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

1. Jan 4, 2013

### Strilanc

I'm trying to understand how optical circuits behave. In particular, if they can interfere "across time". Consider this optical circuit:

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?

2. Jan 4, 2013

### Staff: Mentor

It depends on the coherence length of the photons - and if you check the detectors too precise (if you let the photon interaction time influence their state), you ruin this coherence in time, and don't get interference. If you can check too early, you need detectors with some time-sensitivity.

3. Jan 4, 2013

### Strilanc

Could you explain what you mean by the coherence length of the photons? It seems like just making the long path a lot longer would fix any issue with arrival times not being exact.

4. Jan 4, 2013

### Staff: Mentor

See wikipedia or any physics book about quantum mechanics.
With a light bulb, your experiment will never give interference - as a simplified model, the photons are "too short" to interfere. With a laser, interference in this setup is possible.

5. Jan 4, 2013

### Strilanc

You're saying the states resulting from earlier photon arrivals don't interfere with the states resulting from later photon arrivals, I think. That would mean they must be different states.

Is it simply not possible in reality to make a detector that is truly time-insensitive (until you check it)? I feel like it should be workable in magic idealized abstraction land, unless it violates something fundamental like reversibility. Come to think of it... given the 100% triggered detector you can't go backwards to the photon being emitted. So, maybe it does violate reversibility and the issue is the detectors being impossible.

6. Jan 4, 2013

### Staff: Mentor

The time-reversed process is a photon with long coherence length as well.
It is possible in the mm-range (and probably more) with conventional lasers, I don't see a fundamental problem with the setup. You have to use a laser.

7. Jan 4, 2013

### Strilanc

Yes, it would work with a laser, but with the impossible detectors even a single photon would have interfered with itself.

8. Jan 4, 2013

### Staff: Mentor

I don't see the argument. The experiment is possible, you can get interference. Where is the problem?

9. Jan 5, 2013

### Strilanc

Simulating the system correctly on a computer. I wrote a simple program but it had detectors like what I was describing, so I was seeing interference without a laser.