The von-Neumann equation reads:

$$\frac{d\rho}{dt} = -\frac{i}{\hbar}[H,\rho]$$

The solution:

$$\rho(t)=U\rho U^{\dagger}$$ with $U=e^{-\frac{iHt}{\hbar}}$ is easily obtained when starting from the Schrödinger equation, but I want to obtain the same result starting from the von-Neumann equation.

My approach is to start by integrating both sides:

$$ \rho(t) = \rho_0-\frac{i}{\hbar}\int_0^t[H,\rho(\tau)]d\tau $$

Substituting the left hand side into the right yields:

$$ \rho(t) = \rho_0-\frac{i}{\hbar}\int_0^t[H,\rho_0-\frac{i}{\hbar}\int_0^t[H,\rho(\tau')]d\tau']d\tau $$


$$ \rho(t) = \rho_0-\frac{i}{\hbar}\int_0^t[H,\rho_0]+\frac{i}{\hbar}\int_0^t[H,\frac{i}{\hbar}\int_0^t[H,\rho(\tau')]d\tau']d\tau $$

$$ \rho(t) = \rho_0 + \frac{it}{\hbar}(\rho H - H\rho)+\frac{i}{\hbar}\int_0^t[H,\frac{i}{\hbar}\int_0^t[H,\rho(\tau')]d\tau']d\tau $$

The first order terms are the same as one gets by expanding $U\rho U^{\dagger}$:

$$\rho(t) \approx \Big(1-\frac{iHt}{\hbar}\Big)\rho\Big(1+\frac{iHt}{\hbar}\Big)=\rho_0 + \frac{it}{\hbar}\Big(\rho H - H\rho\Big) $$

However, with the second order terms I run into trouble, because from $U\approx \Big(1-\frac{iHt}{\hbar}+\frac{1}{2}\Big(\frac{iHt}{\hbar}\Big)^2\Big)$ I expect them to be:

$$ -\frac{iHt}{\hbar}\rho\frac{iHt}{\hbar} + \rho \frac{1}{2}\Big(\frac{iHt}{\hbar}\Big)^2 + \frac{1}{2}\Big(\frac{iHt}{\hbar}\Big)^2 \rho $$

but in the recursive expression I wrote down the $n$-th order term is just going to be $n$ nested commutators with $H$, so my 2nd order term ($[H,[H,\rho]]$) is:

$$ -2\frac{iHt}{\hbar}\rho\frac{iHt}{\hbar} + \rho\Big(\frac{iHt}{\hbar}\Big)^2 + \Big(\frac{iHt}{\hbar}\Big)^2 \rho $$

which is a factor of two wrong. The 3rd order term will have the form of the 4th row of pascal's triangle and so on, and the $n$-th order term will be $n!$ too big in my expression. This is where I'm stuck, because I don't see any way of getting a factorial term, but I also don't see anything obviously wrong with my approach.

If anyone could point out where my error is I'd be very grateful.

(Just now, I came across the following identity:

$$ e^A\rho e^A = \rho + [A,\rho] + \frac{1}{2!}[A,[A,\rho]] + \frac{1}{3!}[A,[A,[A,\rho]]] + ... $$

which is exactly what I expect, but I don't know how to get the $\frac{1}{n!}$ factors)

  • $\begingroup$ I see someone added the homework tag, so I just want to clarify that this isn't a homework question. It's just something I thought about in the context of a different problem. $\endgroup$
    – fulis
    Commented Jul 31, 2018 at 17:25
  • 2
    $\begingroup$ Hi fulis. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$
    – Qmechanic
    Commented Jul 31, 2018 at 17:26
  • 1
    $\begingroup$ Your error is right after your "Substituting the left hand side into the right yields". The upper limit of the inside integral is $\tau$, not t. $\endgroup$ Commented Jul 31, 2018 at 19:40
  • $\begingroup$ Looking at it, I think you're right! That would solve it. Thanks. $\endgroup$
    – fulis
    Commented Jul 31, 2018 at 22:02

1 Answer 1



  1. Note that the $n$'th term in the Dyson series is nested integrals over an $n$-simplex integration region, cf. above comment by Cosmas Zachos. If we time-order and normalize with a $1/n!$ factor, we can replace the integration region with an $n$-box.

  2. OP's last equation reads $$e^{{\rm ad}\hat{A}}\hat{B}~=~e^{\hat{A}}\hat{B}e^{-\hat{A}}, \tag{1}$$ where $$ {\rm ad}\hat{A}~\equiv~[\hat{A},\cdot]. \tag{2}$$

    Sketched proof of eq. (1):

    • Replace $\hat{A}\to t\hat{A}$ in eq. (1), where $t\in\mathbb{C}$ is a parameter.

    • Differentiate wrt. $t$.

    • Show that the LHS & the RHS of eq. (1) satisfy the same ODE in $t$.


  • $\begingroup$ 1. I think this is a slightly different problem, because I was trying to derive the solution, not just verify it. 2. Anyway, regarding this answer: does the uniqueness theorem for ODEs apply when we have operator valued functions? I would feel a bit uneasy showing the equality this way. $\endgroup$
    – fulis
    Commented Jul 31, 2018 at 23:21
  • $\begingroup$ 1. I updated the answer. 2. That is a valid mathematical concern. Consider asking it on Mathematics. $\endgroup$
    – Qmechanic
    Commented Aug 1, 2018 at 8:34

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