# Expectation of momentum in the bound state

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).

• I think one can only conclude for non-relativistic, bound, eigenstate $\langle \hat{p} \rangle=0$. Since $$\langle n | p | n \rangle \propto \langle n | [H,x] | n \rangle = \langle n | Hx-xH | n \rangle = E_n ( \langle n| x | n \rangle - \langle n| x | n \rangle) =0$$, here $$H=\frac{p^2}{2m} + V$$. If we relax the state into any bound state $| \rangle$, we have $$\langle | p | \rangle \propto \langle | [H,x] | \rangle = \sum_n c^*_n E_n \langle n | x | \rangle - c_n E_n\langle |x | n \rangle \neq0$$ in general. Feb 22 '14 at 7:52
• @user26143 Why not turn that into an answer?
– user10851
Feb 22 '14 at 9:22
• My reply is a technical note. I am not sure how to answer "What is the physical interpretation of the fact that ⟨p^⟩=0 in an energy eigenstate ψn(x,t) but ⟨p^⟩≠0 in some superposition state" I guess it relates to shape of wavefunction?... Perhaps I will turn it into an answer Feb 22 '14 at 15:45
• if the potential has symmetry along a given direction, then it follows that an eigenstate will move along that direction in the same way regardless of the orientation, hence it must cancel out to zero when averaged Oct 22 '18 at 19:06

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.

• Why won't the argument hold for unbounded states? Jan 31 '17 at 15:27
• @GuillemB In that case x is ill-defined. Dec 29 '20 at 14:26

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

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

• This is certainly not correct: the coherent state $\vert\alpha\rangle$ for Im$(\alpha) \ne 0$ is an easy counter example. Of course, it is a linear combination of bound states, and in general such linear combinations may very well not have $\langle r\rangle=0$ Jul 31 at 13:20
• @ZerotheHour this is only true for eigenstates of the Hamiltonian, which they did mention but could be emphasized better. Jul 31 at 16:01

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.

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.

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:

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