# Density operator commutator in master equation

In Breuer and Petruccione's Theory of Open Quantum Systems Eq. 3.359, they arrive at the master equation for the evolution of a system's density matrix, after interaction with a pointer via a typical measurement interaction Hamiltonian $$H = \theta \hat{A} \hat{Q}$$ where $$\hat{A}$$ acts on the system subspace, $$\hat{Q}$$ is the position of the pointer. The master equation is, \begin{align} \frac{d}{dt}\rho(t) &= - i [ H, \rho(t) ] - \frac{1}{2} \sigma_Q^2 [ A , [ A , \rho(t) ]] \end{align} where $$\sigma_Q^2$$ is the variance of the $$\hat{Q}$$ observable of the pointer.

They associate the second term on the right-hand side to backaction on the system induced by the interaction. I am interested in computing this term for say, a continuous variable system (they only provide an example for qubits). Is this possible to compute in practice? Or is it more of a formal statement in principle?

For example say $$A = X$$, the position of the system and $$\rho(t) = \exp ( - i H t ) \rho(0) \exp ( i H t)$$ where $$H = \hslash \omega a^\dagger a$$ is the free Hamiltonian for the system and we prepare the system in a coherent state $$\rho(0) = | \alpha \rangle\langle \alpha |$$. But simply inspecting the resulting commutator, it looks incredibly painful to compute due to the fact that each of the three operators in $$\rho(t)$$ do not commute with $$X$$. To begin with one can of course use some commutator identities to split $$[X , \rho(t) ]$$ into a sum of three commutators, but this quickly gets out of hand once computing the commutator for e.g. between $$[X , e^{-i H t}]$$ (I used this identity to begin to tackle this).

EDIT: perhaps a sub question of this is, is $$[ X, | \alpha \rangle\langle \alpha | ]$$ (which will be one of the terms in the nested commutator) possible to compute in a closed form? I am new to this business of working with these kinds of evolution equations so help in the right direction is much appreciated.

In general, you will probably want to look into the Gaussian formalism for continuous-variable states and the rotating frame here. But we can solve directly for this particular problem. In the present construction, $$\rho(t)=\exp(-iHt)|\alpha\rangle\langle \alpha|\exp(i Ht)$$ remains pure (note that the time evolution used $$\hbar=1$$ and the Hamiltonian left it explicit, so I will cancel them appropriately): \begin{align} |\psi(t)\rangle&=e^{-i \omega t a^\dagger a}|\alpha\rangle\\ &=e^{-i \omega t a^\dagger a}e^{-|\alpha|^2/2}\sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}}|n\rangle\\ &=e^{-|\alpha|^2/2}\sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}}e^{-i \omega t n}|n\rangle\\ &=|\alpha e^{-i\omega t}\rangle. \end{align} This time evolution simply rotates a coherent state around phase space with angular frequency $$\omega$$.
Then, one must compute $$[X,|\alpha(t)\rangle\langle \alpha(t)|]$$ etc., where $$\alpha(t)=\alpha e^{-i\omega t}$$. This is done by writing $$X$$ in terms of creation and annihilation operators as $$X=(a+a^\dagger)/\sqrt{2}$$ (normalization conventions vary). Then \begin{align} [X,\rho(t)]=\frac{(\alpha(t)+a^\dagger) |\alpha(t)\rangle\langle\alpha(t)|- |\alpha(t)\rangle\langle \alpha(t)|(\alpha(t)^*+a)}{2}. \end{align} You will notice that $$a^\dagger |\alpha\rangle$$ is no longer a coherent state, so things get complicated. It will generally not be true that $$\rho(t)$$ remains pure; that only holds for the unitary evolution.
So, what's a better strategy? Fokker-Planck equations! Your exact evolution is an example on Wikipedia and I bet it's in Breuer and Petruccione somewhere. What one does is replaces the density operator by a function over phase space that retains the same information as the state (a "quasiprobability distribution"), notices that $$a$$ and $$a^\dagger$$ acting on the state from the right and the left lead to particular changes in the phase-space functions, then one has some differential equations that can be solved. That is a sometimes-used method of dealing with commutators like $$[X,\rho]$$ for arbitrary states.
• I agree with the recommendation to look at the Wigner function. However, I wanted to point out that it does not make much sense to plug in $\rho(t) = \exp ( - i H t ) \rho(0) \exp ( i H t)$ into that master equation and try to compute the right hand side, since that does not actually solve the master equation. Commented Jul 23 at 7:26
• Hi @Noiralef thanks for the input. I don't fully understand your comment. Doesn't one have to compute the right hand side to obtain the resulting differential equations for $\rho(t)$ (if one wanted to solve for $\rho(t)$)? In any case, I am less interested in actually solving the equation and moreso just seeing the presence/form of the back action term that arises due to the measurement, and understanding under what scenarios it might vanish. Commented Jul 23 at 12:11
• That makes sense @Quantum Mechanic -- I was a bit puzzled as to what to do with $a^\dagger$ acting on the state e.g. whether I needed to trace or compute some observable (which computing a Wigner function effectively does). Commented Jul 23 at 12:14
• @Noiralef agreed. That's why I said generally $\rho$ does not remain pure. The real usefulness of the rotating frame is to solve the problem in an interaction picture where this rotation is separated from the rest of the problem. Ie, I agree that in general $\rho(t)$ is not merely a unitary update of $\rho(0)$ via the Hamiltonian (that's what the master equation is telling us - things aren't unitary!) Commented Jul 23 at 13:29
• @j.foobles I don't know what you mean by "obtain the resulting differential equations for $\rho(t)$. You already have a differential equation for $\rho(t)$. The problem is that the expression $e^{-iHt} \rho(0) e^{iHt}$ doesn't solve this differential equation, so that expression is not equal to $\rho(t)$, so it is unclear why we would substitute that expression for $\rho(t)$. (A good question, and perhaps what you really meant to ask, is: if $\rho(0)$ is something like $|\alpha\rangle\langle\alpha|$, then what is / how do we evaluate $\dot \rho(0)$ and what does it tell us?) Commented Jul 23 at 13:31