I'm currently trying to solve an exercise for my quantum mechanics class and have run into a bit of a jam:
Suppose we have the following potential :
$V(x) = 0$ if $x > |a/2|$ but $V(x) = V_0$ if $x < |a/2|$.
The Schrödinger equation is the following: $-\hbar^2 / 2m \frac{\partial^2}{\partial x^2} \psi = (E-V(x))\psi$. The ansatz for the three regions are simply a superposition of incoming and reflected plane waves, i.e. $\psi_i(x) = A_iexp(-kx) + B_iexp(kx)$.
One can easily check that this solution solves the Schrödinger equation in all three regions. My question is now the following:
In the solution to this exercise we integrate the Schrödinger equation over a small region around the edges of the three regions, i.e.:
$\psi'(x_0 + \epsilon) - \psi'(x_0 - \epsilon) = \lim_{\epsilon \rightarrow 0} (-\frac{2m}{\hbar^2} \int_{x_0 - \epsilon}^{x_0 + \epsilon} (E-V(x))\psi)$
The sample solution now immediately concluded, that the right side is zero, and therefore both the derivative and thus the function itself are continuous.
I asked an assistant and he told me "it is enough to know that both the potential and the wave function are bounded to show that the right hand side is zero".
Can someone explain to me how exactly that is enough? I mean if $\psi$ were a function without continuity then there could be a jump between $x_0 + \epsilon$ and $x_0 - \epsilon$ and the integral wouldn't vanish?
Hope someone can shed some light on this for me!
Cheers.