# Phase estimation using $U_3$ gate

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!

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 $$\varphi+\lambda$$.

• @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! Jan 26 at 16:00
• Thank you! That helps:)
– IGY
Jan 26 at 16:01