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Starting with the definitions used.

A PVM is a set $\mathcal{P} = \{P_i: P_i^2 = P_i, P_iP_j = \delta_{ij}P_j, \sum{P_i} = \mathbf{I}\}_{i,j=1}^n$, where $n\leq d$ on a Hilbert space $\mathcal{H}^d$ of dimension $d$

A POVM is a set $\mathcal{M} = \{A_i : A_i \geq 0, \sum{A_i }= \mathbf{I}\}_{i=1}^m$ on a Hilbert space $\mathcal{H}^d$ of dimension $d$.

In many articles, I have come across the statement that "Neumark's theorem states that any rank one POVM can be realised as a PVM on a higher-dimensional space". Rank one POVM is where all the operators $A_i$ are of rank one.

$\textbf{Confusion 1:}$ Is the rank one requirement necessary, or is it assumed for the sake of simplicity?

$\textbf{Confusion 2:}$ I have seen the enlargement of the Hilbert space being done in two different ways, by embedding the system in a higher-dimensional space. For eg: A qubit being treated as a qutrit with the third amplitude being zero, and by attaching an ancilla of a suitable size. Are these two approaches equivalent?

$\textbf{Confusion 3:}$ When the Hilbert space is enlarged by attaching an ancilla, to realise the POVM, is the PVM performed on the ancilla alone or on the total system(original system + ancilla)?

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  • $\begingroup$ Do you have some refs of your statement In many articles, ...? $\endgroup$
    – narip
    Apr 23, 2022 at 13:35
  • $\begingroup$ For your confusion 3, the paper here in page 10 states with measurement on ancilla alone but since there is a $U$ in $UI\otimes P U^\dagger$ here, where $P$ is the projective measurement on ancilla, you can also see it as projection on system and ancilla. This link might also be helpful. $\endgroup$
    – narip
    Apr 23, 2022 at 13:40
  • $\begingroup$ @narip of course. R. Jozsa, M. Koashi, N. Linden, S. Popescu, S. Presnell, D. Shepherd, and A. Winter, Quantum Inf. Comput. 3, 405 ?2003. $\endgroup$ Apr 23, 2022 at 18:08

2 Answers 2

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You can assume without loss of generality that the POVM's are rank one because $\sum_i A_i=I$, so it's not necessary just more convenient.

The enlargement of the space in the Naimark dilation theorem comes from the Stinespring dilation theorem. Which says that whenever you have a POVM there is an isometry $V:\mathbb{C}^d\rightarrow \mathbb{C}^{d'}$ with $d'\geq d$ whereby the compression of each projection $P_i$ acting on $\mathbb{C}^{d'}$, is the positive operator $A_i$, i.e. $VP_iV^*=A_i$. The reason that the operators $P_i$ are orthogonal projections is enforced by the requirement that the measure $\mathbb{C}^m\rightarrow M_d(\mathbb{C})$ needs to be a $*$-homomorphism and $\sum_{i=1}^m A_i=I$.

The appearance of an "ancilla space" can be seen as an artifact from the proof of the Stinespring dilation theorem.. However, nice presentations exists without needing to know Stinespring's dilation theorem, such as the presentation in Watrous's text (section 2.3) or the other answer.

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    $\begingroup$ Will decomposing the POVM elements into rank one elements give rise to the same physical scenario? Suppose I have a POVM ith two elemets $P_1$ and $P_2$ where $P_1 = |00 \rangle \langle00| + |11 \rangle \langle11|$ and $P_1 = |01 \rangle \langle01| + |10 \rangle \langle10|$. Breaking this down into rank one operators will just give the projective measurement on two qubits in the computational basis. In what sense are they equialent? $\endgroup$ Apr 19, 2022 at 4:49
  • $\begingroup$ $P_1 + P_2 = |00\rangle \langle00| + |01\rangle \langle01| + |10\rangle \langle10| + |11\rangle \langle11| = \mathbf{I}$ . Am I missing something? $\endgroup$ Apr 19, 2022 at 13:36
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    $\begingroup$ Whoops yeah sorry, you are totally right about it being a POVM, clearly, I should wake up more before commenting. But I think its just equivalence in the sense that we can express $prob(outcome 1)=\langle\psi|P_1|\psi\rangle=\langle\psi|P_1^{|00\rangle}|\psi\rangle+\langle\psi|P_1^{|11\rangle}|\psi\rangle$, where now each one is rank 1. I'm not sure what you mean by physical scenario. $\endgroup$
    – Condo
    Apr 19, 2022 at 13:51
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    $\begingroup$ POVMs just describe probabilities they don't represent anything physical about the system. Everything physical about the scenario is determined by the state. $\endgroup$
    – Condo
    Apr 19, 2022 at 13:59
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    $\begingroup$ perhaps this would also help, arxiv.org/pdf/1104.4886.pdf I did not read the article but the abstract seems applicable. $\endgroup$
    – Condo
    Apr 25, 2022 at 16:06
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Any POVM can be interpreted as a projective measurement in a higher-dimensional space.

