The $\delta(x)$ Dirac delta is not the only "point-supported" potential that we can integrate; in principle all their derivatives $\delta', \delta'', …$ exist also, do they?
If yes, can we look for bound states in any of these $\delta'^{(n)}(x)$ potentials? Are there explicit formulae for them (and for the scattering states)?
To be more precise, I am asking for explicit solutions of the 1D Schroedinger equation with point potential,
$$- {\hbar^2 \over 2m} \Psi_n''(x) + a \ \delta'^{(n)}(x) \Psi(x) \ = E_n \Psi_n(x) $$
I should add that I have read at least of three set of boundary conditions that are said to be particular solutions:
- $\Psi'(0^+)-\Psi'(0^-)= A \Psi(0)$ with $\Psi(0)$ continuous, is the
zero-th derivative case, the "delta potential". - $\Psi(0^+)-\Psi(0^-)= B \Psi'(0)$ with $\Psi'(0)$ continuous, was called "the delta prime potential" by Holden.
- $\lambda \Psi'(0^+)=\Psi'(0^-)$ and $\Psi(0^+)=\lambda\Psi(0^-)$ simultaneusly, was called "the delta prime potential" by Kurasov
The zero-th derivative case, $V(x)=a \delta(x)$ is a typical textbook example, pretty nice because it only has a bound state, for negative $a$, and it acts as a kind of barrier for positive $a$. So it is interesting to ask for other values of $n$ and of course for the general case and if it offers more bound states or other properties. Is it even possible to consider $n$ beyond the first derivative?
Related questions
(If you recall a related question, feel free to suggest it in the comments, or even to edit directly if you have privileges for it)
For the delta prime, including velocity-dependent potentials, the question has been asked in How to interpret the derivative of the Dirac delta potential?
In the halfline $r>0$, the delta is called "Fermi Pseudopotential". As of today I can not see questions about it, but Classical limit of a quantum system seems to be the same potential.
A general way of taking with boundaring conditions is via the theory of self-adjoint extensions of hermitian operators. This case is not very different of the "particle in 1D box", question Why is $ \psi = A \cos(kx) $ not an acceptable wave function for a particle in a box? A general take was the question Physical interpretation of different selfadjoint extensions A related but very exotic question is What is the relation between renormalization and self-adjoint extension? because obviosly the point-supported interacctions have a peculiar scaling
Comments
Of course upgrading distributions to look as operators in $L^2$ is delicate, and it goes worse for derivatives of distributions when you consider its evaluation $<\phi | \rho(x) \psi>$. Consider the case $\rho(x) = \delta'(x) = \delta(x) {d\over dx}$. Should the derivative apply to $\psi$ only, or to the product $\phi^*\psi$?
Best Answer
Ok, I have a solution for $\delta'(x)$ based on a very crude limit. I'm going to neglect factors of $\hbar$, $m$, etc for the sake of eliminating clutter.
Let $V_\epsilon(x)=\frac{\delta(x+\epsilon)-\delta(x)}{\epsilon}$. Then $\lim_{\epsilon\rightarrow 0}V_e(x)=\delta'(x)$. We'll solve the Schrodinger equation for finite $\epsilon$ and then take the limit afterwards.
We have $\Psi''(x) = (E-V_\epsilon(x))\Psi(x)$. If we want a bound solution, we must have $E<0$. Then in the ranges $[-\infty,0),(0,\epsilon),(\epsilon,\infty]$ we must have that $\Psi$ is some exponential function. In other words, $$ \Psi(x) =\left\{\begin{array}{ll} e^{\sqrt{-E}x} &x\in [-\infty,0)\\ Ae^{\sqrt{-E}x}+Be^{-\sqrt{-E}x} &x\in (0,\epsilon)\\ Ce^{-\sqrt{-E}x}&x\in (\epsilon,\infty]\\ \end{array}\right\} $$
From the fact that $\Psi$ must be continuous, we can replace $B$ and $C$ in terms of $A$ to get $$ \Psi(x) =\left\{\begin{array}{ll} e^{\sqrt{-E}x} &x\in [-\infty,0)\\ Ae^{\sqrt{-E}x}+(1-A)e^{-\sqrt{-E}x} &x\in (0,\epsilon)\\ (1-A+Ae^{2\sqrt{-E}\epsilon})e^{-\sqrt{-E}x}&x\in (\epsilon,\infty]\\ \end{array}\right\} $$
We can also write down the derivative of $\Psi$.
$$ \Psi'(x) =\left\{\begin{array}{ll} \sqrt{-E}e^{\sqrt{-E}x} &x\in [-\infty,0)\\ A\sqrt{-E}e^{\sqrt{-E}x}+(A-1)\sqrt{-E}e^{-\sqrt{-E}x} &x\in (0,\epsilon)\\ (A-1-Ae^{2\sqrt{-E}\epsilon})\sqrt{-E}e^{-\sqrt{-E}x}&x\in (\epsilon,\infty]\\ \end{array}\right\} $$
Using the normal method of finding boundary conditions at a $\delta$ function barrier, we have that
$$ \begin{array}{c} \Psi'_{+}(0)-\Psi'_{-}(0) = \Psi(0) \\ \Psi'_{+}(\epsilon)-\Psi'_{-}(\epsilon) = \Psi(\epsilon) \end{array} $$
The first boundary condition gives us
$$ (2A-1)\sqrt{-E} = \frac{1}{\epsilon}$$ or $$ A=\frac{1}{2\epsilon\sqrt{-E}}+\frac{1}{2} $$
The second boundary condition gives us
$$ -2A\sqrt{-E}e^{\sqrt{-E}\epsilon} = -\frac{1}{\epsilon}[Ae^{\sqrt{-E}\epsilon}+(1-A)e^{-\sqrt{-E}\epsilon}] $$ or $$ A=\frac{1}{e^{2\sqrt{-E}\epsilon}(2\epsilon\sqrt{-E}-1)+1} $$
Putting the two conditions together gives us a constraint on $E$.
$$ \frac{1}{2\epsilon\sqrt{-E}}+\frac{1}{2} = \frac{1}{e^{2\sqrt{-E}\epsilon}(2\epsilon\sqrt{-E}-1)+1} $$
We can expand both sides in a Laurent series to first order in $\epsilon$. The left hand side is already expanded. The right hand side becomes (to first order)
$$ \frac{1}{(2\epsilon\sqrt{-E}+1)(2\epsilon\sqrt{-E}-1)+1}=\frac{1}{4\epsilon^2(-E)} $$
The two sides of the equation are then impossible to match in the limit $\epsilon\rightarrow 0$ since they occur at different orders in $\frac{1}{\epsilon}$. Thus, in that limit, no solution for $E$ exists, and so there is no bound state.
I'm sure there's some algebra mistakes in all that mess, but that's the general idea. You could do the same algebra to look at scattering states, if you wanted. One could also apply this method to higher derivatives. For example, $\delta''(0)=\lim_{\epsilon\rightarrow 0}\frac{\delta(x+2\epsilon)-2\delta(x+\epsilon)+\delta(x)}{\epsilon^2}$. Of course, each higher derivative demand one more boundary to account for, so the problem gets correspondingly messier. But doable, in principle.