# Condition for existence of a zero energy eigenvalue for a particle in an infinite well with delta potential

There is this rather elementary question that I can't figure out. Consider a particle of mass $$m$$ in 1D which is confined to the following potential:

$$V(x)=\begin{cases} \lambda\delta(x-a) & \text{for }0

where $$\lambda$$ is a real constant and $$0. Let $$\xi=2m\lambda/\hbar^2$$. Find conditions on $$a$$ and $$\xi$$ such that zero is an energy eigenvalue for this system. Show that this happens when $$\xi \leq \xi_*$$ and determine $$\xi_*$$.

My attempt:

The time-independent Schrodinger equation for $$0 (region $$1$$) and $$a (region $$2$$) gives:

$$\frac{\partial^2\psi}{\partial x^2}=\frac{-2mE}{\hbar^2}\psi$$

For scattering states we should solve for $$E>0$$ which gives complex exponential functions and for bound states we should solve for $$E<0$$ which results in real exponential functions. For $$E=0$$ we get:

$$\frac{\partial^2\psi}{\partial x^2}=0$$

which gives:

$$\psi = Ax+B$$

For each respective region. The boundary conditions give:

$$\psi_1(0)=0 \implies B_1=0, \qquad \psi_2(b)=0 \implies A_2=\frac{-B_2}{b}$$

Therefore the wave function becomes:

$$\psi(x)=\begin{cases} A_1x & \text{for }0

Integrating the time-independent Schrodinger equation around $$a$$ results in another boundary condition:

$$-\frac{\hbar^2}{2m}\frac{\partial \psi}{\partial x}|_{a-\epsilon}^{a+\epsilon} + \lambda = 0$$

in which $$\epsilon$$ is an infinitesimally small number. Substituting the expression for wave function we get:

$$A_2-A_1=\frac{2m\lambda}{\hbar^2}=\xi$$

Now everything can be expressed in terms of $$A_1$$ (or $$A_2$$).

$$\psi(x)=\begin{cases} A_1x & \text{for }0

Continuity of wave function at $$x=a$$ gives:

$$A_1a=A_1(a-b)+\xi(a-b) \implies \xi=\frac{A_1b}{(a-b)}$$

We can find $$A_1$$ by enforcing normalization. But the problem is that, doing so we don't get an inequality. We will get a strict equality.

$$\int_{-\infty}^{+\infty}|\psi(x)|^2 dx=1 = A_1^2\int_0^a x^2dx+(A_1+\frac{A_1b}{a-b})^2\int_a^b (x-b)^2 dx$$

Doing the integrations and substituting $$\xi$$ we get:

$$A_1^2\frac{a^3}{3}-A_1^2(\frac{a}{a-b})^2\frac{(a-b)^3}{3}=1$$

Simplifying and solving for $$A_1$$ we get:

$$A_1 = \sqrt{\frac{3}{a^2b}}$$

Finally the wave function becomes:

$$\psi(x)=\begin{cases} \sqrt{\frac{3}{a^2b}}x & \text{for }0

And the parameter $$\xi$$ becomes:

$$\xi=2m\lambda/\hbar^2=\frac{\sqrt{3b}}{a(a-b)}$$

which puts a restriction to the value of $$\lambda$$ if we were to have a zero energy eigenvalue or restricts $$a$$ and $$b$$ given $$\lambda$$ depending on how you look at the problem. The whole problem is solved and there is no inequality in sight! So I don't know what I'm doing wrong, how this problem can be resolved and how to find $$\xi_*$$.

Another strategy can be solving for a non-zero energy eigenvalue and taking the limit as energy goes to zero. Let's assume $$E>0$$. We will have:

$$\frac{\partial^2\psi}{\partial x^2}=\frac{-2mE}{\hbar^2}\psi=k^2\psi$$

The general solution can be written as:

$$\psi(x)=Asin(kx+\phi)$$

Applying boundary conditions we get:

$$\psi(x)=\begin{cases} A_1sin(kx) & \text{for }0

Continuity at $$x=a$$ gives:

$$A_1sin(ka)=A_2sin(ka-kb)$$

Integrating Schrodinger equation and evaluating around $$x=a$$:

$$A_2cos(ka-kb)-A_1cos(ka)=\frac{\xi}{k}$$

Putting the previous two equations together we get a transcendental equation for the energy eigenvalues:

$$A_1[sin(ka)cot(ka-kb)-cos(ka)]=\frac{\xi}{k}$$

Taking the limit $$E \rightarrow 0$$ and hence $$k \rightarrow 0$$ gives us nothing as you can see. The value of $$A_1$$ can be calculated by enforcing normalization which also doesn't have anything to do with the mentioned inequality.

• You have not yet used continuity at $a$ which gives $A_1a=(A_1+\xi)(a-b)$. Does this help? Jan 31, 2022 at 15:28
• I updated my question. I don't see how it can help. Jan 31, 2022 at 15:49
• What did you get for $A_{1}$? It seems it should be in terms of $b$ and $\xi$. Jan 31, 2022 at 16:42
• I found $A_1$ and edited the question. Jan 31, 2022 at 16:56
• Why don't you solve a general Schrod eq. for a generic eigenvalue and then take the value to 0 see what happens Jan 31, 2022 at 17:16

There is a mistake in your calculations. The correct form of equality obtained by integrating the Schrodinger equation around $$a$$ is $$-\frac{\hbar^2}{2m}\left(\psi'(a+\varepsilon) - \psi'(a-\varepsilon)\right) + \lambda\psi(a) = 0$$ This equation leads to the following relation for $$A_1$$ and $$A_2$$: $$-\frac{\hbar^2}{2m}(A_2-A_1)+\lambda A_1 a = 0.$$ Last equation, together with the continuity condition $$A_1a = A_2(a-b),$$ allows finding required relation between $$\xi$$ and $$a$$: $$\xi = -\frac{b}{a(b-a)}.$$ For a fixed value of $$b$$, the following inequality is valid: $$\xi\leq \xi_{*} = \left. -\frac{b}{a(b-a)}\right|_{a = b/2} =-\frac{4}b.$$