Generally speaking, the description of post-measurement states associated with a POVM seems to always pass through, in some form or another, the formalism of Kraus operators. For example:

  1. Consider a general channel $\Phi$ with Kraus representation $\Phi(\rho)=\sum_a A_a \rho A_a^\dagger$. The collection of operators $A_a^\dagger A_a$ is then always a POVM. Furthermore, via Stinespring, $\Phi$ can always be written as $\Phi(\rho)=\operatorname{Tr}_2[V_\Phi \rho V_\Phi^\dagger]$ with the isometry $V_\Phi$ defined as $$V_\Phi = \sum_a A_a\otimes |a\rangle, \qquad V_\Phi |\psi\rangle = \sum_a A_a |\psi\rangle\otimes |a\rangle,$$ so that $$V_\Phi \rho V_\Phi^\dagger = \sum_{a,b} A_a \rho A_b^\dagger\otimes |a\rangle\!\langle b|.$$ This allows to interpret $\Phi$ as the act of performing the measurement $\{A_a^\dagger A_a\}$, with post-measurement states (up to renormalisation) $A_a \rho A_a^\dagger$, with this operation realised by evolution through the isometry $V_\Phi$, with the "promise" of not accessing the coherences in the second degree of freedom.

  2. The so-called formalism of general measurements, which given a POVM with elements written as $A_a^\dagger A_a$ prescribes post-measurement states of the form $$\frac{A_a \rho A_a^\dagger}{\operatorname{Tr}[A_a^\dagger A_a \rho]},$$ seems to be essentially equivalent to the latter approach.

  3. Let us start now from a POVM $\mu$. There are infinitely many ways to decompose its elements as $\mu_a=A_a^\dagger A_a$ for a collection of Kraus operators $A_a$, and thus multiple ways to define post-measurement states corresponding to measuring $\mu$, as discussed in this answer to a related question. The gist, again, seems to be that we can consider the channel $\Phi$ with $$\Phi(\rho) = \sum_a A_a \rho A_a^\dagger\otimes |a\rangle\!\langle a|,$$ which thus provides a natural way to formalise having both classical and quantum outcome to the measurement (I'm aware this kind of thing is generally referred to as a quantum instrument, but I'm not very familiar with this formalism so I don't know how much more there is to say about it in this context).

On the other hand, given a POVM $\mu$, consider an entanglement-breaking channel $\Phi$ of the form $$\Phi(\rho) = \sum_a \langle \mu_a ,\rho\rangle \sigma_a\equiv \sum_a \operatorname{Tr}(\mu_a\rho) \sigma_a.$$ To some degree, this kind of channel seems to be an even more apt description of a measurement process: we see directly how each outcome $a$ corresponds to a post-measurement state $\sigma_a$. In contrast, the formalism with Kraus operators connects measurement outcomes probabilities and post-measurement states more tightly: $A_a \rho A_a^\dagger$ describes both at the same time, and we can't choose arbitrary post-measurement states to attach to each outcome $a$ (although we do have some freedom, as discussed in the linked post above).


This brings me to the question: why should we consider as the most general description of post-measurement outcomes the formalism with Kraus operators (in one of the shapes outlined above, which I'm considering as all essentially equivalent for the purpose of this discussion)? Why not describe instead post-measurement states via generic entanglement breaking channels attached to the POVM, which would allow to describe more general situations as far as the allowed post-measurement states are concerned?

I have a vague hunch that a possible answer lies in the fact that an entanglement breaking channel of the form outlined above might have a Choi rank larger than the number of outcomes in the corresponding POVM, and thus in some sense would describe a different measurement with a larger number of possible outcomes. I'm not completely sure how to properly formalise this though.

