Delta function proof in QM I'm actually working with some QM problems at the moment but I've hit a wall with a delta potential involved.
The problem asks me to verify that 
$$
 \frac{d \phi_{x=0^{+}}}{dx} -\frac{d \phi_{x=0^{-}}}{dx} = -\frac{2mV_{0}}{ \hbar} \phi_{x=0}
$$
which I believe is a continuity proof.
I know that the sol. of the TISE that relates to $E=-\frac{\hbar k^{2}}{2m} < 0$ is
$$
\phi(x) = \sqrt{k} e^{-k \mid x \mid}
$$
and $k=\frac{mv_{0}}{\hbar}>0$
Here's the thing, I know that even in QM potentials continuity must exist at the potential bounds.  How am I supposed to prove a continuity equation based on a delta potential that is only valid at an infinitesimal point?
 A: There's nothing crazy going on here, you just compute the left and right derivatives. In general, we have that
$$\left. \frac{df}{dx}\right|_{x = a^\pm} = \lim_{h \to a^\pm} \frac{f(a + h) - f(a)}{h}.$$
Let's just apply this to the present problem.
Since $\phi(x) = \sqrt{k}e^{-k|x|}$, we have that
$$\lim_{h \to 0^+} \frac{\sqrt{k}e^{-k|h|} - \sqrt{k}}{h} = \sqrt{k} \lim_{h \to 0^+} \frac{e^{-kh} - 1}{h} = \sqrt{k} \lim_{h \to 0^+} \frac{-ke^{-kh}}{1} = -k\sqrt{k},$$
where we could replace $|h|$ with $h$ since $h$ approaches $0$ through positive values in the above limit. Similarly,
$$\lim_{h \to 0^-} \frac{\sqrt{k}e^{-k|h|} - \sqrt{k}}{h} = \sqrt{k} \lim_{h \to 0^-} \frac{e^{kh} - 1}{h} = \sqrt{k} \lim_{h \to 0^-} \frac{ke^{-kh}}{1} = k\sqrt{k},$$
where we replaces $|h|$ with $-h$ since $h$ is approaching $0$ through negative values in the above limit. Therefore
$$\left. \frac{d\phi}{dx}\right|_{x = 0^+} - \left. \frac{d\phi}{dx}\right|_{x = 0^-} = -2k\sqrt{k}.$$
Now $k = \dfrac{mV_0}{\hbar}$ and $\phi(0) = \sqrt{k}$, so in fact we have
$$\left. \frac{d\phi}{dx}\right|_{x = 0^+} - \left. \frac{d\phi}{dx}\right|_{x = 0^-} = -\frac{2mV_0}{\hbar}\phi(0),$$
as desired.
In the case of the delta potential, the wave function $\phi$ is continuous, but its derivative is not continuous at $x = 0$ because of the $|x|$ in the exponent. The above result shows that the derivative of $\phi$ has a jump at $x = 0$.
