# Proof the solution of von Neumann equation will fluctuate forever if Hamiltonian and initial density matrix does not commute

Given von Neumann equation $$\frac{d}{dt} \rho(t) = -i [H, \rho(t)] = -i e^{-iHt}[H, \rho(0)]e^{iHt}.$$

If we know that $[H, \rho(0)] \neq 0$, how do we prove that the solution will fluctuate forever. (stabilization means the fluctuation approaching to zero, but the derivative can still be large, so here, we want to show that it will endlessly oscillate and never be damped.)

Mathematically, $H$ is an Hermitian matrix and $\rho(t)$ is a square matrix with trace one, positiveness and self-adjoint.

We need a mathematical proof that particularly by using the fact that $[H, \rho(0)] \neq 0$, i.e. $H\rho(0)\neq\rho(0)H$

In the eigenbasis of $H$ the density matrix has elements

$$\rho_{nm}(t)=\rho_{nm}(0)\exp[i(E_m-E_n)t]$$

where $E_n$ are the eigenvalues of $H$; since you assume that $\rho(0)$ does not commute with $H$, there are nonzero matrix elements $\rho_{nm}(0)$ with $n\neq m$. So these matrix elements will oscillate forever, without any damping.

• Thank you for the answer. When it comes to the form with the energy eigenvalues $\exp(-i(E_m - E_n))t$, is it still possible to consider it as an operator such that $U'(t, 0) = \exp(-i(E_m - E_n))t \neq I$, where $I$ denotes the identity operator. Or we can only simply say $\exp(-i(E_m - E_n))t \neq 1$ ? – Xingdong Zuo Aug 22 '14 at 13:30
• $\rho(t)=U(t)\rho(0)U^{\dagger}(t)$ where $U(t)=e^{-iHt}$ has matrix elements $e^{-iE_n t}\delta_{nm}$ in the eigenbasis of $H$. – Carlo Beenakker Aug 22 '14 at 14:21

I'm not sure I fully understand what you want, but here is at least a partial answer. You have $$\det\left(\frac{d}{dt} \rho(t)\right) = (-i)^n \det(e^{-iHt}[H, \rho(0)]e^{iHt}) = (-i)^n \det([H, \rho(0)]),$$ so that the determinant of the derivative is a nonzero constant. Thus $\frac{d}{dt} \rho(t)$ is uniformly bounded away from the zero matrix.

A simple calculation gives $\frac{d^2}{dt^2}\rho(t)=(-i)^2e^{-iHt}[H,[H,\rho(0)]]e^{iHt}$. Since $\frac{d}{dt} \rho(t)$ is bounded away from zero and $\frac{d^2}{dt^2} \rho(t)$ is bounded, $\rho(t)$ cannot have a limit. In fact, you can establish a lower limit on its "oscillation" (assuming this is the kind of oscillation you want).

The oscillation estimate: Suppose $f:\mathbb R\to\mathbb R^N$ is a $C^2$ function so that $|f'|\geq a>0$ and $|f''|\leq A<\infty$ for some constants $a$ and $A$. (This applies in particular to your $\rho$.) By the fundamental theorem of calculus $$f(T+\tau)-f(T) = \tau f'(T)+\int_0^\tau\int_0^sf''(r)drds.$$ Thus for $\tau>0$ $$|f(T+\tau)-f(T)| \geq \tau|f'(T)|-\int_0^\tau\int_0^s|f''(r)|drds \geq a\tau-\frac12A\tau^2.$$ Therefore $|f(T+\tau)-f(T)|$ has a bound from below uniformly in $T$ if $\tau<2a/A$. I could take this behavior as a definition of endless oscillation.

• Thanks for the answer. Yes, we can show that the 'general' derivative is away from zero. Well, but we'd like to know that the solution will fluctuate forever(up and down behaviors, intuitively in graphs), because there can be case that the derivative is nonzero, but the solution is already in an equilibrium (not fluctuate any more). – Xingdong Zuo Aug 22 '14 at 9:07
• @XingdongZuo, can you be more specific what you want? I cannot give a rigorous proof for a vague claim. I understood from your claim that you want the limit $\lim_{t\to\infty}\rho(t)$ not to exist. I extended my answer a bit to show this and more. – Joonas Ilmavirta Aug 22 '14 at 11:12