bei48482_FM

since U0 there. (The total derivative d^2 dx^2 is the same as the partial derivative

^2 x^2 because is a function only of xin this problem.) Equation (5.37) has the

solution

Asin xB cos x (5.38)

which we can verify by substitution back into Eq. (5.37). Aand Bare constants to be

evaluated.

This solution is subject to the boundary conditions that 0 for x0 and for

xL. Since cos 0 1, the second term cannot describe the particle because it does

not vanish at x0. Hence we conclude that B0. Since sin 0 0, the sine term

always yields 0 at x0, as required, but will be 0 at xLonly when

Ln n1, 2, 3,... (5.39)

This result comes about because the sines of the angles , 2, 3,... are all 0.

From Eq. (5.39) it is clear that the energy of the particle can have only certain val-

ues, which are the eigenvalues mentioned in the previous section. These eigenvalues,

constituting the energy levelsof the system, are found by solving Eq. (5.39) for En,

which gives

Particle in a box En n1, 2, 3,... (5.40)

Equation (5.40) is the same as Eq. (3.18) and has the same interpretation [see the

discussion that follows Eq. (3.18) in Sec. 3.6].

Wave Functions

The wave functions of a particle in a box whose energies are Enare, from Eq. (5.38)

with B0,

(^) nA sin x (5.41)
Substituting Eq. (5.40) for Engives
(^) nA sin (5.42)
for the eigenfunctions corresponding to the energy eigenvalues En.
It is easy to verify that these eigenfunctions meet all the requirements discussed in
Sec. 5.1: for each quantum number n, (^) nis a finite, single-valued function of x, and
(^) nand  (^) nxare continuous (except at the ends of the box). Furthermore, the integral
nx
L
 2 mEn
n^2 ^22
2 mL^2
 2 mE
 2 mE
 2 mE
178 Chapter Five
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