I'm trying to understand how optical circuits behave. In particular, if they can interfere "across time". Consider this optical circuit:(adsbygoogle = window.adsbygoogle || []).push({});

Suppose that it is completely isolated from the outside world (until we check it), and that the detectors are time-insensitive (they record IF a photon was received, not WHEN it was received [even implicitly]). Call the time it takes for a photon to follow the bottom branch ##t_{1}## and the time it takes for a photon to follow the top branch ##t_{2}##.

Does the outcome of this circuit depend on whether or not we check it (un-isolate it) before or after ##t_{2}##? From what I understand, it does, but this seems counter-intuitive. Let me go through the math.

At ##t_{1}## the state of the system should be this:

##\frac{-1}{2} RightDetector + \frac{-i}{2} BottomDetector + \sqrt{\frac{1}{2}} i^{MirrorsHit} TopPath##

If we don't un-isolate the system at this point, then at ##t_{2}## the system will have reached this state:

##(\frac{-1}{2} RightDetector + \frac{-i}{2} BottomDetector) + (\frac{-i}{2} BottomDetector + \frac{1}{2} RightDetector)##

Since the detectors aren't time-sensitive, the time the photon arrived doesn't matter and the photon-arriving-at-##t_{1}## states interfere with the photon-arriving-at-##t_{2}## states, giving:

##= -i BottomDetector##

So if we only check after ##t_{2}## then we should see that the bottom detector has gone off 100% of the time.

But, going back to the state at ##t_{1}##:

##\frac{-1}{2} RightDetector + \frac{-i}{2} BottomDetector + \sqrt{\frac{1}{2}} i^{MirrorsHit} TopPath##

If we check the system now, we modify its state into the following:

##\frac{-1}{2} InitiallySawRightDetector + \frac{-i}{2} InitiallySawBottomDetector + \sqrt{\frac{1}{2}} i^{MirrorsHit} TopPath##

and it will evolve into the following state at ##t_{2}##:

##\frac{-1}{2} InitiallySawRightDetector + \frac{-i}{2} InitiallySawBottomDetector + \frac{-i}{2} LaterSawBottomDetector + \frac{1}{2} LaterSawRightDetector##

These states are different, because of our meddling, so they don't interfere. In this case we expect the bottom detector to have gone off 50% of the time by the end of the experiment (25% initially, 25% later) and the right detector to have gone off the other 50% of the time (also 25% initially, 25% later).

Is this analysis correct? Do the right detector states actually get erased after the fact, if we haven't checked to make sure they don't?

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# Can the result of an optical circuit depend on when you check it?

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