2
$\begingroup$

I'm currently working on QRAC and was wondering if there's an encoding protocol in $3 \rightarrow 1$ such that the receiver is able to retrieve any one of the XOR combinations of the bits, along with the original bits itself ( a grand total of 7 functions; if a, b, c are the bit positions, the receiver should be able to guess a, b, c, a+b, b+c, c+a, a+b+c with a probability greater than 1/2).

Is this even possible? My initial guess would be to use some kind of POVM measurements in the decoding part but don't how to proceed with that.

If you have any ideas on this then please let me know. Thank you.

New contributor
Vaisakh M is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
  • $\begingroup$ I'm certain it can be done; for the $2\to1$ QRAC you can use the usual encoding, and measure in the basis $\{\cos\frac\pi8|0\rangle + \sin\frac\pi8|1\rangle, \sin\frac\pi8|0\rangle - \cos\frac\pi8|1\rangle\}$ to get the XOR of the two bits. I guess the analogous will work for the $3\to1$ QRAC; just keep the usual encoding and optimize over the measurement basis to get each of the functions you want. $\endgroup$ – Mateus Araújo Nov 21 at 20:29
  • $\begingroup$ I'm not sure of the scheme you just mentioned. In the usual 2-> 1 QRAC 00 and 11 are encoded using antipodal Bloch vectors. If we take XOR both yield 0. How can a basis give such a result? $\endgroup$ – Vaisakh M Nov 22 at 9:54
  • $\begingroup$ Sorry, you're right, that was a brain fart. You can just perturb the encoding states a little bit, though, so they will not be antipodal anymore, and there will be a basis that works. That's horrible, of course, but should be enough to prove that the $3\to1$ case is possible. $\endgroup$ – Mateus Araújo Nov 22 at 20:55
1
$\begingroup$

I found a way to do it for the $2\to1$ QRAC. I simply guessed that we could leave the measurement bases as they are, $Z$ for the first bit, and $X$ for the second bit, and added $Y$ as the basis with which to extract the XOR of the two bits. From the guess obtaining the optimal encoding states is then easy, we just need to diagonalize the relevant operators. They are \begin{align*} |\psi_{00}\rangle & = \sqrt{p}|0\rangle + \sqrt{1-p}\ e^{\frac{\pi i}4}|1\rangle \\ |\psi_{01}\rangle & = \sqrt{p}|0\rangle + \sqrt{1-p}\ e^{-\frac{3\pi i }4}|1\rangle \\ |\psi_{10}\rangle & = \sqrt{1-p}|0\rangle + \sqrt{p}\ e^{-\frac{\pi i}4}|1\rangle \\ |\psi_{11}\rangle & = \sqrt{1-p}|0\rangle + \sqrt{p}\ e^{\frac{3\pi i}4}|1\rangle, \end{align*} where $p=\frac{3+\sqrt3}{6}$ is also the probability of success in this QRAC. I don't know whether it is optimal, but I guess it is, since it's so nice and symmetrical.

Now for the $3\to1$ case such a simply idea cannot work, as you want to extract 7 different functions, but there are only these 3 mutually unbiased bases in dimension 2. What I would do is to find some set of 7 bases that is in some sense uniformly distributed on the Bloch sphere and see if it works. Another idea is to simply do a see-saw optimization and see what you get.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Hmmm yes, I've thought about it too... I would need to work out the probability for the 4-> 1 case and see if it works. $\endgroup$ – Vaisakh M 2 days ago

Your Answer

Vaisakh M is a new contributor. Be nice, and check out our Code of Conduct.

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.