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I encountered an inconsistency in the one-dimensional delta potential. Suppose we have a one-dimensional infinitely deep square well from $-L$ to $+L$. We know the eigenstates are sine and cosine functions. They are either even or odd.

Now let us add one delta potential $g \, \delta(x)$ in the middle. Here $g\in \mathbb{R}$. By parity consideration, the odd states are not affected at all. The even states are coupled together. This means, the new even-parity eigenstates are linear superpositions of the odd even-parity eigenstates. In turn, this means, the even-parity eigenstates should have zero derivative at $x=0$.

However, if you do this same problem in another standard way, i.e., by integrating the Schrodinger equation across the delta potential, you would get a boundary condition for the right part of the wave function in the form of $a \psi(0) +b \psi'(0) =0 $, where $a$ and $b$ are finite numbers. This boundary condition means $\psi'( 0) \neq 0 $. So the two approaches yield different results! How do we resolve this inconsistency?

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Jiang-min Zhang
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1 Answers1

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The main point that although a pointwise convergent Fourier series of cosine modes is an even function $\psi(-x)=\psi(x)$, it does not have to be differentiable at $x=0$. A pointwise convergent infinite sum of differentiable functions is not necessarily a differentiable function.

More generally, as OP already mentions, the wave function $\psi(x)$ is not necessarily differentiable at the $x$-position of the delta potential and the two well walls. One should allow for a discontinuity in the $\psi^{\prime}(x)$ at these three $x$-positions. See also e.g. my Phys.SE answer here.

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