204
6 T h e Schrodinger Equation
CHAPTER
1925 a n d is n o w k n o w n as the S c h r o d i n g e r e q u a t i o n . Like the classical wave e q u a t i o n , t h e S c h r o d i n g e r e q u a t i o n relates the time a n d space derivatives of the wave function. T h e r e a s o n i n g followed by S c h r o d i n g e r is s o m e w h a t difficult a n d not i m p o r tant for o u r p u r p o s e s . I n any case, we can't derive the S c h r o d i n g e r e q u a t i o n j u s t as we can't derive Newton's laws of motion. T h e validity of any f u n d a m e n t a l equation lies in its a g r e e m e n t with e x p e r i m e n t . A l t h o u g h it would be logical merely to postulate t h e S c h r o d i n g e r e q u a t i o n , it is helpful to get some idea of w h a t to expect by first c o n s i d e r i n g t h e wave equation for p h o t o n s , which is Equation 5-7 with speed v = c a n d with y(x,t) replaced bv the wave function for light, namely, the electric field %(x,t). d% dx
1 c
2
2
2
d % 2
dt
6-1
2
Classical wave equation
As discussed in C h a p t e r 5, a particularly i m p o r t a n t solution of this e q u a t i o n is the h a r m o n i c wave function %(x,t) = (?o cos (kx —
2
2
2
2
2
0
2
= - c-4 OT (i)
= kc
6-2
Using w = E/h a n d p = hk, we have E = pc
6-3
which is t h e relation between the e n e r g v a n d m o m e n t u m of a photon. Let us now use t h e d e Broglie relations for a particle such as an electron to find the relation between o» a n d k for electrons which is a n a l o g o u s to Equation 6-2 for p h o t o n s . We can t h e n use this relation to work backwards a n d see how the wave equation for electrons m u s t differ from Equation 6 - 1 . T h e e n e r g y of a particle of mass m is + V
2m
6-4
w h e r e V is t h e potential e n e r g v . U s i n g the d e Broglie relations we obtain hk + V 2m 2
h (D
2
6-5
T h i s differs from Equation 6-2 for a p h o t o n because it contains t h e potential e n e r g y V a n d because t h e a n g u l a r frequency w d o e s n o t vary linearly with k. N o t e that we get a factor of OJ when we differentiate a h a r m o n i c wave function with respect to time
E nergy-momen tu m relation for photon
SECTION
6-4
Expectation Values and Operators
217
T h e expectation value is t h e same as t h e a v e r a g e value of .v that we would expect to obtain from a m e a s u r e m e n t of t h e positions of a large n u m b e r of particles with t h e s a m e wave function ty(x,<). As we h a v e seen, for a particle in a state of definite energy t h e probability distribution is i n d e p e n d e n t of time. T h e expectation value is t h e n given bv •+x
(x) =
xi/>*(x)t/»(v) dx
6-28
Expectation
value
J -00 F r o m t h e infinite-square-well wave functions, we can see by symmetrv (or bv direct calculation) that (x) is L/2, t h e m i d p o i n t of the well. T h e expectation value of any function / ( x ) is given by
(*)> = I
f{x)$*$dx
6-29
For e x a m p l e , (x ) can b e calculated from t h e wave functions, above, for the infinite s q u a r e well of width L. It is left as an exercise to show that for that case 2
V
L 1 - d i " ! 3 2n IT 2
(x ) 2
=
2
6-30
We should note that we d o n ' t necessarily e x p e c t to m e a s u r e the expectation value. For e x a m p l e , for even n , the probability of m e a s u r i n g x in some r a n g e dx at t h e m i d p o i n t of the well x = L/2 is zero because t h e wave function sin (mrx/L) is zero t h e r e . We get (x) = L/2 because t h e probability function is symmetrical about that point.
Optional
Operators
If we knew the m o m e n t u m p of a particle as a function of x, we could calculate t h e expectation value (p) from Equation 6-29. However, it is impossible in principle to find p as a function of x since, a c c o r d i n g to t h e uncertainty principle, b o t h p a n d x c a n n o t be d e t e r m i n e d at the s a m e time. T o find (p) we n e e d to know the distribution function for m o m e n t u m , which is equivalent to t h e distribution function A(k) discussed in Section 5-4. As discussed t h e r e , if we know i/»(x) we can find A (k) a n d vice versa bv F o u r i e r analysis. Fortunately we n e e d not d o this each time. It can be shown from F o u r i e r analysis that (p) can be found from Expectation value of momentum
»-f>(75i)** Similarly (/>*) can be f o u n d from
6-31
6-5
SECTION
Transitions between Energy States
219
in t e r m s of position a n d m o m e n t u m a n d replace t h e m o m e n t u m variables by t h e a p p r o p r i a t e o p e r a t o r s to obtain the H a m i l t o n i a n o p e r a t o r for t h e system. Questions
4 . For what k i n d of probability distribution would vou expect to get t h e expectation value in a single m e a s u r e m e n t ? 5. Is ( x ) t h e s a m e as (x) ? 2
2
6-5 Transitions between Energy States W e have seen that t h e S c h r o d i n g e r equation leads to e n e r g y quantization for b o u n d systems. T h e existence of these energy levels is d e t e r m i n e d experimentally by observation of t h e e n e r g y emitted o r a b s o r b e d w h e n the system makes a transition from o n e level to a n o t h e r . In this section we shall consider s o m e aspects of these transitions in o n e d i m e n s i o n . T h e results will be readily applicable to m o r e complicated situations. I n classical physics, a c h a r g e d particle radiates w h e n it is accelerated. If t h e c h a r g e oscillates, the frequency of t h e radiation e m i t t e d equals t h e frequency of oscillation. A stationary c h a r g e distribution d o e s not r a d i a t e . C o n s i d e r a particle with c h a r g e q in a q u a n t u m state n d e scribed by the wave function
w h e r e E is t h e e n e r g y a n d <|/„(x) is a solution of t h e timei n d e p e n d e n t S c h r o d i n g e r e q u a t i o n for some potential e n e r g y V(x). T h e probability of finding t h e c h a r g e in dx is ^ f ^ n dx- If we m a k e m a n y m e a s u r e m e n t s on identical svstems (i.e., particles with the s a m e wave function), t h e a v e r a g e a m o u n t of c h a r g e with found in dx will be qtyfity dx. We t h e r e f o r e identify q^t^n t h e charge density p. As we have p o i n t e d out, the probability d e n sity is i n d e p e n d e n t of time if t h e wave function contains a single e n e r g y , so t h e c h a r g e density for this state is also i n d e p e n d e n t of time: n
n
p
n
= qV*(x, y9 (x,t) t
n
= i|»*(xW/(x) = ?
q^mt
W e should t h e r e f o r e expect that this stationary c h a r g e distribution would n o t radiate. (This a r g u m e n t , in tin- case oi the h y d r o g e n a t o m , is t h e q u a n t u m - m e c h a n i c a l e x p l a n a t i o n of Bohr's postulate of n o n r a d i a t i n g orbits.) H o w e v e r , we d o observe t h a t systems m a k e transitions from o n e e n e r g y state to a n o t h e r with t h e emission or a b s o r p t i o n of radiation. T h e cause of the transition is t h e interaction of the electromagnetic field with t T o simplify the notation in this section we shall sometimes omit the functional for y^x.t). d e p e n d e n c e and merely write tlin for 0„(x) and