It is said in a lecture note[1] by John Preskill that,

Equivalently, we may imagine measuring system $B$ in the basis $\{|a\rangle\}$, but failing to record the measurement outcome, so we are forced to average over all the possible post-measurement states, weighted by their probabilities. The result is that the initial density operator $\boldsymbol{\rho} = |\psi\rangle\langle \psi|$ is subjected to a linear map $\mathcal{E}$, which acts as

$$\mathcal{E}(\boldsymbol {\rho}) = \sum_a M_a\boldsymbol{\rho} M^{\dagger}_a, \tag{3.32}$$

where the operators $\{M_a\}$ obey the completeness relation eq.(3.25).

The justification for this name will emerge shortly. Eq.(3.32) is said to be an operator-sum representation of the quantum channel, and the operators $\{M_a\}$ are called the Kraus operators or operation elements of the channel.

It seems that Kraus operators and measurement operators are the same thing. Is that right?

[1]: Lecture Notes for Ph219/CS219: Quantum Information Chapter 3 (John Preskill, 2018)

  • 1
    $\begingroup$ I don't see any mention to "measurement operators" in the text you quote though $\endgroup$
    – glS
    Apr 25, 2019 at 12:59

1 Answer 1


Quantum measurement (without results recording) is just a special case of quantum operation (quantum channel). So, yes, measurement operators (as in general measurement formalism) are indeed Kraus operators. But Kraus operators are more general. For example, they can be "rectangular", while measurement operators can't.

  • $\begingroup$ Why couldn't they be rectangular? In a sufficiently generalized notion of measurement, the post-measurement state can easily live in a different Hilbert space than the input state. In an extreme case, the state can cease to exist, like a photon absorbed by the detector. I simply use Mi = 〈φi| for such cases, understood as maps from H to C. $\endgroup$
    – The Vee
    Oct 30, 2019 at 9:39
  • $\begingroup$ I think most books on quantum information theory assume that a measurement doesn't change the Hilbert space where the state lives. It's not convenient to introduce very general definitions from the start. $\endgroup$
    – Danylo Y
    Oct 30, 2019 at 14:22
  • $\begingroup$ In quantum error correction, are the Kraus operators always square matrix? $\endgroup$
    – quest
    May 17 at 2:48
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    $\begingroup$ @quest I'm not quite sure to what you're referring, but if the input and output dimensions of a quantum operation are the same then it can be described by square Kraus operators. $\endgroup$
    – Danylo Y
    May 17 at 12:46

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