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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.

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

1 Answer 1

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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: $$ \xi\leq \xi_{*} = \left. -\frac{b}{a(b-a)}\right|_{a = b/2} =-\frac{4}b. $$

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  • $\begingroup$ Thank you very much. $\endgroup$
    – Ali Pedram
    Jan 31, 2022 at 20:20

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