x\geq\frac{2}{\pi}x$, so

$\displaystyle\mathfrak{w}(\texttt{G}_{N,\varepsilon})\leq\frac{2}{N}+\frac{2\sqrt{\pi\varepsilon}}{N}\sum_{i=1}^{N/2-1}\frac{1}{\sqrt{\theta_{i}}}.$
(108)

Approximating the sum by an integral,

$\displaystyle\mathfrak{w}(\texttt{G}_{N,\varepsilon})\leq\frac{2}{N}+\sqrt{\pi\varepsilon}\int_{0}^{1}\frac{1}{\sqrt{x}}dx=\frac{2}{N}+2\sqrt{\pi\varepsilon}.$
(109)

∎

##  7 Monogamy games for finite groups

In this section, we introduce the coset monogamy game on non-abelian, but
finite, groups. This serves to bridge the gap between the monogamy game on
$\mathbb{Z}_{2}^{n}$ of [CLLZ21] and the infinite-dimensional monogamy games
of the following sections.

###  7.1 Non-abelian coset states

Let $G$ be an arbitrary finite group. The corresponding Hilbert space
$\mathcal{H}_{G}=\operatorname{span}_{\mathbb{C}}\left\\{{\lvert
g\rangle}\middle|{g\in G}\right\\}$, where the $\lvert g\rangle$ form an
orthonormal basis, is finite-dimensional. Hence, we can, as in the original
$\mathbb{Z}_{2}^{n}$ case, work with a basis of coset states rather than a
measure.

Let $H\leq G$ be a subgroup and fix $g\in G$. In order to have a basis of
coset states, the states on the coset $gH$ need to be an orthonormal basis of
$\operatorname{span}_{\mathbb{C}}\left\\{{\lvert gh\rangle}\middle|{h\in
H}\right\\}$. For an abelian group, it is sufficient to use the characters,
giving coset states of the form

$\displaystyle\lvert gH^{\chi}\rangle=\frac{1}{\sqrt{|H|}}\sum_{h\in
H}\chi(h)\lvert gh\rangle.$ (110)

However, if $H$ is non-abelian, then the irreducible representations are not
one-dimensional and the characters only span the space of class functions, not
the space of all functions on $H$. In this case, the natural generalization of
the irreducible characters are the matrix elements of the irreducible
representations. Let $\gamma:H\rightarrow\mathcal{U}(d_{\gamma})$ be an
irreducible representation, and write the $(m,n)$ matrix element, for $1\leq
m,n\leq d_{\gamma}$ as $\gamma_{m,n}:H\rightarrow\mathbb{C}$. The associated
coset state is

$\displaystyle\lvert
gH^{\gamma}_{m,n}\rangle=\sqrt{\frac{d_{\gamma}}{|H|}}\sum_{h\in
H}\gamma_{m,n}(h)\lvert gh\rangle.$ (111)

The orthonormality of these states is given by the Schur orthogonality
relations: if $\gamma,\gamma^{\prime}$ are either equal or inequivalent
irreducible representations, and $1\leq m,n\leq d_{\gamma}$, $1\leq
m^{\prime},n^{\prime}\leq d_{\gamma^{\prime}}$, then

$\displaystyle\sum_{h\in
H}\overline{\gamma_{m,n}(h)}\gamma_{m^{\prime},n^{\prime}}(h)=\frac{|H|}{d_{\gamma}}\delta_{\gamma,\gamma^{\prime}}\delta_{m,m^{\prime}}\delta_{n,n^{\prime}}.$
(112)

However, unlike the characters, the matrix elements of a representation are
not unique in the sense that the equivalent but unequal representations may
have different matrix elements. This implies that the coset states for
different choices of irreducible representations and coset representatives are
not equal up to global phase, as they are in the abelian case. To remedy that
we need to fix choices of representatives. Let $\textsf{CS}(H)$ be a set of
coset representatives of $G/H$, let $\textsf{Irr}(H)$ be a full set of
inequivalent irreducible representations (irreps) of $H$, and define
$\textsf{IB}(H)=\\{{\gamma_{m,n}}|{\gamma\in\textsf{Irr}(H);1\leq m,n\leq
d_{\gamma}}\\}$. With this, we take the basis of coset states to be

$\displaystyle\left\\{{\lvert
gH^{\gamma}_{m,n}\rangle}\middle|{g\in\textsf{CS}(H),\gamma_{m,n}\in\textsf{IB}(H)}\right\\}.$
(113)

An interesting property of non-abelian coset states is that, whereas abelian
coset states are equal superpositions of the elements of $gH$, non-abelian
states are no longer, as we do not have always that $|\gamma_{m,n}(h)|=1$. We
give a simple example to illustrate that.

Consider $G=D_{N}$, the dihedral group defined with generators and relations
$D_{N}=\langle{r,t};{r^{N}=t^{2}=(tr)^{2}=1}\rangle$, for $N$ odd and its
subgroup $H=\langle{r^{N/p},t}\rangle\cong D_{p}$ for $p$ a divisor of $N$.
There are two one-dimensional irreps of $H$: the trival
$\gamma_{0}(r^{N/p})=\gamma_{0}(t)=1$, and the representation
$\gamma_{-1}(r^{N/p})=1$, $\gamma_{-1}(t)=-1$. The remaining irreps are two-
dimensional, indexed by $k=1,\ldots,\frac{1}{2}(p-1)$:
$\gamma_{k}(r^{N/p})=\exp(\frac{2\pi ik}{p}Z)$, $\gamma_{k}(t)=X$. Further, we
may take $\textsf{CS}(H)=\left\\{{1,r,\ldots,r^{N/p-1}}\right\\}$. Then, the
coset states are

$\displaystyle\begin{split}&\lvert
r^{q}H^{0}\rangle=\frac{1}{\sqrt{2p}}\sum_{j=0}^{p-1}\left\lparen\lvert
r^{Nj/p+q}\rangle+\lvert tr^{Nj/p-q}\rangle\right\rparen\\\ &\lvert
r^{q}H^{-1}\rangle=\frac{1}{\sqrt{2p}}\sum_{j=0}^{p-1}\left\lparen\lvert
r^{Nj/p+q}\rangle-\lvert tr^{Nj/p-q}\rangle\right\rparen\\\ &\lvert
r^{q}H^{k}_{m,n}\rangle=\frac{1}{\sqrt{p}}\sum_{j=0}^{p-1}e^{(-1)^{m}2\pi
i\frac{jk}{p_{I}}}\lvert t^{m+n}r^{Nj/p+(-1)^{m+n}q}\rangle.\end{split}$ (114)

Note in particular that each of the states $\lvert r^{q}H^{k}_{m,n}\rangle$ is
a superposition of only half the elements of $r^{q}H$.

###  7.2 Non-abelian coset monogamy game

######  Definition 7.1.

A _non-abelian coset monogamy game_ is a pair $\texttt{G}=\left\lparen
G,\mathcal{S}\right\rparen$, where $G$ is a finite group and $\mathcal{S}$ is
a finite set of subgroups of $G$.

A _(quantum) strategy_ for a coset measure game G is a tuple
$\texttt{S}=\left\lparen\mathcal{B},\mathcal{C},B,C,\rho\right\rparen$, where
$\mathcal{B}$ and $\mathcal{C}$ are Hilbert spaces,
$B=\\{{B^{H}:\textsf{CS}(H)\rightarrow\mathcal{B}(\mathcal{B})}|{H\in\mathcal{S}}\\}$
and
$C=\\{{C^{H}:\textsf{IB}(H)\rightarrow\mathcal{B}(\mathcal{C})}|{H\in\mathcal{S}}\\}$
are collections of POVMs, and
$\rho\in\mathcal{D}(\mathcal{H}_{G}\otimes\mathcal{B}\otimes\mathcal{C})$ is a
shared density operator.

Let G be a coset game and S be a strategy for it. The _winning probability_ of
S is

$\displaystyle\mathfrak{w}_{\texttt{G}}(\texttt{S})=\underset{H\in\mathcal{S}}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\\!\\!\\!\sum_{\begin{subarray}{c}g\in\textsf{CS}(H)\\\
\gamma_{m,n}\in\textsf{IB}(H)\end{subarray}}\\!\\!\\!\operatorname{Tr}[(\vphantom{gH^{\gamma}_{m,n}}\left\lvert\smash{gH^{\gamma}_{m,n}}\middle\rangle\\!\middle\langle\smash{gH^{\gamma}_{m,n}}\right\rvert\otimes
B^{H}_{g}\otimes C^{H}_{\gamma_{m,n}})\rho],$ (115)

where the expection with respect to $H\in\mathcal{S}$ is uniform.

The winning probability of G is
$\mathfrak{w}(\texttt{G})=\sup_{\texttt{S}}\mathfrak{w}_{\texttt{G}}(\texttt{S})$.

Our main result in this section is an analogous bound to the one of [CV22] on
the winning probability of such a general game.

######  Theorem 7.2.

Let $\texttt{G}=(G,\mathcal{S})$ be a coset monogamy game, and take a set of
mutually orthogonal permutations $\pi_{i}:\mathcal{S}\rightarrow\mathcal{S}$
for $i=1,...,|\mathcal{S}|$, permutations such that $\pi_{i}\circ\pi_{j}^{-1}$
has no fixed points unless $i=j$. Then,

$\displaystyle\mathfrak{w}(\texttt{G})\leq\underset{i}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\max_{\begin{subarray}{c}H\in\mathcal{S}\\\
\gamma\in\textsf{Irr}(H)\end{subarray}}\sqrt{d_{\gamma}\frac{|H\cap\pi_{i}(H)|}{|H|}}\,.$
(116)

First, we will make use of a lemma of [TFKW13] to bound the winning
probability by overlaps of the measurement operators.

######  Lemma 7.3 (Lemma 2 in [TFKW13]).

Let $P_{1},\ldots,P_{n}$ be positive semidefinite operators on a Hilbert
space. Then

$\displaystyle\Big{\|}\sum_{i=1}^{n}P_{i}\Big{\|}$
$\displaystyle\leq\sum_{i=1}^{n}\,\max_{j=1,\ldots,n}\,\Big{\|}\sqrt{\vphantom{P_{\pi_{i}(j)}}P_{j}}\sqrt{P_{\pi_{i}(j)}}\Big{\|}\;,$

where $\pi_{1},\ldots,\pi_{n}$ is any set of mutually orthogonal permutations
of $\\{1,\ldots,n\\}$.

Next, we need to bound the overlaps of particular projectors related to the
coset states.

