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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.

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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.

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

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  • $\begingroup$ 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. $\endgroup$ – suriya00 Jun 16 at 17:54

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