2
$\begingroup$

I am starting with a product state over multiple qubits. That looks like the expression below. $$ |\psi\rangle = \left(\cos\left(\frac{\theta_1}{2}\right)|0\rangle+e^{i\phi_1}\sin\left(\frac{\theta_1}{2}\right)|1\rangle \right)\otimes\left(\cos\left(\frac{\theta_2}{2}\right)|0\rangle+e^{i\phi_2}\sin\left(\frac{\theta_2}{2}\right) |1\rangle\right)\,. $$

After doing some calculations, I obtain a function of $\theta$ and $\phi$. e.g.

$$ f(\theta_1,\theta_2,\phi_1,\phi_2)=\cos ^2\left(\frac{\text{$\theta_1$}}{2}\right) \cos ^2\left(\frac{\text{$\theta_2$}}{2}\right)\,. $$ (This specific example doesn't contain $\phi_1$ or $\phi_2$. But, they are present in general.) I am not sure, how can I integrate $f(\theta_1,\theta_2,\phi_1,\phi_2)$ over Haar random states. I know for single qubit, we can integrate with a measure $\sin(\theta) \,\mathrm{d}\theta$ with $\theta \in [0,\pi]$. My doubt is what should be the way for multiple qubits. Any help is much appreciated. Feel free to ask for any clarification. Thank you in advance.

$\endgroup$
2
  • 1
    $\begingroup$ A Haar-random quantum state on $2$ qubits is parametrized by $3$ amplitudes and $3$ relative phases. Since you only have $2$ of each, does that mean that you want to integrate over tensor products of Haar-random states independently? $\endgroup$
    – Tristan Nemoz
    Commented Oct 25, 2023 at 13:54
  • $\begingroup$ BTW your formula for $\psi$ is missing some basis vectors. Just as a remark: It seems that the function you want to integrate is a polynomial in the state. In this case, there are more elegant ways to integrate which do not involve an explicit parametrization. This is also the way to go in a general $n$-qubit setting. $\endgroup$ Commented Oct 25, 2023 at 14:07

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.