The following general master equation (from this paper 'Dynamical quantum correlations of Ising models on arbitrary lattice and their resilience to decoherence') describes the various types of decoherence relevant to trapped ions, Rydberg atoms and condensed matter systems \begin{equation} \dot{\rho} = i \mathscr{H}(\rho) - \mathscr{L}_{ud}(\rho)-\mathscr{L}_{du}(\rho)-\mathscr{L}_{el}(\rho)~~~~~~~~~~~~~(1) \end{equation} where $$\mathscr{H}(\rho) = [\mathcal{H},\rho],\\ \mathscr{L}_{ud}(\rho) = \frac{\Gamma_{ud}}{2}\sum_{j}(\hat{\sigma}^{+}_j\hat{\sigma}^{-}_{j}\rho + \rho \hat{\sigma}^{+}_{j}\hat{\sigma}^{-}_{j}-2\hat{\sigma}^{-}_{j}\rho \hat{\sigma}^{+}_{j})\\ \mathscr{L}_{du}(\rho) = \frac{\Gamma_{ud}}{2}\sum_{j}(\hat{\sigma}^{+}_j\hat{\sigma}^{-}_{j}\rho + \rho \hat{\sigma}^{+}_{j}\hat{\sigma}^{-}_{j}-2\hat{\sigma}^{-}_{j}\rho \hat{\sigma}^{+}_{j})\\ \mathscr{L}_{el}(\rho) = \frac{\Gamma_{el}}{8}\sum_{j}(2\rho-2\hat{\sigma_j^z}\rho\hat{\sigma}_j^z) $$ The first term involving a commutator describes coherent evolution due to the Ising interaction, and the various terms having subscripts ‘ud’, ‘du’ and ‘el’ correspond respectively to spontaneous relaxation, spontaneous excitation and dephasing. Equation (1) has the formal solution $\rho(t) = \mathscr{U}(t)\rho(0)$, with $$\mathscr{U}(t) = \text{exp}[-t(i\mathscr{H + \mathscr{L}_{ud}+\mathscr{L}_{du} + \mathscr{L}_{el})}]~~~~~~~~~~~~~~~~~(2)$$ An immediate simplification follows from the observation that $$[\mathscr{L}_{el}, \mathscr{H}] = [\mathscr{L}_{el},\mathscr{L}_{du}] = [\mathscr{L}_{el}, \mathscr{L}_{ud}] = 0~~~~~~~~~~~~~~~~~~(3)$$ That the last two commutators vanish is less obvious than the firsts, but physically it has a very clear meaning: spontaneous relaxation/excitation on a site $j$ causes the $j$'th spin to have a well defined value of $\sigma^{z}_j$, and thus to be unentangled with the rest of the system. Since the dephasing jump operator $\hat{\sigma}_{j}^{z}$ changes the relative phase between the states $| \sigma_{j}^{z} = \pm 1 \rangle$, whether spontaneous relaxation/excitation occurs before or after a dephasing event only affects the sign of the overall wave function, which is irrelevant.

As a result, we can write $$\mathscr{U}(t) = e^{-t\mathscr{L}_{el}}e^{-t(i\mathscr{H} + \mathscr{L}_{ud} + \mathscr{L}_{du})}~~~~~~~~~~~~~~(4)$$

Question: I might be missing something simple, but can anyone see why equation (2) reduces to equation (4) as result of the commutativity in equation (3)?


1 Answer 1


It's fundamentally similar to/the same as Baker–Campbell–Hausdorff (BCH).

Generally, in quantum physics, this is most often used (or at least taught) with commuting Hamiltonians: $$e^{-i\left(H_1+H_2\right)t} = \sum_{n=1}^\infty\frac{\left(-it\right)^n}{n!}\left(H_1+H_2\right)^n = e^{-iH_1t}e^{-iH_2t}e^{\frac{1}{2}\left[H_1, H_2\right]t^2}\cdots$$ where the terms of order $e^{t^3}$ and higher (represented by '$\cdots$') all involve the commutator $\left[H_1, H_2\right]$, so if $\left[H_1, H_2\right] = 0$, this becomes $$e^{-i\left(H_1+H_2\right)t} = e^{-iH_1t}e^{-iH_2t} .$$

Similarly, as this is a general mathematical result in Lie algebra (and not at all specific to Hamiltonians), if $$[\mathscr{L}_{el}, \mathscr{H}] = [\mathscr{L}_{el},\mathscr{L}_{du}] = [\mathscr{L}_{el}, \mathscr{L}_{ud}] = 0,$$ then $$[\mathscr{L}_{el}, \mathscr{H}+\mathscr{L}_{du}+\mathscr{L}_{ud}] = 0$$ and we can write \begin{align*}\mathscr{U}(t)\rho &= \exp[-t(i\mathscr{H + \mathscr{L}_{ud}+\mathscr{L}_{du} + \mathscr{L}_{el})}]\rho \\ &= \exp[-t(i\mathscr{H + \mathscr{L}_{ud}+\mathscr{L}_{du}) -\mathscr{L}_{el}t}]\rho \\ &= \exp[-t(i\mathscr{H + \mathscr{L}_{ud}+\mathscr{L}_{du})]\exp[-\mathscr{L}_{el}t}]\exp[\frac{1}{2}\left[i\mathscr H + \mathscr L_{ud} + \mathscr L_{du}, \mathscr L_{el}\right]t^2]\cdots\rho \\ &= \exp[-t(i\mathscr{H + \mathscr{L}_{ud}+\mathscr{L}_{du})]\exp[-\mathscr{L}_{el}t}]\rho\end{align*} and as this is true $\forall\,\rho$, $$\mathscr{U}(t) = e^{-t\mathscr{L}_{el}}e^{-t(i\mathscr{H} + \mathscr{L}_{ud} + \mathscr{L}_{du})} = e^{-t(i\mathscr{H} + \mathscr{L}_{ud} + \mathscr{L}_{du})}e^{-t\mathscr{L}_{el}}$$ (again, due to the same commutation relation, where $e^{-t\mathscr{L}_{el}}$ is doesn't matter). While this may initally look and feel weird as Master equations are superoperators, the underlying maths remains fundamentally unchanged as per e.g. Machnes and Plenio.

  • $\begingroup$ Thanks for a very complete answer. Just one further query, the equation which follows in the paper, namely equation (38) $$e^{-t \mathscr{L}_{el}}\hat{\sigma}^{x,y}_{j}= e^{-\frac{\Gamma_{el}t}{2}}\hat{\sigma}^{x,y}_{j}, ~e^{-t\mathscr{L}_{el}}\hat{\sigma}^{z}_{j}=\hat{\sigma}_{j}^{z}~~~~~~~~(38)$$ is accompanied by the statement "The effect of the time evolution due to $\mathscr{L}_{el}$ can be understood by considering its effect on the Pauli operators". Do you know why would this be sufficient to describe the time evolution? $\endgroup$
    – John Doe
    Commented Sep 13, 2019 at 10:42
  • 1
    $\begingroup$ I assume that, as you can write the density matrix of a two level system as a sum of Pauli matrices, you can do this, calculate the effect on the Paulis (evidently easier than on the entire density matrix), then sum them back together to get the new density matrix $\endgroup$
    – Mithrandir24601
    Commented Sep 13, 2019 at 10:51
  • $\begingroup$ Yeah I see, it's like describing how basis elements transform, as a way to describe how any arbitrary two dimensional vector transforms. $\endgroup$
    – John Doe
    Commented Sep 13, 2019 at 12:10

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