Why does implementing the annihilation operator $\hat a$ not allow action at a distance?

Question

I was reading a paper (link) in which they implement action at a distance using the annihilation operator, and in it they say that "this observation does not imply superluminal signaling because photon annihilation is not a unitary operation, and as such can be realized only probabilistically".

EDIT: A summary of the theoretical part of the paper in a few words: There's an initial state of $|\psi>|{0}>$ entering a Mach-Zehnder inteferometer. After the first beamsplitter, they apply the annihilation operator to one of its output leg (Alice - $\hat{a}_A$). Since it can be written as a linear combination of the input ports, it is proportional to the mode $\hat{a}$ in which $|\psi>$ was introduced. This implies that a local action of the annihilation operator actually affects the total state (unlocal). Then, they say that "the action at a distance of the photon annihilation operator can be made explicit by observing its effect on the mean number of photons in Bob's mode (the other output port - $\hat{a}_B$)". They give the example of $|\psi>$ being a Fock state $|N>$, and once Alice applies $\hat{a}_A$ on the state, the state in mode $\hat{a}$ becomes $|N-1>$, and so the mean number of photons in Bob's mode changes to $\propto|N-1>$. Afterwards, they claim the sentence I'd like to understand.

I'm not really sure what they mean by that and would be happy if someone could explain this statement to me.

Thanks in advance.

Answer 1

The idea is that given any state $|\psi\rangle$, any physical evolution corresponds to applying a unitary operation $U$ to $|\psi\rangle$.

This means that, if $A$ is not a unitary operator, then "applying $A$ to $|\psi\rangle$" is not a physical operation: there is no physical scenario that can send $|\psi\rangle$ to $A|\psi\rangle$ for all $|\psi\rangle$ (essentially, because this would be a non-reversible operation, and thus incompatible with quantum mechanics).

There is, however, a catch to this statement. It is possible to effectively obtain an evolution $|\psi\rangle\mapsto A|\psi\rangle$ for $A$ non-unitary, if you neglect part of the information about the output. In other words, if you "forget" some of the output modes, you can get an effective evolution corresponding to a non-unitary operator $A$.

This is rather easy to see. Consider some state $|\psi\rangle=\sum_k c_k|k\rangle$, and some unitary $U$. If you only consider a subset $S$ of the output modes, you get the effective evolution $|\psi\rangle\mapsto N\mathbb P_S U|\psi\rangle$, where $\mathbb P_S\equiv\sum_{k\in S}|k\rangle\!\langle k|$ projects onto the modes in $S$ and $N$ is a normalisation constant. Such evolution is not unitary in general, essentially because the submatrix of $U$ containing the rows in $S$ will not be unitary/have orthogonal columns.

So going back to the case at hand: there is no way to implement an evolution of the form $|\psi\rangle\mapsto\hat a|\psi\rangle$, as $\hat a$ is not unitary. However, one can implement some unitary evolution on the state, and then just look at the subset of the outputs that corresponds to having one less photon in the state. This is exactly what they do, using the "click" on the detector as witness that the state now has one less photon, and therefore we effectively obtained the mapping $|\psi\rangle\mapsto\hat a|\psi\rangle$.

You cannot "communicate" information with this trick, because you are not really "applying the operation $\hat a$ to the state", if anything, you are waiting for the operation $\hat a$ to happen to $|\psi\rangle$, and then consider only the states obtained when this happens (we usually talk of post-selection in this case). Because you are not actively steering the state, you are not putting into the system any amount of information to be transferred.