# Expectation Value of a Stabilizer

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• Johny Boy

#### Johny Boy

TL;DR Summary
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.

Last edited:

$$S_M = \sum_{i=0}^{n}W_i$$
$$S_Me^{-i\frac{\theta}{2}\sum_{i=0}^{n}X_i}\ =S_M \sum_{k=0}^{\infty} \big(-i\frac{\theta}{2}\sum_{i=0}^{n}X_i\big)^k$$