Equation 43 · Part 3 · The Entry a Relabeling Cannot Write

Symbol W^dagger

\Lambda^\dagger(O_c) = W^\dagger O_c W

W^\dagger

What this part means

$W^d$ agger is an input to the expression that computes the quantity on the left.

Its job in the formula

$W^d$ agger is an input to the expression that computes the quantity on the left.

Full expressionSymbol W^daggerArticle meaning

The passage around this formula

The dual of conjugation by W is conjugation by $W^\dagger$ : from the defining pairing, $\mathrm{Tr}$ [W $\rho$ $W^\dagger$ $O_c$ ] = $\mathrm{Tr}$ [ $\rho$ $W^\dagger$ $O_c$ W] holds for every $\rho$ , so $\Lambda^\dagger(O_c)$ = $W^\dagger$ $O_c$ W . Substituting,

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Learn the underlying idea

An exponent tells how a base is used in multiplication. In x³, x is the base and 3 is the exponent: x³ = x × x × x.

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