How to take derivative of density operator?

Question

I was just trying to confirm to myself that the following density operator

$$\rho(t) = e^{-iHt/\hbar} \rho(0) e^{iHt/\hbar}$$

fulfills the Liouville-von Neumann equation:

$$\frac{d}{dt}\rho(t) = - \frac{i}{\hbar} [H,\rho(t)]$$ where $[H,\rho(t)]$ denotes the commutator of the Hamiltonian with the density operator here. I only have to take the derivative of $\rho(t)$ in order to plug it in, but although the answer is probably trivial, I am struggling at the moment as I don’t really know how to take the derivative of this expression (there are several operators acting here, so I am not sure how to apply the product rule here). Can someone please help me?

Answer 1

The product rule works essentially as you think it might. You just must be careful about the commutation of of operators. For example, $$\frac{d}{dt}\big(AB \big) = \left(\frac{d}{dt}A\right) B + A \left(\frac{d}{dt}B\right) \neq \left(\frac{d}{dt}A\right) B + \left(\frac{d}{dt}B\right)A$$

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If you would like a rigorous description of operator differentiation, we need to be more precise. Let $A:t \mapsto A(t)$ be a family of operators on some (for now, finite-dimensional) Hilbert space $\mathscr H$ which are indexed by a continuous variable $t$. The derivative $A'(t)$ is the operator such that, for arbitrary $\psi\in \mathscr H$,

$$A'(t) \psi = \lim_{\epsilon\rightarrow 0} \frac{A(t+\epsilon)\psi - A(t)\psi}{\epsilon}$$

assuming that this limit exists. Alternatively, $A'(t)$ is the operator such that $$\lim_{\epsilon_{\rightarrow 0}} \left\Vert \big[A(t+\epsilon)- A(t) - \epsilon A'(t)\big]\psi \right\Vert\rightarrow 0$$ for all $\psi\in \mathscr H$. Therefore, as a computational tool we may take an expression which depends on $t$, substitute $t\rightarrow t+\epsilon$, make first order replacements of the form $f(t+\epsilon) = f(t) + \epsilon f'(t)$, and then read off the term proportional to $\epsilon$ at the very end.

For example, $$A(t)B(t) \rightarrow A(t+\epsilon)B(t+\epsilon) \rightarrow \big(A(t) + \epsilon A'(t) \big) \big( B(t) + \epsilon B'(t)\big)$$ $$\rightarrow A(t) B(t) + \epsilon \underbrace{\bigg(A'(t) B(t) + A(t) B'(t)\bigg)}_{\frac{d}{dt} A(t) B(t)} + \epsilon^2 A'(t)B'(t)$$

Answer 2

This is just a product rule.

Take $\rho(t) = e^{-iHt/\hbar} \rho(0) e^{iHt/\hbar}$ and use the product rule:

$$\frac{d}{dt}\rho(t) = \frac{d}{dt}[e^{-iHt/\hbar} \rho(0) e^{iHt/\hbar}]$$

This is just:

$$=\frac{d e^{-iHt/\hbar}}{dt} \rho(0) e^{iHt/\hbar} + e^{-iHt/\hbar}\rho(0)\frac{d e^{iHt/\hbar}}{dt} $$

Which is:

$$ =\frac{-iH}{\hbar}e^{-iHt/\hbar} \rho(0) e^{iHt/\hbar} + e^{-iHt/\hbar}\rho(0)e^{iHt/\hbar}\frac{iH}{\hbar} $$

Which by definition of $\rho(t)$ is just:

$$=\frac{-iH}{\hbar}\rho(t) + \frac{iH}{\hbar}\rho(t) = \frac{-i}{\hbar}[H,\rho(t)]$$

Which is indeed what you wanted!