######  Lemma 7.4.

Let $H,K\leq G$. Then, for any $\gamma_{m,n}\in\textsf{IB}(H)$ and
$q\in\textsf{CS}(K)$,

$\displaystyle\Big{\|}\sum_{g\in\textsf{CS}(H)}\vphantom{gH^{\gamma}_{m,n}}\left\lvert\smash{gH^{\gamma}_{m,n}}\middle\rangle\\!\middle\langle\smash{gH^{\gamma}_{m,n}}\right\rvert\sum_{\varrho_{i,j}\in\textsf{IB}(K)}\vphantom{qK^{\varrho}_{i,j}}\left\lvert\smash{qK^{\varrho}_{i,j}}\middle\rangle\\!\middle\langle\smash{qK^{\varrho}_{i,j}}\right\rvert\Big{\|}\leq\sqrt{d_{\gamma}\frac{|H\cap
K|}{|H|}}\,.$ (117)

###### Proof.

Note that, since the coset states in each of the sums are orthogonal, we’re
dealing with the overlap of two projectors. Next, since $\left\\{{\lvert
qK^{\varrho}_{i,j}\rangle}\middle|{\varrho_{i,j}\in\textsf{IB}(K)}\right\\}$
is an orthonormal basis of $\operatorname{span}_{\mathbb{C}}\left\\{{\lvert
qk\rangle}\middle|{k\in K}\right\\}$, we get

$\displaystyle\sum_{\varrho_{i,j}\in\textsf{IB}(K)}\vphantom{qK^{\varrho}_{i,j}}\left\lvert\smash{qK^{\varrho}_{i,j}}\middle\rangle\\!\middle\langle\smash{qK^{\varrho}_{i,j}}\right\rvert=\sum_{k\in
K}\vphantom{qk}\left\lvert\smash{qk}\middle\rangle\\!\middle\langle\smash{qk}\right\rvert=:\Pi_{qK},$
(118)

a diagonal projector. Then, the overlap is

$\displaystyle\Big{\|}\sum_{g\in\textsf{CS}(H)}\vphantom{gH^{\gamma}_{m,n}}\left\lvert\smash{gH^{\gamma}_{m,n}}\middle\rangle\\!\middle\langle\smash{gH^{\gamma}_{m,n}}\right\rvert\sum_{\varrho_{i,j}\in\textsf{IB}(K)}\vphantom{qK^{\varrho}_{i,j}}\left\lvert\smash{qK^{\varrho}_{i,j}}\middle\rangle\\!\middle\langle\smash{qK^{\varrho}_{i,j}}\right\rvert\Big{\|}=\Big{\|}\Pi_{qK}\sum_{g\in\textsf{CS}(H)}\vphantom{gH^{\gamma}_{m,n}}\left\lvert\smash{gH^{\gamma}_{m,n}}\middle\rangle\\!\middle\langle\smash{gH^{\gamma}_{m,n}}\right\rvert\Pi_{qK}\Big{\|}^{1/2}.$
(119)

Now, since $\Pi_{qK}\lvert gH^{\gamma}_{m,n}\rangle$ is a superposition of
basis vectors from $qK\cap gH$, these states are orthogonal, giving that the
operators
$\Pi_{qK}\vphantom{gH^{\gamma}_{m,n}}\left\lvert\smash{gH^{\gamma}_{m,n}}\middle\rangle\\!\middle\langle\smash{gH^{\gamma}_{m,n}}\right\rvert\Pi_{qK}$
over $g\in\textsf{CS}(H)$ are Hermitian with orthogonal ranges. Since
$\|\sum_{s}X_{s}\|\leq\max_{s}\|X_{s}\|$ for any $X_{i}$ Hermitian with
orthogonal ranges,

$\displaystyle\begin{split}\Big{\|}\sum_{g\in\textsf{CS}(H)}\Pi_{qK}\vphantom{gH^{\gamma}_{m,n}}\left\lvert\smash{gH^{\gamma}_{m,n}}\middle\rangle\\!\middle\langle\smash{gH^{\gamma}_{m,n}}\right\rvert\Pi_{qK}\Big{\|}&\leq\max_{g\in\textsf{CS}(H)}\left\|\Pi_{qK}\vphantom{gH^{\gamma}_{m,n}}\left\lvert\smash{gH^{\gamma}_{m,n}}\middle\rangle\\!\middle\langle\smash{gH^{\gamma}_{m,n}}\right\rvert\Pi_{qK}\right\|\\\
&=\max_{g\in\textsf{CS}(H)}\vphantom{gH^{\gamma}_{m,n}\Pi_{qK}gH^{\gamma}_{m,n}}\left\langle\smash{gH^{\gamma}_{m,n}}\middle\rvert\smash{\Pi_{qK}}\middle\rvert\smash{gH^{\gamma}_{m,n}}\right\rangle.\end{split}$
(120)

We get the result by bounding the inner product

$\displaystyle\vphantom{gH^{\gamma}_{m,n}\Pi_{qK}gH^{\gamma}_{m,n}}\left\langle\smash{gH^{\gamma}_{m,n}}\middle\rvert\smash{\Pi_{qK}}\middle\rvert\smash{gH^{\gamma}_{m,n}}\right\rangle=\frac{d_{\gamma}}{|H|}\sum_{h\in
H\cap g^{-1}qK}|\gamma_{m,n}(h)|^{2}\leq d_{\gamma}\frac{|H\cap
g^{-1}qK|}{|H|}\leq d_{\gamma}\frac{|H\cap K|}{|H|}.$ (121)

∎

Now, we can pass to the proof of the theorem.

###### Proof of Theorem 7.2.

Let $\texttt{S}=\left\lparen\mathcal{B},\mathcal{C},B,C,\rho\right\rparen$ be
a strategy for G. We may assume that $B$ and $C$ are projective. Writing, for
each $H\in\mathcal{S}$,
$\Pi^{H}=\sum_{g,\gamma_{m,n}}\vphantom{gH^{\gamma}_{m,n}}\left\lvert\smash{gH^{\gamma}_{m,n}}\middle\rangle\\!\middle\langle\smash{gH^{\gamma}_{m,n}}\right\rvert\otimes
B^{H}_{g}\otimes C^{H}_{\gamma_{m,n}}$, the winning probability may be
expressed as
$\mathfrak{w}_{\texttt{G}}(\texttt{S})=\underset{H\in\mathcal{S}}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\operatorname{Tr}(\Pi^{H}\rho)\leq\big{\|}\underset{H\in\mathcal{S}}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\Pi^{H}\big{\|}$.
Now, we can use Lemma 7.3:

$\displaystyle\mathfrak{w}_{\texttt{G}}(\texttt{S})\leq\underset{i}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\sup_{H\in\mathcal{S}}\big{\|}\Pi^{H}\Pi^{\pi_{i}(H)}\big{\|},$
(122)

since $(\Pi^{H})^{2}=\Pi^{H}$. Fixing $H,K\in\mathcal{S}$, it remains to
simplify $\left\|\Pi^{H}\Pi^{K}\right\|$. We upper bound

$\displaystyle\begin{split}&\Pi^{H}\leq\sum_{\begin{subarray}{c}g\in\textsf{CS}(H)\\\
\gamma_{m,n}\in\textsf{IB}(H)\end{subarray}}\vphantom{gH^{\gamma}_{m,n}}\left\lvert\smash{gH^{\gamma}_{m,n}}\middle\rangle\\!\middle\langle\smash{gH^{\gamma}_{m,n}}\right\rvert\otimes\mathbb{I}_{B}\otimes
C^{H}_{\gamma_{m,n}}\\\
&\Pi^{K}\leq\sum_{\begin{subarray}{c}q\in\textsf{CS}(K)\\\
\varrho_{i,j}\in\textsf{IB}(K)\end{subarray}}\vphantom{qK^{\varrho}_{i,j}}\left\lvert\smash{qK^{\varrho}_{i,j}}\middle\rangle\\!\middle\langle\smash{qK^{\varrho}_{i,j}}\right\rvert\otimes
B^{K}_{q}\otimes\mathbb{I}_{K},\end{split}$ (123)

and so we can bound the product

$\displaystyle\begin{split}\|\Pi^{H}\Pi^{K}\|&\leq\Big{\|}\sum_{\begin{subarray}{c}g,\gamma_{m,n}\\\
q,\varrho_{i,j}\end{subarray}}\lvert
gH^{\gamma}_{m,n}\rangle\\!\vphantom{gH^{\gamma}_{m,n}qK^{\varrho}_{i,j}}\left\langle\smash{gH^{\gamma}_{m,n}}\middle|\smash{qK^{\varrho}_{i,j}}\right\rangle\\!\langle
qK^{\varrho}_{i,j}\rvert\otimes B^{K}_{q}\otimes
C^{H}_{\gamma_{m,n}}\Big{\|}\\\
&=\max_{q,\gamma_{m,n}}\Big{\|}\sum_{\begin{subarray}{c}g,\varrho_{i,j}\end{subarray}}\lvert
gH^{\gamma}_{m,n}\rangle\\!\vphantom{gH^{\gamma}_{m,n}qK^{\varrho}_{i,j}}\left\langle\smash{gH^{\gamma}_{m,n}}\middle|\smash{qK^{\varrho}_{i,j}}\right\rangle\\!\langle
qK^{\varrho}_{i,j}\rvert\Big{\|},\end{split}$ (124)

since Bob and Charlie’s measurement projectors have orthogonal supports. Now,
we can use Lemma 7.4 and get
$\|\Pi^{H}\Pi^{K}\|\leq\max_{q,\gamma_{m,n}}\sqrt{d_{\gamma}\frac{|H\cap
K|}{|H|}}=\max_{\gamma}\sqrt{d_{\gamma}\frac{|H\cap K|}{|H|}}$. Hence we get
the wanted bound

$\displaystyle\mathfrak{w}_{\texttt{G}}(\texttt{S})\leq\underset{i}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\sup_{H,\gamma}\sqrt{d_{\gamma}\frac{|H\cap\pi_{i}(H)|}{|H|}}.$
(125)

∎

##  8 Monogamy games for abelian topological groups

Here, we generalise the coset monogamy game to arbitrary locally compact
Abelian groups, on which Pontryagin duality provides a generalization of the
Fourier transform. By replacing the coset states with the appropriate
operator-valued measure, we find a family of continuous-variable versions of
this game, and via a similar analysis we can bound the winning probability,
but as a function of not the number of outcomes but the measurement precision.
We also study the version of the game where Alice sends damped coset states
and find that the same winning probability bound applies.

###  8.1 Infinite-outcome measurements and coset measures

In the context of the original coset monogamy game, we take the space of
classical states to be $\mathbb{Z}_{2}^{n}$, the space of $n$-bit strings. We
are interested in extending the space of classical states to an arbitrary
locally compact (Hausdorff) abelian group $G$, for example $U(1)$,
$\mathbb{Z}$, or $\mathbb{R}$. In this way, we will be able to work with games
where the answer sets are non-trivially infinite. The Hilbert space
corresponding to such a set of classical states is $L^{2}(G)$, the space of
square-integrable functions $G\rightarrow\mathbb{C}$, modulo the subspace of
almost-everywhere zero functions. We will write $\lvert\psi\rangle\in
L^{2}(G)$ for a class, and the $\psi:G\rightarrow\mathbb{C}$ for a
representative of this class. Also, we write the Haar measure
$\mu_{G}:\mathscr{B}(G)\rightarrow[0,\infty]$, where $\mathscr{B}(G)$ is the
$\sigma$-algebra of Borel sets, and write the Haar integral as $\int
f(g)d_{G}g$.

The subgroups $H\leq G$ we consider will always be closed, since this is
necessary and sufficient for the quotient space $G/H$ to be Hausdorff as well.
Endowing $H$ with the subspace topology and $G/H$ with the quotient topology,
both $H$ and $G/H$ are locally compact abelian groups, so they have Haar
measures. In fact, fixing Haar measures on $G$ and $H$, there is a unique Haar
measure on $G/H$ such that $\int f(g)d_{G}g=\iint f(gh)d_{H}hd_{G/H}(gH)$.
This holds for a much larger class of topological groups, such as compact
groups [Nac76]. As much as possible, we will assume the Haar measures to be
normalized such that if $G$ is compact $\mu_{G}(G)=1$ and else if $G$ is
discrete $\mu_{G}(\\{1\\})=1$.

