[Donald A. McQuarrie, John D. Simon] Physical Chem(BookZZ.org)

Although we have shown this proportionality only for the case of a vibrating string, it is 
a general result and shows that the intensity of a wave is proportional to the square of the 
amplitude. If we had carried everything through in complex notation instead of sines and 
cosines, then we would have found that En is proportional to I Dn 12 instead of just D~. 
Generally, there are many normal modes present at the same time, and the complete 
solution is (Equation 2.25) 
~ nnx 
u(x, t) = L..- Dn cos(wnt + 4>) sin -
1
-
n=l 
Using the fact that (see Problem 3-16) 
sin -- sm --dx = 0 11 nnx . mnx 0 l l ifm=/=n 
show that 
2 2 00 
E = n v p '\'n2D2 
n 4[ L..- n 
n=l 
3-32. The quantized energies of a particle in a box result from the boundary conditions, or from 
the fact that the particle is restricted to a finite region. In this problem, we investigate the 
quantum-mechanical problem of a free particle, one that is not restricted to a finite region. 
The potential energy V (x) is equal to zero and the Schrodinger equation is 
-OO<X<OO 
Note that the particle can lie anywhere along the x-axis in this problem. Show that the two 
solutions of this SchrOdinger equation are 
and 
P r o b l e m s 
w h e r e 
k = ( 2 m E ) ' I 2 
h 
S h o w t h a t i f E i s a l l o w e d t o t a k e o n n e g a t i v e v a l u e s , t h e n t h e w a v e f u n c t i o n s b e c o m e 
u n b o u n d e d f o r l a r g e x . T h e r e f o r e , w e w i l l r e q u i r e t h a t t h e e n e r g y , E , b e a p o s i t i v e q u a n t i t y . 
W e s a w i n o u r d i s c u s s i o n o f t h e B o h r a t o m t h a t n e g a t i v e e n e r g i e s c o r r e s p o n d t o b o u n d 
s t a t e s a n d p o s i t i v e e n e r g i e s c o r r e s p o n d t o u n b o u n d s t a t e s , a n d s o o u r r e q u i r e m e n t t h a t E 
b e p o s i t i v e i s c o n s i s t e n t w i t h t h e p i c t u r e o f a f r e e p a r t i c l e . 
T o g e t a p h y s i c a l i n t e r p r e t a t i o n o f t h e s t a t e s t h a t 1 / 1 '
1 
( x ) a n d 1 / f
2 
( x ) d e s c r i b e , o p e r a t e o n 
1 / f / x ) a n d 1 / f
2
( x ) w i t h t h e m o m e n t u m o p e r a t o r P ( E q u a t i o n 3 . 1 1 ) , a n d s h o w t h a t 
A d \u2022
1
< 
P l / f =-in-'~'-' = n k l / f 
I d x I 
a n d 
A 
0 
d l / f
2 
P l / f = - 1 h - = - h k l / f 
2 d x 2 
N o t i c e t h a t t h e s e a r e e i g e n v a l u e e q u a t i o n s . O u r i n t e r p r e t a t i o n o f t h e s e t w o e q u a t i o n s i s t h a t 
1 / 1 '
1 
d e s c r i b e s a f r e e p a r t i c l e w i t h f i x e d m o m e n t u m h k a n d t h a t 1 / 1 '
2 
d e s c r i b e s a p a r t i c l e w i t h 
f i x e d m o m e n t u m - h k . T h u s , 1 / 1 '
1 
d e s c r i b e s a p a r t i c l e m o v i n g t o t h e r i g h t a n d 1 / 1 '
2 
d e s c r i b e s 
a p a r t i c l e m o v i n g t o t h e l e f t , b o t h w i t h a f i x e d m o m e n t u m . N o t i c e a l s o t h a t t h e r e a r e n o 
r e s t r i c t i o n s o n k , a n d s o t h e p a r t i c l e c a n h a v e a n y v a l u e o f m o m e n t u m . N o w s h o w t h a t 
h 2 k 2 
E = -
2 m 
N o t i c e t h a t t h e e n e r g y i s n o t q u a n t i z e d ; t h e e n e r g y o f t h e p a r t i c l e c a n h a v e a n y p o s i t i v e 
v a l u e i n t h i s c a s e b e c a u s e n o b o u n d a r i e s a r e a s s o c i a t e d w i t h t h i s p r o b l e m . 
