even. For one odd and one even function,
P (x) = f(x)g(x) — ¯f(—x)g(—x) —
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
- (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
- (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
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.
From Table 5.4, V2(x)
5-27. Show that (p) = 0 and that
= —1101k) 5 l/2
for a harmonic oscillator.
Recall that = —ili— and P Then from Table 5.4,
Y/J2(x) — —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
(p) dx = —ih 1/f2(x) — 1/0 (x) dx
the integrand is seen to be overall an odd function. Consequently, (p) = 0.
1/J2(x) P 2 V2(x) dx
-2h2 —ax2/2 d Saxe—ax 2/2 — 20 2 x 3 e—ax-/2
—2112 50 — I la 2x 2 -h 2cy v
-2112 — 240 3x 4 -l- 2 la 2x 2 —cy.v-
—h2a = —h (11k) 1/2