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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.

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2 Answers 2

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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.

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  • $\begingroup$ Great! How did you find those matrix equations though? Doesn't seem very intuitive to me. :) $\endgroup$
    – user568976
    Jun 4, 2019 at 4:52
  • $\begingroup$ Hi, I edited with ideas about how you can prove that these matrices are essentially unique. I'll leave the algebra to you though :) $\endgroup$ Jun 4, 2019 at 5:07
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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 $U_Y^\dagger = HV^\dagger$.

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