B Is the Momentum Operator Hermitian? A Proof

[hokhani]

TL;DR Summary

How momentum is Hermitian

Momentum operator is $p=-i\frac{d}{dx}$ and its adjoint is $p^\dagger=i\frac{d}{dx}$. So, $p^\dagger=-p$. How is the momentum Hermitian?

 

[Gaussian97]

Remember, the exact definition of an autoadjoint operator is that $\left<\psi\right|\hat{p}\left|\phi\right>=\left<\phi\right|\hat{p}\left|\psi\right>^*$
You can check that with such a definition the momentum is really autoadjoint.
You can prove indeed that $\left(\frac{d}{dx}\right)^\dagger = -\frac{d}{dx}$

 

[etotheipi]

write this\begin{align*}

(f, p g) = -i \hbar \int_D f^*(x) \frac{\partial g}{\partial x}(x)dx &= i\hbar \int_D \frac{\partial f^*}{\partial x}(x) g(x) dx - {\underbrace{\left[ i \hbar f^*(x) g(x) \right]}_{\overset{!}{=} \, 0}}^{\partial D} \\

&= \int_D \left( - i \hbar \frac{\partial f}{\partial x}(x) \right)^* g(x) dx \\

&= (p f, g)

\end{align*}

 

[PeterDonis]

Momentum operator is $p=-i\frac{d}{dx}$ and its adjoint is $p^\dagger=i\frac{d}{dx}$.

No, this is not correct. The correct equation is $p^\dagger = (- i)^* \hbar \left( \frac{d}{dx} \right)^\dagger$. Then:

You can prove indeed that $\left(\frac{d}{dx}\right)^\dagger = -\frac{d}{dx}$

Which means that $p^\dagger = (- i)^* \hbar \left( \frac{d}{dx} \right)^\dagger = i \hbar \left(- \frac{d}{dx} \right) = - i \hbar \frac{d}{dx} = p$.

 

[hokhani]

You can prove indeed that $\left(\frac{d}{dx}\right)^\dagger = -\frac{d}{dx}$

I can't prove it. Could you please help me with that?

 

[etotheipi]

please let $L = \dfrac{d}{dx}$ and re-write $(f, Lg)$ in the form $(Mf, g)$ and then tell what is $M$

 

[PeterDonis]

I can't prove it. Could you please help me with that?

Post #3 is a proof of it. If the presence of the factor of $i\hbar$ in post #3 confuses you, just eliminate it; then you have a straightforward proof. The key to the proof is the sign flip that comes with the integration by parts.

 

[PeterDonis]

Post #3 is a proof of it.

Note, btw, that the proof in post #3 is only valid for functions that vanish at the boundary, i.e., at infinity, so the boundary term in the integration by parts goes away. The usual argument is that any function that can actually be physically realized will have this property; this is what mathematicians often call (somewhat disdainfully) a physicist's level of rigor.