Question

Two particles, each of mass M, are attracted to each other by a potential

V(r) = -(g²/d)  exp (-r/d),

where d = h/mc with mc² = 140 million electron volts (MeV), Mc² = 940 MeV.

(a) Show that for l = 0 the radial Schrodinger equation for this system can be reduced to Bessel's differential equation

by means of the change of variable x = α exp (-\(\beta\)r) for a suitable choice of α and \(\beta\).

(b) Suppose that this system is found to have only one bound state with a binding energy of 2.2 MeV; evaluate g²/hc numerically and state its units.

[Note: a graph of values J_p(x) in x - p plane has been provided with the information at the beginning of the examination (Fig.)].

(c) What would the minimum value of g²/hc have to be in order to have two l = 0 bound states (d and M remaining the same)?.

Answer 1

(a) When I = 0, the radial wave function R(r) = x(r)/r satisfies the equation

the reduced mass being µ = M/2. By the change of variable

we can reduce the Schrodinger equation to Bessel's differential equation of order p

\(\frac{d^2J(x)}{dx^2} + \frac 1x \frac{dJ(c)}{dx} + \left(1 - \frac{p^2}{x^2}\right) J_p(x) = 0\)

Thus the (unnormalized) radial wave function is

(b) For bound states we require that for \(r \to \infty\), R(r) → 0, or J_p remains finite. This demands that p ≥ 0. R(r) must also be finite at r = 0, which means that x(0) = J_p(α) = 0.

This equation has an infinite number of real roots. For E = 2.2 MeV,

Figure shows the contours of J_p(x) for different values (indicated by right and top numbers) of the function in the x - p plane. The lowest zero of J_p(x) for p = 0.65 is 3.3, the next 6.6. Thus for \(\alpha \approx3.3\), the system has one l = 0 bound state, for which

which is a dimensionless constant.

(c) For \(\alpha \approx6.6\), there is an additional 1 = 0 bound state. Thus the minimum value of α for two l = 0 bound states is 6.6, for which