The Liouville von Neumann equation is

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

Now, the hamiltonian is

$$ H= H_0+H_1$$

and I want to transform it into the interaction Picture, so

$$ \rho_I=\exp(i H_0 t) \rho \exp(-i H_0 t)$$.


$$\frac{\partial}{\partial t} \rho_I= i H_0 \rho_I + \exp(i H_0 t) (\frac{\partial }{\partial t} \rho) \exp(-i H_0 t) - i \rho_I H_0$$

Writing it as a commutator and plugging in the Liouville von Neumann equation, this can be written as

\begin{align} \frac{\partial}{\partial t} \rho_I &= i [H_0 ,\rho_I] + \exp(-i H_0 t) (-i [H_0+H_I, \rho(t)])\exp(i H_0 t) \\ &=i[H_0, \rho_I]-i [H_0, \rho_I]-i \exp(i H_0)[H_1, \rho_I] \exp(-i H_0 t) \\ &=-i \exp(i H_0 t) [H_1, \rho] \exp(-i H_0 t) \end{align} Now, if $[H_o, H_1]=0$, this can be cast in the form that I know

$$ \frac{\partial}{\partial t} \rho_I=-i [H_1, \rho_I] \quad (1)$$

But if they don't commute, this is not possible. So why is the Liouville von Neumann equation normally written in the form of equation (1), even though it might be possible that $H_0$ and $H_1$ don't commute? Am I making a mistake in my derivation? Is there a smarter way to derive this equation?

  • 1
    $\begingroup$ In eq1, $H_1$ should be $H_I =e^{iH_0t}H_1e^{-iH_0t}$, maybe your book has a typo, be sure to check it up. $\endgroup$
    – yangcs11
    Feb 16, 2017 at 12:43
  • $\begingroup$ Thanks for your answer! What happens then to the density matrix? Do they actually do it in the following way $-i \exp(i H_0 t )[H_1, \rho]\exp(-i H_0 t )= -i \exp(i H_0 t ) ( H_1 \exp(-i H_0 t ) \exp(i H_0 t ) \rho) \exp(-i H_0 t )+ i \exp(i H_0 t ) (\rho \exp(-i H_0 t ) \exp(i H_0 t ) H_1 )\exp(-i H_0 t )$. In this case, my question would be fully answered! $\endgroup$
    – anonymous
    Feb 16, 2017 at 15:42

1 Answer 1


Yes, it's exactly the way was said in the comments:

$H_0$ and $H_1$ don't need to commute to do the last step and retrieve eq.(1), but instead one inserts the identity of the time evolution operator $\mathbb{1}=U_0(t) U_0(t)^{\dagger}$. Starting from your last line of the derivation and using $U_0(t)=\exp(-iH_0t)$:

\begin{align} \partial_t \rho^{(I)} &= -i U_0^{\dagger} [H_1,\rho] U_0 \\ &= -i U_0^{\dagger} (H_1 \rho - \rho H_1) U_0 \\ &= -i U_0^{\dagger} (H_1 U_0 U_0^{\dagger} \rho - \rho U_0 U_0^{\dagger} H_1) U_0 \\ &= -i [U_0^{\dagger} H_1 U_0, U_0^{\dagger} \rho U_0] \\ &= -i [H_1^{(I)},\rho^{(I)}] \\ \end{align} where $H_1^{(I)} = U_0^{\dagger} H_1 U_0$.

Sorry for the redundancy and late answer, I just had the same problem and hope others will find a definite answer quicker with this.


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