I understand how to show that every POVM is equivalent to a projective measurement on a larger Hilbert space, but I don't understand why the converse is true. The vast majority of explanations of POVM's start by defining a POVM, and then show that given any POVM, you can tensor an appropriate ancilla onto your system and convert that POVM into a projective measurement on the combined system. But I want to know how to go the other direction: that is, given a projective measurement on a larger system, can I reduce it to a POVM on a subsystem? And if so, can I do it in a state-independent way?

For example, suppose I have a pure state whose Hilbert space is a product of two subsystems $A$ and $B$: $| \psi \rangle = \sum_{ab} \psi_{ab} | a \rangle | b \rangle$. I want to make a specified projective measurement $\hat{M} = \sum_m m\, \hat{P}_m$ on the entire combined system, where the $\{\hat{P}_m\}$ are orthogonal projectors. Is there a way to express the expectation value $\langle \psi | \hat{M} | \psi \rangle$ in terms of a POVM on system $A$ alone? If so, does it depend on the state $| \psi \rangle$, or just on $\hat{M}$ and the Hilbert spaces of the systems $A$ and $B$? If it depends on the state $| \psi \rangle$, than this seems like a rather serious limitation, because it means that there's no state-independent way to convert an ordinary projective measurement into a POVM. It seems to me that in an experiment, we might know the details of the measurement we want to make, but not the details of the state we're measuring.

The closest I can find to an explanation of this is at http://arxiv.org/pdf/1110.6815v2.pdf on pgs. 10-11. The author says "any standard measurement involving more than one physical system may be described as a generalized measurement on one of the subsystems," which seems promising. But in the statement of the theorem, he assumes that the measurement is only performed on the ancilla, which seems like a quite restrictive assumption which weakens his claim. (He also assumes that the system and the ancilla are originally unentangled and then undergo arbitrary unitary evolution. But if you were to start from an arbitrary experimental state, there is no state-independent unitary operator that unentangles $A$ and $B$, so again this setup seems quite state-dependent.)

Edit: Perhaps I misunderstood the point of the POVM formulation. The Wikipedia article on POVM says "In rough analogy, a POVM is to a PVM what a density matrix is to a pure state ... POVMs on a physical system are used to describe the effect of a projective measurement performed on a larger system." I took this to mean that a POVM measurement is a way of restricting the effect of an arbitrary projective measurement of the purified state onto just the original system, but perhaps this is incorrect.

The standard proof I've seen shows that an arbitrary POVM measurement is equivalent to a very specific type of projective measurement on a composite system. How do we know that a more complicated/general projective measurement on a composite system (e.g. a joint measurement on both the original system and any added ancilla) can be expressed as a POVM measurement?


1 Answer 1


What you are proposing cannot work: You cannot replace a (projective) measurement on a general composite system AB by a (POVM) measurement on part A only. To see this, simply consider the case where the joint state $\vert\psi\rangle$ is of the form $$ \vert\psi\rangle_{AB} = \vert0\rangle_A\vert\vartheta\rangle_B\ . $$ The reduced state of A is $\vert0\rangle\langle0\vert$ and thus completely independent of $\vert\vartheta\rangle$. No measurement on A will thus be able to reveal any information about $\vert\vartheta\rangle$.

However, you are misunderstanding the "POVM <-> projective measurement on a larger system" relation. The statement is that any POVM on a system A is equivalent to (i) adding an ancilla B in a well-defined state (say, $\vert0\rangle_B$), (ii) performing a specific unitary on AB, and (iii) carrying out a projective measurement on AB. In that case, the state of AB after step (ii) carries exactly the same information as the state A before step (i), and everything works out fine.

Note also that this is a standard textbook result (in both directions), did you check some textbooks/lecture notes (e.g. Nielsen+Chuang, Preskill, ...)?

  • $\begingroup$ (a) Re your statement "You cannot replace a (projective) measurement on a general composite system AB by a (POVM) measurement on part A only": the linked article says "any standard [i.e. projective] measurement involving more than one physical system may be described as a generalized measurement on one of the subsystems." Does this not contradict your claim? $\endgroup$
    – tparker
    Dec 29, 2015 at 16:33
  • 2
    $\begingroup$ The statement is that a POVM on A is equivalent to adding an ancilla B in a well-defined state (=> unentangled with A!), and then performing a projective measurement on the joint system (which is equivalent to doing a unitary and measuring B, if B has been chosen sufficiently large). $\endgroup$ Dec 29, 2015 at 23:19
  • 1
    $\begingroup$ With respect to the "this is in most textbooks" note at the end: textbooks are a lot harder to access and search than stack exchange sites. $\endgroup$ Dec 30, 2015 at 5:45
  • 1
    $\begingroup$ NOW I understand the equivalence that you, Paris, and Preskill are trying to explain. The point that I wasn't getting is that the ancilla is under the experimenter's control and in a known state - in which case as you say, it can't be initially entangled with the system to be measured - and that entangling the ancilla to the system is part of the measurement process. Sorry I was being dense, and thank you very much for your patience. $\endgroup$
    – tparker
    Dec 30, 2015 at 8:18
  • 1
    $\begingroup$ @tparker Indeed, ancillas are supposed to be in a well-defined state (after all, they are there to help). $\endgroup$ Dec 30, 2015 at 9:30

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.