# What observable(s) does a mirror operate on?

I'm reading this introduction to the Mach-Zehnder interferometer, and I'm confused about what the mirrors are doing.

Suppose we're describing the path (upper / lower) and polarization (vertical / horizontal) of a particle in the basis $$|UV\rangle$$, $$|UH\rangle$$, $$|LV\rangle$$, $$|LH\rangle$$. They describe the mirrors' operation as:

$$-\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix} = -(I_{pos}\otimes I_{pol}) = -I_{pos}\otimes I_{pol} = I_{pos}\otimes -I_{pol}$$

A couple of questions:

1. Many resources online describe a mirror as "phase shifting" the photon, but they don't give a precise explanation of what this means. From the above, I gather it's not a relative phase shift (which is what the phrase normally means in QIT) but a global one. Is this correct? (We could say that each mirror introduces a phase shift relative to the other path, but the pair introduces a global phase shift. Another way of saying this is that if there were only one path, the phase shift would not be relative to anything.)

2. It seems like a global phase shift cannot be specific to a particular observable: $$-|\psi\rangle \otimes |\phi\rangle = |\phi\rangle \otimes -|\psi\rangle$$ after all. Even worse, this means it isn't specific to a specific particle: it's equivalent to phase shifting the universal wave function, so it doesn't seem to make sense to ask "where" the shift got applied.

Clearly I'm misunderstanding something fundamental, and any tips would be appreciated.

• a gate $U$ is physically identical to the gate $-U$, or to any $e^{i\phi}U$ for that matter. This follows from the quantum states being defined up to their global phase, and noting that $e^{i\phi}U|\psi\rangle$ differs from $U|\psi\rangle$ only by a global phase. The "phase shifting due to reflection" is not of this form, as it only happens only in reflection, see e.g. physics.stackexchange.com/q/32122/58382 and links therein. As a gate this is something like a $iY$ gate with $Y$ the Pauli matrix, but redefining the modes this is equivalent to just using $Z$ – glS Nov 5 '19 at 13:05
• @glS Thanks for commenting. What's unintuitive for me is this: a plate of glass introduces a relative phase shift to the position observable, but if it's wide enough that an entire wavepacket passes through, then this is effectively a global phase shift. And a global phase shift is not specific to any observable (if it can be said to have happened at all). Also, a waveplate introduces a relative polarization phase shift, and there's no way to make it global by moving or extending it: position eigenstates are by definition separated in space, while the polarization eigenstates aren't. – A_P Nov 5 '19 at 15:27
• "but if it's wide enough that an entire wavepacket passes through, then this is effectively a global phase shift" I think you are confusing something here. The size of the plate is irrelevant, we are always assuming that the whole of the beam impinges on it (otherwise the description would get much more complicated). Part of the light is transmitted, and part of it is reflected. The reflected part gets a phase shift, while the other does not. This is an observable phase shift, because you can then make reflected and transmitted components interact with each other – glS Nov 5 '19 at 15:55
• also note that two operations differing only by a global phase model the same exact physical operation, just as $|\psi\rangle$ and $-|\psi\rangle$ are the same exact physical state, just written in two different ways – glS Nov 5 '19 at 15:57
• @glS You're talking about half-silvered mirrors, right? I meant for a simple plate of glass or a fully-silvered mirror. – A_P Nov 5 '19 at 16:10

If the full state is being reflected, then the introduced phase is not observable because it's a global phase. If, however, only part of the beam is reflected by a mirror while some other part is not, then this phase shift is not global, and can be observed for example by making the parts of the beam interfere with each other.

And a global phase shift is not specific to any observable (if it can be said to have happened at all). Also, a waveplate introduces a relative polarization phase shift, and there's no way to make it global by moving or extending it: position eigenstates are by definition separated in space, while the polarization eigenstates aren't.

There is complete symmetry between position and polarisation in this regard.

If a photon is in a superposition of spatial modes, say something of the form $$|\text{left path}\rangle+|\text{right path}\rangle$$, you can introduce a phase shift between the two modes using for example a mirror, which might send this state into one of the form $$|\text{left path}\rangle-|\text{right path}\rangle$$.

Similarly, a photon can be in a superposition of different polarisation states, e.g. $$\lvert\uparrow\rangle+\lvert\downarrow\rangle$$, and a phase between the two polarisation modes can be introduced using a waveplate, which might produce something like $$\lvert\uparrow\rangle-\lvert\downarrow\rangle$$.

It is true that you can think of reflecting both spatial modes of the photon via a mirror, and then this would amount to an irrelevant global phase. You could then think that the analogous operation for the polarisation should be to "apply a waveplate to both polarisation states of the photon", which is not really a meaningful physical operation. That is not overly surprising though: different physical carriers of information can be handled in different ways. You could still do something similar as the "mirror on one side" operation on the polarisation by using a PBS to spatially separate the two polarisation modes, then put a mirror on one of the two paths, and use another PBS to join the paths together again.

Another difference that you might be wondering about is the following: we can understand the phase shift introduced by the waveplate as due to a "rotation" in the polarisation space. But in the case of the mirror, it's hard to see where this "rotation" is happening. The mirror is operating on a single spatial mode, there would seem to be no "auxiliary space" on which the mirror is acting. The answer is that there actually is such a space: the photon interacts with the mirror in a highly complex way, that will involve the atoms making up the mirror etc. A rotation is happening in this much larger space, but things are set up in such a way that the overall result is simply a phase shift, so you don't need to worry about the describing the whole thing, and can just use the effective description in which the mirror is modelled as simply "applying a phase shift to the state".

• Thanks very much for the detailed answer! For the PBS, it still has to spatially separate the polarization components to let us operate on them separately, right? Here's another way of stating my initial confusion: to change a mirror's behavior (wrt position modes) you can simply move it, whereas for a waveplate you must modify it (e.g., thicken it). Similarly: a "position phase shifter" is commonly just called a "phase shifter," whereas a "polarization phase shifter" is not. I've asked physicists before which observable a "phase shifter" operates on, and they don't understand the question. – A_P Nov 9 '19 at 18:35
• Anyway, I think my confusion is mostly resolved. Thanks! I still have to digest your final paragraph. – A_P Nov 9 '19 at 18:35
• indeed, there is no "observable associated with a phase shift". A phase shift between two states can be introduced, roughly speaking, by any operation that treats those two states differently. Note that phase shifts introduced by different operations might be indistinguishable (for example a phase shift on a photon due to a mirror or due to a waveplate). The phase between two states is really a way to quantify how those two states will interfere with each other. Also, different physical devices operate on different degrees of freedoms in different ways, that is to be expected I would say – glS Nov 9 '19 at 18:39
• Consider the device simply known as a "phase shifter" in quantum optics. Surely we can say that the relevant observable there is position, yes? It cannot introduce a phase shift w.r.t any other observable. Part of what's been hard about getting to that answer is that if we place it just right (so that a whole beam is incident), it's no longer a "phase shifter" at all (at least, w.r.t. that beam). In a sense, it's our choice of positioning that makes it a "position phase shifter," yet it still seems to be intrinsically a position phase shifter. – A_P Nov 9 '19 at 18:51
• @A_P that's right, how you place it does matter, because if you introduce the same phase shift on all of the position modes, it's the same as doing nothing. I don't know if this will help, but if you think of a photon in a superposition of two different spatial modes, then a phase shifter in only one of the two modes is essentially a "controlled-shift operation", in the notation of quantum circuits, because it only applies a phase shift conditionally to the photon being in a specific position state. From this pov, it is a "more complex" operation than a polarisation waveplate – glS Nov 9 '19 at 19:00