Proof that the linear momentum operator is hermitian

[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

 

[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 $\left.\psi_b^*\psi_a\right|_{-\infty}^\infty=0$.

 

[micromass]

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".

 

[Paul Black]

thank you for your answers
if i take a special function which satisfy the property then is my solution ok?

 

[Matterwave]

Yea, it looks OK to me.

 

[Paul Black]

ok thank you very much