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In many papers about quantum optics and interferometry, it's assumed or said that "it's well known" that linear optics commutes with uniform losses. In particular if we have a beam splitter and the two input modes are affected by the same loss, we can always move them to the output modes (see e.g. Michał Oszmaniec and Daniel J Brod 2018 New J. Phys. 20 092002 for details).

However, when attempting to perform explicit calculations, I encounter difficulties in grasping how this process actually functions in practice.

Lossy optical modes are commonly modeled through a beam splitter coupling them to external environmental modes in the vacuum state: the trace over the environment returns the map acting on the mode. If I label the input modes with "$1$" and "$2$" and the environmental modes with "$0$" and "$3$" respectively, in the case of lossy input I have two beam splitters coupling $0$-$1$ and $2$-$3$ (namely $$\hat{U}_{01} = \hat{U}_{23} = \hat{W}$$ in the uniform case) and then the proper optical gate $\hat{U}_{12}$. Finally, I operate the partial trace over the modes $0$ and $3$ (or equivalently I can consider the Kraus operators $$\{\langle \mu |\hat{W}| 0 \rangle\}_{\mu}$$ for each lossy mode).

I do not get why, within this formalism, $W$ or the related Kraus operators commute with any $\hat{U}_{12}$.

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  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Jul 14, 2023 at 14:52

1 Answer 1

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The answer is easiest with input/output relations. If linear optics does $a_i\to \sum_j U_{ij}a_j$ and loss does $a_i\to \eta_i a_i$, then if all the loss parameters $\eta_i$ are the same then the overall evolution is $$a_i\to \eta \sum_j U_{ij}a_j$$ regardless of the order in which the loss and linear optical channels are applied.

You might argue that the loss channel needs another operator for normalization, really evolving as $a_i\to \eta a_i+\sqrt{1-\eta^2}v_i$, but you'll find the evolution to still be the same once you ignore the vacuum modes:

$$a_i\to \sum_j U_{ij}(\eta a_j+\sqrt{1-\eta^2}v_j)=\sqrt{1-\eta^2}v_i^\prime+ \eta\sum_j U_{ij}a_j$$ versus $$a_i\to \eta (\sum_j U_{ij}a_j)+\sqrt{1-\eta^2} v_i.$$ The only difference is that one order of operators has vacuum operators $v_i^\prime=\sum_j U_{ij} v_j$ and the other has $v_i$, but since we can do arbitrary linear mode transformations of the vacuum without changing any physics then the results of either order of transformations will always be the same.

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  • $\begingroup$ Thanks for the answer. I know that with alternative formulations (mode operators, in this case) the description might be simpler but what surprises me is that this fact is not at all obvious dealing with dynamical maps explicitly. I don't see a way to obtain the same without some assumptions on $U$ and $W$. Or I'm probably missing something. $\endgroup$
    – D.C
    Jul 14, 2023 at 19:23
  • $\begingroup$ @D.C fair, but there is no crime in using the simplest description! You could use the same evolution for the creation operators making the state and the input/output relations would tell you how the state evolves, where again you'd find that the state evolves in the same way while the environment might change with the order of operations $\endgroup$ Jul 14, 2023 at 20:08
  • $\begingroup$ I definitely agree, but checking that all the descriptions return the same result might be of interest. That's why I specified the formalism (maps) in my question without resorting to input-output relation. Thanks again for your reply. $\endgroup$
    – D.C
    Jul 14, 2023 at 20:37

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