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I am looking to translate interferometric processes onto a quantum circuit model and am running into issues regarding feedback and the reuse of circuit elements. My question is framed in terms of a Sagnac interferometer but applies to translating more general problems onto quantum circuits.

At its heart, a Sagnac interferometer applies some unitary $U$ to a target system if the control is in some state $|0\rangle$ and the opposite unitary $U^\dagger$ to the target system if the control is in state $|1\rangle$. This is easy to implement with a quantum circuit if we have an individual gate implementing each of $U$ and $U^\dagger$; then we only need two control gates. But the crux of Sagnac interferometry is that $U^\dagger$ arises from implementing $U$ backwards: the control qubit governs the direction in which the target system experiences the unitary $U$. Can this be represented by a quantum circuit?

The first step is straightforward: if control is in $|0\rangle$, implement $U$, otherwise ignore. I could envision the next step to be: if control is in $|1\rangle$, reflect backward through the circuit. Is this allowed? Is there a standard method for doing this?

Perhaps there is a way of implementing some controlled gate that says "if the control is in some state, perform a complex conjugation operation on the next gate" can that be done, instead? And, if so, can that be done in arbitrary dimensions?

At the end of the day, Sagnac interferometry needs the two processes to be able to reinterfere with each other, so an ideal solution would allow for that, but we can postpone that problem for now.

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    $\begingroup$ This doesn't really answer your question (hence the comment) but might help with perspective. The quantum circuit is a completely general specification of an experiment in terms of time-ordered operations. It's up to you how to physically implement it. Some implementations translate that time ordering into space ordering. If that's the case, you gain the ability to loop things, run them backwards etc. But it's a specific advantage of those implementations which I don't think can be translated back to the more general circuit model except by using new elements: $U^\dagger$ or repetition. $\endgroup$
    – DaftWullie
    Jul 7 at 7:10
  • $\begingroup$ Ok, so the answer is probably just "no"! In this paper's fig 1 they distinguish between a quantum circuit and a "functional circuit" - that's probably the distinction I'm looking for. $\endgroup$ Jul 7 at 13:29
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the control qubit governs the direction in which the target system experiences the unitary U. Can this be represented by a quantum circuit?

Sure. For example, you can do this:

enter image description here

Or this:

enter image description here

Or lots of other things that are equivalent to "if control is off apply $U^{-1}$ otherwise apply $U$".

In particular, note that if you have an explicit circuit for $U$ then you can often make it conditionally run backwards by controlling just a few key operations inside of it. For example, an increment operation can be reversed into a decrement by, before and after the increment, applying a NOT gate to all of the qubits it operates on. This is much cheaper than the alternative.

I could envision the next step to be: if control is in |1⟩, reflect backward through the circuit. Is this allowed?

No, it's not allowed. Quantum circuits are a series of instructions to apply one after another. They are ordered in time, not in space, so you can't go backwards.

I think the key idea you are missing is that you are trying to re-use circuit elements when instead you should just repeat them. You don't put a gate in a circuit and then apply it to a qubit multiple times by literally drawing a circular loop in the circuit, you instead unroll the winding circle into a line and repeat the gate multiple times.

For example, a quantum circuit representing a sagnac interferometer could have one qubit to indicate the direction of the photon in the loop, and a handful of qubits forming a register to indicate the photon's position in the loop. You would then have a "time step" subroutine that incremented the position register if the direction was clockwise and decremented otherwise, and if the position was 0 (at the half silvered mirror) then do something there possibly involving other qubits used to represent states outside the loop, including possibly some measurements watching for the simulated photon hitting a simulated detector. You would then repeat this subroutine some number of times and interpret the measurement results you got as the result of the experiment.

Here's my attempt at that in Quirk:

enter image description here

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  • $\begingroup$ This directly avoids the question, unfortunately. The entire goal is to reuse the same circuit element - of course copying the unitary will work. $\endgroup$ Jun 11 at 18:15
  • $\begingroup$ "you should just repeat them" is exactly the opposite of interferometry, in which the goal is to measure parameters of some unknown unitary in a clever way $\endgroup$ Jun 11 at 18:16
  • $\begingroup$ Your answer is simply repeating "This is easy to implement with a quantum circuit if we have an individual gate implementing each of $U$ and $U^\dagger$; then we only need two control gates." $\endgroup$ Jun 11 at 18:18
  • $\begingroup$ @QuantumMechanic If you are going to execute a $U$ in a quantum circuit once, you must have already decomposed it into gates. Therefore you can execute it twice by executing the gates twice, or execute a controlled version by executing controlled versions of each of the gates. The only obstacle is that you dislike the idea of running it twice. But in general it is in fact necessary to run it twice to implement the "backward vs forward" effect that you want. Maybe it's not "real" interferometry but it's simulating the same unitary so you can conclude how the interferometry will behave. $\endgroup$ Jun 11 at 19:07
  • $\begingroup$ That's the crux of my problem here. So your answer is that quantum circuits cannot be used to describe interferometry, only to simulate an already-known setup? If I call $U$ an oracle that you can only access once would you change your mind? A "no" answer is disappointing but I can accept it. $\endgroup$ Jun 11 at 19:43

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