In order to work with games with infinitely many answers, we need to be able
to handle quantum measurements with infinitely many outcomes. If the outcomes
are discrete, then it is possible to handle the measurement as a POVM in the
usual way. However, the set of measurement outcomes will not in general be
discrete, in which case problems arise. One way to handle this is to use a
basis of unnormalizable states to represent the measurement basis. For
example, we can take and unnormalizable basis of $L^{2}(G)$ as
$\left\\{{\lvert g\rangle}\middle|{g\in G}\right\\}$, where each $\lvert
g\rangle$ is a Dirac delta function such that $\langle h|g\rangle=0$ when
$g\neq h$ and
$\vphantom{gh}\left\langle\smash{g}\middle|\smash{h}\right\rangle=\delta_{G}(g^{-1}h)$.
Of course, none of these are elements of $L^{2}(G)$, but $\lvert\psi\rangle\in
L^{2}(G)$ can nonetheless be represented as
$\lvert\psi\rangle=\int\psi(g)\lvert g\rangle d_{G}g$. The drawback of this
approach is the difficulty of handling convergence of such expressions, which
can make the handling of the probability of a measurement outcome ambiguous.
To remedy this, we generalise the measurement with respect to a basis not by a
generalised basis, but by an operator-valued generalization of a probability
measure. For instance, in the previous example, for a measurable subset
$E\subseteq G$, we see the probability of measuring some $g\in E$ on some
state $\lvert\psi\rangle\in L^{2}(G)$ is
$\int_{E}\left\lvert\psi(g)\right\rvert^{2}d_{G}g=\vphantom{\psi\int_{E}\vphantom{g}\left\lvert\smash{g}\middle\rangle\\!\middle\langle\smash{g}\right\rvert
d_{G}g\psi}\left\langle\smash{\psi}\middle\rvert\smash{\int_{E}\vphantom{g}\left\lvert\smash{g}\middle\rangle\\!\middle\langle\smash{g}\right\rvert
d_{G}g}\middle\rvert\smash{\psi}\right\rangle$.

######  Definition 8.1.

Let $X$ be a measurable space, $\mathcal{H}$ be a Hilbert space, and
$\mathscr{S}_{X}$ be a $\sigma$-algebra on $X$. An _operator-valued measure_
is a map $P:\mathscr{S}_{X}\rightarrow\mathcal{B}(\mathcal{H})$ that is weakly
countably additive, i.e., for all countable collections of disjoint measurable
sets $\\{E_{i}\\}_{i=1}^{\infty}\subseteq\mathscr{S}_{X}$ and
$\lvert\phi\rangle,\lvert\psi\rangle\in\mathcal{H}$,

$\displaystyle\vphantom{\phi
P\Big{\lparen}\bigcup_{i=1}^{\infty}E_{i}\Big{\rparen}\psi}\left\langle\smash{\phi}\middle\rvert\smash{P\Big{\lparen}\bigcup_{i=1}^{\infty}E_{i}\Big{\rparen}}\middle\rvert\smash{\psi}\right\rangle=\sum_{i=1}^{\infty}\vphantom{\phi
P(E_{i})\psi}\left\langle\smash{\phi}\middle\rvert\smash{P(E_{i})}\middle\rvert\smash{\psi}\right\rangle.$
(126)

Additionally, $P$ is

  * •

a _POVM_ if $P(E)$ is positive for all $E\in\mathscr{S}_{X}$ and
$P(X)=\mathbb{I}$;

  * •

_projective_ (or spectral) if $P(E\cap F)=P(E)P(F)$ for all
$E,F\in\mathscr{S}_{X}$;

  * •

a _PVM_ if it is a projective POVM.

Let $P$ be a POVM measure. For any $\rho\in\mathcal{D}(L^{2}(G))$,
$E\mapsto\operatorname{Tr}(P(E)\rho)$ is a probability measure, representing
the probability of measuring an outcome in $E$. POVM measures constitute a
physically natural generalization of usual finite POVMs because of the role
they play in the spectral theorem for general Hermitian operators. For any
Hermitian operator, $A\in\mathcal{B}(\mathcal{H})$, there exists a spectral
measure $P:\mathbb{R}\rightarrow\mathcal{B}(\mathcal{H})$ such that
$A=\int\lambda dP(\lambda)$ [BB03]; as such, since any observable in quantum
mechanics is represented by a Hermitian operator, the distribution of
measurement outcomes is naturally represented by a POVM measure. The role of
operator-valued measures in quantum theory has been studied from a variety of
perspectives [Hol11, Mor17].

Before moving on to the particular measure of interest here, we state a
general result on POVM measures that allows us to dilate any POVM measure to a
PVM measure.

######  Theorem 8.2 (Naimark [Pau02]).

Let $X$ be a measurable space, $\mathcal{H}$ be a Hilbert space, and
$Q:\mathscr{S}_{X}\rightarrow\mathcal{B}(\mathcal{H})$ be a POVM measure.
Then, there exists a Hilbert space $\mathcal{K}$, an isometry
$V:\mathcal{H}\rightarrow\mathcal{K}$, and a PVM measure
$P:\mathscr{S}_{X}\rightarrow\mathcal{B}(\mathcal{H})$ such that, for all
measurable $E\subseteq X$,

$\displaystyle Q(E)=V^{\dagger}P(E)V.$ (127)

We go into detail about integration with respect to POVM measures in Appendix
A.

The original coset monogamy game, given a subspaces $H\leq\mathbb{Z}_{2}^{n}$,
utilizes bases indexed by
$\mathbb{Z}_{2}^{n}/H\times\mathbb{Z}_{2}^{n}/H^{\perp}$. $H^{\perp}$ is
defined using the dot product, which can be seen as the family of
homomorphisms $\mathbb{Z}_{2}^{n}\rightarrow S^{1}=\\{{z\in\mathbb{C}}|{\lvert
z\rvert=1}\\}$ indexed by $a\in\mathbb{Z}_{2}^{n}$ defined
$x\mapsto(-1)^{a\cdot x}$. For a general abelian topoloogical group, this
natural family of homomorphisms is given not by $G$ itself but by the dual
group $\hat{G}$, which the group of continuous homomorphisms $G\rightarrow
S^{1}$, with product $(\gamma\eta)(g)=\gamma(g)\eta(g)$. $G$ is isomorphic to
$\hat{G}$ if $G$ is finite. If $G$ is locally compact abelian, then so is
$\hat{G}$. Then, the natural generalization of

$\displaystyle H^{\perp}=\left\\{{\gamma\in\hat{G}}\middle|{\gamma(h)=1\forall
h\in H}\right\\},$ (128)

and we have that $\hat{G}/H^{\perp}\cong\hat{H}$ homeomorphically by
Pontryagin duality. Thus, the basis of coset states should generalise to an
operator-valued measure on $G/H\times\hat{H}$. An important tool for working
with the function spaces on these groups is the Fourier transform.

######  Definition 8.3.

Let $G$ be a locally compact abelian group. The _Fourier transform_ is the
operator $\mathcal{F}_{G}\in\mathcal{B}(L^{2}(G),L^{2}(\hat{G}))$ defined on
$\lvert\psi\rangle\in L^{2}(G)$ continuous with compact support as

$\displaystyle(\mathcal{F}_{G}\lvert\psi\rangle)(\gamma)=\int\psi(g)\overline{\gamma(g)}d_{G}g,$
(129)

and extended uniquely to all $L^{2}(G)$ by continuity.

There exists a unique scaling of the Haar measure on $\hat{G}$, called the
dual measure, that makes the Fourier transform an isometry. Using that
measure, the inverse Fourier transform on $\lvert\psi\rangle\in
L^{2}(\hat{G})$ continuous with compact support is
$(\mathcal{F}_{G}^{-1}\lvert\psi\rangle)(g)=\int\psi(\gamma)\gamma(g)d_{\hat{G}}g$,
which again extends to the whole space by continuity.

Finally, we need the concept of a Fourier transform with respect to a closed
subgroup $H\leq G$. For $\lvert\psi\rangle\in L^{2}(G)$, we can simply take
$\mathcal{F}_{H}\lvert\psi\rangle=\mathcal{F}_{H}\lvert\psi|_{H}\rangle:\hat{H}\rightarrow\mathbb{C}$.
In general, $\lvert\psi|_{H}\rangle$ is not always in $L^{2}(H)$, but in the
cases we consider, this will be true almost everywhere.

######  Definition 8.4.

Let $G$ be a locally compact abelian group and $H\leq G$ be a closed subgroup.
The _coset operator measure_ is the map
$A^{H}:\mathscr{B}(G/H)\otimes\mathscr{B}(\hat{H})\rightarrow\mathcal{B}(L^{2}(G))$
defined, for $E\subseteq G/H\times\hat{H}$ measurable and
$\lvert\phi\rangle,\lvert\psi\rangle\in L^{2}(G)$ as

$\displaystyle\vphantom{\phi
A^{H}(E)\psi}\left\langle\smash{\phi}\middle\rvert\smash{A^{H}(E)}\middle\rvert\smash{\psi}\right\rangle=\int_{E}\overline{(\mathcal{F}_{H}\lvert\phi\circ
g\rangle)(\gamma)}(\mathcal{F}_{H}\lvert\psi\circ
g\rangle)(\gamma)d_{{}_{G/H\times\hat{H}}}(gH,\gamma),$ (130)

where $\lvert\psi\circ g\rangle(h)=\psi(gh)$.

Note that the integral defining $\vphantom{\phi
A^{H}(E)\psi}\left\langle\smash{\phi}\middle\rvert\smash{A^{H}(E)}\middle\rvert\smash{\psi}\right\rangle$
is well-defined, since we have that $\mathcal{F}_{H}(\lvert\psi\circ
gh\rangle)(\gamma)=\gamma(h)\mathcal{F}_{H}(\lvert\psi\circ g\rangle)(\gamma)$
for all $h\in H$, so $\overline{\mathcal{F}_{H}(\lvert\phi\circ
g\rangle)(\gamma)}\mathcal{F}_{H}(\lvert\psi\circ g\rangle)(\gamma)$ does not
depend on the choice of coset representative. Also, as $\int f(g)d_{G}g=\iint
f(gh)d_{H}hd_{G/H}(gH)$, $\lvert\psi\circ g|_{H}\rangle$ is in $L^{2}(H)$
$\mu_{G/H}$-almost everywhere, so the integrand is in fact integrable. From
the definition, it is direct that $\vphantom{\phi
A^{H}(E)\psi}\left\langle\smash{\phi}\middle\rvert\smash{A^{H}(E)}\middle\rvert\smash{\psi}\right\rangle$
is in fact linear in $\lvert\psi\rangle$ and antilinear in
$\lvert\phi\rangle$. We can show that it is bounded and that it is a PVM
simultaneously.

######  Theorem 8.5.

$A^{H}$ is a PVM measure.

