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?

  • $\begingroup$ Reading all the comments you have written, I think the best would be to derive the product rule yourself. See e.g. the linked answer (somewhere in the comments). $\endgroup$ Dec 2, 2023 at 12:58

2 Answers 2


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

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

  • $\begingroup$ Thank you for that clarification. However, how would it work (step -by-step) in this case? $\endgroup$
    – Physchem16
    Dec 2, 2023 at 1:25
  • $\begingroup$ @Physchem16 To turn the question around, is there a particular step you feel like you're struggling with? The result is almost immediate, so if something seems wrong then there may be a more fundamental misunderstanding at play. $\endgroup$
    – J. Murray
    Dec 2, 2023 at 4:10
  • $\begingroup$ @J.Murray If A and B do not commute AB ≠ BA → A'B+AB' ≠ B'A + BA'. Could it look like this too? You seem to have replaced B'A with A'B. Does it not matter in difference in showing the non-equality? $\endgroup$ Dec 2, 2023 at 5:19
  • 1
    $\begingroup$ @Proscionexium No, it doesn't matter. Sometimes people learn the product rule as $(fg)' = f'g + g' f$, my point was merely that with non-commuting operators this is not the right ordering. $\endgroup$
    – J. Murray
    Dec 2, 2023 at 5:24
  • 1
    $\begingroup$ @Physchem16 1) It doesn't matter, because $[H, e^{iHt/\hbar}]=0$. 2) From a heuristic standpoint, the composition of operators is analogous to ordinary multiplication (other than the general lack of commutivity). If you pretend that all of these quantities are numbers rather than operators (while preserving the correct ordering), then you will get the correct answer. $\endgroup$
    – J. Murray
    Dec 2, 2023 at 16:46

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!

  • $\begingroup$ Thank you for that elaborate answer. Two things I have to ask about that: $\endgroup$
    – Physchem16
    Dec 2, 2023 at 12:48
  • $\begingroup$ First, in the last line of your answer, I think \rho and H should be swapped (\rho should come before H, i.e. \rho should act on H, in order to yield the commutator), shouldn’t it? Secondly, why exactly can I just take the derivative of the second exponential? Isn’t there an operator (\rho(0)) still acting on that function? That’s essentially the step that is not clear to me. $\endgroup$
    – Physchem16
    Dec 2, 2023 at 12:56
  • $\begingroup$ @Physchem16 Your comment is correct, $\rho$ and $H$ should be swapped. However, because you've mentioned this several times I think it's worth emphasizing that it's probably best not to think of the operators acting on each other, but rather being composed with one another such that they act sequentially on some arbitrary vector. This can be an important distinction - when in doubt, put a test vector on the right-hand side. $\endgroup$
    – J. Murray
    Dec 2, 2023 at 19:12
  • $\begingroup$ @J.Murray Thanks a lot, now it got clear to me! $\endgroup$
    – Physchem16
    Dec 3, 2023 at 16:11

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