In 1d, for $V(x) = g\delta(x)$, integrating the TISE yields (assuming that $\psi$ is bounded$^\dagger$, so as to suppress the term containing $E$) $$ -\frac{\hbar^2}{2m} \left( \psi'(\varepsilon) - \psi'(-\varepsilon) \right) + g\psi(0) = 0 $$ for $\varepsilon\to0^+$.

Now, this doesn't at all imply that $\psi$ has to be continuous (even though it is bounded) at $x=0$ (unlike the case when $V$ is bounded). But still, everywhere I've seen, it is implicitly assumed that $\psi$ is continuous at $x=0$, and eigenstates are derived using this.

I can see that in the above equation, if $\psi$ is discontinuous at $x=0$, then we'll have trouble writing $\psi(0)$—I could define it anything if it isn't continuous. Moreover, I am doubtful is the following equality for $\varepsilon>0$ (which was used in getting the above equation) is valid even when $\psi$ is discontinuous at $x=0$. $$ \int_{-\varepsilon}^\varepsilon \delta(x)\psi(x)\; dx \stackrel{?}{=} \psi(0) $$

Question: If the above questioned equation is valid only for continuous $\psi$, is it "our desire" to exploit this equality so that we end up restricting the continuity of $\psi$ "by hand?" If so, what is the guarantee that there are not other kinds of solutions?

$^\dagger$This is "put in by hand."

Note: A similar question was posted five years ago, but was never satisfactorily answered. So I post this again, with more details, hoping that it'll be clearly answered this time.

  • 1
    $\begingroup$ maybe you can clarify what parts of those answers in the linked post are not satisfying? $\endgroup$ Nov 15 '20 at 20:02
  • $\begingroup$ My answer to a related question adopts a perspective that may be of interest. $\endgroup$
    – J. Murray
    Nov 17 '20 at 0:10

Were $\psi(x)$ not continuous at $x=0$ then $\psi''(x)$ would contain the derivative of a $\delta$-function, and there is nothing else in the equation $H\psi=E\psi$ that could cancel it, so a discontinuity is not allowed.

  1. Let us assume the integral form $$ \psi(x)~=~ \frac{2m}{\hbar^2} \int^{x}\mathrm{d}y \int^{y}\mathrm{d}z\ (V(z)-E)\psi(z) \tag{1}$$ of the time independent 1D Schrödinger equation (TISE), where the potential $$V(x)~=~V_0\delta(x)\tag{2}$$ is a Dirac delta potential.

  2. Let us assume that the wavefunction $\psi\in {\cal L}_{\rm loc}^1(\mathbb{R})$ is a locally integrable function, so that the integral $\int^{y}\mathrm{d}z\ E\psi(z)$ in eq. (1) is well-defined.

  3. Note in particular that we do not assume that $\psi$ is continuous, cf. OP's title question. The following crucial question arises: How should we define the integral $$\int^{y}\mathrm{d}z\ V(z)\psi(z)\tag{3}$$ in eq. (1)?

    • Should we define (3) to be equal to $$V_0\theta(y)\psi(0)+C,\tag{3'}$$ where $C$ is an integration constant?

    • Or if the left and right limits $\psi(0^-)$ and $\psi(0^+)$ exist, should we define (2) to be equal to $$V_0\theta(y)\frac{\psi(0^-)+\psi(0^+)}{2}+C~?\tag{3''}$$

  4. Whatever definition we propose for the Dirac delta distribution, it seems likely that (3) will become a locally integrable function of $y$.

    A mathematical bootstrap argument involving the TISE (1) then shows that $\psi\in C(\mathbb{R})$ is continuous, cf. OP's title question. See also e.g. my related Phys.SE answer here.

  • $\begingroup$ Thanks! Your answer definitely expands my understanding. But this argument works only because one assumes that $\psi$ is integrable. It won't explain scattering states. Right? $\endgroup$
    – Atom
    Nov 16 '20 at 5:45
  • $\begingroup$ I updated the answer to include scattering states. $\endgroup$
    – Qmechanic
    Nov 16 '20 at 9:09
  • $\begingroup$ Shouldn't there be a lower limit in the integral in $(1)$, affecting the LHS? $\endgroup$
    – Atom
    Nov 16 '20 at 10:58
  • $\begingroup$ It is an indefinite integral with an implicit integration constant $C$. $\endgroup$
    – Qmechanic
    Nov 16 '20 at 11:15

You can think of the delta potential $V(x)=g\delta(x)$ as representing the limiting form of a potential barrier (or well) of height $\Lambda$ and width $w$, with $w\Lambda=g$.

For any finite $\Lambda$, $\psi(x)$ and $\psi’(x)$ must be continuous at both edges of the barrier. Within the barrier, Schrödinger’s equation implies that $\psi’’(x)\sim\Lambda$, so the total change in $\psi’(x)$ across the barrier is $\sim g$.

This means that, in the limit $\Lambda\rightarrow\infty$ with $g$ fixed, $\psi’(x)$ is finite everywhere within the barrier, but changes by a finite amount from one side of the barrier to the other. Since $\psi’(x)$ is finite everywhere, $\psi(x)$ is continuous.


The other answers are very good. I want to address one mis-think in the question.

"... if $\psi$ is discontinuous at $x=0$, then we'll have trouble writing $\psi(0)$ ..."

Not always. As a vast oversimplification, suppose that you have obtained $\psi$ as a Fourier transform, which has $\lim_{x \rightarrow 0^-} \psi(x) = -1$ and $\lim_{x \rightarrow 0^+} \psi(x) = 1$. Then we know $\psi(0) = 0$. At a step discontinuity, a Fourier series converges to the midpoint between the values to the left and right. (This is part of the Dirichlet conditions. There are mild technical hypotheses. The cited reference describes the real-valued version; the complex-valued version is essentially the same.)


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