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I am intrigued about two points given in an answer to a similar question (https://physics.stackexchange.com/a/38198/262985).

On one hand, it is stated that wavefunctions inside the well (excluding its boundaries) are $C^1$, and may have a discontinuity at the well boundaries (which I take to mean that either the function, or its derivative may be discontinuous there). Is there a simple/published proof?

On the other hand, an example of a discontinuous wavefunction is given as $Sign(x)$ inside the well. It is later dismissed as having an infinite kinetic energy. The second derivative (proportional to the kinetic energy operator) is $2 delta'(x)$ - twice the derivative of the Dirac distribution. If I integrate the product of this second derivative of the wavefunction with the wavefunction itself, the result, which I take to be proportional to the expectation of the kinetic energy, is zero (according to Wolfram calculator), i.e., not unbounded. what is the catch?

Qmechanic
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Gérard Nihous
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2 Answers2

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For the first part of your question:

Start with \begin{align} -\frac{\hbar^2}{2m}\psi^{\prime\prime}(x)=\left(E-V(x)\right)\psi(x)\, , \end{align} integrate from $-\epsilon$ to $\epsilon$: \begin{align} -\frac{\hbar^2}{2m}\left(\psi^\prime(\epsilon)-\psi^\prime(-\epsilon)\right) &=\int_{-\epsilon}^\epsilon dx (E-V(x))\psi(x)\, , \tag{1} \end{align} and take $\lim \epsilon\to 0$.

Next, recall that, if $m$ is the minimum of $(E-V(x))\psi(x)$ on the interval, and $M$ is the maximum of $(E-V(x))\psi(x)$ on the interval, then \begin{align} \lim_{\epsilon\to 0} 2\epsilon m \le \lim_{\epsilon\to 0} \int_{-\epsilon}^\epsilon dx (E-V(x))\psi(x)\le \lim_{\epsilon\to 0} 2M\epsilon \end{align}

Thus, unless $m\to -\infty$ or $M\to \infty$, we have by the sandwich theorem \begin{align} \lim_{\epsilon\to 0} \left(\psi^\prime(\epsilon)- \psi^{\prime}(-\epsilon)\right)\to 0 \end{align} i.e. the derivative is continuous.

The proof fails if $(E-V(x))\psi(x)$ blows up somewhere on the interval. Since $\psi(x)$ is expected on physical grounds to be continuous and finite, and since $E$ is also expected on physical grounds to be finite, this leaves the possibility of $V(x)$ having some infinite discontinuity somewhere. This is precisely the case at the edges of the infinite well, or if $V(x)=\Omega \delta(x)$, i.e proportional to a $\delta$-function. In such cases $\psi^\prime$ is NOT continuous across the discontinuity, and the size of the jump in the derivative is related to the integral on the RHS of (1).

ZeroTheHero
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I certainly was careless to simply look at Sign(x) as the example given. It is not even square integrable, and has nothing to do with Schrodinger's equation. Instead, the example proposed was Sign(x) only in the interval between -a and a. Mea culpa.

It is relatively easy to show that the expected kinetic energy corresponding to this wavefunction would be unbounded by checking the smoothness of its autocorrelation function R(Tau). R(Tau) is only C0, having three points where the derivative is discontinuous. In momentum space, this means that the probability density (the Fourier transform of R(Tau)) asymptotically behaves like 1/p^2. Hence, the expected kinetic energy in momentum space clearly diverges.

There are other ways, but the autocorrelation method avoids possible traps with directly handling distributions and their derivatives at discontinuities. It can simply be used, for example, in the infinite potential well case, to discard the cosine (discontinuous) solution candidates.

Gérard Nihous
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