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

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?

• Please use MathJax for properly typesetting mathematical expressions and use the appropriate tags. Go through How to write a good question?. I've edited it this time. – Sanchayan Dutta Feb 12 at 15:38
• What are $P$ and $M$ in your question? That should add some clarity to the question. Also what do you mean by why do they equal each other? Adding a reference to which page of the Nielsen and Chuang are you reading will also be helpful. – Josu Etxezarreta Martinez Feb 12 at 15:48
• I assume it's equation 4.64 (Box 4.1: Approximating quantum circuits) that you're referring to? (page 195 in the one I've got) – Mithrandir24601 Feb 12 at 21:39
• @Mithrandir24601 Yes. Can you please clarify this equality? – bilanush Feb 12 at 22:25
• Do you want absolute values on both sides? Neither? – AHusain Feb 13 at 15:28

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.

• Thanks. But is there any intuitive/ mathematical way to understand this equality p(m)=⟨ψ|M|ψ⟩.??? Or any Link explaining why this is the connection between povm to normal measurements? – bilanush Feb 13 at 0:16
• I edited the answer to include what you asked now. I assume that by normal measurements you refer to what I was stating. – Josu Etxezarreta Martinez Feb 13 at 13:51
• Thank you very much. There are still a few things I don't understand here. But you helped me. – bilanush Feb 13 at 17:45
• Why when you multply ⟨ψU|M |ψU⟩ you take the dagger of the first state and U?? The book says this is just a notation for inner product |ψ> with M |ψ⟩. So why did you take the dagger of the first? You referred to it as a bra instead of a ket? When the book says it's ket by ket in general – bilanush Feb 13 at 22:23
• Yeah, it is indeed a ket by ket inner product, but the standard inner product in linear algebra is defined as $x \cdot y=x^\dagger y$ if $x$ and $y$ are complex column vectors. That inner product between column vectors is translated then to the bra-ket notation as $\langle\phi|\psi\rangle$ if $|\phi\rangle$ and $|\psi\rangle$ are quantum states. Remenber that the bra-ket is just a notation of standard linear algebra for QM. – Josu Etxezarreta Martinez Feb 14 at 14:24