Constancy of Wronskian when potential has Finite Discontinuities I am working on a problem where for the 1-dimensional Hamiltonian $\hat{H}=(-1/2m)(d^2/dx^2)+\mathcal{W}(x)$ with $\mathcal{W}$ assumed to be smooth and real and $\Psi$ as solution to $\hat{H}\Psi=E\Psi$, I need to prove that the Wronskian, $W=\overline{\Psi(x)}(d\Psi(x)/dx)+(d\overline{\Psi(x)}/dx)\Psi(x)$, does not depend on $x$ if $\mathcal{W}$ has finite discontinuities and $\Psi,\Psi'$ being continuous.
I do not know how to approach this problem. I have proven that the Wronskian does not depend on $x$ via taking the derivative of $W$ with respect to $x$ and showing that $\overline{\Psi(x)}(d^2\Psi(x)/dx^2)-(d^2\overline{\Psi(x)}/dx^2)\Psi(x)=0$ using the difference of $\hat{H}\overline{\Psi}\Psi=E\overline{\Psi}\Psi$ and $\hat{H}\Psi\overline{\Psi}=E\Psi\overline{\Psi}$.
Since $\overline{\Psi(x)}(d^2\Psi(x)/dx^2)-(d^2\overline{\Psi(x)}/dx^2)\Psi(x)=0$ regardless of the value of $\mathcal{W}$ shouldn't this be sufficient as a proof? I am asking as in the next step I somehow have to use this information to show that $|R(k)|^2+|T(k)|^2=1$ in the square-well problem.
 A: Hints:

*

*The main point of the problem seems to be that with the given assumptions, i.e.

*

*that the wavefunction $\Psi\in C(\mathbb{R})$ is continuous, and


*that the potential $V$ is piecewise continuous with finite jumps/discontinuities,
one can use a bootstrap argument to prove that $\Psi$ is in fact of class $C^1$ and piecewise twice differentiable, cf. e.g. my Phys.SE answer here.


*Similar for $\Psi^{\prime}$.


*Therefore the Wronskian $W(\Psi,\Psi^{\prime})$ is a continuous piecewise differentiable function.


*Now differentiate, and use the TISE twice (assuming the same energy $E$) to conclude that the Wronskian $W(\Psi,\Psi^{\prime})$ is a constant.
A: Starting from your equation:
$${\Psi^*(x)}\,(d^2\Psi(x)/dx^2)-(d^2{\Psi^*(x)}/dx^2)\,\Psi(x)=0$$
You are just one-step away from the answer:
\begin{align}
0 &= {\Psi^*(x)}\,(d^2\Psi(x)/dx^2)-(d^2{\Psi^*(x)}/dx^2)\,\Psi(x)\\
&= \frac{d}{dx} \left\{ {\Psi^*(x)}\,(\Psi(x)/dx)-(d{\Psi^*(x)}/dx)\,\Psi(x) \right\}
\end{align}
Thus,
$$
\left\{ {\Psi^*(x)}\,(\Psi(x)/dx)-(d{\Psi^*(x)}/dx)\,\Psi(x) \right\} = \text{constant}.
$$
