Momentum operator for stationary solution?

Question

Okay, so I'm having some difficulty understanding where I'm going wrong with this problem. I'm trying to show that $$ \langle \vec{P} \rangle_{\psi} = \langle \psi | \vec{P} | \psi \rangle = 0 $$ where $\psi$ is a stationary solution of $H$ (i.e. satisfies $H | \psi \rangle = E | \psi \rangle$) and $\vec{P} = \frac{\hbar}{i}\frac{\partial}{\partial x}$.

So what I have so far is since $\psi$ is a stationary solution of $H$, it has the form $\psi(x,t) = \psi(x) e^{-\frac{i}{\hbar} E t}$ so I get

$$ \langle \vec{P} \rangle_{\psi} = \langle \psi | \vec{P} | \psi \rangle = \int_{-\infty}^{\infty} dx \: \psi^*(x,t) \frac{\hbar}{i}\frac{\partial}{\partial x} \psi(x,t) $$ $$ = \int_{-\infty}^{\infty} dx \: \psi^*(x,t) \frac{\hbar}{i} e^{-\frac{i}{\hbar} E t} \frac{\partial}{\partial x} \psi(x) = \frac{\hbar}{i} e^{-\frac{i}{\hbar} E t} \int_{-\infty}^{\infty} dx \: \psi^*(x,t) \frac{\partial}{\partial x} \psi(x) $$

My problem is I'm not following how this is suppose to be equalling $0$, as I know it should? I've been told this gives a real function but I'm not sure how that helps. Any insight would be appreciated.

Answer 1

$\newcommand{\bk}[1]{\left<\psi\middle|#1\middle| \psi\right>}$Assuming that the Hamiltonian has the usual form $H = p^2/2m + V(x)$ (I don't write operators with hats, nor do I put vector arrows!). You can look at the commutator $[x,H] = [p^2,x]/2m =- i\hbar p/m$. Hence, for a state $H \left|\psi\right> = E\left|\psi\right> $ we have:

$$\bk {[x,H]} = -i\hbar/m \bk{p}$$

on the other hand: $$\bk{[x,H]} = \bk{xH} - \bk{Hx} = E\big(\bk{x}-\bk{x}\big)=0 $$

combining them we get: $$-i\hbar/m\bk p =0 \implies \bk p =0$$

As a general rule of thumb you should look at commutators of kind $[\mathcal A,H]\propto \mathcal O$ for some hermitian operator $\mathcal A$ if you want to show that $\bk{\mathcal O}$ is zero for an observable $\mathcal O$ and energy eigensate $\left|\psi\right>$.

Answer 2

For bound states, first write $\psi(x,t)^*=e^{+i Et/\hbar}\psi(x)^*$ so as to cancel the exponentials and remove the time dependence. Next if $\psi(x)$ is real $$ -i\hbar \int_{-\infty}^{\infty} dx\, \psi(x)^* \frac{d}{dx}\psi(x) $$ will be purely imaginary or $0$. Since the average momentum $\langle P\rangle$ must be a real quantity, it cannot be purely imaginary and thus be must $0$.

As an alternate derivation, $\langle P\rangle= m\frac{d}{dt}\langle x\rangle $ but $$ \langle x\rangle = \int_{-\infty}^{\infty} dx \,x \psi(x)^* \psi(x) $$ is time-independent for a stationary solution (almost be definition) so immediately $\frac{d}{dt}\langle x\rangle=0$ and thus so is $\langle p\rangle$.