How do you implement a square root of NOT gate in an optical quantum computing circuit? What optical element or combination of elements would you use - especially, if you want to manipulate polarization qubits?

The square root of not gate has the following matrix representation: $$ \sqrt{X} = \sqrt{NOT} = \frac{1}{2} \begin{bmatrix} 1+i & 1-i \\ 1-i & 1+i \end{bmatrix}. $$

  • $\begingroup$ It would perhaps be best if you posted this on quantumcomputing.stackexchange.com. Our quantum computing SE site. thanks. $\endgroup$
    – joseph h
    Jun 22 at 2:52
  • 1
    $\begingroup$ I’m voting to close this question because it is better suited to quantumcomputing.stackexchange.com. $\endgroup$
    – joseph h
    Jun 22 at 2:54
  • 3
    $\begingroup$ I think this question is better suited for physics than quantum computing stack exchange because the answer will involve theoretical and experimental aspects of optics. It does not matter for the answer the application for the question is quantum computing. $\endgroup$
    – Jagerber48
    Jun 22 at 3:04
  • $\begingroup$ @josephh Quantum Computing questions are completely on topic here. Not to mention the quantum optics part. $\endgroup$ Jun 22 at 9:58

For a polarization qubit, you need to rotate the polarization. This is done with a waveplate. A $\lambda/2$ plate implements a NOT gate. Thus, a $\lambda/4$ plate will implement a square root of the NOT gate.

  • $\begingroup$ Thanks! Just to confirm whether my understanding is correct, that would be a quarter-wave plate with a 45-degree angle between the fast axis and whatever polarization corresponds to the input (e.g., |0> encoded in horizontal polarization)? $\endgroup$
    – triclope
    Jun 23 at 2:42

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.