###### Proof.

First, we have that, for any measurable $E\subseteq G/H\times\hat{H}$ and
$\lvert\psi\rangle\in L^{2}(G)$,

$\displaystyle\vphantom{\psi
A^{H}(E)\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(E)}\middle\rvert\smash{\psi}\right\rangle=\int_{E}\left\lvert(\mathcal{F}_{H}\lvert\psi\circ
g\rangle)(\gamma)\right\rvert^{2}d_{G/H\times\hat{H}}(gH,\gamma)\geq 0,$ (131)

so $A^{H}(E)$ is positive. Next, if $E\subseteq E^{\prime}$, $\vphantom{\psi
A^{H}(E)\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(E)}\middle\rvert\smash{\psi}\right\rangle\leq\vphantom{\psi
A^{H}(E^{\prime})\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(E^{\prime})}\middle\rvert\smash{\psi}\right\rangle$.
In particular, $\vphantom{\psi
A^{H}(E)\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(E)}\middle\rvert\smash{\psi}\right\rangle\leq\vphantom{\psi
A^{H}(G/H\times\hat{H})\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(G/H\times\hat{H})}\middle\rvert\smash{\psi}\right\rangle$.
We have, by Plancherel’s theorem, that

$\displaystyle\begin{split}\vphantom{\psi
A^{H}(G/H\times\hat{H})\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(G/H\times\hat{H})}\middle\rvert\smash{\psi}\right\rangle&=\int_{G/H}\int_{\hat{H}}\left\lvert(\mathcal{F}_{H}\lvert\psi\circ
g\rangle)(\gamma)\right\rvert^{2}d_{\hat{H}}\gamma d_{G/H}(gH)\\\
&=\int_{G/H}\int_{H}\left\lvert\lvert\psi\circ
g\rangle(h)\right\rvert^{2}d_{H}hd_{G/H}(gH)\\\
&=\int_{G/H}\int_{H}\left\lvert\psi(gh)\right\rvert^{2}d_{H}hd_{G/H}(gH)\\\
&=\int_{G}|\psi(g)|^{2}d_{G}g=\vphantom{\psi}\left\langle\smash{\psi}\middle|\smash{\psi}\right\rangle.\end{split}$
(132)

Thus, $A^{H}(G/H\times\hat{H})$ is bounded, and therefore $A^{H}(E)$ is
bounded.

To get that $A^{H}$ is a POVM, we need next that it is weakly countably
additive. Let $\\{E_{i}\\}_{i=1}^{\infty}$ be a countable collection of
disjoint measurable sets in $G/H\times\hat{H}$. Then, by monotone convergence

$\displaystyle\begin{split}\vphantom{\psi
A^{H}\Big{\lparen}\bigcup_{i=1}^{\infty}E_{i}\Big{\rparen}\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}\Big{\lparen}\bigcup_{i=1}^{\infty}E_{i}\Big{\rparen}}\middle\rvert\smash{\psi}\right\rangle&=\int_{\bigcup_{i=1}^{\infty}E_{i}}\left\lvert(\mathcal{F}_{H}\lvert\psi\circ
g\rangle)(\gamma)\right\rvert^{2}d_{G/H\times\hat{H}}(gH,\gamma)\\\
&=\int\sum_{i=1}^{\infty}\chi_{E_{i}}(gH,\gamma)\left\lvert(\mathcal{F}_{H}\lvert\psi\circ
g\rangle)(\gamma)\right\rvert^{2}d_{G/H\times\hat{H}}(gH,\gamma)\\\
&=\sum_{i=1}^{\infty}\int\chi_{E_{i}}(gH,\gamma)\left\lvert(\mathcal{F}_{H}\lvert\psi\circ
g\rangle)(\gamma)\right\rvert^{2}d_{G/H\times\hat{H}}(gH,\gamma)\\\
&=\sum_{i=1}^{\infty}\vphantom{\psi
A^{H}(E_{i})\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(E_{i})}\middle\rvert\smash{\psi}\right\rangle.\end{split}$
(133)

Using the polarization identity gives the full form of weak countable
additivity.

Finally, we want to show that $A^{H}$ is projective. To see that, note
$(A^{H}\lvert\psi\rangle)(g)=\int_{(E)_{gH}}\left\lparen\mathcal{F}_{H}\lvert\psi\circ
g\rangle\right\rparen(\gamma)d_{\hat{H}}\gamma$ almost everywhere, where
$(E)_{gH}=\\{{\gamma\in\hat{H}}|{(gH,\gamma)\in E}\\}$ is the cross-section of
$E$ at $gH$. In fact, for any $\lvert\phi\rangle\in L^{2}(G)$ continuous with
compact support, it follows that

$\displaystyle\begin{split}\vphantom{\phi
A^{H}(E)\psi}\left\langle\smash{\phi}\middle\rvert\smash{A^{H}(E)}\middle\rvert\smash{\psi}\right\rangle&=\int_{E}\int_{H}\overline{\phi(gh)}\gamma(h)d_{H}h(\mathcal{F}_{H}\lvert\psi\circ
g\rangle)(\gamma)d_{{}_{G/H\times\hat{H}}}(gH,\gamma)\\\
&=\int_{\hat{H}}\chi_{(E)_{gH}}(\gamma)\int_{G/H}\int_{H}\overline{\phi(gh)}(\mathcal{F}_{H}\lvert\psi\circ
gh\rangle)(\gamma)d_{H}hd_{G/H}gHd_{\hat{H}}\gamma\\\
&=\int_{(E)_{gH}}\int_{G}\overline{\phi(g)}(\mathcal{F}_{H}\lvert\psi\circ
g\rangle)(\gamma)d_{G}gd_{\hat{H}}\gamma\\\
&=\int_{G}\overline{\phi(g)}\int_{(E)_{gH}}\left\lparen\mathcal{F}_{H}\lvert\psi\circ
g\rangle\right\rparen(\gamma)d_{\hat{H}}\gamma d_{G}g.\end{split}$ (134)

Thus, for any measurable $E,F$, we get that
$(A^{H}(E)A^{H}(F)\lvert\psi\rangle)(g)=\int_{(E)_{gH}}\left\lparen\mathcal{F}_{H}(A^{H}(F)\lvert\psi\rangle\circ
g)\right\rparen(\gamma)d_{\hat{H}}\gamma$ and since, for $h\in H$,

$\displaystyle\begin{split}(A^{H}(F)\lvert\psi\rangle\circ
g)(h)&=(A^{H}(F)\lvert\psi\rangle)(gh)=\int_{(F)_{ghH}}(\mathcal{F}_{H}\lvert\psi\circ
gh\rangle)(\gamma)d_{\hat{H}}\gamma\\\
&=\int_{(F)_{gH}}\gamma(h)(\mathcal{F}_{H}\lvert\psi\circ
g\rangle)(\gamma)d_{\hat{H}}\gamma\\\
&=\mathcal{F}_{H}^{-1}\lparen\chi_{(F)_{gH}}\mathcal{F}_{H}\lvert\psi\circ
g\rangle\rparen(h),\end{split}$ (135)

if $\mathcal{F}_{H}\lvert\psi\circ g\rangle$ is continuous with compact
support. By Fourier inversion, this set is dense, so this identity holds for
all $\lvert\psi\rangle$. As such,

$\displaystyle\begin{split}(A^{H}(E)A^{H}(F)\lvert\psi\rangle)(g)&=\int_{(E)_{gH}}\left\lparen\mathcal{F}_{H}\left\lparen\mathcal{F}_{H}^{-1}\lparen\chi_{(F)_{gH}}\mathcal{F}_{H}\lvert\psi\circ
g\rangle\rparen\right\rparen\right\rparen(\gamma)d_{\hat{H}}\gamma\\\
&=\int_{(E)_{gH}}\chi_{(F)_{gH}}(\gamma)\lparen\mathcal{F}_{H}\lvert\psi\circ
g\rangle\rparen(\gamma)d_{\hat{H}}\gamma\\\ &=\int_{(E\cap
F)_{gH}}\lparen\mathcal{F}_{H}\lvert\psi\circ
g\rangle\rparen(\gamma)d_{\hat{H}}\gamma=(A^{H}(E\cap
F)\lvert\psi\rangle)(g).\end{split}$ (136)

∎

###  8.2 The coset measure game

######  Definition 8.6.

An _abelian coset measure monogamy game_ is a tuple $\texttt{G}=\left\lparen
G,\mathcal{S},E,F\right\rparen$, where $G$ is a locally compact Hausdorff
abelian group, $\mathcal{S}$ is a finite set of closed subgroups of $G$, and
$E\subseteq G$ and $F\subseteq\hat{G}$ are symmetric neighborhoods of
identity, _i.e._ $E=E^{-1}$ and $1\in E$ and the same for $F$.

A _(quantum) strategy_ for a coset measure game G is a tuple
$\texttt{S}=\left\lparen\mathcal{B},\mathcal{C},B,C,\rho\right\rparen$, where
$\mathcal{B}$ and $\mathcal{C}$ are Hilbert spaces,
$B=\\{{B^{H}:\mathscr{B}(G/H)\rightarrow\mathcal{B}(\mathcal{B})}|{H\in\mathcal{S}}\\}$
and
$C=\\{{C^{H}:\mathscr{B}(\hat{H})\rightarrow\mathcal{B}(\mathcal{C})}|{H\in\mathcal{S}}\\}$
are collections of POVM measures, and
$\rho\in\mathcal{D}(L^{2}(G)\otimes\mathcal{B}\otimes\mathcal{C})$.

Let G be a coset measure game and S be a strategy for it. The _winning
probability_ of S is

$\displaystyle\mathfrak{w}_{\texttt{G}}(\texttt{S})=\underset{H\in\mathcal{S}}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\operatorname{Tr}\left[(A^{H}\otimes
B^{H}\otimes C^{H})(E_{H})\rho\right],$ (137)

where $E_{H}=\\{{(gH,\gamma|_{H},egH,(\varphi\gamma)|_{H})}|{g\in
G,\gamma\in\hat{G},e\in E,\varphi\in F}\\}\in G/H\times\hat{H}\times
G/H\times\hat{H}$ and the expection with respect to $H\in\mathcal{S}$ is
uniform.

The winning probability of G is
$\mathfrak{w}(\texttt{G})=\sup_{\texttt{S}}\mathfrak{w}_{\texttt{G}}(\texttt{S})$.

