Find Transcendental Equation for Triple Delta Potential Energy

[mtjs]

Homework Statement

I'm looking for the bound energy of a triple delta potential:

V(x) = -w \left [ \delta(x-a) + \delta(x) + \delta(x+a) \right ]

What is the correct transcendental equation for kappa?

Homework Equations

My wave function is \psi_1(x) = A e^{\kappa x} for x < -a, \psi_2(x) = B \cosh(\kappa(x+a/2) for -a < x < 0, \psi_3(x) = C \cosh(\kappa(x-a/2)) for 0< x < a, \psi_4(x) = D e^{-\kappa x} forx > a.

We use this continuity formula \psi&#039;( z+\epsilon) - \psi&#039;(z-\epsilon) = -\frac{2mw}{\hbar^2} \psi(z)

The Attempt at a Solution

Calculating the continuity formula at x = 0 gives \kappa \tanh(\frac{\kappa a}{2}) = \frac{m w}{\hbar^2}

This means, you get the same bound energy as for one delta potential if a is very large, and something weird for small a?

 

[mfb]

Very large a just mean your particle is in the middle and the other parts are not important.

Shouldn't there be a second equation from the second potential? And I think your approach for the inner wavefunctions is not general enough to find all solutions.

 

[mtjs]

Is my equation for x = 0 correct? For x = a or x = -a, you get the same transcendental equation as for the double-delta potential?

1 + \tanh(\frac{\kappa a}{2}) =\frac{2 m w}{\hbar^2 \kappa}

 

[mfb]

Is my equation for x = 0 correct?

For your chosen wavefunctions, it looks fine.

For x = a or x = -a, you get the same transcendental equation as for the double-delta potential?

1 + \tanh(\frac{\kappa a}{2}) =\frac{2 m w}{\hbar^2 kappa}

Up to factors of two maybe, but you have to check your ansatz for the wavefunctions.

 

[mtjs]

So after all, you get two different bound energy values. One for the delta at x = 0 and one for two deltas at x = -a and x = a.

 

[mfb]

That just shows your wavefunction is not general enough. The bound energy states have to satisfy both equations at the same time.

 

[mtjs]

Now I see where I did wrong.

The correct wave function is:
\psi_1(x) = A e^{\kappa x}
for x < -a,

\psi_2(x) = B e^{\kappa x} + C e^{-\kappa x}
for -a < x < 0,

\psi_3(x) = D e^{\kappa x} + E e^{-\kappa x}

for 0< x < a,

\psi_4(x) = F e^{-\kappa x}
for x > a.

 

[mfb]

That is better, right.

I would use $\psi_1(x) = A e^{\kappa (x+a)}$ and similar expressions to make the coefficients easier to evaluate, but that is a detail.