Modern Physics (Tipler)

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6 T h e Schrodinger Equation1925 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

Transcript of Modern Physics (Tipler)

204 C H A P T E R 6 The Schrodinger Equation

1925 and is now known as the Schrod inger equat ion. Like the classical wave equat ion, the Schrod inger equat ion relates the t ime and space derivatives of the wave function. T h e reasoning followed by Schrodinger is somewhat difficult and not impor­tant for o u r purposes . In any case, we can't der ive the Schrodinger equat ion jus t as we can't der ive Newton 's laws of motion. T h e validity of any fundamenta l equat ion lies in its a g r e e m e n t with exper iment . Al though it would be logical merely to postulate the Schrodinger equat ion, it is helpful to get some idea of what to expect by first consider ing the wave equa­tion for pho tons , which is Equation 5-7 with speed v = c and with y(x,t) replaced bv the wave function for light, namely, the electric field %(x,t).

d2% dx2

1 d 2 %

c2 dt2 6-1 Classical wave equation

As discussed in Chap te r 5, a particularly impor tan t solution of this equat ion is the ha rmonic wave function %(x,t) = (?o cos (kx — <at). Differentiating this function twice we obtain d2%/dt2 = - w 2 g 0 c o s ( & c - tat) = -<02%(x,t) and d2%/dx2 = — k2%(x,t). Substitution into Equation 6-1 then gives

= - 4 c-

OT

(i) = kc 6-2

Using w = E/h and p = hk, we have

E = pc 6-3

which is the relat ion between the energv and m o m e n t u m of a pho ton .

Let us now use the de Broglie relations for a particle such as an electron to find the relation between o» and k for electrons which is analogous to Equat ion 6-2 for pho tons . We can then use this relat ion to work backwards and see how the wave equat ion for electrons must differ from Equation 6 -1 . T h e energy of a part icle of mass m is

2 m + V 6-4

where V is the potential energv. Using the de Broglie relations we obtain

E nergy-momen tu m relation for photon

h (D h2k2

2 m + V 6-5

Th i s differs from Equation 6-2 for a pho ton because it contains the potential energy V and because the angular frequency w does not vary linearly with k. Note that we get a factor of OJ when we differentiate a ha rmon ic wave function with respect to t ime

S E C T I O N 6-4 Expectation Values and Operators 217

T h e expectat ion value is the same as the average value of .v that we would expect to obtain from a m e a s u r e m e n t of the positions of a large n u m b e r of particles with the same wave function ty(x,<). As we have seen, for a particle in a state of definite en­ergy the probability distr ibution is i n d e p e n d e n t of t ime. T h e ex­pectation value is then given bv

•+x

(x) = xi/>*(x)t/»(v) dx 6-28 Expectation value J -00

From the infinite-square-well wave functions, we can see by sym-metrv (or bv direct calculation) that (x) is L/2, the midpoin t of the well.

T h e expectat ion value of any function / ( x ) is given by

</(*)> = I f{x)$*$dx 6-29

For example , (x2) can be calculated from the wave functions, above, for the infinite square well of width L. It is left as an exer­cise to show that for that case

L2 12

(x2) = V - d i " ! 6-30 3 2n IT

We should note that we don ' t necessarily expect to measu re the expectat ion value. For example , for even n , the probability of measur ing x in some r ange dx at t he midpoin t of the well x = L/2 is zero because the wave function sin (mrx/L) is zero there . We get (x) = L/2 because the probability function is symmetrical about that point.

Optional

Expectation value of momentum

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 the expectat ion value (p) f rom Equation 6-29. However , it is impossible in principle to find p as a function of x since, according to the uncer ta inty principle, both p and x cannot be de t e rmined at the same t ime. T o find (p) we need to know the distr ibution function for m o m e n t u m , which is equiva­lent to the distr ibution function A(k) discussed in Section 5-4. As discussed there , if we know i/»(x) we can find A (k) and vice versa bv Four ier analysis. Fortunately we need not d o this each t ime. It can be shown from Four ier analysis that (p) can be found from

» - f > ( 7 5 i ) * * Similarly (/>*) can be found from

6-31

S E C T I O N 6-5 Transitions between Energy States 219

in te rms of position and m o m e n t u m and replace the m o m e n t u m variables by the app rop r i a t e opera to rs to obtain the Hamil tonian o p e r a t o r for the system.

Quest ions

4. For what kind of probability distr ibution would vou expect to get the expectat ion value in a single measurement?

5. Is ( x 2 ) the same as (x) 2 ?

6-5 Transitions between Energy States

We have seen that the Schrodinger equat ion leads to energy quant izat ion for b o u n d systems. T h e existence of these en­ergy levels is de t e rmined experimental ly by observation of the energy emit ted o r absorbed when the system makes a transi­tion from one level to ano the r . In this section we shall consider some aspects of these transit ions in one d imension. T h e results will be readily applicable to m o r e complicated situations.

In classical physics, a charged particle radiates when it is accelerated. If t he charge oscillates, the frequency of the radia­tion emit ted equals the frequency of oscillation. A stationary charge distr ibution does not radia te .

Consider a particle with charge q in a q u a n t u m state n de­scribed by the wave function

where En is the energy and <|/„(x) is a solution of the time-i n d e p e n d e n t Schrodinger equat ion for some potential energy V(x). T h e probability of finding the charge in dx is ^ f ^ n dx- If we make many measu remen t s on identical svstems (i.e., particles with the same wave function), the average a m o u n t of charge found in dx will be qtyfity n dx. We there fore identify q^t^n with the charge density p. As we have poin ted out , the probability den­sity is i n d e p e n d e n t of t ime if the wave function contains a single energy, so the charge density for this state is also i n d e p e n d e n t of t ime:

pn = qV*(x,ty9n(x,t) = ?i|»*(xW/(x) = q^mt

We should therefore expect that this stationary charge distribu­tion would n o t radiate. (This a rgumen t , in tin- case oi the hydro­gen a tom, is the quantum-mechanica l explanat ion of Bohr 's postulate of nonrad ia t ing orbits.) However , we d o observe that systems m a k e transit ions from one energy state to ano the r with the emission or absorpt ion of radiat ion. T h e cause of the transit ion is the interaction of the electromagnetic field with

t T o simplify the notation in this section we shall sometimes omit the functional dependence and merely write tlin for 0„(x) and for y^x.t).