# Homework Help: Proof that d/dx is anti-hermitian by integration by parts

1. Jul 7, 2011

### HotMintea

The attempt at a solution

$$\begin{equation*} \begin{split} \ -\ i\int\psi^* \frac{\partial{\psi}}{\partial x}= \\ -i(\psi^*\psi\ - \int\psi \frac{\partial\psi^*}{\partial x})\space\ (?) \end{split} \end{equation*}$$

I thought $\psi^*\psi\ = \ constant\neq\ 0.$ However, it vanishes in the correct proof.

2. Jul 7, 2011

### tiny-tim

Hi HotMintea!

Not sure what this is , but doesn't [constant]ab = 0 ?

3. Jul 7, 2011

### Matterwave

As tiny-tim pointed out, you need limits of integration.

However, psi*psi is not a constant. It is only a constant when integrated over all space, but it is not integrated in this case.

The real reason the first term vanishes is that for limits at plus or minus infinity, the wave-function must decay to 0 to be normalizable. For periodic boundary conditions, the two limits are the same point so they must subtract to 0.

4. Jul 7, 2011

### HotMintea

Hi tiny-tim and Matterwave!
For a decaying wave function, is the following correct?:
$$\begin{equation*} \begin{split} \ -\ i\int_{-\infty}^{\infty} \psi^*\frac{\partial \psi}{\partial x}\ =\\-\ i\ (\psi^*\psi\ |_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \psi\frac{\partial \psi^*}{\partial x})\ = \\ i\int_{-\infty}^{\infty}\psi\frac{\partial\psi^*}{\partial x} \end{split} \end{equation*}$$
Thus, $\psi\ = \ 0 \ at \ x\ = -\infty\ \ and \ \infty$?
For periodic boundary conditions, $\ -\ i\int_{a}^{a} \psi^*\frac{\partial \psi}{\partial x}\ = \ i\int_{a}^{a}\psi\frac{\partial\psi^*}{\partial x}\ = \ 0$ ?
For $\psi\ = \ Ae^{i(k x-\omega t)}$ where A is an arbitrary constant, $\psi^* \psi\ = \ A^2 \ = \ constant.$ Is $\psi^* \psi\ = \ constant$ not true in general? (Example taken from: http://en.wikipedia.org/wiki/Normalizable_wave_function#Example_of_normalization)

Last edited: Jul 8, 2011
5. Jul 8, 2011

### tiny-tim

Hi HotMintea!
Yes.
No, $\psi$ is the same (at each boundary), so the [] is zero, not the ∫.
That's a featureless plane wave, so of course $\psi^* \psi$ is constant.

Not all waves are so uninteresting.

6. Jul 8, 2011

### HotMintea

!
For the periodic boundary conditions, is the following correct?:
$$\begin{equation*} \begin{split} \ -\ i\int_{a}^{b} \psi^*\frac{\partial \psi}{\partial x}\ = -\ i\ (\psi^*\psi\ |_{a}^{b} - \int_{a}^{b}\psi\frac{\partial\psi^*}{\partial x})\ \end{split} \end{equation*}$$
However, $\psi(a)\ = \psi(b).$ Moreover, conjugates $\psi^*(a)\ = \psi^*(b).$ Thus, $\psi^*\psi\ |_{a}^{b}\ = \ 0.$

Therefore,
$$\begin{equation*}\begin{split} -\ i\int_{a}^{b} \psi^*\frac{\partial \psi}{\partial x}\ = \ i\int_{a}^{b}\psi\frac{\partial\psi^*}{\partial x} \end{split}\end{equation*}$$

Last edited: Jul 8, 2011
7. Jul 8, 2011

Yes!

8. Jul 8, 2011