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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?

Qmechanic
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Jiang-min Zhang
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  • I have a similar doubt. Even without the delta potential, you can't apply the derivative boundary condition in 1-d well. The derivative is non-zero at boundary inside the well, and zero outside. But we simply choose to ixnore that condition. Why? – Cheeku Oct 21 '14 at 13:34
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    Related: https://physics.stackexchange.com/q/81574/25794 – doetoe Oct 21 '14 at 13:38
  • I do not get why on the delta do you expect $\psi(0^+)=-\frac ba \psi'(0^+)$. And if b and a are real numbers, that implies that no probability crosses the barrier. – arivero Aug 23 '15 at 20:20

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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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Qmechanic
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    There must be some mathematical way to get around it. But, the point is that, now i do not dare to use delta potential anymore. At least for the concrete problem above, i am not sure the two approaches will yield the same result. – Jiang-min Zhang Oct 21 '14 at 14:57