######  Theorem 8.7.

Let $\texttt{G}=(G,\mathcal{S},E,F)$ be a coset measure game for second-
countable $G$, and let $\pi^{i}:\mathcal{S}\rightarrow\mathcal{S}$ for
$i=1,\ldots,|\mathcal{S}|$ be a set of orthogonal permutations, _i.e._
$\pi_{i}\circ\pi_{j}^{-1}$ has no fixed points unless $i=j$. Then, the winning
probability of G is bounded above as

$\displaystyle\mathfrak{w}(\texttt{G})\leq\underset{i}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\sup_{H\in\mathcal{S},g\in
G}\sqrt{\mu_{H}\big{\lparen}H\cap
gE\pi_{i}(H)\big{\rparen}\mu_{\hat{H}}\big{\lparen}F\big{\rparen}}.$ (138)

######  Lemma 8.8.

Let $\mathcal{H}$ be a Hilbert space and $\mathcal{K}$ be a separable Hilbert
space, $X$ be a measurable space,
$P:\mathscr{S}_{X}\rightarrow\mathcal{B}(\mathcal{H})$ be a POVM measure, and
$F:X\rightarrow\mathcal{B}(\mathcal{K})$ be a bounded weakly measurable
function. Then,

$\displaystyle\Big{\|}\int F\otimes dP\Big{\|}\leq\sup_{x\in X}\|F(x)\|.$
(139)

###### Proof.

Let $\varepsilon>0$. Using Lemma A.5, there exists a sequence of simple
functions $\lparen F_{n}=\sum_{l}M^{n}_{l}\chi_{E^{n}_{l}}\rparen$ that
converges strongly pointwise to $F$ and
$\|F_{n}(x)\|\leq\sup_{y}\|F(y)\|+\varepsilon$. Since this implies
$\sup_{n}\left\|\int F_{n}\otimes dP\right\|<\infty$, Theorem A.9 gives that
$\int F_{n}\otimes dP\rightarrow\int F\otimes dP$ weakly. We know that there
exist $\lvert u\rangle,\lvert v\rangle\in\mathcal{K}\otimes\mathcal{H}$ with
$\|\lvert u\rangle\|,\|\lvert v\rangle\|\leq 1$ such that
$\left\lvert\vphantom{u\int F\otimes
dPv}\left\langle\smash{u}\middle\rvert\smash{\int F\otimes
dP}\middle\rvert\smash{v}\right\rangle\right\rvert>\left\|\int F\otimes
dP\right\|-\varepsilon$. Then, using the weak convergence, there exists
$N\in\mathbb{N}$ such that $\left\lvert\vphantom{u\int(F_{n}-F)\otimes
dPv}\left\langle\smash{u}\middle\rvert\smash{\int(F_{n}-F)\otimes
dP}\middle\rvert\smash{v}\right\rangle\right\rvert<\varepsilon$ $\forall n\geq
N$, so

$\displaystyle\begin{split}\Big{\|}\int F\otimes
dP\Big{\|}&<\Big{\lvert}\vphantom{u\int F\otimes
dPv}\left\langle\smash{u}\middle\rvert\smash{\int F\otimes
dP}\middle\rvert\smash{v}\right\rangle\Big{\rvert}+\varepsilon\\\
&<\Big{\lvert}\vphantom{u\int F_{n}\otimes
dPv}\left\langle\smash{u}\middle\rvert\smash{\int F_{n}\otimes
dP}\middle\rvert\smash{v}\right\rangle\Big{\rvert}+2\varepsilon\\\
&\leq\Big{\|}\sum_{l}M^{n}_{l}\otimes P(E^{n}_{l})\Big{\|}+2\varepsilon\\\
&\leq\sup_{l}\|M^{n}_{l}\|+2\varepsilon\\\ &<\sup_{x\in
X}\|F(x)\|+3\varepsilon.\end{split}$ (140)

As this holds for all $\varepsilon>0$, we get $\Big{\|}\int F\otimes
dP\Big{\|}\leq\sup_{x\in X}\|F(x)\|$ as wanted. ∎

######  Lemma 8.9.

Let $H,K\leq G$ be closed subgroups, and let $E\subseteq G$ and
$F\subseteq\hat{G}$ be symmetric neighborhoods of the identity. Then, for any
$\eta\in\hat{G}$ and $qK\in G/K$

$\displaystyle\left\|A^{H}\big{\lparen}G/H\times(F\eta)|_{H}\big{\rparen}A^{K}\big{\lparen}EqK/K\times\hat{K}\big{\rparen}\right\|\leq\sup_{g\in
G}\sqrt{\mu_{H}\big{\lparen}H\cap
gEK\big{\rparen}\mu_{\hat{H}}\big{\lparen}F\big{\rparen}}$ (141)

###### Proof.

First, we use Fourier inversion as a continuous-valued version of
orthogonality of the characters to simplify
$A^{K}\big{\lparen}EqK/K\times\hat{K}\big{\rparen}$. For any
$\lvert\psi\rangle\in L^{2}(G)$ continuous with compact support and
$\|\lvert\psi\rangle\|=1$, and $g\in G$, since
$(EqK/K\times\hat{K})_{gK}=\hat{K}$ if $g\in EqK$ and $\varnothing$ otherwise,

$\displaystyle\begin{split}\left\lparen
A^{K}\big{\lparen}EqK/K\times\hat{K}\big{\rparen}\lvert\psi\rangle\right\rparen(g)&=\chi_{EqK}(g)\int_{\hat{K}}(\mathcal{F}_{K}\lvert\psi\circ
g\rangle)(\gamma)d_{\hat{K}}\gamma\\\
&=\chi_{EqK}(g)(\mathcal{F}_{K}^{-1}\mathcal{F}_{K}\lvert\psi\circ
g\rangle)(1)\\\ &=\chi_{EqK}(g)\lvert\psi\circ
g\rangle(1)=\chi_{EqK}(g)\psi(g).\end{split}$ (142)

Write $A^{K}\big{\lparen}EqK/K\times\hat{K}\big{\rparen}=\Pi_{EqK}$, the
projector onto the subspace of $L^{2}(G)$ where $\psi(g)=0$ if $g\notin EqK$.
Then, using the Cauchy-Schwarz inequality,

$\displaystyle\begin{split}\|A^{H}(G/H\times(F\eta)|_{H})\Pi_{EqK}\lvert\psi\rangle\|^{2}&=\int_{G/H\times(F\eta)|_{H}}\left\lvert(\mathcal{F}_{H}(\Pi_{EqK}\lvert\psi\rangle)\circ
g)(\gamma)\right\rvert^{2}d_{{}_{G/H\times\hat{H}}}(gH,\gamma)\\\
&=\int_{G/H\times(F\eta)|_{H}}\left\lvert\int_{H}\chi_{EqK}(gh)\psi(gh)\overline{\gamma(h)}d_{H}h\right\rvert^{2}d_{{}_{G/H\times\hat{H}}}(gH,\gamma)\\\
&\leq\int_{G/H\times(F\eta)|_{H}}\int_{H}\left\lvert\psi(gh)\right\rvert^{2}d_{H}h\int_{H}\left\lvert\chi_{EqK}(gh)\overline{\gamma(h)}\right\rvert^{2}d_{H}hd_{{}_{G/H\times\hat{H}}}(gH,\gamma)\\\
&=\int_{G/H\times(F\eta)|_{H}}\int_{H}\left\lvert\psi(gh)\right\rvert^{2}d_{H}h\mu_{H}(H\cap
g^{-1}EqK)d_{{}_{G/H\times\hat{H}}}(gH,\gamma)\\\
&\leq\int_{G/H}\int_{H}\left\lvert\psi(gh)\right\rvert^{2}d_{H}hd_{G/H}gH\sup_{g\in
G}\mu_{H}(H\cap g^{-1}EqK)\mu_{\hat{H}}{(F\eta)|_{H}}\\\
&\leq\|\lvert\psi\rangle\|^{2}\sup_{g\in G}\mu_{H}(H\cap
g^{-1}EqK)\mu_{\hat{H}}\left\lparen(F\eta)|_{H}\right\rparen.\end{split}$
(143)

Finally, we note by Haar invariance that
$\mu_{\hat{H}}\left\lparen(F\eta)|_{H}\right\rparen=\mu_{\hat{H}}\left\lparen
F(\eta|_{H})\right\rparen=\mu_{\hat{H}}\left\lparen F\right\rparen$, giving
the result. ∎

###### Proof of Theorem 8.7.

Let $\texttt{S}=\left\lparen\mathcal{B},\mathcal{C},B,C,\rho\right\rparen$ be
a strategy for $\texttt{G}=\left\lparen G,\mathcal{S},E,F\right\rparen$. Due
to Naimark’s theorem Theorem 8.2, we may assume that $B$ and $C$ are PVMs.
Writing, for each $H\in\mathcal{S}$, $\Pi^{H}=(A^{H}\otimes B^{H}\otimes
C^{H})(E_{H})$, the winning probability may be expressed as
$\mathfrak{w}_{\texttt{G}}(\texttt{S})=\underset{H\in\mathcal{S}}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\operatorname{Tr}(\Pi^{H}\rho)$.
Since the overlap result Lemma 7.3 doesn’t depend on the dimension of the
underlying space, we apply it to upper bound the winning probability of the
game, getting

$\displaystyle\mathfrak{w}_{\texttt{G}}(\texttt{S})\leq\Big{\|}\underset{S\in\mathcal{S}}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\Pi^{H}\Big{\|}\leq\underset{i}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\sup_{H\in\mathcal{S}}\left\|\Pi^{H}\Pi^{\pi_{i}(H)}\right\|,$
(144)

since $(\Pi^{H})^{2}=\Pi^{H}$. Fixing $H,K\in\mathcal{S}$, it remains to
simplify $\left\|\Pi^{H}\Pi^{K}\right\|$. Given that

$\displaystyle\begin{split}E_{H}&\subseteq\left\\{{(gH,\gamma|_{H},g^{\prime}H,(\varphi\gamma)|_{H})}\middle|{g,g^{\prime}\in
G;\gamma\in\hat{G},\varphi\in F}\right\\}\\\
&=\bigcup_{\gamma\in\hat{G}}G/H\times\\{\gamma|_{H}\\}\times
G/H\times(F\gamma)|_{H},\end{split}$ (145)

we get that

$\displaystyle\begin{split}\Pi^{H}&\leq\int_{\hat{H}}(A^{H}\otimes
B^{H})(G/H\times(F\eta)|_{H}\times G/H)\otimes dC^{H}(\eta)\\\
&=\int_{\hat{H}}A^{H}(G/H\times(F\eta)|_{H})\otimes\mathbb{I}_{B}\otimes
dC^{H}(\eta).\end{split}$ (146)

Similarly, $E_{K}\subseteq\bigcup_{g\in G}\\{gK\\}\times\hat{K}\times
EgK/K\times\hat{K}$, so $\Pi^{K}\leq\int_{G/H}A^{K}(EgK/K\times\hat{K})\otimes
dB^{K}(gK)\otimes\mathbb{I}_{C}$. Thus, we may bound the norms

$\displaystyle\begin{split}\left\|\Pi^{H}\Pi^{K}\right\|&\leq\Big{\|}\int\\!A^{H}(G/H\times(F\eta)|_{H})\otimes\mathbb{I}_{B}\otimes
dC^{H}(\eta)\int\\!A^{K}(EgK/K\times\hat{K})\otimes
dB^{K}(gK)\otimes\mathbb{I}_{C}\Big{\|}\\\ &=\Big{\|}\int
A^{H}(G/H\times(F\eta)|_{H})A^{K}(EgK/K\times\hat{K})\otimes d(B^{K}\otimes
C^{H})(gK,\eta)\Big{\|}.\end{split}$ (147)

Using Lemma 8.8, $\left\|\Pi^{H}\Pi^{K}\right\|\leq\sup_{gK\in
G/K,\eta\in\hat{H}}\left\|A^{H}(G/H\times(F\eta)|_{H})A^{K}(EgK/K\times\hat{K})\right\|$;
and then using Lemma 8.9. $\left\|\Pi^{H}\Pi^{K}\right\|\leq\sup_{g\in
G}\sqrt{\mu_{H}(H\cap gEK)\mu_{\hat{H}}(F)}$. Thus, putting it together,

$\displaystyle\mathfrak{w}_{\texttt{G}}(\texttt{S})\leq\underset{i}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\sup_{H\in\mathcal{S},g\in
G}\sqrt{\mu_{H}(H\cap gE\pi_{i}(H))\mu_{\hat{H}}(F)},$ (148)

and since this holds for every strategy, we have the wanted result. ∎

Note that it follows directly from the proof that we also have the bound

$\displaystyle\mathfrak{w}(\texttt{G})\leq\underset{i}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\sup_{H\in\mathcal{S},g\in
G}\min\left\\{{1,\sqrt{\mu_{H}\big{\lparen}H\cap
gE\pi_{i}(H)\big{\rparen}\mu_{\hat{H}}\big{\lparen}F\big{\rparen}}}\right\\},$
(149)

which may be better in the case that the Haar measure of one of the groups is
not $1$.