  • $\begingroup$ The Kraus representation is always useful because it comes from the general posterior state formula $\widehat{\rho|_P} = \frac{P\rho P}{\mathrm{Tr}(\rho P)}$ where $P$ is a Boolean measurement. Your question partially boils down to where the posterior state formula comes from. The short answer is that it is motivated from the simpler posterior state precept for vector states. BTW, I'm not sure what you mean by EBT maps being more general than the Kraus operator formalism -- it seems from Theorem 4 here that EBT maps do have Kraus representations. $\endgroup$ Aug 8, 2022 at 11:36
  • $\begingroup$ @SanchayanDutta yes, I'm saying that formula is essentially equivalent to interpreting a Kraus representation of a channel in a specific way. Every channel has a Kraus representation, of course, I'm referring here to whether it is meaningful to use the decomposition characteristic of an entanglement-breaking channel as a way to specify a measurement, rather than (one of) its Kraus decomposition(s). This is also what the last paragraph is about btw: the Choi rank is the minimum number of elements in a Kraus decomposition of the channel. $\endgroup$
    – glS
    Aug 8, 2022 at 12:51
  • $\begingroup$ As far as I understand, the decomposition characteristic of an EB channel is what Horodecki et al. call the Holevo form. Under the channel-state duality, an EB channel corresponds to a separable state. You can't always write general POVM measurements (specified with Kraus operators $E_i = A_iA_i^\dagger$) in the Holevo form, because there's a theorem in finite dimensions stating that a channel is EB iff it has a Holevo (measure-and-prepare) form iff the corresponding state is separable. General POVM measurements are not of this form. $\endgroup$ Aug 8, 2022 at 19:02
  • $\begingroup$ @SanchayanDutta yea, that's a good point. There's plenty of channels that are not EB, but can still be interpreted as a measurement in the above sense. Difference being then I guess in the amount of coherence left after the measurement (which also connects nicely with the answer in the linked post). EB channels are a particular case b/c they describe situations where enough info is extracted from states that there is no a priori relation between pre- and post-measurement states anymore. Whereas general ch allow to describe "weaker" measurements, which might perturbe states less. Makes sense $\endgroup$
    – glS
    Aug 9, 2022 at 14:02

1 Answer 1


TL;DR: There are three possible fates that befall quantum information$^1$ subjected to measurement: it may be converted to classical information, it may be lost in the environmental degrees of freedom or it may be retained in the system under study. Entanglement-breaking channels describe processes in which quantum information is either lost or converted to classical information. Therefore, this type of channel fails to account for the measurements that preserve some of the original quantum information. Moreover, in a world where all measurements are described by entanglement-breaking channels, standard scheme for quantum error correction doesn't work.

Naimark dilation

The above idea is formalized in Naimark dilation theorem which states that any measurement may be realized as a unitary on the Hilbert space $\mathcal{H}_A\otimes\mathcal{H}_E\otimes\mathcal{H}_M$ describing the system under study $A$, its environment $E$ and a measurement apparatus $M$, followed by a projective measurement on $M$ and a partial trace over $E$. The partial trace can be thought of as a projective measurement whose results have been lost due to our lack of control over the environmental degrees of freedom.

This allows us to classify measurements in terms of quantum information they preserve in $A$, the quantum information they lose in $E$ and the classical information they produce in $M$. For example, a unitary on $A$ preserves all quantum information in $A$, loses nothing in $E$ and obtains no classical information in $M$. As another example, a fully depolarizing channel $\mathcal{D}(\rho)=\frac{I}{N}$ loses all input quantum information while obtaining no classical information.

Partial measurements

Similarly, a projective measurement obtains some classical information while losing no quantum information to $E$, i.e. it keeps all information within $A$ and $M$. If the projective measurement is degenerate then the measurement outcome does not uniquely identify the post-measurement state so some information remains in $A$. If the measurement is non-degenerate then the outcome completely determines the post-measurement state and thus all information passes from $A$ to $M$. In both cases, the measurement is efficient in the sense that no information is lost in $E$.

Suppose we measure $|\psi\rangle=\alpha|00\rangle+\beta|01\rangle+\gamma|10\rangle+\delta|11\rangle$ using $P_0=\mathrm{diag}(1, 0, 0, 1)$ and $P_1=\mathrm{diag}(0, 1, 1, 0)$ and obtain outcome $0$. The post-measurement state is $\frac{\alpha|00\rangle+\delta|11\rangle}{\sqrt{|\alpha|^2+|\delta|^2}}$. Note that the set of all possible post-measurement states associated with this outcome is infinite. Thus, the knowledge of the measurement outcome does not uniquely identify the post-measurement state. In fact, the set contains an infinite number of pure states. Therefore, the measurement leaves some "uncollapsed" quantum information behind. At the same time, the classical information it obtains is only partial - it yields the parity of the two qubits but fails to distinguish between $|00\rangle$ and $|11\rangle$.

