# Finite potential well; finding the energy “in a limit”

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 $$-a/2 where $$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.

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}

• 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 $\delta(x)$? – user234775 Jun 17 '19 at 16:38
• 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. – Thomas Fritsch Jun 17 '19 at 17:22
• 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... :-) – user234775 Jun 17 '19 at 17:44
• – Thomas Fritsch Jun 17 '19 at 17:50