Irodov – Problems in General Physics

6.84. Let us integrate the Schrodinger equation over a small

interval of the coordinate x within which there is a discontinuity

in U (x), for example at the point x = 0:

+6

84) p r

—07 + 6)— ai (— 6) = )

2m

(E — U) dx.

Since the discontinuity U is finite the integral tends to zero as

6 —* 0. What follows is obvious.

6.85. (a) Let us write the Schrodinger equation for two regions

0 < x < 1, + k 2 11) 1 = 0, k 2 = 2mE/h 2 ,

x > /, 1); x2,T2 ___ 0, x2^ 2m (U 0 — E)1h 2.

Their common solutions

(x) = a sin (kx a), '11) 2 (x) = be -xx cemx

must satisfy the standard and boundary conditions. From the condi-

tion Y 1 (0) = 0 and the requirement for the finiteness of the wave

Fig. 45.

function it follows that a = 0 and c = O. And finally, from the

continuity of 1) (x) and its derivative at the point x = 1 we obtain

tan kl = —k/x, whence

sin kl = t kl h 2 /2m/zUo.

Plotting the left-hand and right-hand sides of the last equation
(Fig. 45), we can find the points at which the straight line crosses
the sine curve. The roots of the equation corresponding to the eigen-
values of energy E are found from those intersection points (kl)i^
for which tan (kl) < i 0, i.e. the roots of that equation are located
in the even quadrants (these segments of the abscissa axis are shown
heavy in the figure). It is seen from the plot that the roots of the
equation, i.e. the bound states of the particle, do not always exist.
The dotted line indicates the ultimate position of the straight line.
(b) 1 (1 2 U - 0,1 min = 7t 2 h 2 18M, ( 12 U0)n min = (2n — 1) n2h2/8m.