Graduate Expectation Value of a Stabilizer

[Johny Boy]

TL;DR

Interested in the form of the expectation value of a stabilizer operator used in a quantum information paper on graph states.

Given that operator $S_M$, which consists entirely of $Y$ and $Z$ Pauli operators, is a stabilizer of some graph state $G$ i.e. the eigenvalue equation is given as $S_MG = G$ (eigenvalue $1$).

In the paper 'Graph States as a Resource for Quantum Metrology' (page 3) it states that the expectation value is given by

\begin{align*}
\langle S_M \rangle &= \text{Tr}(e^{i\frac{\theta}{2}\sum_{i=0}^{n}X_i}S_M e^{-i\frac{\theta}{2}\sum_{i=0}^{n}X_i}G) \\
&= \text{Tr}(e^{i \theta \sum_{i=0}^{n}X_i}G).
\end{align*}

It seems that they are working in the Heisenberg picture and the above equations imply that
$$
S_M e^{-i\frac{\theta}{2}\sum_{i=0}^{n}X_i} = (S_M e^{-i\frac{\theta}{2}\sum_{i=0}^{n}X_i})^{\dagger} = e^{i\frac{\theta}{2}\sum_{i=0}^{n}X_i^\dagger}S_{M}^{\dagger} = e^{i\frac{\theta}{2}\sum_{i=0}^{n}X_i}S_{M},
$$
but in order to do this I assumed that $S_M e^{-i\frac{\theta}{2}\sum_{i=0}^{n}X_i}$ is Hermitian. We only know that $S_M$ is Hermitian and unitary (being Pauli operators) and $e^{-i\frac{\theta}{2}\sum_{i=0}^{n}X_i}$ is unitary. What am I missing that allows the above simplification?

Thanks for any assistance.

 

[tnich]

This was giving fits until I realized that $X_i$ is meant to be a Pauli $X$ operator. With that it's easy: $YX = -XY$ and $ZX = -ZY$. Assuming that "$S_M$ is entirely $Y$ and $Z$ operators" means
$$S_M = \sum_{i=0}^{n}W_i$$
where $W_i$ is $X_i$ or $Z_i$, then
$$S_Me^{-i\frac{\theta}{2}\sum_{i=0}^{n}X_i}\
=S_M \sum_{k=0}^{\infty} \big(-i\frac{\theta}{2}\sum_{i=0}^{n}X_i\big)^k$$
For even values of $k$, $(\sum_{i=0}^{n}X_i)^k$ is the identity operator. So $S_M$ and $(\sum_{i=0}^{n}X_i)^k$ commute for even values of $k$ and anti-commute for odd values.