# Liouville von Neumann equation in interaction picture

The Liouville von Neumann equation is

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

Now, the hamiltonian is

$$H= H_0+H_1$$

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

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

Then

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

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 how come that the Liouville von Neumann equation is 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?

• 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. Feb 16 '17 at 12:43
• 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! Feb 16 '17 at 15:42

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 $$\mathbb{1}=U(t)^{\dagger}U(t)$$. Starting from your last line of the derivation and using $$U(t)=\exp(iH_0t)$$:

\begin{align} \partial_t \rho_I(t) &= -i U(t)[H_1,\rho]U(t)^{\dagger} \\ &= -i U(t)H_1\rho U(t)^{\dagger} -i U(t)\rho H_0U(t)^{\dagger} \\ &= -i U(t)H_1 U(t)^{\dagger}U(t)\rho U(t)^{\dagger} -i U(t)\rho U(t)^{\dagger}U(t)H_0 U(t)^{\dagger} \\ &= -i [H_1(t),\rho_I(t)] \end{align} where $$H_1(t)=U(t)H_1 U(t)^{\dagger}.$$

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