Stretched Harmonic Oscillator

Use WKB in one dimension to calculate the eigenvalues of a particle of mass m in the following potential (see Figure 1.1):

(i)

Figure 1.1

SOLUTION

The turning point C is where the argument of k(x) = 0. For the present potential the turning point is

(1)

(2)

In the interval –a < x < a then V(x) = 0, k(x) is a constant, and the integral is just 2ak. The potential V(x) is nonzero in the two intervals –C < x < -a and a < x < C. Since the WKB integral is symmetric, we get

(3)

To evaluate the second integral, change variables to y = x – a:

(4)

The last integral equals d²π/4. Writing K = mω², we find

(5)

(6)

We have to determine E. Equation (5) is a quadratic equation for the variable  Solving the quadratic by the usual formula gives the final result:

(7)

(8)