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## Homework Statement

Consider [itex]V(x)=-aV_{0}δ(x)[/itex]. Show that it admits a bound state of energy [itex]E=-ma^2V_{0}/2\hbar^{2}[/itex].Are there any other bound states?

**Hint**:solve Schrodinger's equation outside the potential for E<0, and keep only the solution that has the right behavior at infinity and is continuous at x=0.

Draw the wave function and see how there is a cusp, or a discontinuous change of slope at x=0. Calculate the change in the slope at equate it to [itex]\int_{-ε}^{ε}( \frac{d^2\psi}{dx^2}) dx[/itex]

(where ε is infinitesimal) determined from Shrodinger's equation.

## Homework Equations

Shrodinger equation in X basis : [itex]\frac {d^2\psi}{dx^2}+(2m/\hbar)(E-V)\psi=0[/itex].

## The Attempt at a Solution

First how can we prove that [itex]E=-ma^2V_{0}/2\hbar^{2}[/itex] admits a bound state, shall I substitute it in the shrodinger equation then prove that the obtained wave function goes to zero when x goes to infinity?

For are there any other bound states?

I tried to solve the Shrodinger's equation with E<0 and outside the potential, what I got is this :

Let E=-A where A is a positive integer.

[itex]\frac {d^2\psi}{dx^2}+(2m/\hbar)(E)\psi=0[/itex]

[itex]\frac {d^2\psi}{dx^2}-(2m/\hbar)(A)\psi=0[/itex]

[itex]\psi=Ae^{-\sqrt{2mA/\hbar^2}x}+Be^{\sqrt{2mA/\hbar^2}x}[/itex]

[itex]Be^{\sqrt{2mA/\hbar^2}x}[/itex] blows up to zero when x goes to infinity, so B=0.

what we finally get is [itex]\psi=Ae^{-\sqrt{2mE/\hbar^2}x}[/itex]. I reached here and I don't know how to complete.

If anybody can help.Thanks.

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