# Expectation Value of a Stabilizer

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• Johny Boy
In summary: Therefore$$S_M e^{-i\frac{\theta}{2}\sum_{i=0}^{n}X_i} \= S_M + i\frac{\theta}{2}\sum_{i=0}^{n}X_iS_M + \frac{\theta^2}{4}\sum_{i=0}^{n}X_i^2S_M + ...$$The ##X_i^2## terms all commute with each other and with ##S_M##, and so on. So S_M e^{-i\frac{\theta}{2}\sum_{i=0}^{n}X_i} = e^{-i \theta \sum_{i=0}^{n}X 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},
$$but in order to do this I assumed that ##S_M e^{-i\frac{\theta}{2}\sum_{i=0}^{n}X_i}## is Hermitian. We only know that ##S_M## is Hermitian and unitary (being Pauli operators) and ##e^{-i\frac{\theta}{2}\sum_{i=0}^{n}X_i}## is unitary. What am I missing that allows the above simplification? Thanks for any assistance. Last edited: This was giving fits until I realized that ##X_i## is meant to be a Pauli ##X## operator. With that it's easy: ##YX = -XY## and ##ZX = -ZY##. Assuming that "##S_M## is entirely ##Y## and ##Z## operators" means$$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.

## What is the expectation value of a stabilizer?

The expectation value of a stabilizer is a measure of the average value of a physical quantity (such as position, momentum, or energy) for a given quantum state. It is calculated by taking the inner product of the stabilizer operator with the state vector, and then taking the absolute value squared.

## How is the expectation value of a stabilizer used in quantum mechanics?

In quantum mechanics, the expectation value of a stabilizer is used to make predictions about the outcomes of measurements on a quantum system. It allows us to determine the most likely values of physical quantities for a given quantum state.

## What is the relationship between the expectation value of a stabilizer and uncertainty principle?

The expectation value of a stabilizer is related to the uncertainty principle in that it represents the minimum possible uncertainty in the measurement of a physical quantity. This means that the expectation value of a stabilizer is the most accurate prediction we can make about the value of a physical quantity for a given quantum state.

## Can the expectation value of a stabilizer be negative?

No, the expectation value of a stabilizer cannot be negative. This is because it is calculated by taking the absolute value squared of the inner product, which always results in a positive value.

## How is the expectation value of a stabilizer affected by changes in the quantum state?

The expectation value of a stabilizer can change depending on the quantum state. For example, if the state is an eigenstate of the stabilizer operator, the expectation value will be equal to the eigenvalue. However, if the state is not an eigenstate, the expectation value will be a combination of all possible eigenvalues weighted by their respective probabilities.

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