Proof that d/dx is anti-hermitian by integration by parts

[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.

 

[tiny-tim]

Hi HotMintea!

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

 

[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.

 

[HotMintea]

Hi tiny-tim and Matterwave!

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

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

For a decaying wave function, is the following correct?:
$$\begin{equation*} \begin{split} \ -\ i\int_{-\infty}^{\infty} \psi^*\frac{\partial \psi}{\partial x}\ =\\-\ i\ (\psi^*\psi\ |_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \psi\frac{\partial \psi^*}{\partial x})\ = \\ i\int_{-\infty}^{\infty}\psi\frac{\partial\psi^*}{\partial x} \end{split} \end{equation*}$$

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.

Thus, $\psi\ = \ 0 \ at \ x\ = -\infty\ \ and \ \infty$?

For periodic boundary conditions, the two limits are the same point so they must subtract to 0.

For periodic boundary conditions, $\ -\ i\int_{a}^{a} \psi^*\frac{\partial \psi}{\partial x}\ = \ i\int_{a}^{a}\psi\frac{\partial\psi^*}{\partial x}\ = \ 0$ ?

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.

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)

 

[tiny-tim]

Hi HotMintea!

Thus, $\psi\ = \ 0 \ at \ x\ = -\infty\ \ and \ \infty$?

Yes.

For periodic boundary conditions, $\ -\ i\int_{a}^{a} \psi^*\frac{\partial \psi}{\partial x}\ = \ i\int_{a}^{a}\psi\frac{\partial\psi^*}{\partial x}\ = \ 0$ ?

No, $\psi$ is the same (at each boundary), so the [] is zero, not the ∫.

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?

That's a featureless plane wave, so of course $\psi^* \psi$ is constant.

Not all waves are so uninteresting. ​

 

[HotMintea]

Not all waves are so uninteresting.

​

!

No, $\psi$ is the same (at each boundary), so the [] is zero, not the ∫.

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

 

[tiny-tim]

Yes!

 

[HotMintea]

Thanks for your help!