Expectation of momentum in the bound state

Question

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).

---

NOTE The already exist many excellent answers to this post. However, I am particularly interested in the answer provided by user26143. His argument seems quite robust. I do not understand why his calculation fails for unbound or scattering states.

Answer 1

Is it logically correct to assert that the expectation of the momentum $\langle p \rangle=0$ for any bound state because it is bound to some finite region?

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 any 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 into any bound 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.

Answer 2

$\newcommand{ket}{\left| #1 \right>}$ $\newcommand{bra}{\left< #1 \right|}$ $\newcommand{\bk}[3]{\left< #1| #2 |#3\right>}$ 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:

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

Answer 3

I would adjust Gonenc's answer a bit, because it's not always true that one can choose a bound energy eigenstate to have a real position-space representation. There are 1D systems with degeneracy (e.g. the "isotonic oscillator" discussed here), and in those cases an energy eigenstate is of the form $\alpha \chi_1 + \beta \chi_2$ with $|\alpha|^2 + |\beta|^2 = 1$ and $\chi_{1,2}$ real functions. In general such a state can't be rotated into a purely real one. (In systems without degeneracy we have $\psi = e^{i \alpha} \psi^*$ for any energy eigenstate, since both $\psi$ and $\psi^*$ are solutions to the time-independent Schrödinger equation with the same energy and then Gonenc's answer goes through.)

Nevertheless, writing $\psi = \psi_R + i\psi_I$ for an energy eigenstate $\psi$, some algebra shows $$\langle p \rangle_\psi \propto \int \mathrm{d}x \, W[\psi_R,\psi_I] \,,$$ where $W[\chi,\varphi] = \chi \varphi' - \chi' \varphi$ is the Wronskian. In the algebra I've assumed $\psi \rightarrow 0$ as $|x| \rightarrow \infty$ i.e. we're in a bound state. Now for the Wronskian, it's easy to see that $W' = 0$ everywhere from the Schrödinger equation. To get $W = 0$ everywhere we should assume that both products $\psi_R \psi_I'$ and $\psi_R' \psi_I$ vanish at infinity. This is not always the case even if $\psi_{R,I}$ vanish at infinity. A sufficient condition is that $V(x) - E > M^2$, for all $x > x_0$, for some numbers $x_0,M$, where $E$ is the energy of the bound state in question, since then both the wave function and its derivative decay exponentially at infinity.

Answer 4

$\renewcommand{\Re}{\operatorname{Re}}\renewcommand{\Im}{\operatorname{Im}}$Another way to phrase the derivation of @Latrace and @Gonenc, maybe more physically, is to recognize that the Wronskian which appear is just the probability current. $$j = \frac{\hbar}{2mi}(\psi^*\partial_x \psi -(\partial_x\psi)^*\psi) = \frac{\hbar}{m}\operatorname{Im}(\psi^*\partial_x\psi)$$

Note that this is $j = \frac{\hbar}{2mi}W[\psi^*,\psi] = \frac{\hbar}{m}W[\Re\psi,\Im\psi]$.

Then, we have, in a general state $\psi$ : $$\langle \hat p\rangle_\psi = m\int_{-\infty}^{+\infty} j(x)\text dx \tag 1$$

As probability is conserved, we have, for a general solution $\psi(t)$ of the Schrödinger equation : $$\partial_t |\psi|^2 + \partial_x j = 0$$ For a stationary state, $|\psi|^2$ is constant everywhere constant in time, so we have $\partial_x j = 0$ (This can also be checked explicitly from the time-independent Schrödinger equation).

For a bound stationary state, the fact that the wavefunction must falloff at infinity implies that $j= 0$ : the probability current vanishes (as we expect for a bound state). From $(1)$, we then have $\langle \hat p\rangle_\psi =0$.

For a scattering state, there is no normalizability condition, and indeed we can have $j\neq 0$. More precisely, for any $E>0$, we have two solutions with $j = \pm \sqrt{\frac{2E}{m}}$ (corresponding to plane waves $e^{\pm i x\sqrt{2mE}/\hbar}$ near $x \to -\infty$). Then, $\langle \hat p\rangle_\psi$ diverges.

This is the flaw in @user26143's reasoning : scattering state are not normalizable, and therefore we should'nt expect $\langle \psi|\hat p|\psi\rangle$ to be finite. If it is not finite, then the calculation does not work.

Answer 5

A hint on this could be the fact that a superposition of stationary states of different energies is NOT a stationary state, because you can not express the wave function as the product of a single time-dependent exponential tiames a spatial function.

Answer 6

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:

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