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.


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

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?

  • $\begingroup$ 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. $\endgroup$ – Sanchayan Dutta Feb 12 at 15:38
  • 3
    $\begingroup$ 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. $\endgroup$ – Josu Etxezarreta Martinez Feb 12 at 15:48
  • 1
    $\begingroup$ 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) $\endgroup$ – Mithrandir24601 Feb 12 at 21:39
  • 1
    $\begingroup$ @Mithrandir24601 Yes. Can you please clarify this equality? $\endgroup$ – bilanush Feb 12 at 22:25
  • $\begingroup$ Do you want absolute values on both sides? Neither? $\endgroup$ – 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:


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:

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


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.

  • $\begingroup$ 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? $\endgroup$ – bilanush Feb 13 at 0:16
  • 1
    $\begingroup$ I edited the answer to include what you asked now. I assume that by normal measurements you refer to what I was stating. $\endgroup$ – Josu Etxezarreta Martinez Feb 13 at 13:51
  • $\begingroup$ Thank you very much. There are still a few things I don't understand here. But you helped me. $\endgroup$ – bilanush Feb 13 at 17:45
  • $\begingroup$ 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 $\endgroup$ – bilanush Feb 13 at 22:23
  • 1
    $\begingroup$ 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. $\endgroup$ – Josu Etxezarreta Martinez Feb 14 at 14:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.