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!!!

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

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

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

  • 2
    $\begingroup$ why do you say $|+\rangle\!\langle+|\neq(|+\rangle\!\langle+|)^2$? $\endgroup$
    – glS
    Jul 5, 2021 at 23:02
  • $\begingroup$ 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 $P^2$ = $P$." $\endgroup$ Jul 5, 2021 at 23:08
  • 2
    $\begingroup$ yes. I'm asking why you think that particular inequality is true. What calculation did you make? $\endgroup$
    – glS
    Jul 5, 2021 at 23:08
  • $\begingroup$ oh boy... must be my monday-brain... $\endgroup$ Jul 5, 2021 at 23:14
  • $\begingroup$ so I guess the Pauli matrices are projective measurements then? $\endgroup$ Jul 5, 2021 at 23:15

1 Answer 1

  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.


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