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I've been trying to make a simple puzzle game based on quantum optical circuits. Problem is, I don't actually know the physics that well (this is a hobby project with the goal of increasing that understanding). In particular, I'm not sure how to deal with cycles or a photon arriving at a detector at different times.

For example, start with a Sagnac interferometer and flip the half-silvered mirror.

Modified Sagnac interferometer

Presumably this creates a situation where the photon 'decays' out of the loop exponentially. I think the state of the system evolves like this:

[note: s = 1/sqrt(2), i = sqrt(-1)]
1 |PhotonLeavingSource>
is |PhotonHeadingToDetector> + s |PhotonHeadingToBottomRight>
is |PhotonDetected> + is |PhotonHeadingToTopRight>
is |PhotonDetected> + -s |PhotonHeadingToTopLeft>
is |PhotonDetected> + -is |PhotonHeadingToBottomLeft>
is |PhotonDetected> + -i/2 |PhotonHeadingToDetector> + 1/2 |PhotonHeadingToBottomRight>
i(s-1/2) |PhotonDetected> + 1/2 |PhotonHeadingToBottomRight>
i(s-s^2-s^3-s^4-...) |PhotonDetected> = -i |PhotonDetected>

I'm pretty sure I've made a mistake in there. The final state has an amplitude of 1, but some of the intermediate state don't (they aren't unitary). Am I supposed to be normalizing as I go? Should the detected states be parametrized by arrival time and thereby not interfere with each other? Does that happen conditionally based upon the type of detector?

Any insights would be appreciated.

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up vote 1 down vote accepted

Assuming a balanced beam splitter (BS) and denoting $|k\rangle$ state corresponding to $k$ passages of the photon through the loop, the state can be found iteratively as follows:

  1. After the first passage through the BS, we get $$\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle).$$

  2. The state $|1\rangle$ needs to be expanded when it goes through the BS second time; this gives $$|1\rangle\to\frac{1}{\sqrt{2}}(|1\rangle-|2\rangle).$$ Note that due to unitarity of the BS operation, there has to be a phase shift between light coming from the left reflected down and the light coming from upwards and reflected to the right. The complete state after the second BS passage is then $$\frac{1}{\sqrt{2}}|0\rangle+\frac{1}{2}(|1\rangle-|2\rangle).$$

  3. You can continue this for as long you want. Now $|2\rangle$ gets transformed the same way as $|1\rangle$ above, and so on.

I'm not sure I understood your approach correctly but I hope this will help you to understand what is going on.

As far as interference is concerned the events of the photon arriving at the detector at different times are distinguishable and do not interfere. Using these so called time-bins is, in fact, one possible way of encoding qubits in quantum optics. This wouldn't be possible if these states interfered.

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My approach was 'multiply by i when reflecting' and 'propagate each component of the superposition independently of the others'. Each new line is a further evolution of the state. The fact that there are time-bins basically resolves the question for me, I think. Are there any cases where time-bins interfere with each other? – Craig Gidney Jan 3 '13 at 21:26
The photon cannot interfere with itself at different times; that would contradict causality so time-bins cannot interfere with each other. You could, however, run into difficulties if your detector didn't have sufficient temporal resolution so that you couldn't resolve individual time-bins. But then you would just have to add the probabilities, not the amplitudes. – Ondřej Černotík Jan 3 '13 at 21:34
Well, as I understand it, anytime two components of a superposition 'evolve into each other' they interfere. So if the detector reacted the same way whenever a photon arrived (no marking down the arrival time) and stayed in the same state (no counting up the elapsed time) then all of the arrival times would interfere because they'd correspond to the same final state. I don't think this breaks causality, although I'd guess that in reality there's always some inherent 'counting up the elapsed time', due to how the underlying wave equation works, that prevents interference. – Craig Gidney Jan 3 '13 at 21:41

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