1
$\begingroup$

I recently was watching these 2 videos on Coursera which show how to build a simple quantum computer that can implement the simplest case of the Deutsch-Jozsa algorithm (which uses only 2 qubits).

https://www.coursera.org/lecture/quantum-computing-algorithms/quantum-computer-prototype-diy-dCKRO

https://www.coursera.org/lecture/quantum-computing-algorithms/quantum-computer-prototype-solving-the-deutschs-problem-7EuD2

In it, the professor mentioned that the photon represented 2 qubits: one was the polarization qubit and one was the path/position qubit. Later, the professor stated that a half-wave plate affected the path/position qubit (not the polarization qubit). I found this really strange as I thought half waveplates only affected polarization.

How can a half-wave plate affect the position qubit and not the polarization qubit?

$\endgroup$
  • $\begingroup$ 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 Deutsch algorithm and the presented implementation, there are only 4 options for the location of the half-wave plates and and in none of them there is a situation to affect only on the position qubit but not on the polarization qubit ... $\endgroup$ – Aleksey Zhuravlev Aug 8 at 5:07
1
$\begingroup$

The waveplates can be understood 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 single path, the net effect is "if the photon traveled down that path, do x to it". Hence, it affects the position component.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.