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

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.

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    $\begingroup$ 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
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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 $-i$.

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