Density operator commutator in master equation

Question

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.

Answer 1

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.