Energy eigenvalues for a particle in Dirac delta potential
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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.
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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.
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 ...
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