Expectation Value of Stabilizer

Question

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 $S_MG = G$.

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 help.

Answer 1

The way that I approach the calculation is quite different. I think of $$ e^{i\frac{\theta}{2}\sum_n X_i}=\prod_ne^{i\frac{\theta}{2}X_i}. $$ So, this means that you can consider the term $$ e^{i\frac{\theta}{2}\sum_n X_i}S_Me^{-i\frac{\theta}{2}\sum_n X_i} $$ on a qubit-by-qubit basis. You'll have terms like $$ e^{i\frac{\theta}{2}X}Ze^{-i\frac{\theta}{2}X}, $$ where $Z$ might be replaced by $Y$ (doesn't make any difference to the forthcoming analysis). Recall that $e^{i\frac{\theta}{2}X}=\cos\frac{\theta}{2} I+i\sin\frac{\theta}{2} X$. This lets us use the anti-commutation of $Z$ with $X$: $$ Z(\cos\frac{\theta}{2} I-i\sin\frac{\theta}{2} X)=(\cos\frac{\theta}{2} I+i\sin\frac{\theta}{2} X)Z. $$ This means that our per-qubit term has become $$ e^{i\frac{\theta}{2}X}Ze^{-i\frac{\theta}{2}X}=e^{i\frac{\theta}{2}X}e^{i\frac{\theta}{2}X}Z=e^{i\theta X}Z. $$ So, overall we have $$ \text{Tr}\left(e^{i\theta\sum_n X_i}S_MG\right), $$ at which point we just use that $S_MG=G$.