New answers tagged photonics
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I ask to excuse me, but I can not add comments.
Could you indicate the exact time on the video when the professor claims "a half-wave plate affected the path/position qubit (not the polarization qubit)"?
Or a literal quote from the text under the video? (I tried to find but did not find such a statement...)
It is very interesting since based on the ...
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The waveplates can be undertsood in terms of polarisation, but the change that they implement is independent of what polarisation the photon is in. It is simply "if the photon passes through this waveplate, x happens to it" where x might be "apply phase $e^{i\pi}$".
Because there are two different paths, and you put a waveplate on a ...
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As for any platform, one has to choose a suitable $d$-dimensional "computational" subspace. Suitability depends on your application, but generally it means that one should be able to perform operations on that subspace and couple it to other qudits. In practice, these operations will couple the qudit to degrees of freedom outside of the subspace ...
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You start with a polarisation filter. This does nothing to the path of your photon and, effectively, measures the polarisation of the photon, meaning that you prepare the "second" qubit in the fixed state determined by what polarisation the filter is detecting. So, at this point, you have
$$
|0\rangle|-\rangle
$$
Then, you input to a beamsplitter. ...
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Not exactly.
You are correct, there is a single photon; qubit 1 is its path, qubit 2 is its polarization.
The waveplates implement an oracle (one from 4 possible).
You are wrong about the beam splitters; beam splitters do not affect polarization, so they act on the qubit 1 only as Hadamard gates.
The $|-\rangle$ state of qubit 2 is created by the ...
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