# Proof that the linear momentum operator is hermitian

1. May 23, 2014

### Paul Black

hello
i have to proof that Px (linear momentum operator ) is hermitian or not
i have added my solution in attachments

please look at my solution and tell me if its correct

thank you all

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• ###### CCI05232014_00001.jpg
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2. May 23, 2014

### Matterwave

There's a sign error in the last expression of the first page, but you "fixed" it in the next expression.

This looks OK to me, but of course you should explain why $\left.\psi_b^*\psi_a\right|_{-\infty}^\infty=0$.

3. May 23, 2014

### micromass

Staff Emeritus
What is the domain of the linear momentum operator that you're looking at? You can't just look at arbirary wave functions, because then $\psi_b^*\psi_a\vert_{-\infty}^{+\infty}$ will not be zero. You need to restrict to special wave function which do have that property.

Another solution is that you use "distributional differentiation".

4. May 24, 2014

### Paul Black

if i take a special function which satisfy the property then is my solution ok?

5. May 24, 2014

### Matterwave

Yea, it looks OK to me.

6. May 24, 2014

### Paul Black

ok thank you very much