# unitary that transform $\sigma^x \pm \alpha \sigma^z$ into $\sigma^x$ and $\sigma^z$

Consider $$\sigma^x \pm \sigma^z$$, where $$\sigma^x$$ and $$\sigma^z$$ are Pauli $$X$$ and $$Z$$ matrices. Let unitary $$U$$ be a $$\pi/4$$ rotation matrix around $$Y$$-axis. Then, $$$$U(\sigma^x + \sigma^z)U^\dagger \propto \sigma^x, \quad U(\sigma^x - \sigma^z)U^\dagger \propto \sigma^z,$$$$ where $$\propto$$ refers to the equality sign up to a constant factor.

I want to generalize this problem. Specifically, what would be a unitary $$U$$ that satisfies: $$$$U(\sigma^x + \alpha \sigma^z)U^\dagger \propto \sigma^x, \quad U(\sigma^x - \alpha \sigma^z)U^\dagger \propto \sigma^z,$$$$ where $$\alpha \in \mathbb{R}$$ is a fixed non-zero constant? Does this unitary even exist in the first place?

• $U$ is simply a change in basis for these operators. The eigenvectors of $X+\alpha Z$ gets mapped to the eigenvalues of $X$, which should give most of the structure $U$ immediately. Commented Jul 1 at 21:52

Short answer: Such a unitary $$U$$ exists if and only if $$\alpha=1$$ or $$\alpha=-1$$
Long answer: Because $$\sigma_x-\alpha\sigma_z$$ are traceless matrices with determinant $$-\alpha^2-1$$ their eigenvalues are $$-\sqrt{1+\alpha^2}$$. Thus any unitary $$U$$ which diagonalizes $$\sigma_x-\alpha\sigma_z$$ has to satisfy$${}^1$$ $$U(\sigma_x-\alpha\sigma_z)U^\dagger=\sqrt{1+\alpha^2}\sigma_z\,.\tag1$$ This is our starting point: let $$U$$ be a general element of $$\mathsf{SU}(2)$$, i.e. $$U=\begin{pmatrix} \cos(\phi)e^{i\xi}&\sin(\phi)e^{i\omega}\\ -\sin(\phi)e^{-i\omega}&\cos(\phi)e^{-i\xi} \end{pmatrix}$$ for some $$\phi,\xi,\omega\in\mathbb R$$. We have to tune the parameters such that, equivalently to (1), $$\sqrt{1+\alpha^2}U^\dagger\sigma_z U=\sigma_x-\alpha\sigma_z=\begin{pmatrix}-\alpha&1\\1&\alpha\end{pmatrix}.\tag2$$ A straightforward computation shows that $$\sqrt{1+\alpha^2}U^\dagger\sigma_z U=\begin{pmatrix} \sqrt{\alpha^2+1} \cos (2 \phi ) & \sqrt{\alpha^2+1} e^{i (\omega -\xi )} \sin (2 \phi ) \\ \sqrt{\alpha^2+1} e^{-i (\omega -\xi )} \sin (2 \phi ) & -\sqrt{\alpha^2+1} \cos (2 \phi ) \end{pmatrix}$$ meaning that $$\sqrt{\alpha^2+1} \cos (2 \phi )$$ has to equal $$-\alpha$$. Thus we find $$\cos(2\phi)=\frac{-\alpha}{\sqrt{\alpha^2+1}}$$, that is, $$\phi=\frac12\arccos\Big(\frac{-\alpha}{\sqrt{\alpha^2+1}}\Big)\in[0,\pi]$$ (which is well defined because $$\frac{-\alpha}{\sqrt{\alpha^2+1}}\in(-1,1)$$). With this choice of $$\phi$$ $$\sqrt{\alpha^2+1} \sin (2 \phi )=\sqrt{\alpha^2+1} \sqrt{1-(\cos (2 \phi ))^2}=\sqrt{\alpha^2+1}\sqrt{1-\frac{\alpha^2}{\alpha^2+1}}=1$$ so $$\sqrt{1+\alpha^2}U^\dagger\sigma_z U=\begin{pmatrix} -\alpha & e^{i (\omega -\xi )} \\ e^{-i (\omega -\xi )} &\alpha \end{pmatrix};$$ comparing with (2) shows $$e^{i \omega}=e^{i\xi }$$. But with these choices of $$\phi,\omega$$ one readily verifies that $$U(\sigma_x+\alpha\sigma_z)U^\dagger= \frac1{\sqrt{\alpha^2+1}} \begin{pmatrix} 1-\alpha^2 & -2 \alpha e^{2 i \omega} \\ -2 \alpha e^{-2 i \omega}&\alpha^2-1 \end{pmatrix}$$ which is proportional to $$\sigma_x$$ (in particular: has zero diagonal) if and only if $$\alpha^2=1$$.
1: Actually, $$U(\sigma_x-\alpha\sigma_z)U^\dagger=-\sqrt{1+\alpha^2}\sigma_z$$ would also be allowed but that case is handled analogously.
Let $$A$$ and $$B$$ be your two initial matrices. You have $$\text{Tr}(AB)=2(1-\alpha^2).$$ After the unitary, you have $$\text{Tr}(UAU^\dagger UBU^\dagger)=\text{Tr}(AB)=2(1-\alpha^2).$$ Your aim is to have this value be 0. That can only happen if $$\alpha=\pm1$$.