It was recently floated in chat that a particle in a box with a delta-function spike inside it, with hamiltonian $$ H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V_0\delta(x-a) $$ and with the box walls at $x=0$ and $x=L$, might make an interesting example to teach perturbation theory if it is exactly solvable, since it would allow for a full-blown solution of both the full potential as well as the perturbation theory, similarly to the model in this recent question.

Is this model solvable? If so, what are the eigenfunctions and the energy eigenvalues?


There's a lot you can do, but the short of it is that the eigen-energy quantization condition still boils down to a transcendental equation that must be solved numerically.

Start with the wavefunction Ansatz $$ \psi(x) = \begin{cases} A \sin(kx) & 0<x<a \\ B \sin(k(L-x)) & a<x<L \end{cases} $$ where $E = \hbar^2 k^2/2m$.

To find the effect of the delta-function potential, start with the Schrödinger equation, $$ \left[-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V_0\delta(x-a)\right] \psi(x) = E\psi(x), $$ and integrate it between $a-\epsilon$ and $a+\epsilon$ to get, in the $\epsilon\to0$ limit, $$ -\frac{\hbar^2}{2m}(\psi'(a^+)-\psi'(a^-)) + V_0 \psi(a) = 0. $$ When putting in our Ansatz, this reads $$ -\frac{\hbar^2}{2m}(-Bk\cos(k(L-a))-Ak\cos(ka)) + V_0 \frac12 (A\sin(ka) + B\sin(k(L-a)) = 0, $$ which is further supplemented by the continuity equation $$ A\sin(ka) = B\sin(k(L-a)) $$ to produce the coupled set of equations \begin{align} A\sin(ka) - B\sin(k(L-a)) &= 0 \\ \left[\frac{\hbar^2}{2m}k\cos(ka) + \frac12V_0\sin(ka)\right]A + \qquad\qquad\qquad\qquad\qquad\qquad & \\ +\left[\frac{\hbar^2}{2m}k\cos(k(L-a))+\frac12V_0\sin(k(L-a))\right] B &= 0, \end{align} or a bit more simply \begin{align} A\sin(ka) - B\sin(k(L-a)) &= 0 \\ \left[\frac{\hbar^2}{2m}k\cos(ka) + V_0\sin(ka)\right]A +\frac{\hbar^2}{2m}k\cos(k(L-a)) B &= 0, \end{align} which are best expressed in matrix form, as $$ \begin{pmatrix} \sin(ka) & - \sin(k(L-a)) \\ \frac{\hbar^2}{2m}k\cos(ka) +V_0\sin(ka) & \frac{\hbar^2}{2m}k\cos(k(L-a)) \end{pmatrix} \begin{pmatrix}A \\ B\end{pmatrix} = \begin{pmatrix} 0\\0 \end{pmatrix}. $$ Since we're looking for a nonzero solution, we require this system to be singular, which means that we require the determinant to vanish, $$ \det\mathopen{} \begin{pmatrix} \sin(ka) & - \sin(k(L-a)) \\ \frac{\hbar^2}{2m}k\cos(ka) +V_0\sin(ka) & \frac{\hbar^2}{2m}k\cos(k(L-a)) \end{pmatrix} \mathclose{} = 0, $$ and this is what forms the quantization condition.

To see its explicit form, we start by unpacking the determinant: $$ \sin(ka)\frac{\hbar^2}{2m}k\cos(k(L-a))+ \sin(k(L-a))\left(\frac{\hbar^2}{2m}k\cos(ka) +V_0\sin(ka)\right) = 0, $$ or a bit more cleanly, in terms of the characteristic wavevector $\kappa = 2mV_0/\hbar^2$, $$ k\sin(ka)\cos(k(L-a))+ \sin(k(L-a))\left(k\cos(ka) +\kappa\sin(ka)\right) = 0, $$ and it seems that this is about as far as you can go in simplifying this quantization condition.

You can make a little bit of extra headway if you assume that the delta-function potential is in the middle of the well, so $a=L-a = L/2$. In this case you have $$ k\sin(kL/2)\cos(kL/2)+ \sin(kL/2)\left(k\cos(kL/2) +\kappa\sin(kL/2)\right) = 0, $$ so things factorize, $$ \sin(kL/2) \left[ 2 k\cos(kL/2) +\kappa\sin(kL/2)\right] = 0. $$ This means that you either have the fully unperturbed $$ \sin(kL/2) = 0 $$ (which is natural - the eigenfunctions of the original problem where $kL/\pi$ is even will have a zero at the center, so they don't see the delta-function potential), or you have $$ 2 k\cos(kL/2) +\kappa\sin(kL/2) = 0, $$ which simplifies to $$ \cot(kL/2) = -\frac{\kappa}{2k}, $$ but this is still not quite within exactly-solvable territory. This can be easily solved numerically, but it's still outside of the range of what you can deal with using analytical methods.

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    $\begingroup$ The perturbation theory is in deed beautiful here all $V$ get 1 and simply $$E_{nk_i} = n^2 - k_i^2$$ (using notation from here (open the "show" link)). I think it cant get simpler than that. Would be interesting to see if there is a closed form for the $n$th-term. $\endgroup$ – Rudi_Birnbaum Mar 8 '19 at 8:40

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