quantum mechanics - Solving a time independent Schrödinger equation with a given potential

I'm trying to rework this old homework problem, and I am having problems arriving at the same solution on the answer sheet:

Let

$$V(x)=\begin{cases}\infty &\text{ if } x < 0\\ \alpha\delta(x-a) &\text{ if } x \geq 0 \end{cases}$$ where $a$ and $\alpha$ are of the appropriate units.

A particle starts out in the well $0 < x < a$, but because of tunneling its wave function gradually leaks out through the delta-function barrier.

(a) Solve the time independent Schrödinger equation for this potential; impose appropriate boundary conditions, and determine the energy $E$. An implicit equation will do.

Here is what I know:

$$\psi(x)=\begin{cases}Ae^{ikx}+Be^{-ikx} &\text{ if } 0 \leq x \leq a\\ Fe^{ikx} &\text{ if } x > a \end{cases}.$$

$\psi(0)=A+B$ and this implies $-A=B$. I also understand that $\psi$ is continuous at $a$. What I am stuck on is why $\psi'(a)$ is discontinuous. Specifically, why is it that:

$$ikFe^{ika}-ikA(e^{ika}+e^{-ika})=\frac{2m\alpha}{\hbar^2}\psi(a).$$

Answer

Note: I have tried to make my answer a little more general, with detail, so that it will be useful for more people.

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The question is what boundary conditions do we apply to our wavefunction either side of a Dirac delta function?

In your example we have the potential

$$V(x)=\begin{cases}\infty &\text{ if } x < 0\\ \alpha~\delta(x-a) &\text{ if } x \geq 0 \end{cases}$$ We are interested in the boundary conditions either side of $x=a$. What information do we have? Well, due to the probabilistic interpretation of the wavefunction we require continuity of the wavefunction. That is, our first boundary condition is $$(1)~~~~~~~~~~~~\boxed{\psi_{-}(a) = \psi_{+}(a)}$$ where the $\pm$ subscripts represent the right and left sides of $x=a$ respectively. What other conditions can we set? Well, usually we would ask that the first derivative is also matched either side of $x=a$ (you should be asking yourself why do we do this?), but in this case this is not the right condition to impose. Let's see why.

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Where does the boundary condition on $\frac{\partial \psi}{\partial x}$ come from?

Our wavefunction is a solution to the 1-D time-independent Schrödinger Equation:

$$H~\psi(x) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)+V(x)\psi(x) = E~\psi(x)$$ Looking at this equation, we see that we can get a boundary condition on $\frac{\partial \psi}{\partial x}$ at any point $a$ by integrating it w.r.t $x$ over the region $[a-\epsilon,a+\epsilon]$, taking $\epsilon\rightarrow 0$: $$\lim_{\epsilon\rightarrow 0}\left[-\int^{a+\epsilon}_{a-\epsilon}dx\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)+\int^{a+\epsilon}_{a-\epsilon}dx~V(x)\psi(x)\right] = E~\lim_{\epsilon\rightarrow 0}~\int^{a+\epsilon}_{a-\epsilon}dx~\psi(x)\\ \implies -\frac{\hbar^2}{2m}\left[\frac{\partial \psi_{+}(a)}{\partial x}-\frac{\partial \psi_{-}(a)}{\partial x}\right] + \lim_{\epsilon\rightarrow 0}\int^{a+\epsilon}_{a-\epsilon}dx~V(x)\psi(x) = 0$$ where we have used the continuity of $\psi$ to evaluate the RHS as zero. Note that for any $V(x)$ the second term doesn't necessarily vanish. But, if $V(x)$ is continuous, this term will vanish for the same reason the RHS did, and in those cases we yield the usual boundary condition $$\boxed{\frac{\partial \psi_{+}(a)}{\partial x}=\frac{\partial \psi_{-}(a)}{\partial x}}~~~~~~~(\mbox{when $V(a)$ is continuous at $x=a$})$$ We note that the for example given here, $V(x)$ is definitely not continuous at $x=a$, where it diverges to infinity. So this second term doesn't vanish, in fact $$\lim_{\epsilon\rightarrow 0}\int^{a+\epsilon}_{a-\epsilon}dx~V(x)\psi(x) = \alpha\lim_{\epsilon\rightarrow 0}\int^{a+\epsilon}_{a-\epsilon}dx~\delta(x-a)\psi(x) = \alpha~\lim_{\epsilon\rightarrow 0}~\psi(a) = \alpha~\psi(a)$$ so rearranging our results, the second boundary condition for the problem is $$(2)~~~~~~~~~~~~\boxed{\frac{\partial \psi_{+}(a)}{\partial x}-\frac{\partial \psi_{-}(a)}{\partial x} = \frac{2m\alpha}{\hbar^2}\psi(a)}$$

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Explicitly using the form of your wavefunctions (using $B=-A$ to eliminate $B$)

$$\psi_{+}(x) = Fe^{ikx},~~~~~~\psi_{-}(x) = A\left(e^{ikx}-e^{-ikx}\right)$$ this boundary condition becomes $$\boxed{ikFe^{ika}-ik\left(Ae^{ika}+e^{-ika}\right) = \frac{2m\alpha}{\hbar^2}\psi(a)}$$ as required.