In these notes, on page 13, there is a claim:

we have $$ \sum_{\mu, a} N_\mu M_a |\psi \rangle \langle \psi|M_a^\dagger N_\mu^\dagger = |\psi \rangle \langle \psi|. $$ Since the left-hand side is a sum of positive operators, the equation can only hold if each of these terms is proportional to $|\psi\rangle \langle \psi|$, hence $$ N_\mu M_a = \lambda_{\mu a} I. $$

I do not understand how it follows from positivity (and the form of the terms) that they are all proportional to the RHS.

  • 4
    $\begingroup$ Hint: What happens if you apply each side to a state orthogonal to $|\psi\rangle$ $\endgroup$ Jan 31, 2018 at 14:52
  • 2
    $\begingroup$ @BySymmetry you get 0 on RHS and on left-hand side sum of $| \langle v|N_\mu M_a |\psi \rangle |^2$ terms which are all $\geq 0$ so that they each must be zero. Therefore, $|v \rangle$ is orthogonal to $N_\mu M_a |\psi \rangle$ for each $\mu, a$ and since $|v \rangle$ was any orthogonal state it implies that $N_\mu M_a |\psi \rangle$ is proportional to $|\psi \rangle$. Is this correct? Also, I think that positivity is implicitly used due to the form of the LHS terms but that it itself is not enough. Is this true? Thank you for the hint. $\endgroup$
    – user110503
    Jan 31, 2018 at 15:47


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