###  8.3 State-sending version of the game

In this section, we additionally assume that, for the subgroups $H\leq G$ we
consider, the Haar measures on $G/H$ and $\hat{H}$ are $\sigma$-finite, that
is the sets may be written as the countable union of sets of finite measure.
This holds for all our groups of interest here, in particular
$\mathbb{R}^{n}$, $\mathbb{C}$, and $U(1)$, which appear above.

We use approximate maximally entangled states to be able to interpret the
coset monogamy game as a game where Alice samples a random pair $(gH,\gamma)$,
prepares a damped version of the associated unnormalizable coset states, and
then sends to it to Bob and Charlie through an adversarially-chosen channel.
This will allow us to generalise the original interpretation of the coset
monogamy game as a quantum encoding of classical messages as in [CLLZ21].

######  Definition 8.10.

Let $H\leq G$ be a closed subgroup, let $c:L^{2}(G)\rightarrow L^{2}(G)$ be a
complex conjugate, and $(\Delta_{n})$ be a damping sequence in
$\mathcal{B}(L^{2}(G))$ that $\mu_{G/H}\times\mu_{\hat{H}}$-damps $A^{H}$. By
Lemma B.5, there exists measurable
$\rho_{n}:G/H\times\hat{H}\rightarrow\mathcal{D}(L^{2}(G))$ and integrable
$\pi_{n}:G/H\times\hat{H}\rightarrow[0,\infty)$ such that
$\Delta_{n}^{\dagger}A^{H}(E)\Delta_{n}=\int\rho_{n}(gH,\gamma)\pi_{n}(gH,\gamma)d(gH,\gamma)$.
The _state associated to $H$_ is the measurable function

$\displaystyle\begin{split}&\sigma^{H}_{n}:gH\times\hat{H}\rightarrow\mathcal{D}(L^{2}(G))\\\
&\sigma^{H}_{n}(gH,\gamma)=c\rho_{n}(gH,\gamma)c;\end{split}$ (150)

and the _probability measure associated to $H$_ is

$\displaystyle\begin{split}&\mu^{H}_{n}:\mathscr{B}(G/H)\otimes\mathscr{B}(\hat{H})\rightarrow[0,1]\\\
&\mu^{H}_{n}(E)=\frac{1}{\|\Delta_{n}\|_{2}^{2}}\int_{E}\pi_{n}(gH,\gamma)d(gH,\gamma).\end{split}$
(151)

$\mu^{H}_{n}$ is in fact a probability measure, as

$\displaystyle\mu^{H}_{n}(G/H\times\hat{H})=\frac{1}{\|\Delta_{n}\|_{2}^{2}}\int\operatorname{Tr}(\rho_{n})\mu_{n}d\mu=\frac{\operatorname{Tr}(\Delta_{n}^{\dagger}A^{H}(G/H\times\hat{H})\Delta_{n})}{\|\Delta_{n}\|_{2}^{2}}=\frac{\operatorname{Tr}(\Delta_{n}^{\dagger}\Delta_{n})}{\|\Delta_{n}\|_{2}^{2}}=1,$
(152)

and we have
$\frac{1}{\|\Delta_{n}\|_{2}^{2}}c\Delta_{n}^{\dagger}A^{H}(E)\Delta_{n}c=\int_{E}\sigma^{H}_{n}d\mu^{H}_{n}$.

It is also important to note that, due to the definition of $A^{H}$ in terms
of the measure $\mu_{G/H}\times\mu_{\hat{H}}$, _any_ damping sequence
$\mu_{G/H}\times\mu_{\hat{H}}$-damps it.

######  Definition 8.11.

Let $\texttt{G}=\left\lparen G,\mathcal{S},E,F\right\rparen$ be a coset
monogamy game, $c:L^{2}(G)\rightarrow L^{2}(G)$ be a complex conjugate, and
$(\Delta_{n})$ be a damping sequence in $\mathcal{B}(L^{2}(G))$ that
$\mu_{G/H}\times\mu_{\hat{H}}$-damps $A^{H}$ for each $H\in\mathcal{S}$. The
_state-sending version_ of G is the sequence of tuples
$\texttt{G}_{n}=(G,\mathcal{S},E,F,\Delta_{n},c)$.

A _strategy_ for $\texttt{G}_{n}$ is a tuple
$\texttt{S}=(\mathcal{B},\mathcal{C},B,C,\Phi)$, where $\mathcal{B}$ and
$\mathcal{C}$ are the Hilbert spaces held by Bob and Charlie,
$B=\\{{B^{H}:\mathscr{B}(G/H)\rightarrow\mathcal{B}(\mathcal{B})}|{H\in\mathcal{S}}\\}$
and
$C=\\{{C^{H}:\mathscr{B}(\hat{H})\rightarrow\mathcal{B}(\mathcal{C})}|{H\in\mathcal{S}}\\}$
are collections of POVM measures, and
$\Phi:\mathcal{T}_{1}(L^{2}(G))\rightarrow\mathcal{T}_{1}(\mathcal{B}\otimes\mathcal{C})$
is a completely positive trace preserving map.

The _winning probability_ of S is

$\displaystyle\mathfrak{w}_{\texttt{G}_{n}}(\texttt{S})=\underset{H\in\mathcal{S}}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\int\operatorname{Tr}\left[(B^{H}\otimes
C^{H})(EgH\times(F\gamma)|_{H})\Phi(\sigma^{H}_{n}(gH,\gamma))\right]d\mu^{H}_{n}(gH,\gamma).$
(153)

The winning probability of $\texttt{G}_{n}$ is
$\mathfrak{w}(\texttt{G}_{n})=\sup_{\texttt{S}}\mathfrak{w}_{\texttt{G}_{n}}(\texttt{S})$.

Now, we bound the winning probability of the state-sending version by the
winning probability monogamy version.

######  Lemma 8.12.

Let $\mathcal{H}$ be a Hilbert space, $(\Delta_{n})$ in
$\mathcal{B}(\mathcal{H})$ be damping sequence,
$c:\mathcal{H}\rightarrow\mathcal{H}$ be a complex conjugate, and
$(\lvert\Psi_{n}\rangle)$ be the associated approximate maximally entangled
state. Writing
$\operatorname{Tr}_{1}:\mathcal{T}_{1}(\mathcal{H}\otimes\mathcal{H})\rightarrow\mathcal{T}_{1}(\mathcal{H})$
the partial trace on the first copy on $\mathcal{H}$, we have that for any
$A\in\mathcal{B}(\mathcal{H})$,

$\displaystyle\operatorname{Tr}_{1}\left[(A\otimes\mathbb{I})\vphantom{\Psi_{n}}\left\lvert\smash{\Psi_{n}}\middle\rangle\\!\middle\langle\smash{\Psi_{n}}\right\rvert\right]=\frac{1}{\|\Delta_{n}\|_{2}^{2}}c\Delta_{n}^{\dagger}A^{\dagger}\Delta_{n}c.$
(154)

###### Proof.

Using the singular-value decomposition
$\Delta_{n}=\sum_{i=1}^{\infty}s_{n,i}\vphantom{\phi_{n,i}\chi_{n,i}}\left\lvert\smash{\phi_{n,i}}\middle\rangle\\!\middle\langle\smash{\chi_{n,i}}\right\rvert$,
the associated approximate maximally entangled state
$\lvert\Psi_{n}\rangle=\frac{1}{\|\Delta_{n}\|_{2}}\sum_{i=1}^{\infty}s_{n,i}\lvert\phi_{n,i}\rangle\otimes
c\lvert\chi_{n,i}\rangle$. Then,

$\displaystyle\begin{split}\operatorname{Tr}_{1}\left[(A\otimes\mathbb{I})\vphantom{\Psi_{n}}\left\lvert\smash{\Psi_{n}}\middle\rangle\\!\middle\langle\smash{\Psi_{n}}\right\rvert\right]&=\frac{1}{\|\Delta_{n}\|_{2}^{2}}\sum_{i,j}s_{n,i}s_{n,j}\vphantom{\phi_{n,j}A\phi_{n,i}}\left\langle\smash{\phi_{n,j}}\middle|\smash{A\phi_{n,i}}\right\rangle\vphantom{c\chi_{n,i}c\chi_{n,j}}\left\lvert\smash{c\chi_{n,i}}\middle\rangle\\!\middle\langle\smash{c\chi_{n,j}}\right\rvert\\\
&=\frac{1}{\|\Delta_{n}\|_{2}^{2}}\sum_{i,j}s_{n,i}s_{n,j}\vphantom{cA\phi_{n,i}c\phi_{n,j}}\left\langle\smash{cA\phi_{n,i}}\middle|\smash{c\phi_{n,j}}\right\rangle\vphantom{c\chi_{n,i}c\chi_{n,j}}\left\lvert\smash{c\chi_{n,i}}\middle\rangle\\!\middle\langle\smash{c\chi_{n,j}}\right\rvert\\\
&=\frac{1}{\|\Delta_{n}\|_{2}^{2}}\sum_{i,j}s_{n,i}s_{n,j}\vphantom{c\chi_{n,i}c\phi_{n,i}}\left\lvert\smash{c\chi_{n,i}}\middle\rangle\\!\middle\langle\smash{c\phi_{n,i}}\right\rvert(cAc)^{\dagger}\vphantom{c\phi_{n,j}c\chi_{n,j}}\left\lvert\smash{c\phi_{n,j}}\middle\rangle\\!\middle\langle\smash{c\chi_{n,j}}\right\rvert.\end{split}$
(155)

To simplify this, note that for any
$\lvert\phi\rangle,\lvert\psi\rangle\in\mathcal{H}$,

$\displaystyle\sum_{i}s_{n,i}\vphantom{\psi
c\chi_{n,i}}\left\langle\smash{\psi}\middle|\smash{c\chi_{n,i}}\right\rangle\\!\\!\vphantom{c\phi_{n,i}\phi}\left\langle\smash{c\phi_{n,i}}\middle|\smash{\phi}\right\rangle=\sum_{i}s_{n,i}\vphantom{c\phi\phi_{n,i}}\left\langle\smash{c\phi}\middle|\smash{\phi_{n,i}}\right\rangle\\!\\!\vphantom{\chi_{n,i}c\psi}\left\langle\smash{\chi_{n,i}}\middle|\smash{c\psi}\right\rangle=\vphantom{c\phi\Delta_{n}c\psi}\left\langle\smash{c\phi}\middle|\smash{\Delta_{n}c\psi}\right\rangle=\vphantom{c\Delta_{n}c\psi\phi}\left\langle\smash{c\Delta_{n}c\psi}\middle|\smash{\phi}\right\rangle=\vphantom{\psi(c\Delta_{n}c)^{\dagger}\phi}\left\langle\smash{\psi}\middle|\smash{(c\Delta_{n}c)^{\dagger}\phi}\right\rangle,$
(156)

so
$\operatorname{Tr}_{1}\left[(A\otimes\mathbb{I})\vphantom{\Psi_{n}}\left\lvert\smash{\Psi_{n}}\middle\rangle\\!\middle\langle\smash{\Psi_{n}}\right\rvert\right]=\frac{1}{\|\Delta_{n}\|_{2}^{2}}(c\Delta_{n}c)^{\dagger}(cAc)^{\dagger}(c\Delta_{n}c)$.
Finally, note that conjugation by a complex conjugate commutes with the
adjoint: for any $T\in\mathcal{B}(\mathcal{H})$,

