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:
$V(x)=\begin{cases}
\lambda\delta(x-a) & \text{for }0<x<b\\
+\infty & \text{otherwise}
\end{cases}$
where $\lambda$ is a real constant and $0<a<b$. 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 $0<x<a$ (region $1$) and $a<x<b$ (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:
$\psi(x)=\begin{cases}
A_1x & \text{for }0<x<a\\
A_2(x-b) & \text{for }a<x<b\\
0 & \text{otherwise}
\end{cases}$
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 $A_2$).
$\psi(x)=\begin{cases}
A_1x & \text{for }0<x<a\\
(A_1+\xi)(x-b) & \text{for }a<x<b\\
0 & \text{otherwise}
\end{cases}$
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:
$\psi(x)=\begin{cases}
\sqrt{\frac{3}{a^2b}}x & \text{for }0<x<a\\
\frac{\sqrt{3/b}}{a-b}(x-b) & \text{for }a<x<b\\
0 & \text{otherwise}
\end{cases}$
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:
$\psi(x)=\begin{cases}
A_1sin(kx) & \text{for }0<x<a\\
A_2sin(kx-kb) & \text{for }a<x<b\\
0 & \text{otherwise}
\end{cases}$
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.
Best Answer
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: $$ \xi\leq \xi_{*} = \left. -\frac{b}{a(b-a)}\right|_{a = b/2} =-\frac{4}b. $$