This possibility is the reason why general measurement cannot be described using generic entanglement breaking channels. Those channels correspond to an extreme special case of the measurement process. Namely, the case where all quantum information is either lost or converted to classical information. This can be shown more rigorously by noting that on one hand the output of any entanglement breaking channel is a convex combination of a finite set of pure states and on the other hand the set of post-measurement states of a partial measurement, like the parity measurement above, may contain an infinite number of pure states.

Proof by quantum error correction

There is an interesting connection to quantum error correction. If entanglement-breaking channels of the form $$ \Phi(\rho) = \sum_i\mathrm{tr}(\mu_i\rho) \sigma_i\tag1 $$ were sufficient to describe general measurements, then the standard scheme for quantum error correction - which entails syndrome measurements for error diagnostics followed by active error correction$^2$ - would be impossible.

To see this, let $\mathcal{E}:L(\mathbb{C}^2)\to\mathcal{H}$ denote an operation that encodes a logical qubit into a quantum system with Hilbert space $\mathcal{H}$, let $\mathcal{N}:\mathcal{H}\to\mathcal{H}$ represent the noise in the system, let the set of non-trace-preserving operations $\mathcal{S}_a:\mathcal{H}\to\mathcal{H}$ represent the syndrome measurements labeled by syndromes $a$, let $\mathcal{R}_a:\mathcal{H}\to\mathcal{H}$ represent the unitary recovery operations conditional on the classical syndrome information $a$ and finally let $\mathcal{D}:\mathcal{H}\to L(\mathbb{C}^2)$ denote the decoding operation. Quantum error correction is successful if $$ \mathcal{D}(\mathcal{R}_a(\mathcal{S}_a(\mathcal{N}(\mathcal{E}(\rho)))))=p(a|\rho)\cdot\rho\tag2 $$ where $p(a|\rho)=\mathrm{tr}(\mathcal{S}_a(\mathcal{N}(\mathcal{E}(\rho))))$ is the probability of measuring syndrome $a$.

Now, suppose that for some $a$, the measurement operation $\mathcal{S}_a$ can be written as $\mathcal{S}_a(\rho) = \sum_i\mathrm{tr}(\mu_i\rho)|\psi_i\rangle\langle\psi_i|$ for some finite set of pure states $\{|\psi_i\rangle\}$. In this case, $(2)$ fails, because the left-hand side is a convex combination of a finite set of states $\{\mathcal{D}(\mathcal{R}_a(|\psi_i\rangle\langle\psi_i|))\}$ while for any pure $\rho$ the right-hand side is an extreme point of the set of single-qubit states and thus not a convex combination of any finite set of states that doesn't include $\rho$. The case of $\mathcal{S}_a$ of the form $\mathcal{S}_a(\rho) = \sum_i\mathrm{tr}(\mu_i\rho)\sigma_i$ for arbitrary $\sigma_i$ can be reduced to the preceding case by expanding $\sigma_i$ as a convex combination of pure states.

Thus, the fact that general quantum measurement allows for the possibility of obtaining partial information about a quantum state while leaving behind a state collapsed to a subspace whose dimension is greater than one is key to the feasibility of quantum error correction.

$^1$ Here, we use the term "information" informally as "the thing that information processing devices process". Similarly, "quantum information" is understood informally as "the stuff that quantum information processing devices process". In particular, here the latter term does not denote von Neumann entropy.
$^2$ In practice, active error correction is avoided and replaced by classical bookkeeping using the Pauli frame.

  • $\begingroup$ I like this perspective, but I'm not sure about saying that "projective measurements lose no quantum information to the environment". That's only true within their degenerate eigenspaces. An IC PVM will of course lose all quantum information (and correspond to some EB channel). $\endgroup$
    – glS
    Sep 1, 2022 at 20:38
  • $\begingroup$ Thanks! In the first paragraphs I used the term "information" rather informally to give some intuition behind the more rigorous argument later in the post. I just added some clarification about what I meant. In the terms I use in the answer, any non-degenerate PVM looses all information in $A$, but that information "goes" to $M$, not $E$. This way of talking about the process is meant to convey the fact that we learn the measurement outcome and hence have no uncertainty about the post-measurement state. This is in contrast to an EB channel that looses that information yielding a mixture. $\endgroup$ Sep 1, 2022 at 21:16

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