# What's the difference between Kraus operators and measurement operators?

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)

• I don't see any mention to "measurement operators" in the text you quote though – glS Apr 25 at 12:59