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I'm trying to understand how to implement quantum phase estimation (QPE) for a generic single-qubit Hamiltonian. The general time-evolution could be simulated using $U_3$ gate, in Qiskit documentation,

$$ U_3(\theta,\phi,\lambda)= \begin{pmatrix} \cos(\theta/2) & -\mathrm{e}^{i\lambda}\sin(\theta/2) \\ \mathrm{e}^{i\phi}\sin(\theta/2) & \mathrm{e}^{i(\phi+\lambda)}\cos(\theta/2) \end{pmatrix} $$

If I've already figured out the 3 Euler angles, in quantum phase estimation, am I only estimating $\theta$ ? In this example on Qiskit textbook, we can confirm the angle of $T$ gate as well as other phase angles $\theta$. My question is if we want to apply QPE to a general rotation, are we only interested in $\theta$? Is that because only this angle encodes information about energy? Thanks!

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If you were to apply QPE to this unitary, what you will get, assuming you start with a proper eigenvector $|\Lambda\rangle$, is an estimation of $x$, if the associated eigenvalue $\Lambda$ is written as $\mathrm{e}^{2\mathrm{i}\pi x}$. Thus, to know what you will measure, you have to know what the eigenvalues of this unitary are.

There may be a faster way of doing this, but what you can do is to find the root of the characteristic polynomial $\chi$ associated to this matrix: $$\chi = \begin{vmatrix}X-\cos\left(\frac\theta2\right)&\mathrm{e}^{\mathrm{i}\lambda}\sin\left(\frac\theta2\right)\\-\mathrm{e}^{\mathrm{i}\varphi}\sin\left(\frac\theta2\right)&X-\mathrm{e}^{\mathrm{i}(\varphi+\lambda)}\cos\left(\frac\theta2\right)\end{vmatrix}=X^2-2\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)\mathrm{e}^{\mathrm{i}\frac{\varphi+\lambda}{2}}+\mathrm{e}^{\mathrm{i}(\varphi+\lambda)}.$$ The roots of this polynomial are: $$\Lambda=\mathrm{e}^{\mathrm{i}\frac{\varphi+\lambda}{2}}\left(\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)\pm\mathrm{i}\sqrt{1-\cos^2\left(\frac\theta2\right)\cos^2\left(\frac{\varphi+\lambda}{2}\right)}\right)$$ We can easily see that these roots lie, as expected, on the unit circle. We are thus interested by their argument. There are three cases to be dealt with:

  1. If $\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)>0$, then the eigenvalues are $$\exp\left(\mathrm{2i\pi\frac{\frac{\varphi+\lambda}{2}+\arctan\left(\frac{\sqrt{1-\cos^2\left(\frac\theta2\right)\cos^2\left(\frac{\varphi+\lambda}{2}\right)}}{\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)}\right)}{2\pi}}\right)$$ and $$\exp\left(\mathrm{2i\pi\frac{2\pi+\frac{\varphi+\lambda}{2}-\arctan\left(\frac{\sqrt{1-\cos^2\left(\frac\theta2\right)\cos^2\left(\frac{\varphi+\lambda}{2}\right)}}{\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)}\right)}{2\pi}}\right).$$
  2. If $\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)=0$, then the eigenvalues are $\exp\left(2\mathrm{i}\pi\frac{\varphi+\lambda+\pi}{4\pi}\right)$ and $\exp\left(2\mathrm{i}\pi\frac{\varphi+\lambda-\pi}{4\pi}\right)$.
  3. If $\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)<0$, then the eigenvalues are $$\exp\left(\mathrm{2i\pi\frac{\frac{\varphi+\lambda+\pi}{2}+\arctan\left(-\frac{\sqrt{1-\cos^2\left(\frac\theta2\right)\cos^2\left(\frac{\varphi+\lambda}{2}\right)}}{\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)}\right)}{2\pi}}\right)$$ and $$\exp\left(\mathrm{2i\pi\frac{\pi+\frac{\varphi+\lambda}{2}-\arctan\left(-\frac{\sqrt{1-\cos^2\left(\frac\theta2\right)\cos^2\left(\frac{\varphi+\lambda}{2}\right)}}{\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)}\right)}{2\pi}}\right).$$

Thus, the value you get isn't only dependent on $\theta$, but also on $\varphi+\lambda$.

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    $\begingroup$ @IGY Well, $\chi$ is a polynomial, so I used $X$ to denote it, even though its roots are denoted $\Lambda$. You're right that I forgot to put it on the bottom-right entry of the determinant though! $\endgroup$ Jan 26 at 16:00
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    $\begingroup$ Thank you! That helps:) $\endgroup$
    – IGY
    Jan 26 at 16:01

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