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I have a question about section 5 in 'Statistical mechanics' (Pathria).

According this book, the density matrix (operator) should satisfy the following identity, which describes the time evolution of density matrix: $$i\hbar \dot{\hat{\rho}} =\left[\hat H,\hat \rho \right]$$ After this equation, the book says, if the density matrix is stationary ($\dot{\hat \rho}=0$),

  1. density matrix operator should commute with Hamiltonian and

  2. Hamiltonian has no time dependence.

I think that the condition is too strong. Can we show that (1) & (2) is 'necessary and sufficient condition' for stationary condition of density matrix? That is, the stationary condition ($\dot{\hat \rho}=0$) always leads to (1)&(2)?

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  • $\begingroup$ I updated the last line at the main post, my question was: the stationary condition always leads to (1)&(2)? $\endgroup$
    – momo
    Commented Nov 15, 2022 at 10:55
  • $\begingroup$ I don't think that if $\rho$ commutes with $H(t)$ then $H$ is time-independent. Indeed, if $\rho(t)=\mathbb I/d = \rho(t_0)$ where $d:=\mathrm{dim}\,h$, where $h$ is the (finite-dimensonal, complex) Hilbert space in question, then $\rho$ commutes with every operator and in particular with $H(t)$, whatever this then may be. However, if the equation in the question holds, then if $\dot \rho =0$, it follows that $[\rho,H(t)]=0$ for all $t$. Conversely, if $\rho$ commutes with $H(t)$ at all times, then $\dot \rho =0$. $\endgroup$
    – Tobias Fünke
    Commented Nov 15, 2022 at 12:01
  • $\begingroup$ Related if not duplicate $\endgroup$
    – Tobias Fünke
    Commented Nov 15, 2022 at 12:13
  • $\begingroup$ I think so, too. Thank you for your comments. $\endgroup$
    – momo
    Commented Nov 15, 2022 at 12:50

1 Answer 1

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I'd say the book is correct. Given that $i\hbar \partial_t{\hat{\rho}} =\left[\hat H,\hat \rho \right]$ and $\partial_t{\hat{\rho}} =0$, we have:

  1. $\left[\hat H,\hat \rho \right]=0$ so the Hamiltonian and the density matrix commute.

  2. One sufficient condition to ensure the commutation is that the density is a functional of the Hamiltonian, $\rho=\rho[\hat H]$.

  3. Then you can check by the chain rule that $\partial_t{\hat{\rho}[\hat H]} =0$ implies that $\partial_t\hat H =0$ (this is, the Hamiltonian does not depend explicitly on time).

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  • $\begingroup$ But your condition $2.$ is an extra assumption, no? See for example my comment under the question. $\endgroup$
    – Tobias Fünke
    Commented Nov 15, 2022 at 11:55
  • $\begingroup$ @TobiasFünke, I am not sure, but wouldn't $\left[\hat H,\hat \rho \right]=0$ imply that $\rho$ is a functional of $\hat H$? Just as any other relation $f(\hat H,\hat \rho)=0$ would. I am more used to working with classical SM, where this kind of reasoning is standard in Liouville's theorem. Sorry, I don't get your example in the comment section. $\endgroup$
    – Javi
    Commented Nov 15, 2022 at 13:17
  • $\begingroup$ I don't understand that reasoning at all. For example, $[X,S_z]=0$ - does this imply that $X$ a function(al) of $S_z$?! Where by an abuse of notation I mean $X\sim X\otimes \mathbb I_{\mathbb C^2}$ and $S_z \sim \mathbb I_{L^2} \otimes S_z$ on the Hilbert space $L^2(\mathbb R)\otimes \mathbb C^2$. Or even easier: on $L^2(\mathbb R^2)$ we have $[X,P_y]=0$; is $X$ a functional of $P_y$? I don't see it, but of course I can be very wrong here, so I really mean it as a question. $\endgroup$
    – Tobias Fünke
    Commented Nov 15, 2022 at 13:25
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    $\begingroup$ Regarding the example/comment under the question. With $\rho(t_0)=\mathbb I/d$ the initial value problem is solved by $\rho(t)=\rho(t_0)$ for all $t$ (easy to check). Now this does not, in any way, put a constraint on $H(t)$. In particular, it can be time-dependent. So your argument does not work here, as far as I can see. $\endgroup$
    – Tobias Fünke
    Commented Nov 15, 2022 at 14:35
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    $\begingroup$ Regarding our discussion, cf. this. $\endgroup$
    – Tobias Fünke
    Commented Nov 15, 2022 at 16:17

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