Why does $|P_U − P_V |$ equal $\langle \psi |U^{\dagger} M U|\psi\rangle −\langle \psi |V^{\dagger} M V |\psi\rangle$?

Question

In QC and QI by Chuang and Nielsen, they state that the $P_U$ of operation $U$ acting on $\psi$ can be reached by $\langle \psi |U^{\dagger} M U |\psi\rangle$.

Where $P_U$ (or $P_V$) is the probability of obtaining the corresponding measurement outcome if the operation $U$ (or $V$). And $M$ is a POVM measurement element.

Then

$$|P_U − P_V | = \langle \psi |U^{\dagger} M U|\psi\rangle −\langle \psi |V^{\dagger} M V |\psi\rangle.$$

This equality appears in the book on page 195 (Box 4.1: Approximating quantum circuits; equation 4.64).

I don't understand it. Can anyone explain it? And why do they equal each other?

Answer 1

The probability that outcome $m$ associated with POVM measurement $M$ comes out after measuring state $|\psi\rangle$ can be calculated by:

$p(m)=\langle\psi|M|\psi\rangle$.

The box in the Isaac and Chuang book says that $P_U$ is the probability of such outcome if $U$ operation is applied, and $P_V$ if $V$ is applied. Consequently, we want to calculate such probabilities for states:

• $|\psi_U\rangle=U|\psi\rangle$
• $|\psi_V\rangle= V|\psi\rangle$

Applying the definition for calculating such probabilities that I presented at the beginning, then you can obtain what you need:

• $P_U=\langle\psi_U|M|\psi_U\rangle=(U|\psi\rangle)^\dagger MU|\psi\rangle=\langle\psi|U^\dagger MU|\psi\rangle$
• $P_V=\langle\psi_V|M|\psi_V\rangle=(V|\psi\rangle)^\dagger MV|\psi\rangle=\langle\psi|V^\dagger MV|\psi\rangle$

EDIT:

To follow the question you gave in the comment to the answer. Postulate 3 of quantum mechanics states that those are described by a collection of measurement operators $\{M_m\}$ related with each of the outcomes $m$ that the quantum state $|\psi\rangle$ can have. Such postulate does also state that the probability to get outcome $m$ is given by

$p(m)=\langle\psi|M_m^\dagger M_m|\psi\rangle$.

POVM measurements are given by a collection of positive operators $E_m$ that fullfil that $\sum_m E_m=I$. Such operators can be related with the measument operators like

$E_m\equiv M_m^\dagger M_m$.

All this is stated in the Isaac and Chuang book on quantum computation and information that seems that you are using, so refer there for more complete details.