2014-02-22 07:04:19 8 Comments

Is it logically correct to assert that the expectation of the momentum $$\langle \hat p \rangle=0$$ for any bound state because it is bound to some finite region? What is the physical interpretation of the fact that $$\langle \hat p \rangle=0$$ in an energy eigenstate $\psi_n(x,t)$ but $$\langle \hat p \rangle\neq0$$ in some superposition state $$\psi(x,t)=c_m\psi_m(x,t)+c_n\psi_n(x,t)~?$$ Here $\psi_n(x,t)$ the eigenstates of the Hamiltonian, for example, in the problem of particle in a box (say).

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## 4 comments

## @dsm 2018-10-22 07:52:48

I know this has already been answered, but I think there is a nice way to see this that hasn't been mentioned. If the state in question is a stationary state (energy eigenstate), then we know

$$ H|\Psi\rangle=E|\Psi\rangle $$

which means that

$$ \langle[x,H]\rangle=\langle\Psi|xH-Hx|\Psi\rangle=E\Big(\langle\Psi|x|\Psi\rangle-\langle\Psi|x|\Psi\rangle\Big)=0, $$

and since $\hat{x}$ has no explicit time dependence we have the simple differential equation for the $\langle x\rangle :$

$$ \frac{d\langle x\rangle}{dt}=\frac{1}{i\hbar}\langle [x,H]\rangle =0. $$

Now recall the immensely useful commutation rule

$$ [x,F(p)] = i\hbar\frac{\partial F}{\partial p}. $$

Since the potential only depends on position, it commutes with $x$, so the above time derivative can be written

$$ 0=\frac{d\langle x\rangle}{dt}=\frac{1}{i\hbar}\langle [x,H]\rangle = \frac{1}{i\hbar}\langle [x,\frac{p^2}{2m}]\rangle = \frac{1}{2m}\langle\frac{\partial}{\partial p}p^2\rangle = \frac{\langle p\rangle}{m}. $$

So we see that the expectation of potential for an energy eigenstate is zero:

$$ \langle p\rangle = 0. $$

## @Gonenc 2015-04-08 20:43:29

$\newcommand{ket}{\left| #1 \right>}$ $\newcommand{bra}{\left< #1 \right|}$ $\newcommand{\bk}[3]{\left< #1| #2 |#3\right>}$ In a one dimensional problem $\langle \hat p \rangle$ is always zero. $$\langle \hat p \rangle = \bk{\psi}{\hat p}{\psi}=\int \mathrm{d}x \,\psi^* \hat p \, \psi \propto \int \mathrm{d}x \, \psi^* \psi' \overset{(1)}{=}-\int \mathrm{d}x \, (\psi^*)' \psi \overset{(2)}{=} -\int \mathrm{d}x \, \psi' \psi^* $$

Where in (1) I integrated by parts and assumed that $\psi \to 0$ as $x \to \infty$ and in (2) I used the fact that you can always choose a bounded energy eigenstate to be real, which implies that I can take the complex conjugate at no cost. Notice that we have the following:

$$ \bk{\psi}{p}{\psi} \propto - \int \mathrm{d}x \, \psi' \psi^* = \int \mathrm{d}x \, \psi' \psi^* \iff \bk{\psi}{p}{\psi} = 0$$

## @Yuman 2015-04-08 19:11:55

A hint on this could be the fact that a superposition of stationary states of different energies is

NOTa stationary state, because you can not express the wave function as the product of a single time-dependent exponential tiames a spatial function.## @user26143 2014-02-22 19:12:29

Bound state means the particles are bounded somewhere. Its wavefunction will vanish at the asymptotic limit. A bound state could be a superposition of a finite number of bound eigenstates. For instance, the superposition of the ground and first excited-state wavefunction of particle-in-box will still vanish at far limit.

I think one can only conclude for non-relativistic, bound, eigenstate (not

bound state) $\langle \hat{p} \rangle=0$. Since $$\langle n | p | n \rangle \sim \langle n | [H,x] | n \rangle = \langle n | Hx-xH | n \rangle = E_n ( \langle n| x | n \rangle - \langle n| x | n \rangle) =0 $$. If we relax the state intoanybound state $| \rangle$, we have $$\langle | p | \rangle \sim \langle | [H,x] | \rangle = \sum_n c^*_n E_n \langle n | x | \rangle - c_n E_n\langle |x | n \rangle \neq0 $$ in general.any## @GuillemB 2017-01-31 15:27:55

Why won't the argument hold for unbounded states?