Homework Help: Triple delta potential

1. Apr 17, 2014

mtjs

1. The problem statement, all variables and given/known data

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?

2. Relevant 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 $$\psi'( z+\epsilon) - \psi'(z-\epsilon) = -\frac{2mw}{\hbar^2} \psi(z)$$

3. 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?

Last edited: Apr 17, 2014
2. Apr 17, 2014

Staff: Mentor

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.

3. Apr 18, 2014

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?

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

Last edited: Apr 18, 2014
4. Apr 18, 2014

Staff: Mentor

For your chosen wavefunctions, it looks fine.

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

5. Apr 18, 2014

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.

Last edited: Apr 18, 2014
6. Apr 18, 2014

Staff: Mentor

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

7. Apr 18, 2014

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.

8. Apr 18, 2014

Staff: Mentor

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.