# Counterfactual communication - Photons absorbed by Bob make it non-counterfactual?

I have found this paper claiming that information can be transmitted between two parties without needing to send any physical particles (or energy). This is called quantum "counterfactual" communication.

A similar thing can be achieved classically; this article makes the connection:

If there were a major crime being committed the Bat-Signal would appear in the sky, and so the Bat-Signal’s absence counterfactually communicates to Bruce Wayne that all is well. Whenever we receive information from a sign’s absence we are being signalled to counterfactually (e.g. the signal that an engine’s components are functioning as they should is that the warning light is off).

The quantum variant of this needs no energy to send either a 0 bit or a 1 bit, which is astonishing.

But looking at the paper, I don't quite understand why it is claiming to be counterfactual.

This diagram is being used to represent the protocol. Bob can "send" Alice a 0 bit by blocking their communication channel, and can send a 1 bit by keeping it open.

I have two primary concerns with this:

1. Bob, by blocking the communication channel, absorbs the photons being sent through it. Why does this not count as an energy transfer?
2. The photons are being sent to Bob through the channel. He is just quickly reflecting them with his mirror, isn't he?

I haven't fully read the paper yet; the topic goes a bit above what I'm used to. If I'm completely wrong, I'd appreciate if someone could correct me. Thanks for any help.

I would recommend you to have a look at the paper by Salih et al. in which this particular protocol was first proposed. I'm going to use a figure from that paper to explain the counterfactuality of the protocol.

In subfigure (a) you have a sequence of concatenated Mach-Zehnder interferometers. The reflectivity of the beam-splitters of these interferometers is chosen to be

$$R = \cos^2 \frac{\pi}{2N},$$ where $$N$$ is the number of beam splitters. The unitary transformation of a single beam-splitter can then be expressed as

$$U = \mathrm{exp}\Bigl(i\frac{\pi}{2N}\sigma_x\Bigr),$$

where

$$\sigma_x = \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}.$$

Assuming the interferometric phase of each sub-interferometer is zero, the transformation of the total chain of MZIs is then simply the beam-splitter transformation applied $$N$$ times:

$$U^N = \mathrm{exp}\Bigl(i\frac{\pi}{2N}\sigma_x\Bigr)^N = \mathrm{exp}\Bigl(i\frac{\pi}{2}\sigma_x\Bigr) = i\sigma_x.$$

This is a swap operation, and it means that a photon entering the interferometer chain from the top left will always be detected in $$\mathrm{D}_2$$. If Bob blocks the mirrors in his laboratory, then the photon returns to Alice if it successfully reflects off all the beam-splitters, which happens with probability $$R^N$$, and in the limit of infinitely many beam-splitters this approaches 1. So, in an idealised scenario Bob can send the photon back to Alice without interacting with it, but when he doesn't block his mirrors and the photon enters his lab and the protocol wouldn't be counterfactual if the photon returns to Alice.

The idea of Salih's protocol is to address this problem by wrapping the interferometer chain in a larger interferometer. To understand how this works, think of the inner interferometer chain as being an effective beam-blocker when Bob doesn't block his mirrors; as we just saw, in this case the photon is always detected in $$\mathrm{D}_2$$ in (a), or equivalently in $$\mathrm{D}_3$$ in (b). That means that in (b), if the photon enters the transmission channel and Bob doesn't block his mirrors the photon will never return to Alice. However, the beam-splitters $$\mathrm{BS}_{\mathrm{M}}$$ are also chosen to have a high reflectivity, such that in all likelihood the photon simply reflects off all of them and returns to Alice in $$\mathrm{D}_1$$

When Bob does block his mirrors the photon will reflect off all the mirrors in the inner interferometer chain, and will interfere with the outer photon path, such that the photon is detected in $$D_2$$.

To summarise, if Bob doesn't block his mirrors a photon entering the transmission channel will always be detected in one of the detectors labelled $$\mathrm{D}_3$$, but this happens with a vanishingly small probability due to the high reflectivity of $$\mathrm{BS}_{\mathrm{M}}$$. If Bob does block his mirrors, the photon also clearly cannot travel back from the transmission channel to Alice. In this case it reflects off all the beam-splitters, and interference between the inner and outer paths occurs.

To address your first question explicitly, if Bob absorbs a photon that is indeed a transfer of energy, but as explained above: in the ideal case the probability of this approaches zero. Furthermore, Bob is sending information to Alice, so depending on your definition of counterfactuality this may or may not violate it.

There has been some debate about the counterfactuality of the protocol, but in this answer I gave the justification put forth in the original paper.