Derivation of the dilated representation

More precisely, let $\{\mu(a)\}_a\subset\operatorname{Pos}(\mathcal X)$ be a POVM actin in some (finite-dimensional) Hilbert space $\mathcal X$. This means in particular that $\sum_a\mu(a)=I$. Consider now the operator $V:\mathcal X\to\mathcal X\otimes\mathcal Y$ defined as $$Vu = \sum_a (\sqrt{\mu(a)}u)\otimes |a\rangle, \qquad u\in\mathcal X,$$ for some orthonormal basis $\{|a\rangle\}_a$ for $\mathcal Y$. You can verify that this $V$ is an isometry, and its Hermitian conjugate acts on basis vectors as $$V^\dagger(u\otimes v) = \sum_a \langle a,v\rangle \sqrt{\mu(a)} u, \qquad u\in\mathcal X,\, v\in\mathcal Y.$$ This expression can be obtained from the inner product $$\langle v\otimes w, Vu\rangle = \sum_a \langle v,\sqrt{\mu(a)}u\rangle \langle w,a\rangle = \langle V^\dagger(v\otimes w),u\rangle.$$

With these, observe that $$V^\dagger(I\otimes |a\rangle\!\langle a|) V \,u = \mu(a) u.$$ In other words, we can write the elements of the POVM as $$\mu(a) = V^\dagger(I\otimes |a\rangle\!\langle a|) V.$$ This is essentially the statement at hand: $V$ is an isometry that maps the states into a larger space, where a projective measurement in the basis $|a\rangle$ is performed.

The probabilities produced by the POVM then read $$\langle \mu(a),\rho\rangle = \langle I\otimes |a\rangle\!\langle a|, V\rho V^\dagger\rangle \equiv \langle a| \operatorname{Tr}_1[V\rho V^\dagger]|a\rangle,$$ which is again directly interpreted as evolving $\rho$ through the isometry $V$, and then performing a projective measurement on the ancillary degree of freedom.

See e.g. Watrous' book, section 2.3, for more info.

Examples

You can find a few examples of POVMs in this other answer. Consider here the somewhat trivial single-qubit two-outcome POVM with $\mu(1)=\mu(2)=\frac12 I$. This has elements that have rank 2, and following the procedure above we find dilations of the form $$V = \frac1{\sqrt2}\sum_{a=1}^2 I\otimes |a\rangle.$$ In matrix form, this can be represented as $$V = \frac1{\sqrt2}\begin{pmatrix}1&0 \\ 0&1 \\ 1& 0\\0&1\end{pmatrix}$$ I should point out that the way you represent $V$ (as any linear operator really) depends on the choice of basis. The matrix representation I'm using here would arguably be the one more naturally attached to $\frac12\sum_a |a\rangle\otimes I$ rather than $V$ itself, but as long as one knows how things are being represented there is no harm in using this representation. Note also that the dilation isometry is not unique. It depends on our choice of basis in the ancillary space.

So, given an initial state $|\psi\rangle$, we get $V|\psi\rangle=|\psi\rangle\otimes|+\rangle$, and thus the outcome probabilities are $p(i) = \| (I\otimes \langle i|)(|\psi\rangle\otimes|+\rangle)\|^2=\frac12$, for $i=0,1$. Which is compatible with the POVM being such that $\langle\mu(i),\rho\rangle=\frac12$. Admittedly, this is a completely useless measurement which extracts no information from the state, I only use it to illustrate the procedure. Similar calculations can be performed in less trivial cases though, e.g. something like $\mu(1)=I/2+\epsilon$ and $\mu(2)=I-\mu(1)$ for $\epsilon>0$ small enough.

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  • $\begingroup$ Can the projective measurements be rank one projectors on the total Hilbert space of the system + ancilla or they have to be projectors on the ancilla alone? $\endgroup$ Apr 19, 2022 at 4:58
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    $\begingroup$ @AbhishekBanerjee btw, I think that's an interesting question, but I'd suggest making a post asking specifically about only that point. It's better to have each post laser-focused, on a specific issue, as that makes it easier to reuse and reference information in the future, and gives you better chances at good answers. You could focus this post on the issue on the first or first two issues you brought up, and ask the other one separately $\endgroup$
    – glS
    Apr 19, 2022 at 8:48
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    $\begingroup$ @AbhishekBanerjee No, the dimension of the Hilbert space on which the elements of the PVM act. $\endgroup$ Apr 19, 2022 at 14:07
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    $\begingroup$ I'm sorry, I'm not aware of a reference that discusses this, the only thing I know is the above mentioned book by Peres. $\endgroup$ Apr 20, 2022 at 5:55
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    $\begingroup$ @narip well, yes, nothing prevents that in the definition. Though of course such an example describes a completely useless type of measurement (where you get no information about the state) $\endgroup$
    – glS
    Apr 23, 2022 at 18:52

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