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More specifically, how do you physically prepare the superposition of two beams of polarized light?

Suppose beam A and beam B, the two input beams, each have unknown polarizations.
Beam C, the output beam, must be the superposition of beams A and B.
How do you do it?

You can create an overlap of two beams with a half silvered mirror at 45%.
You can also creat a localized overlap by crossing the beams with a small angle.
In both cases you get a mixture, not a superposition.

How do you get the superposition?

Bonus Points: How do you prepare $ a|A \rangle + b|B\rangle $?

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Two 'different' beams can not be superimposed (in sense of QM). –  user10001 Jul 21 '12 at 16:21

3 Answers 3

up vote 2 down vote accepted

Your problem is ill-posed, becaus the state of a beam is not described by a state vector $\psi$, but by the corresponding density matrix $\psi\psi^*$, which determines $\psi$ only up to a phase. Thus it is meaningless to talk about ''the'' superposition of two beams.

If the two beams are known to have a fixed, but unknown, relative phase, the question becomes well-defined. The only way to gain this knowledge is to check that the beams have been coherently generated from a common source. In this case, a half-silvered mirror will produce the required superposition, and one can change the relative phase by a rotator (as described in the book by Mandel and Wolf).

For preparing arbitrary states of a collection of beams from a single coherent beam see my paper http://de.arxiv.org/abs/quant-ph/0306123. See also
M.Reck, A. Zeilinger, H.J. Bernstein and P. Bertani, Experimental realization of any discrete unitary operator, Phys. Rev. Lett. 73 (1994), 58–61.
These papers specify how to create appropriate transformations; to prepare a particular state, find a transformation that creates it from a state you know how to prepare, and then implement this transformation.

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OK, suppose beams A and B are in unknown pure states. Does this make the question well-defined? (Thanks for an unexpected answer. I am primarily interested in how one can prepare an arbitrary superposition from specified resources. TIA) –  Jim Graber Jul 22 '12 at 1:10
The question is still ill-defined, as the value of $\psi+\psi'$ depends on which arbitrary phases you choose to describe A and B. The question would make sense if you'd ask: Given two state vectors $|A>$ and $|B>$ in some Hilbert space (rather than attached to some particular beams), can one prepare a beam with state $|A>+|B>$ (yes!) --- or: Given two beams in a pure state, can one transform it into a beam whos state is in the state space spanned by both beams, but different from the state of A and B? (This space is well determined, while a fixed pure state selected from it is ill-determined.) –  Arnold Neumaier Jul 22 '12 at 13:47
Overall comment: You seem to be talking in mathematical terms. I was originally trying to speak only of physical objects such as beams, mirrors, polarizers, etc. Nevertheless, thank you for your answers. I will try to switch into a more mathematical mode. To make my original question still more simple ,(and still more precise,) the two beams may be thought of as having a fixed, but unknown, relative phase. Does this make the question well-defined? –  Jim Graber Jul 23 '12 at 12:16
The first formulation was mathematical, the second physical. - Your new assumptions means that your beams must have been coherently generated from a common source (there is no other way to set up this situation). In this case, a half-silvered mirror will produce the required superposition, and one can change the relative phase by a rotator (as described in the book by Mandel and Wolf). I edited my answer to account for that. –  Arnold Neumaier Jul 23 '12 at 13:11
@JimGraber: I updated my answer, hoping that the new information is closer to what you hope for. –  Arnold Neumaier Jul 23 '12 at 13:26

If you combine two beams, when do you get a superposition versus a mixture? You get a superposition if the beams are mutually-coherent (their relative phase does not change, or at least does not change appreciably over the course of your measurement). You get a mixture if the beams are mutually-incoherent (their relative phase changes so fast that it is for all intents and purposes random).

In practice, if you want a superposition you should start with a single light source--ideally a laser--then split it to get the two beams, then recombine the two beams. If you want a mixture you should take two independent light sources: They are almost guaranteed to randomly differ by many gigahertz or terahertz, so their relative phase is essentially random.

Note that different frequencies within a beam (even a laser beam is not purely monochromatic) may have different phase ... and different (spatial) parts of a beam can also have different phase. So even if you start with a single light source, it is possible to wind up with a mixture rather than a superposition. You can also accidentally get a mixture if your table is vibrating (for example), because that randomly changes the path length difference that the two beams are traveling.

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I had not realized this was so delicate. –  Jim Graber Jul 23 '12 at 23:15

The assumption is not really true, a half-silvered mirror does create a superposition. The key is that in both the output directions, the photons could come from either input but end up in (roughly?) the same mode, thus (almost?) indistinguishable.

We usually observe the superposition (e.g. in a Mach-Zender interferometer setting, where the two paths are led back together and interfere again) when the two beams come from the same source which has been split using another half-silvered mirror and then manipulated. However, it can equally well arise for different sources, but the coherence is so short-lived (possibly less than one period of the oscillation) that it cancels out to a mixed state for all practical purposes.

Remark: I assumed that you meant superposition states in individual photons of the two beams. A "superposition of light beams" might also be interpreted as something much more complicated, e.g. a state where all the light goes in one beam or the other. This, despite being perfectly physically possible, would be extremely hard to achieve. You can look up "N00N states" of light to get an image.

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I am looking up N00N states. –  Jim Graber Jul 22 '12 at 1:12
+1 for Mach-Zehnder and N00N states. But I am not clear on how you would use N00N states to combine two arbitrary beams. –  Jim Graber Jul 23 '12 at 10:44
I would not, that was just to give an image of what a superposition of beams may theoretically look like and how extremely non-classical and hard to prepare it would be. As far as I know, no one was able to prepare one with $N$ high enough to be eligible to be called a "beam" yet and probably never will be. Lacking knowledge of particular experimental techniques to produce them in known states, I don't dare to propose any way of doing so for a priori unknown ones. Anyway, thanks for the +1! –  Vašek Potoček Jul 23 '12 at 11:47
There is also something called two-mode optical Schrödinger cat states, which is a concept similar to N00N states but with coherent (i.e. classical) states in the modes instead of Fock states (which are very nonclassical on their own). These might be somewhat easier to obtain. But I have not thought of getting them in two arbitrary modes, either. Maybe someone doing experimental quantum optics would have an idea. –  Vašek Potoček Jul 23 '12 at 11:54

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