L a s t , s h o w t h a t 1 / f ; ( x ) l / f
1
( x ) = A~A
1 
= J A
1
1
2 
= c o n s t a n t a n d t h a t 1 f J ' ; ( x ) l / f
2
( x ) = 
A ; A
2 
= I A i = c o n s t a n t . D i s c u s s t h i s r e s u l t i n t e r m s o f t h e p r o b a b i l i s t i c i n t e r p r e t a t i o n 
o f 1 / f * l / f . A l s o d i s c u s s t h e a p p l i c a t i o n o f t h e U n c e r t a i n t y P r i n c i p l e t o t h i s p r o b l e m . W h a t 
a r e a - P a n d a - x ? 
3 - 3 3 . D e r i v e t h e e q u a t i o n f o r t h e a l l o w e d e n e r g i e s o f a p a r t i c l e i n a o n e - d i m e n s i o n a l b o x b y 
a s s u m i n g t h a t t h e p a r t i c l e i s d e s c r i b e d b y s t a n d i n g d e B r o g l i e w a v e s w i t h i n t h e b o x . 
3 - 3 4 . W e c a n u s e t h e U n c e r t a i n t y P r i n c i p l e f o r a p a r t i c l e i n a b o x t o a r g u e t h a t f r e e e l e c t r o n s 
c a n n o t e x i s t i n a n u c l e u s . B e f o r e t h e d i s c o v e r y o f t h e n e u t r o n , o n e m i g h t h a v e t h o u g h t 
t h a t a n u c l e u s o f a t o m i c n u m b e r Z a n d m a s s n u m b e r A i s m a d e u p o f A p r o t o n s a n d 
A - Z e l e c t r o n s , t h a t i s , j u s t e n o u g h e l e c t r o n s s u c h t h a t t h e n e t n u c l e a r c h a r g e i s + Z . S u c h 
a n u c l e u s w o u l d h a v e a n a t o m i c n u m b e r Z a n d m a s s n u m b e r A . I n t h i s p r o b l e m , w e w i l l 
u s e E q u a t i o n 3 . 4 1 t o e s t i m a t e t h e e n e r g y o f a n e l e c t r o n c o n f i n e d t o a r e g i o n o f n u c l e a r s i z e . 
T h e d i a m e t e r o f a t y p i c a l n u c l e u s i s a p p r o x i m a t e l y 1 0 -
1 4 
m . S u b s t i t u t e a = 1 0 -
1 4 
m i n t o 
E q u a t i o n 3 . 4 1 a n d s h o w t h a t a - P i s 
f Y ; ; ; : : : 3 X 1 0 -
2 0 
k g · m · S - l 
p 
1 0 3 
104 
Show that 
Chapter 3 I The Schrodinger Equation and a Particle In a Box 
a2 
E = ____!?__ = 5 X 10-10 J 
2m 
~ 3000MeV 
where millions of electron volts (MeV) is the common nuclear physics unit of energy. It 
is observed experimentally that electrons emitted from nuclei as f3 radiation have energies 
of only a few MeV, which is far less than the energy we have calculated above. Argue, 
then, that there can be no free electrons in nuclei because they should be ejected with much 
higher energies than are found experimentally. 
3-35. We can use the wave functions of Problem 3-29 to illustrate some fundamental symmetry 
properties of wave functions. Show that the wave functions are alternately symmetric and 
antisymmetric or even and odd with respect to the operation x --+ - x, which is a reflection 
through the x = 0 line. This symmetry property of the wave function is a consequence of 
the symmetry of the Hamiltonian operator, as we now show. The Schrodinger equation may 
be written as 
Reflection through the x = 0 line gives x --+ - x and so 
H(-x)l/f (-x) = E 1/f (-x) 
n n n 
Now show that H(x) = H( -x) (i.e., that His symmetric) for a particle in a box, and so 
show that 
H(x)l/f (-x) = E 1/f (-x) 
n n n 
Thus, we see that 1/fn (-x) is also an eigenfunction of H belonging to the same eigen-
value En. Now, if only one eigenfunction is associated with each eigenvalue (the state is 
nondegenerate), then argue that 1/fn(x) and 1/f/-x) must differ only by a multiplicative 
constant [i.e., that 1/f.(x) = cl/fn (-x) ]. By applying the inversion operation again to this 
equation, show that c = ±1 and that all the wave functions must be either even or odd with 
respect to reflection through the x = 0 line because the Hamiltonian operator is symmetric. 
Thus, we see that the symmetry of the Hamiltonian operator influences the symmetry of 
the wave functions. A general study of symmetry uses group theory,