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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.

1Surely this doesn’t work? The trace is invariant under cyclic permutations so in the first equation you move the last term to the front, and combine it with the first term to give the identity? – DaftWullie – 2019-11-15T06:35:39.873

@DaftWullie There was a typo there, apologies. The last term is the density operator $G$ (graph state) in the first equation. – John Doe – 2019-11-15T06:39:36.113

And X_i are Pauli X matrices on qubit i? – DaftWullie – 2019-11-15T06:58:55.590

@DaftWullie That's correct. – John Doe – 2019-11-15T07:00:31.157