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. $\endgroup$ Feb 12, 2019 at 15:38
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    $\begingroup$ 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. $\endgroup$ Feb 12, 2019 at 15:48
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    $\begingroup$ 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, 2019 at 21:39
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    $\begingroup$ @Mithrandir24601 Yes. Can you please clarify this equality? $\endgroup$
    – bilanush
    Feb 12, 2019 at 22:25
  • $\begingroup$ Do you want absolute values on both sides? Neither? $\endgroup$
    – AHusain
    Feb 13, 2019 at 15:28

1 Answer 1


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, 2019 at 0:16
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    $\begingroup$ I edited the answer to include what you asked now. I assume that by normal measurements you refer to what I was stating. $\endgroup$ Feb 13, 2019 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, 2019 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, 2019 at 22:23
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    $\begingroup$ 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. $\endgroup$ Feb 14, 2019 at 14:24

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