It is well-known that a POVM $\boldsymbol\mu\equiv (\mu_a)_{a\in\Sigma}$ describes outcome probabilities, but not post-measurement outcomes, which in many scenarios exist and are of interest. To actually describe post-measurement outcomes associated to a given measurement, the most natural approach seems to me to be use a channel. As far as I can tell, the most general possible way to write a channel $\Phi$ that corresponds to a measurement $\boldsymbol\mu$ is via the formalism of quantum instruments. That is, to consider a channel of the form $$\Phi(\rho) = \sum_{a\in\Sigma} \Phi_a(\rho)\otimes \mathbb{P}_a, \qquad \mathbb{P}_a\equiv |a\rangle\!\langle a|,$$ with $\Phi_a$ CP trace-non-increasing maps. For $\Phi$ to be compatible with the POVM we need $\operatorname{tr}(\Phi_a(\rho))=\langle \mu_a,\rho\rangle$ for all $\rho$, which should be equivalent to $\Phi_a^\dagger(I)=\mu_a$ with $\Phi_a^\dagger$ adjoint channel of $\Phi_a$.

As discussed in Given a state $\rho$ and operator $0\le \Lambda\le I$, what does $\sqrt\Lambda \rho \sqrt\Lambda$ represent?, and What is the most general way to describe post-measurement states?, it is possible to have channels corresponding to a given POVM that preserve different amounts of coherence. Entanglement-breaking channels would be an extreme case where al of the coherence in the original states is destroyed. At the opposite side of the spectrum, it seems we have "gentle measurements", that is, the channels of the form $$\Phi(\rho) = \sum_{a\in\Sigma} V_a \sqrt{\mu_a} \rho \sqrt{\mu_a}V_a^\dagger \otimes \mathbb{P}_a.$$ for any set of isometries $V_a$.

Is there a way to prove that this choice is indeed the one that optimally preserves the coherence in the input states (assuming this to be the case)? In fact, this question probably boils down to what is a good way to quantify what precisely "preserves the coherence" means here. Given the gentle operator lemma, my guess is that this should mean the channel such that the post-measurement outcomes have minimal (averaged?) trace-distance, but I'm not sure how exactly to do this.

  • $\begingroup$ If I remember correctly, one can show that the TNICP maps in the quantum instrument can always be decomposed as first applying the Lüder's rule $\rho \mapsto \sqrt{\mu_a} \rho \sqrt{\mu_a}$ and then applying a quantum channel $\Phi_a$. Perhaps then you can argue that the Lüder's update is somehow less disturbing via a data-processing inequality argument. $\endgroup$
    – Rammus
    Jan 9, 2023 at 18:12
  • $\begingroup$ @Rammus interesting. Have you got a reference? Is that only a result for quantum instruments, or a way to characterise generic TNICP maps? As in, every TNICP map can be written as $\Phi=\tilde\Phi\circ\Psi$ with $\Psi(\rho)=\sqrt{\Phi^\dagger(I)}\rho \sqrt{\Phi^\dagger(I)}$? $\endgroup$
    – glS
    Jan 9, 2023 at 22:43
  • $\begingroup$ It would just be for instruments and no I don't have a reference in mind but I can try to find one. Also at the moment the question is a bit vague, can you make precise how you measure disturbance? $\endgroup$
    – Rammus
    Jan 9, 2023 at 22:47
  • $\begingroup$ @Rammus I'm not sure tbh. The intuition would be to find the set of post-measurement outcomes corresponding to the least amount of disturbance to the state. For projective measurements, the natural choice is $P_a \rho P_a$ renormalized. The formalism of "general measurements" also uses $\sqrt{\mu_a}\rho\sqrt{\mu_a}$ as a rule to define post-measurement states. But these seem to skim over the fact that multiple channels can be considered compatible with a POVM. So I'm trying to understand if there's a more formally precise way to pinpoints these particular ways people define post-meas states $\endgroup$
    – glS
    Jan 9, 2023 at 23:05
  • $\begingroup$ a plausible candidate would be the set of post-measurement outcomes from which one can get as close as possible to the original state via some other channel, as this would quantify the amount of information lost to the measurement. But I don't know how easy following this route would be $\endgroup$
    – glS
    Jan 9, 2023 at 23:07

1 Answer 1


I'll write down the conclusions reached in the comments of the question. Consider a generic quantum instrument compatible with a POVM $\mu$: $$\Phi(\rho) = \sum_a \Phi_a(\rho) \otimes \mathbb{P}_a,$$ where $\Phi_a$ are trace-non-increasing CP maps with $\Phi_a^\dagger(I)=\mu_a$,or equivalently, $\operatorname{tr}(\Phi_a(\rho))=\langle\mu_a,\rho\rangle$.

Observe that for all $a$, the map $\Psi_a(X)\equiv \Phi_a(\mu_a^{-1/2}X\mu_a^{-1/2})$ is CPTP: it's clearly CP if $\Phi_a$ is, and furthermore $$\operatorname{tr}(\Psi_a(X))=\langle\mu_a, \mu_a^{-1/2}X\mu_a^{-1/2}\rangle=\operatorname{tr}(X).$$ We can now write $\Phi_a(\rho) = \Psi_a(\sqrt{\mu_a} \rho \sqrt{\mu_a})$ for all $a$ and $\rho$, and therefore we finally obtain the decomposition $$\Phi(\rho) = \sum_a \Psi_a(\sqrt{\mu_a} \rho \sqrt{\mu_a})\otimes \mathbb{P}_a$$ for some collection of channels $\Psi_a$. This tells us that $\sqrt{\mu_a} \rho \sqrt{\mu_a}$ are the post-measurement states that optimally preserve coherence, as clearly the least-perturbing channels $\Psi_a$ we can have are isometric channels, and we thus find the "gentle measurement channels" as the ones corresponding to the least amount of perturbation on post-measurement states.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.