I am trying to complete this exercise regarding noisy channels. I need to measure a density matrix to get the Kraus operators. However, if I measure, I only get scalars. Can someone please explain how I would measure this density matrix to get the Kraus operators?

$$ \begin{pmatrix} \alpha^2 (1-p) + \beta^2 p & 2 \beta \alpha \sqrt{p}\sqrt{(1-p)} \\ 2 \beta \alpha \sqrt{p} \sqrt{(1-p)} & \beta^2 (p) + \alpha^2 p \end{pmatrix} $$

Here is the exercise:

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It is from this CMU paper: https://quantum.phys.cmu.edu/QCQI/qitd412.pdf

  • $\begingroup$ where does it say there that you are "measuring a density matrix"? The exercise says "measuring $f$", and in Figure 3 $f$ is markedly not $\rho$ $\endgroup$
    – glS
    Commented May 24 at 8:46
  • $\begingroup$ f is just a qubit right? And I can represent its state as a density matrix right? and to get that density matrix I just trace subsystem A out of the whole system? I'm not sure though. $\endgroup$ Commented May 24 at 17:49
  • $\begingroup$ Yes. Thought I don't know how that's related to the question at hand $\endgroup$
    – glS
    Commented May 25 at 9:56

1 Answer 1



If you measure a qubit in the computational basis without getting the result of the measurement the corresponding channel can be described by the Kraus operators $K_0 = |0\rangle \langle 0|$ and $K_1 = |1\rangle \langle 1|$ where these projectors act in the subspace of the measured qubit. They do not depend on the initial state of your qubit.

Some more details

Your question looks very similar to this one: Is an unobserved measurement represented by a mixed state? There, you will find a description on how to handle the general case of any measurement being described by Kraus operators.

You are right that if you measure you get scalars. They are the eigenvalues of the corresponding operators that stand for the measurement. The Kraus representation does something different. It describes how your quantum state changes under a given channel. That channel can be a measurement (with known or with unknown outcome), but it does not have to be a measurement.

  • $\begingroup$ true, but I think in the case at end the exercise refers instead to a circuit where an ancillary qubit is measured, and the Kraus operators describe how measuring the ancilla "transforms" the (non-measured) state. $\endgroup$
    – glS
    Commented May 24 at 16:29

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