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


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 $\phi^*\psi$?

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    $\begingroup$ It's difficult to come up with boundary conditions on the barrier. Integrating both sides of the equation infintesimally gives $-\frac{\hbar^2}{2m}[\Psi^{'}_{+}(0)-\Psi^{'}_{-}(0)]=a\Psi^{'}(0)$, but I'm not sure how to interpret that. $\endgroup$ Aug 20, 2015 at 5:02
  • $\begingroup$ @JahanClaes I am not sure of the $a=0$ case, because of the $0 \ \infty$ indeterminacy, but yes for $\delta'^{(0)}$ and $a \neq 0$ this is the usual argument in textbooks, that the equation amounts to require boundary conditions $\Psi'(0^+)-\Psi'(0^-) \propto \Psi(0)$. For $ n > 0$, if the first derivative is continuous then your formula drives to a condition $\Psi'^{(n)}=0$, and I agree that it is unclear how to interpret, and if it is the most general solution. $\endgroup$
    – arivero
    Aug 20, 2015 at 16:04
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    $\begingroup$ Yeah, but of course I'm not sure we can require that $\Psi^{'}$ is continuous at 0, since it isn't for a $\delta$-function potential. $\endgroup$ Aug 20, 2015 at 17:24
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    $\begingroup$ Are you sure the hack with the "boundary conditions" can be made to work for general distributional potentials? I mean, the wavefunctions are technically functions in $L^2$, which are functions only defined up to a zero-measure set, so you can't evaluate them at points. Then again, a delta function can't even properly act on $L^2$ functions. So, what is the space of functions this Schrödinger equation is supposed to be operating on? I guess you could try to sanitize the $\delta$-case by representing it as a limit of sharply peaked Gaussians, but can you represent the derivatives in such a way? $\endgroup$
    – ACuriousMind
    Aug 22, 2015 at 22:40
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    $\begingroup$ @ACuriousMind At the very least, we know any solution has to be a sinusoid/exponential away from $x=0$, so all that we need are boundary conditions. This is not to say, of course, that the same trick will work to get the boundary conditions. $\endgroup$ Aug 22, 2015 at 23:13

3 Answers 3


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.

  • $\begingroup$ Kind of Dirac hairpin instead of Dirac comb. I like the approach, specially for higher derivatives of the delta... it is good that they are more boundary conditions because as one increases $n$, one of the puzzles was how to stay only with 2nd order conditions. $\endgroup$
    – arivero
    Aug 23, 2015 at 3:15
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    $\begingroup$ I have started to review the bibliography and there was an attempt to solve the $\epsilon \to 0$ conflict by using the "Strong Resolvent Limit" of the sequence of hamiltonians. Whatever it is, the result was not encouraging, it amounts to Dirichlet conditions. $\endgroup$
    – arivero
    Aug 23, 2015 at 19:14
  • $\begingroup$ +1, but one can really just take the "normal" delta derivatives to realize there is no way to match higher derivatives of two exponentials falling off on both sides (aka bound states). This also makes physical sense when you think of the deltas in terms of multipolar expansions; the absence of the mere $\delta$ means that the source has no "overall monopole", i.e. that it has no "overal binding potential". $\endgroup$
    – Void
    Aug 23, 2015 at 19:22
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    $\begingroup$ @JahanClaes When you investigate the sources of the usual Laplace equation, you will find that the source of a monopole is a delta, a dipole a delta-derivative and so on. Now, by a loose hand-waivy argument, the Schrödinger equation in $x$-representation is somehow a weird Laplace equation (at least it is linear and second order) and we can represent the effect of the "source" $V(x)$ by a sequence of $\delta, \delta',...$ very similarly to the expansion of the source $\rho(x)$ in the Laplace equation. However, I am not sure about the precise applicability of such an expansion. $\endgroup$
    – Void
    Aug 23, 2015 at 20:52
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    $\begingroup$ @JahanClaes Have you looked into the effect of the 1/$\epsilon$ factor in the potential? I was playing around with the second derivative as a square well surrounded by two barriers. The heights of the barriers are proportional to the cube of the widths. In that case the wells become impenetrable as the width goes to zero. I must add that I haven't completed the analysis. $\endgroup$ Aug 24, 2015 at 14:51

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.

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:enter image description here

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.

  • $\begingroup$ Hmm physics.stackexchange.com/questions/143630/… given that if $\int V(x) < 0$ we have granted the existence of bound states, I am under the impression that we are with some kind of borderline case... perhaps with a E=0 state? $\endgroup$
    – arivero
    Aug 26, 2015 at 1:11
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    $\begingroup$ I am not saying that every function of the limiting process does not have bound states, it does, and that would be easy to prove. It's just that the full $\delta$-derivative does not, because the $\psi_+,\psi_-$ state can essentially do only a cusp corresponding to a $\delta$ potential. (Also note that unlike the theorem you cite, the limiting $V(x)$ is not non-negative, $\int V(x) = 0$ which is not $<0$, and we are dealing with distributions here, so statements such as $V(x)><0$ have no good meaning.) $\endgroup$
    – Void
    Aug 26, 2015 at 8:39

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 $n>1$

  • $\begingroup$ And yet, probability leaking, or non-unitary evolution, could be useful to describe unstable situations. So the deltas produce an interesting family of one-point sinks $\endgroup$
    – arivero
    Aug 29, 2015 at 16:26

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