# 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.

Hints:

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

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

2. 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.

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, $$\left\{ {\Psi^*(x)}\,(\Psi(x)/dx)-(d{\Psi^*(x)}/dx)\,\Psi(x) \right\} = \text{constant}.$$

• Perhaps I wasn't clear but this is how I proved that the Wronskian, $W$, does not depend on $x$, this would part a) of the exercise, part b) is the part I am stuck with where I need to prove that the same is true if $\mathcal{W}$ has finite discontinuities and $\Phi,\Phi'$ are continuous. This is the part I am not sure how to approach as just like you, the equation that I hinged the proof for $W$ not depending on $x$ is $0$ for all values of $\mathcal{W}$. I am not sure if this is sufficient for what I require though which is what I am asking. – suriya00 Jun 16 at 17:54