Bound states of the $V(x)=\pm \delta'^{(n)}(x)$ potential?

Question

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 $\phi^*\psi$?

Answer 1

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.

Answer 2

Let's simplify things, $\hbar=m=1$ and we put in the $\delta^{(n)}$ as $V(x)=V_0\delta^{(n)}/2$. Then the problem is $$-\psi'' - V_0\delta^{(n)}\psi = E \psi$$ Apart from $x=0$ the equation gives $$\psi'' = -E \psi $$ we want a bound state $E<0$ which falls off at infinity, so we get a solution $\psi_+ = A \exp(-k x)$ for $x>0$ and $\psi_- = A \exp(kx)$ for $x<0$ with $k = \sqrt{-E}$. Now we need to sew the solution around $x=0$. The stated unique nontrivial solution has $\psi \sim$ cusp, $\psi' \sim$ jump, $\psi'' \sim -\delta$, $\psi''' \sim -\delta'$,... around $x=0$. When plugging in the non-trivial $\psi_-,\psi_+$ solution into our dynamical equation we find $$2k^2 A \delta - A V_0 \delta^{(n)} = 0$$ which gives us

1. No solution for $n \neq 0$
2. A single bound state $E=-V_0/2$ for $n=0$ (see that the 1/2 in the definition of the potential strikes back!); $A$ is determined by wave-function normalization.

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Now you see that there is really no solution for $\delta$-derivative potentials, at least in one dimension. As already somehow touched upon in the comments, this can be also seen from the fact that all $\delta$-derivatives look like "multi-peaks" of some kind, without any "overall" binding.

To better understand what I mean, consider $\delta'$, which can be obtained by a limiting process of the derivative of a gaussian:

It is then somehat physically intuitive that the even though there are bound states in the well on the right, they are somehow eliminated by the infinite squeezing of this double-peak.

Answer 3

The article of D J Griffiths assumes that the delta interaction can be approximated by a sequence of even functions and then infers two boundary conditions: $$ \Psi'(0^+)-\Psi'(0^-)= (-1)^n {m c \over \hbar^2} (\Psi'^{(n)}(0^+)+\Psi'^{(n)}(0^-))$$ $$ \Psi(0^+)-\Psi(0^-)= (-1)^{n-1} {m c \over \hbar^2} n (\Psi'^{(n-1)}(0^+)+\Psi'^{(n-1)}(0^-))$$

My own take travels differently. Integrating $$- {\hbar^2 \over 2m} \Psi''(x) + a \ V(x) \Psi(x) \ = \lambda \Psi (x) $$ from $-\epsilon < 0$ to $u$ we get

$$- {\hbar^2 \over 2m} (\Psi'(u) -\Psi'(-\epsilon)) + a \int_{-\epsilon}^u \ V(x) \Psi(x) \ = \int_{-\epsilon}^u \lambda \Psi (x) $$

Integrating again $u$ from $-\epsilon < 0$ to $v$ $$- {\hbar^2 \over 2m} (\Psi(v)-\Psi(-\epsilon) - (v+\epsilon) \Psi'(-\epsilon)) + a \int_{-\epsilon}^v du \int_{-\epsilon}^u dx \ V(x) \Psi(x) \ = \int_{-\epsilon}^v \int_{-\epsilon}^u \lambda \Psi (x) $$

the first integration gives the boundary condition in the limit $$- {\hbar^2 \over 2m} (\Psi'(0^+) -\Psi'(0^-)) + a <\rho |\Psi(x)> \ = 0 $$

For the second equation, instead of going across multiple integration by parts for each derivative, I think we can use it just one time:

$$ {d\over du} ( u \int_{-\epsilon}^u dx \ V(x) \Psi(x)) = \int_{-\epsilon}^u dx \ V(x) \Psi(x) + u V(u) \Psi(u) $$

$$ ( u \int_{-\epsilon}^u dx \ V(x) \Psi(x))|^v_{-\epsilon} = \int_{-\epsilon}^v\int_{-\epsilon}^u dx \ V(x) \Psi(x) +\int_{-\epsilon}^v u V(u) \Psi(u) $$

