Computing the Bloch sphere representation of an arbitrary operator in $U(2)$

Question

Computing the Matsumoto-Amano normal form of an operator in $U(2)$ involves finding the Bloch sphere representation of said operator, see Remarks on Matsumoto and Amano’s normal form for single-qubit Clifford+T operators, Theorem 4.1.

I've been trying to perform that computation for a variety of operators but the general (detailed) procedure still eludes me. For instance, in trying to compute $TYT^\dagger$, I end up with a matrix of the form $\begin{pmatrix} 0 & -ie^{-\frac{i\pi}{4}} \\ ie^{\frac{i\pi}{4}} & 0 \end{pmatrix}$ which I can't reduce to a multiplication of the form $n \times P$ where $n$ is a scalar and $P \in \{X, Y, Z\}$, a Pauli matrix.

How would one find the Bloch sphere representation of, say, the $T$ gate and/or an arbitrary rotation about the $X$ axis?

I'm essentially looking for the actual procedure via an example or two.

Answer 1

Looking at the right hand side Eq (12) of the linked paper, you should not assume that $$ U Y U^\dagger = nP\,, \, P \in \{ X, Y, Z \} \,. $$ Such an equality should hold true only when the second column of the Bloch sphere representation $\hat{U}$ of the operator $U$ has only one nonzero element. But in Eq (13) we see that this does not hold for $\hat{T}$. Instead we should solve $$ T(xX + yY + zZ)T^\dagger = x'X + y'Y + z'Z $$ simultaneously to get $x'$, $y'$ and $z'$. Since $TZT^\dagger = Z$ we immediately see $z'=z$, which is why the bottom right entry of $\hat{T}$ in Eq. (13) is one. For the other two equations we get $$ x' + iy' = \, e^{i\pi/4}(x+iy) $$ $$ x' - iy' = - e^{-i\pi/4}(x-iy) $$ which need to be solved for x' and y'. Finally, you just fill in the matrix elements of $\hat{T}$ such that $$ \hat{T} \left( \begin{array}{l} x \\ y \\ z \end{array} \right) = \left( \begin{array}{l} x' \\ y' \\ z' \end{array} \right) $$ recovers the solutions for $x'$ and $y'$ .