# How to calculate the unitaries satisfying $U_YXU_Y^\dagger=Y$ and $U_ZXU_Z^\dagger=Z$?

These are the Pauli $$X$$, $$Y$$ and $$Z$$ matrices respectively:

$$X=\begin{bmatrix}0&1\\1&0\end{bmatrix},\ Y=\begin{bmatrix}0&-i\\i&0\end{bmatrix} \text{ and } Z = \begin{bmatrix}1&0\\0&-1\end{bmatrix}.$$

I'm trying to find the $$2\times 2$$ unitary matrices $$U_Y$$ and $$U_Z$$ that satisfy

$$U_YXU_Y^\dagger = Y \text{ and } U_ZXU_Z^\dagger = Z.$$

What would be a quick algorithmic method to calculate $$U_Y$$ and $$U_Z$$? This is the context.

Another approach is to keep in mind that these are similarity relations.

For example, $$U_Z X U_Z^\dagger = Z$$ is the same as $$X=U_Z^\dagger Z U_Z$$, where $$Z$$ is diagonal. This means that this expression is essentially the eigendecomposition of $$X$$, and that the elements of $$Z$$ are the eigenvalues of $$X$$.

Thus, $$U_Z$$ must be the set of eigenvectors of $$X$$. More precisely, the columns of $$U_Z^\dagger$$ are the eigenvectors of $$X$$. Note that via this simple observation you immediately get that $$U_Z$$ must be Hadamard $$H=(X+Z)/\sqrt2$$, as expected.

You can follow similar ideas for the other case. $$U_Y X U_Y^\dagger=Y$$ is equivalent to $$X=U_Y^\dagger Y U_Y$$, but also $$Y=VZV^\dagger$$ where $$V=\frac{1}{\sqrt2}\begin{pmatrix}1 & 1 \\ i & -i\end{pmatrix},$$ which you know immediately if you know the eigenvalues and eigenvectors of $$Y$$. Then, $$X=U_Y^\dagger VZV^\dagger U_Y$$. But then again, this means that $$U_Y^\dagger V=H$$ by the first argument in this answer. You conclude that $$U_Y^\dagger = HV^\dagger$$.

It is easy to check that the following matrix equations hold true:

$$\frac{X+Z}{\sqrt{2}}X\frac{X+Z}{\sqrt{2}}=Z~~,~~\frac{X+Y}{\sqrt{2}}X\frac{X+Y}{\sqrt{2}}=Y$$

Thus algorithmically it suffices to know how to apply the Hadamard- and $$\pi/2$$-gates to your qubits, since up to an unimportant phase

$$H=\frac{X+Z}{\sqrt{2}}~~,~~ S^7Y=e^{-i\pi/4}(\frac{X+Y}{\sqrt{2}})$$

where $$S=\pmatrix{1&0\\0&i}$$.

Usually in quantum computing constructions it is assumed that one can perform an arbitrary rotation of the 1-qubit state, or at least a set of rotations that is universal, so it can approximate an arbitrary rotation with arbitrary accuracy, thus S is considered to be a given, along with H, which is physically realizable by measuring a qubit in the x-axis basis.

EDIT:

In this simple $$2\times2$$ case one can find all the matrices that solve the equations $$U^{\dagger}RU=R'$$ by substituting in the most general form of a unitary matrix, namely:

$$U=\pmatrix{a&b^*\\-b&a^*}$$

which can be derived by imposing the restriction $$U^\dagger U=1$$ and $$|\det U|=1$$ on an arbitrary 2-d matrix. Here a and b are arbitrary complex variables constrained by the condition $$|a|^2+|b|^2=1$$. The proof that the matrices mentioned are the only ones satisfying the equations, modulo an arbitrary but again, uninteresting in quantum computing phase factor, is left to the reader.

• Great! How did you find those matrix equations though? Doesn't seem very intuitive to me. :) – S.D. Jun 4 '19 at 4:52
• Hi, I edited with ideas about how you can prove that these matrices are essentially unique. I'll leave the algebra to you though :) – DinosaurEgg Jun 4 '19 at 5:07