So that the limit of the second integration produces the boundary condition

$$- {\hbar^2 \over 2m} (\Psi(0^+) -\Psi(0^-)) - a <\rho |x \Psi(x)> \ = 0 $$

It is immediately visible that for $\rho = \delta'^{(n)}(x)$ my result differs which differs from Griffiths in the sign alternance $(-1)^n$. This is the only discrepancy (the $n$ in the second condition appears naturally, given that $< \delta'^{(n)}(x) | x f(x)>$ is just the delta over $x f'^{(n)} + n f'^{(n-1)}$) so it could be simply some issue on the definition of the n-th derivative of the delta.

In any case, they are more evident objections against the result: it requires to keep track of the regularisation to be sure that all the distributions apply to n-th derivatives by averaging left and right; it does not cover all the possible boundary conditions of a point interaction -well, it was not expected to cover all-, and worse of it, to me, it introduces boundary conditions with derivatives greater than the first, when we are simply solving a second-order differential equation. Question: are such boundary conditions compatible with a self-adjoint hamiltonian? I would think they are not.

Lets look now at the scattering matrix. We apply the boundary conditions to a function $\psi_k(x<0)=e^{ikx}+R e^{-ikx}, \psi_k(x>0)=T e^{ikx}$, so that $$\psi'^{(n)}_k(0^-)=(ik)^n (1+(-1)^n R), \psi_k^{(n)}(0^+)=T (ik) ^n$$

The BC, for $n \geq 1$, solve to:

$$- {\hbar^2 \over 2m} ik (T-1+R) + \frac a2 (ik)^n (T+1+(-1)^n R) \ = 0 $$ $$- {\hbar^2 \over 2m} (T-1-R) + \frac a2 n (ik)^{n-1} (-T-1+(-1)^n R) \ = 0 $$

while for the usual delta, $n=0$,

$$- {\hbar^2 \over 2m} ik (T-1+R) + \frac a2 (T+1+ R) \ = 0 $$ $$- {\hbar^2 \over 2m} (T-1-R) \ = 0 $$

In this case if we solve for the transmission coefficient:

$$ T = {ik \over ik- {a m /\hbar^2 } } $$

and see that it has a pole, at $ik= \frac a2 {2m \over \hbar^2}$, corresponding to the bound state. Not all the poles are bound states, but all the bound states are poles, so this technique can be useful to extract information also in the $n>1$ case.

Ah, note that conservation of probability current implies the extra requirement $T^2+R^2=1$. This could be used to check if the solution is consistent. It is easy to check that this condition will not work for $n>1$, as in this case we can pivot over $ik$ in both equations to produce the extra constraint

$$n (T-1+R)(-T-1+(-1)^n R) = (T-1-R)(T+1+(-1)^n R) $$

which reduces to: $$n(-T^2+(-1)^n TR+1-(-1)^n R-TR-R+(-1)^n R^2) = (+T^2+(-1)^n TR-1-(-1)^n R-RT-R-(-1)^n R^2) $$

for $n$ even

$$n(-T^2+1-2R+ R^2) = (+T^2-1-2R- R^2) $$

for $n$ odd $$n(-T^2- 2 TR+1- R^2) = (+T^2- 2 TR-1+ R^2) $$

To see the incompatibility with preservation of probability we impose simultaneously this condition, sum of $R^2$ plus $T^2$ equal to 1 and we get

• for $n$ even: $n(-1+ R)R = (-1- R)R $ and then $R$ and $T$ should be independent of $k$, which they can not be if $n>1$
• for $n$ odd: $nTR = TR $ and either $RT=0$ or $n=1$, but either $R$ or $T$ equal to zero fixes the other to 1, and then again they are independent of $k$, which they can not be if $n>1$

In conclusion, Griffiths approach allows to give sense to all the derivatives of the delta, but the resulting boundary conditions leak probability for $n>1$