I've come a across the following variant of the finite-potential-well-problem in quantum mechanics: The potential is given by $V(x)=0$ for $|x| \geq a/2$ and $V(x)=-V_0/a$ for $-a/2<x<a/2$ where $a,V_0>0$.The task is to find an eigenfunction and an eigenvalue for the Hamiltonian "in the limit $a \to 0$".

My problem is that I am not sure what "in the limit $a \to 0$" should mean in this context. My first idea was to use the energy formula for the finite well, plugging in $V_0/a$ and $a/2$ for the potential resp. the width of the box and then take the limit $a \to 0$. However, this does not give a meaningful result.

The energy formula I am working with is $$K\tan(K\frac{a}{2})=S$$ where $$K=\sqrt{\frac{2m(E+V_0/a)}{h^2}}$$ $$S=\sqrt{\frac{-2mE}{h^2}}$$ and $h$ is Planck's constant.

Maybe someone has an idea what "in the limit $a \to 0$" should mean here.


1 Answer 1


The given potential energy in the finite-potential-well-problem is $$V(x)=\begin{cases} 0, & |x| > \frac{a}{2} \\ -\frac{V_0}{a}, & -\frac{a}{2} < x < \frac{a}{2} \end{cases}$$

In the limit $a \to 0$ this potential is an infinitely narrow and deep peak. Therefore it can most conveniently be written by using the Dirac delta function. Its size (height$\cdot$width) is $-\frac{V_0}{a} \cdot a = -V_0$, hence $$V(x)=-V_0 \delta(x)$$

Then Schrödinger's time-independent equation becomes $$-\frac{\hbar^2}{2m}\psi''(x) - V_0 \delta(x) \psi(x) = E \psi(x) \tag{1}$$

This differential equation can be solved with standard calculus methods. Below I only sketch how to find the bound-state solution. Finding the solutions for unbound states is left to you.

Inspired by the solution of the finite-potential-well-problem, we make the following ansatz for eigenfunction and eigenvalue (with a still unknown $\alpha$). $$\begin{align} \psi(x) &= A e^{-\alpha|x|} \tag{2a} \\ E &= - \frac{\hbar^2\alpha^2}{2m} \tag{2b} \end{align}$$

From (2a) and with the help of $|x|'' = 2\delta(x)$ (see this question) we can calculate the second derivative of $\psi$ $\psi''(x) = \alpha^2 Ae^{-\alpha|x|} - 2\alpha A\delta(x) \tag{3}$

When plugging (2a), (2b) and (3) into Schrödinger's equation (1), we see that most terms cancel out, and get a single solution for $\alpha$: $$\alpha = \frac{mV_0}{\hbar^2} \tag{4}$$ and hence there is only one bound state.

Using this $\alpha$ in the ansatz (2) finally gives the solution $$\begin{align} \psi(x) &= A e^{-mV_0|x|/\hbar^2} \\ E &= -\frac{mV_0^2}{2\hbar^2} \end{align} \tag{5}$$

  • $\begingroup$ Thank you for your answer. I can "mimic" the computation you outlined but I do not understnad what I am doing. (I am a math student.) So, what exactly is $\delta(x)$? I know some basics about distributions and - as far as I understand - $\delta(x)$ is an abuse of notation. So, what would be the correct notation for $\delta(x)$ or for the equation involving $\delta(x)$? $\endgroup$
    – user234775
    Commented Jun 17, 2019 at 16:38
  • $\begingroup$ Well, $\delta(x)$ is not a function in the sense of rigorous mathematics. I know, mathematicians call it a distribution. But I never needed to learn distribution theory to understand and use it. Looosely speaking, you can imagine $\delta(x)$ as a peaked "function" which is $=0$ for all $x\neq 0$, and $=\infty$ for $x=0$, in such a way that $\int_{-\infty}^{+\infty} \delta(x) dx = 1$. I know mathematicians feel uncomfortable with this explanation, but for physicists (including me) this is just enough to know. $\endgroup$ Commented Jun 17, 2019 at 17:22
  • $\begingroup$ Ok. Thank you, Thomas. I asked my question about the exact meaning of $\delta(x)$ on math stackexchange. The title of my question is "Schrödinger equation involving the Dirac-Delta". Maybe they know more... :-) $\endgroup$
    – user234775
    Commented Jun 17, 2019 at 17:44
  • $\begingroup$ Link: Schrödinger equation involving the Dirac-Delta $\endgroup$ Commented Jun 17, 2019 at 17:50

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