Why are the $H$-Gate and the $S^{\dagger}$-Gate used for measurements with the $\{|+\rangle,|-\rangle\}$-basis, etc

Question

In the context of quantum tomography we measure first at the computational basis $\{|0\rangle,|1\rangle\}$. Here we have a Circuit $C$ and measure: $C|0\rangle\rightarrow measure$

The second step is to measure at the basis: $\{|+\rangle,|-\rangle\}$. For that we use at the very first the $H$-Gate, and just before the measurement too: $HCH|0\rangle\rightarrow measure$

The third step is to measure at the basis $\{|i\rangle,|-i\rangle\}$. For that we use at the very first the $S\cdot H$ (I am not sure about $S$), and just before the measurement $H \cdot S^{\dagger}$: $H \cdot S^{\dagger}\cdot C\cdot S\cdot H |0\rangle\rightarrow measure$

My Question is: What is the relation between $U_1=H, U_2=H \cdot S^{\dagger}$ and the three bases?

Answer 1

We'll use the following facts:

• A quantum state on a qubit is uniquely defined by its position on the Bloch sphere;
• The cartesian coordinates $\begin{pmatrix}x\\y\\z\end{pmatrix}$ of such a quantum state $\rho$ on the Bloch sphere are $\newcommand\tr[1]{\mathrm{tr}\left[#1\right]} \begin{pmatrix}\tr{\rho X}\\\tr{\rho Y}\\\tr{\rho Z}\end{pmatrix}$
• When we measure a state in the computational basis, we actually measure in the $Z$ basis, meaning that it's almost like we were computing $\tr{\rho Z}$. More precisely, if $p_0$ is the probability to measure $|0\rangle$ when measuring in the computational basis, then $\tr{\rho Z}=2p_0-1$.

Now, how can we evaluate $\tr{\rho X}$? Well note that we have $X=HZH$, so the quantity we want to compute is actually $\tr{\rho HZH}=\tr{H\rho HZ}$. We can see this as measuring the state $H\rho H$ in the computational basis, which is nothing but the state $\rho$ onto which we have applied an $H$ gate!

Similarly, we have $Y=(SH)Z(SH)^\dagger$, so in order to evaluate $\tr{\rho Y}=\tr{(SH)^\dagger\rho (SH) Z}$, we simply have to apply a $(SH)^\dagger=HS^\dagger$ gate before measuring.

All in all, the relation you are looking for is the one that allows to transform a Pauli $X$ or $Y$ into a $Z$.

Answer 2

M has the stabilizer flow Z -> output.

H has the stabilizer flow X -> Z

S_DAG has the stabilizer flow Y -> X.

So H then M has the flow X -> Z -> output (i.e. measures X)

So S_DAG then H then M sends Y -> X -> Z -> output (i.e. measures Y)