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I read the section in the attached picture about the stabilizer formalism and was wondering about the last sentence in the pic. It says that all operators of the stabilizer group are hermitian, because the negative unity operator is not included, but isn't that operator hermitian as well?

Section about stabilizer formalism

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    $\begingroup$ please be so kind as to properly typeset the equations rather than using images. $\endgroup$ – ZeroTheHero Jan 17 at 2:13
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Notice that an element $A$ of the Pauli group is either Hermitian and then $A^2=I$, or anti-Hermitian and then $A^2=-I$.

If the stab' group included a non-Hermitian element, it would also include it's square, $-I$. Hence if it doesn't include $-I$ it cannot include non-Hermitian elements.

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  • $\begingroup$ ok i get your first point, but why would the stabilizer group also include the square of the non-hermitian element? $\endgroup$ – Serwj Jan 19 at 23:45
  • $\begingroup$ @Serwj Because groups are closed under group-multiplication. $\forall a,b \in G: ab \in G $ so in particular $\forall a \in G: a^2 \in G $ $\endgroup$ – Tal Sheaffer Jan 20 at 2:56

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