$$ \hat U_d = \exp\Big( -i\underbrace{\frac{Q}{2}\int_0^T A(t^\prime)\mathop{dt^\prime}}_{\Theta(t)} \hat\sigma_x \Big) $$

This equation is a Unitary transformation applied to a qubit from time $t = 0$ to time $t=T$. I want the underlined envelope function $\Theta(t)$ to equal $\pi$ to result in an $X$ gate. What code should I write to accomplish this? Should I do frequency sweeps of multiple parameters? I know that a Gaussian pulse should be involved.

  • 1
    $\begingroup$ What do you mean by ...to equal π to result in an X gate...? $\endgroup$ Oct 13 at 17:26
  • $\begingroup$ I want to get the underlined part to be equal to pi. If this happens, then the result will be an X gate. $\endgroup$ Oct 14 at 2:13
  • $\begingroup$ Sorry but I still do not understand. How the result could be a gate? I would expect a quantum state to be the result. $\endgroup$ Oct 14 at 6:21
  • $\begingroup$ Maybe the result is a quantum state that had an X gate performed on it. $\endgroup$ Oct 14 at 14:41

I will try to provide you with some hints how to implement your unitary operation.

Let's firstly rewrite it as $U_d = \mathrm{exp}(-i\Theta(t)\sigma_x)$ or $U_d = \mathrm{exp}(-i\Theta(t)X)$ (i.e. I replaced the notation $\sigma_x$ for Pauli $X$ gate by symbol $X$).

This means that $U_d$ is in fact $x$ rotation given by formula $$ Rx(\theta)= \mathrm{exp}\Big(-i\frac{\theta}{2}X\Big)= \begin{pmatrix} \cos(\theta/2) & -i\sin(\theta/2)\\ -i\sin(\theta/2) & \cos(\theta/2) \end{pmatrix}. $$ Setting $\theta/2 = \pi/2$ (or $\theta = \pi$) we get $$ Rx(\pi) = \begin{pmatrix} \cos(\pi/2) & -i\sin(\pi/2)\\ -i\sin(\pi/2) & \cos(\pi/2) \end{pmatrix} = \begin{pmatrix} 0 & -i\\ -i & 0 \end{pmatrix} = -iX, $$ or in other words $x$ rotation with the rotation angle equal to $\pi$ is equivalent to Pauli $X$ gate up to global phase $-i$.

So, if you are able to compute your definite integral $\Theta(t)$, then the operation $U_d$ is simply $Rx(\theta)$ gate with the angle $\theta = 2\Theta(t)$. If $2\Theta(t) = \pi$ then you will get $X$ gate up to global phase $-i$.


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.