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Griffiths gives a derivation of the energy states of a particle in a potential well of the form: \begin{equation} V(x)=-k\cdot(\delta(x-a)+\delta(x+a));\qquad k>0 \end{equation}

This makes schrodinger's equation: $$-\frac{\hbar^2}{2m}\psi'' -k(\delta(x-a)+\delta(x+a))=E\psi.$$

Here's a picture of the two wavefunctions they get:

double delta potential well, even and odd solution

Naive reasoning can get you to the even solution. How the odd solution could possibly come about? The second (weak) derivative of the odd solution ends up being something of the form:

$$\psi_{\text{odd}}'' = (\text{exponentials}) + \delta(x+a) - \delta(x-a).$$

This doesn't appear to satisfy the Schrodinger equation; the first delta has the wrong sign! Where is my interpretation wrong, and how do you make sense of, or arrive at the odd solution?

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You have the time independent Schodinger's equation wrong. It's $H|\psi\rangle = E|\psi\rangle$. So your second equation should read: $$-\frac{\hbar^2}{2m}\psi'' - k [\delta(x-a) + \delta(x+a)] \psi = E\psi.$$ When you take the two derivatives of $\psi$, the exponentials will multiply the $\delta$ functions, too. When you factor everything, if you do it correctly, you'll see that the equation is obeyed.

In detail, you need to use the following identities: $$\begin{align} f(x) & \equiv A \operatorname{e}^{-k |x-a|} \Rightarrow\\ f'(x) & = -k A \operatorname{e}^{-k |x-a|} \operatorname{sgn}(x-a),\ \mathrm{and} \\ f''(x) & = k^2 A \operatorname{e}^{-k | x-a|} - 2 k A \operatorname{e}^{-k |x-a|} \delta(x-a).\end{align}$$

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