$\displaystyle\vphantom{\psi(cTc)^{\dagger}\phi}\left\langle\smash{\psi}\middle\rvert\smash{(cTc)^{\dagger}}\middle\rvert\smash{\phi}\right\rangle=\vphantom{cTc\psi\phi}\left\langle\smash{cTc\psi}\middle|\smash{\phi}\right\rangle=\vphantom{c\phi
Tc\psi}\left\langle\smash{c\phi}\middle|\smash{Tc\psi}\right\rangle=\vphantom{T^{\dagger}c\phi
c\psi}\left\langle\smash{T^{\dagger}c\phi}\middle|\smash{c\psi}\right\rangle=\vphantom{\psi
cT^{\dagger}c\phi}\left\langle\smash{\psi}\middle|\smash{cT^{\dagger}c\phi}\right\rangle,$
(157)

and therefore
$\operatorname{Tr}_{1}\left[(A\otimes\mathbb{I})\vphantom{\Psi_{n}}\left\lvert\smash{\Psi_{n}}\middle\rangle\\!\middle\langle\smash{\Psi_{n}}\right\rvert\right]=\frac{1}{\|\Delta_{n}\|_{2}^{2}}c\Delta_{n}^{\dagger}ccA^{\dagger}cc\Delta_{n}c=\frac{1}{\|\Delta_{n}\|_{2}^{2}}c\Delta_{n}^{\dagger}A^{\dagger}\Delta_{n}c$.
∎

######  Theorem 8.13.

Let $\texttt{G}_{n}=(G,\mathcal{S},E,F,\Delta_{n},c)$ be the state-sending
version of a coset monogamy game G, and $(\lvert\Psi_{n}\rangle)$ be the
approximate maximally entangled state associated to $(\Delta_{n})$. For any
strategy $\texttt{S}=(\mathcal{B},\mathcal{C},B,C,\Phi)$ for $\texttt{G}_{n}$,
write
$\texttt{S}_{n}=\left\lparen\mathcal{B},\mathcal{C},B,C,(\mathbb{I}\otimes\Phi)(\vphantom{\Psi_{n}}\left\lvert\smash{\Psi_{n}}\middle\rangle\\!\middle\langle\smash{\Psi_{n}}\right\rvert)\right\rparen$,
which is a strategy for G. We have that

$\displaystyle\mathfrak{w}_{\texttt{G}_{n}}(\texttt{S})=\mathfrak{w}_{\texttt{G}}(\texttt{S}_{n}).$
(158)

In particular, this implies that
$\mathfrak{w}(\texttt{G}_{n})\leq\mathfrak{w}(\texttt{G})$. However, we do not
necessarily have that
$\lim_{n\rightarrow\infty}\mathfrak{w}(\texttt{G}_{n})=\mathfrak{w}(\texttt{G})$,
as this does not even always hold in the case of finite information.

###### Proof.

First, we may write

$\displaystyle\begin{split}\mathfrak{w}_{\texttt{G}}(\texttt{S}_{n})&=\underset{H\in\mathcal{S}}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\operatorname{Tr}\left[(A^{H}\otimes
B^{H}\otimes
C^{H})(E_{H})(\mathbb{I}\otimes\Phi)(\vphantom{\Psi_{n}}\left\lvert\smash{\Psi_{n}}\middle\rangle\\!\middle\langle\smash{\Psi_{n}}\right\rvert)\right]\\\
&=\underset{H\in\mathcal{S}}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\operatorname{Tr}\left[\left\lparen\int
dA^{H}(gH,\gamma)\otimes(B^{H}\otimes
C^{H})(EgH\times(F\gamma)|_{H})\right\rparen(\mathbb{I}\otimes\Phi)(\vphantom{\Psi_{n}}\left\lvert\smash{\Psi_{n}}\middle\rangle\\!\middle\langle\smash{\Psi_{n}}\right\rvert)\right].\end{split}$
(159)

Fix some $H\in\mathcal{S}$ and let $(F_{k}=\sum_{i}M^{k}_{i}\chi_{E^{k}_{i}})$
be a sequence of simple functions
$G/H\times\hat{H}\rightarrow\mathcal{B}(\mathcal{B}\otimes\mathcal{H})$ that
converges strongly pointwise to $(gH,\gamma)\mapsto(B^{H}\otimes
C^{H})(EgH\times(F\gamma)|_{H})$. Then.

$\displaystyle\begin{split}\operatorname{Tr}\left[\left\lparen\int
dA^{H}\otimes
F_{k}\right\rparen(\mathbb{I}\otimes\Phi)(\vphantom{\Psi_{n}}\left\lvert\smash{\Psi_{n}}\middle\rangle\\!\middle\langle\smash{\Psi_{n}}\right\rvert)\right]&=\sum_{i}\operatorname{Tr}\left[\left\lparen
A^{H}(E^{k}_{i})\otimes
M^{k}_{i}\right\rparen(\mathbb{I}\otimes\Phi)(\vphantom{\Psi_{n}}\left\lvert\smash{\Psi_{n}}\middle\rangle\\!\middle\langle\smash{\Psi_{n}}\right\rvert)\right]\\\
&=\sum_{i}\operatorname{Tr}\left[M^{k}_{i}\Phi\left\lparen\operatorname{Tr}_{1}\left[(A^{H}(E^{k}_{i})\otimes\mathbb{I})\vphantom{\Psi_{n}}\left\lvert\smash{\Psi_{n}}\middle\rangle\\!\middle\langle\smash{\Psi_{n}}\right\rvert\right]\right\rparen\right]\\\
&=\sum_{i}\frac{1}{\|\Delta_{n}\|_{2}^{2}}\operatorname{Tr}\left[M^{k}_{i}\Phi(c\Delta_{n}^{\dagger}A^{H}(E^{k}_{i})\Delta_{n}c)\right]\\\
&=\sum_{i}\int_{E^{k}_{i}}\operatorname{Tr}\left[M^{k}_{i}\Phi(\sigma^{H}_{n})\right]d\mu^{H}_{n}\\\
&=\int\operatorname{Tr}\left[F_{k}(gH,\gamma)\Phi(\sigma^{H}_{n}(gH,\gamma))\right]d\mu^{H}_{n}(gH,\gamma).\end{split}$
(160)

By weak convergence of the integral, we get that

$\displaystyle\begin{split}\mathfrak{w}_{\texttt{G}}(\texttt{S}_{n})&=\lim_{k\rightarrow\infty}\underset{H\in\mathcal{S}}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\int\operatorname{Tr}\left[F_{k}(gH,\gamma)\Phi(\sigma^{H}_{n}(gH,\gamma))\right]d\mu^{H}_{n}(gH,\gamma)\\\
&=\underset{H\in\mathcal{S}}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\int\operatorname{Tr}\left[(B^{H}\otimes
C^{H})(EgH\times(F\gamma)|_{H})\Phi(\sigma^{H}_{n}(gH,\gamma))\right]d\mu^{H}_{n}(gH,\gamma)\\\
&=\mathfrak{w}_{\texttt{G}_{n}}(\texttt{S}).\end{split}$ (161)

∎

##  9 Monogamy games for compact groups

In this section, we extend the bound to games on compact groups – groups that
may be both non-abelian and infinite. However, the compactness is required
here, unlike in the abelian case of Section 8, in order to be able to make use
of the Peter-Weyl theorem when considering the representations. This provides
a class of groups that, although infinite, are conceptually more similar to
the finite groups of Section 7.

###  9.1 Compact coset measure

Let $G$ be a compact Hausdorff group. The space of quantum states on $G$ is
again $L^{2}(G)$ with inner product given by the Haar integral. We know, by
the Peter-Weyl theorem, that the matrix elements of the finite-dimensional
irreducible representations of $G$ are orthogonal, and span $L^{2}(G)$ as a
Hilbert space. We can again choose a full set of finite-dimensional irreps
$\textsf{Irr}(G)$, by making use of the axiom of choice, and their matrix
elements $\textsf{IB}(G)$. Then, for
$\gamma_{m,n},\gamma^{\prime}_{m^{\prime},n^{\prime}}\in\textsf{IB}(G)$, we
get the orthogonality relation extending Schur orthogonality

$\displaystyle\left\langle\gamma_{m,n},\gamma^{\prime}_{m^{\prime},n^{\prime}}\right\rangle_{G}:=\int\overline{\gamma_{m,n}(g)}\gamma^{\prime}_{m^{\prime},n^{\prime}}(g)d_{G}g=\frac{1}{d_{\gamma}}\delta_{\gamma,\gamma^{\prime}}\delta_{m,m^{\prime}}\delta_{n,n^{\prime}}.$
(162)

Also, since we will be summing over elements of $\textsf{IB}(G)$, we can
trivially endow it with the $\sigma$-algebra of all sets
$\mathscr{P}(\textsf{IB}(G))$ and the counting measure.

Let $H\leq G$ be a closed subgroup. Unlike the abelian case, $G/H$ is not
necessarily a group, so we will need to fix a set of coset representatives
$\textsf{CS}(H)$, again using the axiom of choice. For $g\in G$, write
$[g]_{H}\in\textsf{CS}(H)$ for its representative, dropping the subscript when
$H$ is obvious. As the Haar measures of $G$ and $H$ induce a measure on $G/H$,
they induce a measure on $\textsf{CS}(H)$. The $\sigma$-algebra is the Borel
algebra of the quotient topology
$\mathscr{B}(\textsf{CS}(H))=\left\\{{E\subseteq\textsf{CS}(H)}\middle|{EH\in\mathscr{B}(G)}\right\\}$,
and the corresponding measure $\mu_{\textsf{CS}(H)}(E)=\mu_{G}(EH)$. We denote
$d[g]=d\mu_{\textsf{CS}(H)}([g])$. This measure interacts well with the Haar
measures on $G$ and $H$.

