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Energy eigenvalues for a particle in Dirac delta potential

A particle of mass m moves in the potential

V(x) = -g*delta(x) x>-a
infinity x<-a

(delta (x) = dirac delta function)
a. Without worrying about continuity or boundary conditions, write down the general solution of the Schrodinger equation for a bound state (energy E<0) in regions( -a less than x less than 0) and x>0.

b. In terms of your solution in part (a), write down the boundary conditions at
x=-a and continuity conditions at x=0.

c. From the conditions in part (b), derive a transcendental equation determining the bound state energy E. Solve the equation in the limit as a approaches infinity.

Solution Preview

The Hamiltonian is

H= -h-bar^2/(2m)d^2/dx^2 - g delta(x) (1)

The energy eigenstates satisfy:

H psi = E psi (2)

If we stay away from x = 0, then we can omit the delta function in (1) and we obtain:

-h-bar^2/(2m)d^2/dx^2 psi = E psi (3)

The solutions of (3) are:

psi(x) = A Exp[-i k x] + B Exp[i k x] (4)

where k = squareroot[2 m E] / h-bar (5)

If E < 0 then k is imaginary and we can rewrite (4) and (5) as:

psi(x) = A Exp[- u x] + B Exp[u x] (6)

where u = squareroot[-2 m E] / h-bar (7)

psi must thus be of the form (6) in all the regions, but the coefficients A and B can different. To solve the problem you must impose the correct boundary conditions. Loosely speaking, you can reason as follows. An infinite jump in the potential like at x=-a or at x = 0, means that
d^2 psi/dx^2 will also have an infinite jump, but the derivative of ...

Solution Summary

An equation is derived for the bound state energy eigenvalue of a particle in the potential:

V(x) = -g*delta(x) x>-a
infinity x<-a

The equation is solved in the limit a --> infinity.

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