Consider a system of identical particles (bosons or fermions) with field operator $\hat{\psi}(x)$. The particle density operator is $\hat{\psi}^\dagger(x)\hat{\psi}(x)$.

Suppose that the particle density is constant everywhere - that is to say that: $$\rho(x,t)=\langle\Psi(t)|\hat{\psi}^\dagger(x)\hat{\psi}(x)|{\Psi(t)}\rangle=c(x)$$ where $c(x)$ is some spatial function that is constant in time.

Does this necessarily imply that the system state is an eigenstate of the Hamiltonian? Or are there states with no fluctuation in particle density that aren't eigenstates of the Hamiltonian? How would I prove this in general, or come up with a counterexample? All my attempts thus far have failed.

I strongly suspect that this is true in the single-particle case, but struggled to prove even that. Does it hold in that limit?

I am considering a non-relativistic Hamiltonian of the form: $$\hat{H} = \int dx \hat{\psi}^\dagger(x)\left(-\frac{\hbar^2}{2m}\nabla^2_x+V(x)\right)\hat{\psi}(x)+\frac{g}{2}\int dx \hat{\psi}^\dagger(x)\hat{\psi}^\dagger(x)\hat{\psi}(x)\hat{\psi}(x)$$

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    $\begingroup$ If superpositions of different particle numbers $N$ are allowed, then here's a counterexample: Any sum of a 1-particle state with energy $E_1$ and a 2-particle state with energy $E_2\neq E_1$ gives a non-stationary state with a time-indep $c(x,t)$, because the cross-terms are zero. If sup'ns of different $N$ are not allowed, then take $g=0$ and take $V(x)$ to have two spatially-separated wells with infinitely high walls. Consider a superposition of single-particle states of different energies, each with the particle localized in a different well. Then the cross-terms are zero as before. $\endgroup$ Sep 29 '19 at 13:37
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    $\begingroup$ @ChiralAnomaly You should convert this comment to an answer. $\endgroup$
    – tparker
    Oct 6 '19 at 1:25

Does $\langle \psi | \hat{x} | \psi\rangle = c$ mean that $\hat{x} | \psi \rangle = c | \psi \rangle$?

In case of free particles, QFT Hamiltonian: $$ H = \int \frac{d^3p}{(2 \pi)^3} \; \sqrt{p^2 + m^2} a^\dagger_\vec{p} a_\vec{p} $$ is not even proportional to the particle density.

So the state with mean particle density being equal to $c$ is not necessarily an eigenstate of the Hamiltonian.


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