- #1

Johny Boy

- 10

- 0

- 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

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.

*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: