# In what sense are Pauli matrices measurement operators?

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, $$|0\rangle\langle 0|$$ and $$|1\rangle\langle 1|$$, 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 $$|+\rangle\langle +|$$ and $$|-\rangle\langle -|$$ which do not satisfy one of the properties of projectors. As in, $$|+\rangle\langle +| \neq (|+\rangle\langle +|)^2$$. 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!!!

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

$$|+\rangle\langle +| = \begin{bmatrix} \frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}$$

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

• why do you say $|+\rangle\!\langle+|\neq(|+\rangle\!\langle+|)^2$?
– glS
Jul 5, 2021 at 23:02
• it is a property of projectors that $P = P^2$. Seen in exercise 2.16 of Neilson and Chuang: "Show that any projector $P$ satisfies the equation $P^2$ = $P$." Jul 5, 2021 at 23:08
• yes. I'm asking why you think that particular inequality is true. What calculation did you make?
– glS
Jul 5, 2021 at 23:08
• oh boy... must be my monday-brain... Jul 5, 2021 at 23:14
• so I guess the Pauli matrices are projective measurements then? Jul 5, 2021 at 23:15

1. Given any unit vector, $$v\equiv |v\rangle\in\mathcal X$$ for some finite-dimensional complex vector space $$\mathcal X$$, the operator $$vv^\dagger\equiv|v\rangle\!\langle v|$$ defined by $$|v\rangle\!\langle v|\equiv v v^\dagger\in \operatorname{Lin}(\mathcal X), \\ (vv^\dagger)(w)\equiv (|v\rangle\!\langle v|)(|w\rangle) \equiv v \langle v,w\rangle,$$ 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) $$A\in\mathrm{Lin}(\mathcal X)$$, 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 $$|\pm\rangle$$, the corresponding ket-bras $$|\pm\rangle\!\langle\pm|$$ are rank-one projections, and sum to the identity for the corresponding two-dimensional space.