Thermodynamics, Statistical Physics, and Quantum Mechanics

QUANTUM MECHANICS 307

The integral on the left equals The easiest way to see this result

is to use the change of variables and the integrand becomes

between 0 and (Actually, just note that this is the area of

a quadrant of a disk of radius C). We get

b) The constraint that there be only one bound state is that and

This gives the following constraints on the last constant in the

energy expression:

5.55 Stretched Harmonic Oscillator (Tennessee)

We use (S.5.54.1) and (S.5.54.2) as the basic equations. The turning point
C is where the argument of For the present potential the turning
point is

The integral in (S.5.54.1) has three regions. In the interval then
is a constant, and the integral is just The potential
is nonzero in the two intervals and Since
the WKB integral is symmetric, we get

To evaluate the second integral, change variables to