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In what sense are Pauli matrices measurement operators?

Quantum Computing Asked on July 15, 2021

Neilson and Chuang’s textbook shows a nice example of measuring in the $Z$ basis on page 89 in section 2.2.5. The Hermitians for measuring in the $Z$ basis, $|0ranglelangle 0|$ and $|1ranglelangle 1|$, satisfy the definition of being a projective measurement. However, if we measure in the $X$ basis (i.e. using the $X$ Pauli operator), by the same logic we get the Hermitians $|+ranglelangle +|$ and $|-ranglelangle -|$ which do not satisfy one of the properties of projectors. As in, $|+ranglelangle +| neq (|+ranglelangle +|)^2$. In the textbook, it says the Hermitians $P_m$ making up the projective measurement operator $M$ must be projectors, but in the $X$ basis they are not projectors.

Am I doing something wrong here?…

EDIT: I was wrong with my calculations!!!

$|+rangle = begin{bmatrix} frac{1}{sqrt{2}} frac{1}{sqrt{2}}
end{bmatrix}$

$|+ranglelangle +| = begin{bmatrix} frac{1}{2} & frac{1}{2} frac{1}{2} & frac{1}{2}
end{bmatrix}$

$(|+ranglelangle +|)^2 = begin{bmatrix} frac{1}{2} & frac{1}{2} frac{1}{2} & frac{1}{2}
end{bmatrix}$

One Answer

  1. Given any unit vector, $vequiv |vrangleinmathcal X$ for some finite-dimensional complex vector space $mathcal X$, the operator $vv^daggerequiv|vrangle!langle v|$ defined by $$|vrangle!langle v|equiv v v^daggerin operatorname{Lin}(mathcal X), (vv^dagger)(w)equiv (|vrangle!langle v|)(|wrangle) equiv v langle v,wrangle,$$ is a rank-one projection. This means, in particular, that $(vv^dagger)^2=vv^dagger$. I'm including here both the bra-ket and the more standard linear algebraic way to denote these objects, for better clarity.

  2. Given any normal matrix (and thus in particular any Hermitian matrix) $Ainmathrm{Lin}(mathcal X)$, there is an orthonormal set of eigenvectors of $A$ that is a basis for $mathcal X$.

  3. Given any orthonormal basis of vectors $v_k$ for a finite-dimensional vector space $mathcal X$, the corresponding projections $v_k v_k^dagger$ sum to the identity (and thus form a projective measurement).

So, for example, taking $A=X$, taking any orthonormal basis of eigenvecors of $X$, such as $|pmrangle$, the corresponding ket-bras $|pmrangle!langlepm|$ are rank-one projections, and sum to the identity for the corresponding two-dimensional space.

Correct answer by glS on July 15, 2021

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