# 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).

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.

• 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, 2014 at 7:52
• @user26143 Why not turn that into an answer?
– user10851
Feb 22, 2014 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, 2014 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, 2018 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, 2017 at 15:27
• @GuillemB In that case x is ill-defined. Dec 29, 2020 at 14:26
• @GuillemB I also have the same question.
– SRS
Sep 9, 2023 at 7:50

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

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

• 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, 2021 at 13:20
• @ZerotheHour this is only true for eigenstates of the Hamiltonian, which they did mention but could be emphasized better. Jul 31, 2021 at 16:01

$$\renewcommand{\Re}{\operatorname{Re}}\renewcommand{\Im}{\operatorname{Im}}$$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.

• Why do you say that $\langle p\rangle$ is not finite in a scattering state $|\psi\rangle$? Calculation (with regularization) gives a finite answer and it is zero. @SolubleFish
– SRS
Sep 17, 2023 at 13:00
• If you regularize on a finite length $L$ with periodic boundary conditions, there are states with non-zero expectation value $\frac{\langle \psi | \hat p |\psi\rangle}{\langle \psi | \psi\rangle}$. However, as you let the length $L$ go to infinity, we switch to the continuous normalization for the scattering eigenstates, so $\langle \psi|\psi\rangle \sim L$ and $\langle \psi | \hat p |\psi\rangle$ diverges unless it was zero in the first place. Sep 18, 2023 at 16:20
• Another way to see this is that you can surely construct wavepackets out of superposition of scattering eigenstates with a finite non-zero value for $\langle \hat p \rangle$. This is coherent with the assertion that $\langle\psi|\hat p|\psi\rangle$ diverges (as the diagonal value of a distribution). Sep 18, 2023 at 16:23
• Still confused! If you box normalize, $\psi=\frac{1}{\sqrt{L}}e^{ikx}$, and this state, $\langle p\rangle=\frac{1}{L}\int_{-L/2}^{L/2}e^{-ikx}(-i\hbar \frac{d}{dx}) e^{ikx}dx=\frac{1}{L}(\hbar k)\int_{-L/2}^{L/2}e^{-ikx}e^{ikx}dx=\hbar k$. Since the integral is independent of the regulator $L$, the answer is finite in the limit $L\to\infty$. I don't know why you say that it is infinite. Moreover, even if it is infinite, how does that make user26143 's calculation invalid? He never made use of the finiteness of $\langle p\rangle$. Right? @SolubleFish
– SRS
Sep 20, 2023 at 4:21
• The limit you are considering is indeed finite as $L\to +\infty$, but it does not go to $\langle \hat p \rangle$. This is because of the normalization : the momentum eigenstate in the infinite line is normalized according to (up to a finite multiplicative constant) $\langle p |p'\rangle = \int\text d x e^{ix(p'-p)} = \delta(p'-p)$. This means that, to recover $|p\rangle$ from the finite volume state $|p\rangle_L$, you need to take the limit $\sqrt{L}|p\rangle_L \overset{L\to\infty}{\longrightarrow} |p\rangle$. Sep 21, 2023 at 9:48

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.

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 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:

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