# Given a state $\rho$ and operator $0\le \Lambda\le I$, what does $\sqrt\Lambda \rho \sqrt\Lambda$ represent?

An expression that is found in a good number of results is $$\sqrt\Lambda\rho\sqrt\Lambda$$, for some pair of positive semidefinite operators $$\rho,\Lambda\ge0$$. For example, in the gentle operator lemma one considers an operator $$0\le \Lambda\le I$$, which thus can represent some measurement outcome, and a state $$\rho$$, and shows that, roughly speaking, if $$\operatorname{Tr}(\Lambda\rho)\ge1-\epsilon$$ then $$\sqrt\Lambda\rho_x\sqrt\Lambda\simeq\rho_x$$ (on average on $$x$$, in trace distance).

In such cases, it does not seem far-fetched to think of $$\sqrt\Lambda\rho\sqrt\Lambda$$ as something representing the post-measurement state corresponding to measuring $$\rho$$ and finding the outcome $$\Lambda$$. At least the trace of this operator will equal the corresponding outcome probability, and such interpretation would be consistent with the result above: the probability of finding $$\Lambda$$ can only be very close to unity if $$\rho$$ is essentially in the range of $$\Lambda$$.

At the same time, given a POVM $$\sum_a \Lambda_a=I$$, it's not clear whether it is meaningful to interpret the operators $$\sqrt{\Lambda_a}\rho\sqrt{\Lambda_a}$$ as post-measurement outcomes. Especially because different $$\Lambda_a$$ can have overlapping ranges, and thus different $$\sqrt{\Lambda_a}\rho\sqrt{\Lambda_a}$$ also not be orthogonal. I can see that one could define the channel $$\Phi(\rho) = \sum_a \sqrt{\Lambda_a}\rho\sqrt{\Lambda_a},$$ which sends $$\rho$$ into an ensemble with weights the outcome probabilities. Still, I don't know if there is a natural way to interpret the particular states associated to such ensembles.

For concreteness, say $$\Lambda_1=\begin{pmatrix}1-a&0\\0&0\end{pmatrix}$$ and $$\Lambda_2=\begin{pmatrix}a&0\\0&1\end{pmatrix}$$, for some (small enough) $$a>0$$. Then $$\sqrt{\Lambda_1}\rho\sqrt{\Lambda_1} = (1-a)\begin{pmatrix}\rho_{11}&0\\0&0\end{pmatrix}, \qquad \sqrt{\Lambda_2}\rho\sqrt{\Lambda_2} = \begin{pmatrix}a \rho_{11} & \sqrt a \rho_{12} \\ \sqrt a \rho_{21} & \rho_{22}\end{pmatrix}.$$

Is there a better way to interpret $$\sqrt{\Lambda_a}\rho\sqrt{\Lambda_a}$$, as something related to some post-measurement outcomes, or otherwise?

• The post-measurement outcomes of a generalised measurement need not be orthogonal. In general, they will not be (otherwise, you'd pretty much be reduced to the case of projective measurements). The specific example you give, for instance, is basically the set of measurements for deterministic LOCC conversion between entangled states. Aug 5, 2022 at 7:38
• @DaftWullie uh, I'm not sure I follow. I mean, ok I agree that post-measurement states need not be orthogonal. The $\Phi$ in my question is one such example (when $\Lambda_a$ are not orthogonal). But given a POVM post-measurement outcomes are not specified at all, so what makes the particular ones corresponding to $\Phi$ "special" or noteworthy? I don't really understand your comment about deterministic LOCC conversion I'm afraid
– glS
Aug 5, 2022 at 8:28
• There's nothing special about these. There's a non-uniqueness about going from POVMs to the measurement operators as that is not a direction you should ever be going. Aug 5, 2022 at 12:25
• My comment about deterministic LOCC conversion is simply to point out that the operators $\sqrt{\Lambda_i}$ do have a genuine operational usage as measurements because, look, here's an example. Aug 5, 2022 at 12:26
• yes, but I mean I don't understand the example. Deterministic LOCC conversion between what states?
– glS
Aug 5, 2022 at 12:28

TL;DR: Yes, we can give meaning to $$\sqrt{\Lambda}\rho\sqrt{\Lambda}$$: it is the post-measurement state up to an unknown change of basis (the correct basis being exactly the piece of information that POVMs are notorious for forgetting).

## Summary

POVM is an incomplete description of quantum measurement that "remembers" the outcome probability distribution, but "forgets" the post-measurement states. However, it only forgets the eigenbasis - and not the eigenvalues - of the post-measurement density matrix. In general, the state $$\sqrt{\Lambda}\rho\sqrt{\Lambda}$$ is not the post-measurement state, but it is a state that necessarily has the same eigenvalues as the actual post-measurement state.

## Generalized measurements

Generalized quantum measurement is described by a collection of operators $$M_m$$, one for each measurement outcome $$m$$, such that $$\sum_mM_m^\dagger M_m=I$$. This description enables us to obtain the subnormalized post-measurement state $$\rho_m:=M_m\rho M_m^\dagger$$ from which we can then obtain both the outcome probabilities $$p(m|\rho)=\mathrm{tr}(\rho_m)=\mathrm{tr}(\rho M_m^\dagger M_m)$$ and the normalized post-measurement states $$\sigma_m=\frac{\rho_m}{p(m|\rho)}$$.

## POVMs are forgetful...

A POVM measurement, given by a collection of positive operators $$\Lambda_m$$ with $$\sum_m\Lambda_m=I$$, may be thought of as an incomplete description of a quantum measurement that facilitates the computation of measurement outcome probabilities $$p(m|\rho)=\mathrm{tr}(\rho\Lambda_m)$$, but is ambiguous about the precise post-measurement state. Equivalently, we can think of a POVM as a class of measurements that all give the same output distributions for any given input state, but differ in the precise post-measurement states.

