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    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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    https://brainmass.com/physics/energy/energy-eigenvalues-for-a-particle-in-dirac-delta-potential-77512

    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.

    $2.19