Recent content by 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.

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.

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}

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) =...