Confusion regarding Neumark's/Naimark's extension of POVM

Question

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)?

Answer 1

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.

Answer 2

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.