Mathematically, this corresponds to the fact that a generalized quantum measurement described by operators $$M_m$$ gives rise to a POVM via

$$\Lambda_m=M_m^\dagger M_m.\tag1$$

Clearly, any other quantum measurement given by operators $$N_m=U_m M_m$$ for some unitary$$^1$$ $$U_m$$ gives rise to the same POVM. One could say that POVM "forgets" the unitaries $$U_m$$. The unitaries $$U_m$$ cancel out in the computation of outcome probabilities, but not in the computation of post-measurement states. This is why POVMs enable the former, but not the latter.

## ...but not completely forgetful

This raises the question: is there anything at all we can learn about the post-measurement states from a POVM? The answer is yes!

A quantum state is described by a density operator. We can think of such an operator as consisting of two pieces of information: eigenbasis and eigenvalues. It turns out that POVM's failure to enable the computation of the (subnormalized) post-measurement state $$\rho_m$$ is only partial. It cannot tell us the eigenbasis of $$\rho_m$$, but it does tell us its eigenvalues. Namely, the eigenvalues are the same as those of $$\sqrt{\Lambda_m}\rho\sqrt{\Lambda_m}$$. To see this, note that by polar decomposition $$M_m=V_m\sqrt{M_m^\dagger M_m}=V_m\sqrt{\Lambda_m}$$, so the subnormalized post-measurement state $$\rho_m$$ for outcome $$m$$ of any of the generalized measurements consistent with the POVM specified by $$\Lambda_m$$ is $$\rho_m=V_m\sqrt{\Lambda_m}\rho\sqrt{\Lambda_m}V_m^\dagger$$ which has the same eigenvalues as $$\sqrt{\Lambda_m}\rho\sqrt{\Lambda_m}$$ by similarity.

Equivalently, just as we can think of a POVM as designating a class of generalized measurements up to a unitary, we can also think of the state $$\sqrt{\Lambda_m}\rho\sqrt{\Lambda_m}$$ as designating a class of post-measurement states up to a change of basis.

## Examples

For example, in the case described at the end of the question we see right away that the post-measurement state corresponding to the outcome $$\Lambda_1$$ is always a pure state (albeit we have no way of knowing which one) regardless of the input state. Moreover, $$\det\left(\sqrt{\Lambda_2}\rho\sqrt{\Lambda_2}\right)=a\det(\rho)$$, so we can also conclude that the post-measurement state associated with the outcome $$\Lambda_2$$ is a pure state if and only if the input state was pure. Finally, if for a given $$\rho$$ we find that $$\sqrt{\Lambda_2}\rho\sqrt{\Lambda_2}$$ is the maximally mixed state then we know that the actual post-measurement state is also the maximally mixed state.

$$^1$$ Note that unitaries $$U_m$$ may depend on measurement outcome. Consequently, unlike in projective measurement, in generalized quantum measurement, post-measurement states corresponding to different outcomes may fail to be orthogonal.

• Very nice answer. If I remember correctly your post measurement update result can even be extended to quantum instruments. I.e. any quantum instrument can be seen as first applying some Lüder's rule update $\rho\mapsto \sqrt{\Lambda_m} \rho \sqrt{\Lambda_m}$ followed by a quantum channel whose choice can depend on the outcome of the measurement. You recover the update rules in your answer by restricting to unitary channels. Aug 7, 2022 at 11:26
• can't you find measurements with arbitrary post-measurement states, essentially by considering entanglement-breaking channels associated to the POVM? E.g. If I define $\Psi(\rho)=\langle\Lambda_1,\rho\rangle \sigma_1 + \langle\Lambda_2,\rho\rangle \sigma_2$ with $\sigma_1=|0\rangle\!\langle0|$ and $\sigma_2=|1\rangle\!\langle 1|$ I get a measurement whose outcomes are both pure. You seem to reach your conclusion assuming measurements to correspond to Kraus operators of a channel, but how do we know that assumption to be correct?
– glS
Aug 7, 2022 at 13:32
• @glS This would fall under the extension to quantum instruments. You could just apply replacement channels after the Lüders update rule. Aug 7, 2022 at 16:50
• Thanks, @Rammus! I agree that we can generalize to cover the case like entanglement-breaking channels by replacing unitary $V_m$ with more general quantum operations $\mathcal{E}_m$. However, in this case, $\sqrt{\Lambda_m}\rho\sqrt{\Lambda_m}$ won't generally be similar to $\mathcal{E_m}(\sqrt{\Lambda_m}\rho\sqrt{\Lambda_m})$. My guess is that if $\mathcal{E}_m$ are "nearly unitary" then the post-measurement state's eigenvalues would still be close to those of $\sqrt{\Lambda_m}\rho\sqrt{\Lambda_m}$, but for "highly non-unitary" $\mathcal{E}_m$, only the sum of the eigenvalues is preserved. Aug 7, 2022 at 23:57
• @gIS I chose this particular description of measurement because it endows $\sqrt{\Lambda_m}\rho\sqrt{\Lambda_m}$ with meaning related to the post-measurement state. I'd argue that the situation is still fairly general, e.g. it is more general than the projective measurement. That said, it is interesting to ask why there are limits to the above interpretation. One way to answer this question is to look at Naimark's construction where $M_m$ are "embedded" in a unitary on a higher-dimensional Hilbert space formed by adjoining an auxiliary system followed by discarding of the input system. (...) Aug 8, 2022 at 0:30