######  Lemma 9.1.

  * •

The measurable functions $\textsf{CS}(H)\rightarrow\mathbb{C}$ may be
identified with the measurable functions $G\rightarrow\mathbb{C}$ that are
constant on the cosets of $H$. If $f:\textsf{CS}(H)\rightarrow\mathbb{C}$ is
integrable

$\int_{\textsf{CS}(H)}f([g])d[g]=\int_{G}f([g])d_{G}g.$

  * •

If $f:G\rightarrow\mathbb{C}$ is integrable,

$\int_{G}f(g)d_{G}g=\int_{\textsf{CS}(H)}\int_{H}f([g]h)d_{H}hd[g].$

  * •

If $E\in\mathscr{B}(G)$, then

$\mu_{G}(E)=\int_{\textsf{CS}(H)}\mu_{H}([g]^{-1}E\cap H)d[g].$

###### Proof.

Let $f:\textsf{CS}(H)\rightarrow\mathbb{C}$ be measurable. So for each Borel
set of $B\subseteq\mathbb{C}$, $f^{-1}(B)\in\mathscr{B}(\textsf{CS}(H))$
$\iff$ $f^{-1}(B)H\in\mathscr{B}(G)$. Letting
$f^{\prime}:G\rightarrow\mathbb{C}$ be $f^{\prime}(g)=f([g])$,
$(f^{\prime})^{-1}(B)=\\{g\in G|f([g])\in B\\}=f^{-1}(B)H$, so $f^{\prime}$ is
measurable. Conversely, if $f^{\prime}:G\rightarrow\mathbb{C}$ is a measurable
function constant on the cosets, let
$f^{\prime\prime}=f^{\prime}|_{\textsf{CS}(H)}$. Then,
$(f^{\prime\prime})^{-1}(B)H=((f^{\prime})^{-1}B\cap\textsf{CS}(H))H=(f^{\prime})^{-1}(B)\in\mathscr{B}(G)$.
So, $f\leftrightarrow f^{\prime}$ provides the bijective correspondence we
were looking for. Let $f:\textsf{CS}(H)\rightarrow\mathbb{C}$ is integrable.
There exists a sequence of simple functions that $(f_{n})$ that converges
pointwise to $f$. Then, $(f_{n}^{\prime})$ converges to $f^{\prime}$. For some
$n$, writing $f_{n}=\sum_{i}c_{i}\chi_{E_{i}}$,

$\int
f_{n}([g])d[g]=\sum_{i}c_{i}\mu_{\textsf{CS}(H)}(E_{i})=\sum_{i}c_{i}\mu_{G}(E_{i}H)=\int
f_{n}^{\prime}(g)d_{G}g.$

Therefore,

$\int f([g])d[g]=\lim_{n\rightarrow\infty}\int
f_{n}([g])d[g]=\lim_{n\rightarrow\infty}\int f_{n}^{\prime}(g)d_{G}g=\int
f^{\prime}(g)d_{G}g=\int f([g])d_{G}g.$

Let $f:G\rightarrow\mathbb{C}$ be integrable. Consider the function $G\times
H\rightarrow\mathbb{C}$, $(g,h)\mapsto f(gh)$. As the product is continuous,
this is a composition of measurable functions, so measurable. Also, by
invariance of the Haar measure on $G$,

$\int_{H}\int_{G}|f(gh)|d_{G}gd_{H}h=\int_{H}\int_{G}|f(g)|d_{G}gd_{H}h=\int_{G}|f(g)|d_{G}g,$

which means by Tonelli’s theorem that the map is integrable on the product
measure space. So, the integral $\int f(gh)d_{H}h$ exists for almost every
$g\in G$ and the map $g\mapsto\int f(gh)d_{H}h$ is measurable with

$\int_{G}\int_{H}f(gh)d_{H}hd_{G}g=\int_{H}\int_{G}f(gh)d_{G}gd_{H}h=\int_{G}f(g)d_{G}g$

by Fubini’s theorem. Finally, by invariance of the Haar measure on $H$,
$\int_{H}f(gh)d_{H}h$ is constant on the cosets of $H$, so

$\int_{G}f(g)d_{G}g=\int_{G}\int_{H}f(gh)d_{H}hd_{G}g=\int_{\textsf{CS}(H)}\int_{H}f([g]h)d_{H}hd[g].$

For the last point, take $f:G\rightarrow\mathbb{C}$ to be $f=\chi_{E}$. Then,
$f$ is bounded and measurable, so integrable, giving

$\mu_{G}(E)=\int_{G}\chi_{E}(g)d_{G}g=\int_{\textsf{CS}(H)}\int_{H}\chi_{E}([g]h)d_{H}hd[g]=\int_{\textsf{CS}(H)}\mu_{H}([g]^{-1}E\cap
H)d[g].$

∎

Now, similarly to Definition 8.4, we define the coset operator measure for
$H$.

######  Definition 9.2.

Let $G$ be a compact Hausdorff group and $H\leq G$ be a closed subgroup. The
_coset operator measure_ is the map
$A^{H}:\mathscr{B}(\textsf{CS}(H))\otimes\mathscr{P}(\textsf{IB}(H))\rightarrow\mathcal{B}(L^{2}(G))$
defined, for
$E=\bigcup_{\gamma_{m,n}}E_{\gamma_{m,n}}\times\\{\gamma_{m,n}\\}\subseteq\textsf{CS}(H)\times\textsf{IB}(H)$
measurable and $\lvert\phi\rangle,\lvert\psi\rangle\in L^{2}(G)$, as

$\displaystyle\vphantom{\phi
A^{H}(E)\psi}\left\langle\smash{\phi}\middle\rvert\smash{A^{H}(E)}\middle\rvert\smash{\psi}\right\rangle=\sum_{\gamma_{m,n}}d_{\gamma}\int_{E_{\gamma_{m,n}}}\left\langle\phi\circ[g],\gamma_{m,n}\right\rangle_{H}\left\langle\gamma_{m,n},\psi\circ[g]\right\rangle_{H}d[g],$
(163)

where $[g]$ is seen as the left shift $[g](h)=[g]h$.

It is direct to see that $A^{H}(E)$ is linear, where it is defined, by
linearity of the integral. Next, it is well-defined as a function on
$L^{2}(G)$ as it does not depend on the choice of representative of
$\lvert\psi\rangle$. In fact, if $\lvert\psi\rangle=0$, then
$\int_{H}|\psi([g]h)|^{2}d_{H}h=0$ for almost all $[g]\in\textsf{CS}(H)$, so
$\langle\gamma_{m,n},\psi\circ[g]\rangle_{H}=0$ almost everywhere, and
therefore $A^{H}(E)\lvert\psi\rangle=0$. Positivity is due to the fact that

$\displaystyle\vphantom{\psi
A^{H}(E)\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(E)}\middle\rvert\smash{\psi}\right\rangle=\sum_{\gamma_{m,n}}d_{\gamma}\int_{E_{\gamma_{m,n}}}\lvert\langle\gamma_{m,n},\psi\circ[g]\rangle_{H}\rvert^{2}d[g]\geq
0,$ (164)

and that $A^{H}(E)$ is Hermitian, given that

$\displaystyle\vphantom{\varphi
A^{H}(E)\psi}\left\langle\smash{\varphi}\middle\rvert\smash{A^{H}(E)}\middle\rvert\smash{\psi}\right\rangle=\sum_{\gamma_{m,n}}d_{\gamma}\int_{E_{\gamma_{m,n}}}\overline{\langle\gamma_{m,n},\varphi\circ[g]\rangle_{H}\langle\psi\circ[g],\gamma_{m,n}\rangle_{H}}d[g]=\overline{\vphantom{\psi
A^{H}(E)\varphi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(E)}\middle\rvert\smash{\varphi}\right\rangle}.$
(165)

To show boundedness, it follows from the polarization identity that we need
only show that the collection of $\vphantom{\psi
A^{H}(E)\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(E)}\middle\rvert\smash{\psi}\right\rangle$
for all $\|\lvert\psi\rangle\|=1$ is bounded. First, note that by monotonicity
of the integral, if $E\subseteq E^{\prime}$, we have
$E_{\gamma_{m,n}}\subseteq E^{\prime}_{\gamma_{m,n}}$ for all $\gamma_{m,n}$,
and therefore

$\displaystyle\begin{split}\vphantom{\psi
A^{H}(E)\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(E)}\middle\rvert\smash{\psi}\right\rangle&=\sum_{\gamma_{m,n}}d_{\gamma}\int_{E_{\gamma_{m,n}}}\lvert\langle\gamma_{m,n},\psi\circ[g]\rangle_{H}\rvert^{2}d[g]\\\
&\leq\sum_{\gamma_{m,n}}d_{\gamma}\int_{E^{\prime}_{\gamma_{m,n}}}\lvert\langle\gamma_{m,n},\psi\circ[g]\rangle_{H}\rvert^{2}d[g]=\vphantom{\psi
A^{H}(E^{\prime})\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(E^{\prime})}\middle\rvert\smash{\psi}\right\rangle.\end{split}$
(166)

So, $\vphantom{\psi
A^{H}(E)\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(E)}\middle\rvert\smash{\psi}\right\rangle\leq\vphantom{\psi
A^{H}(\textsf{CS}(H)\times\textsf{IB}(H))\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(\textsf{CS}(H)\times\textsf{IB}(H))}\middle\rvert\smash{\psi}\right\rangle$,
and using Tonelli’s theorem

$\displaystyle\begin{split}\vphantom{\psi
A^{H}(\textsf{CS}(H)\times\textsf{IB}(H))\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(\textsf{CS}(H)\times\textsf{IB}(H))}\middle\rvert\smash{\psi}\right\rangle&=\int_{\textsf{CS}(H)}\sum_{\gamma_{m,n}}d_{\gamma}\langle\psi\circ[g],\gamma_{m,n}\rangle_{H}\langle\gamma_{m,n},\psi\circ[g]\rangle_{H}d[g]\\\
&=\int_{\textsf{CS}(H)}\langle\psi\circ[g],\psi\circ[g]\rangle_{H}d[g]\\\
&=\int_{\textsf{CS}(H)}\int_{H}|\psi([g]h)|^{2}d_{H}hd[g]\\\
&=\int_{G}|\psi(g)|^{2}d_{G}g=1.\end{split}$ (167)

Thus, $\vphantom{\psi
A^{H}(E)\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(E)}\middle\rvert\smash{\psi}\right\rangle\leq
1$ and the positivity implies $\|A^{H}(E)\|\leq 1$.

######  Lemma 9.3.

$A^{H}$ is a PVM measure.

###### Proof.

Most importantly, we need that $A^{H}$ is weakly countably additive. Let
$E^{1},E^{2},\ldots\subseteq\textsf{CS}(H)\times\textsf{IB}(H)$ be a countable
disjoint collection of measurable sets. We have that, for each
$\gamma_{m,n}\in\textsf{IB}(H)$,
$E^{1}_{\gamma_{m,n}},E^{2}_{\gamma_{m,n}},\ldots$ are disjoint. So, using
monotone convergence,

$\displaystyle\begin{split}\vphantom{\psi
A^{H}\Big{\lparen}\bigcup_{i}E^{i}\Big{\rparen}\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}\Big{\lparen}\bigcup_{i}E^{i}\Big{\rparen}}\middle\rvert\smash{\psi}\right\rangle&=\sum_{\gamma_{m,n}}d_{\gamma}\int\chi_{\bigcup_{i}E^{i}_{\gamma_{m,n}}}([g])\lvert\langle\gamma_{m,n},\psi\circ[g]\rangle_{H}\rvert^{2}d[g]\\\
&=\sum_{\gamma_{m,n}}d_{\gamma}\int\sum_{i=1}^{\infty}\chi_{E^{i}_{\gamma_{m,n}}}([g])\lvert\langle\gamma_{m,n},\psi\circ[g]\rangle_{H}\rvert^{2}d[g]\\\
&=\sum_{i=1}^{\infty}\sum_{\gamma_{m,n}}d_{\gamma}\int\chi_{E^{i}_{\gamma_{m,n}}}([g])\lvert\langle\gamma_{m,n},\psi\circ[g]\rangle_{H}\rvert^{2}d[g]\\\
&=\sum_{i=1}^{\infty}\vphantom{\psi
A^{H}(E^{i})\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(E^{i})}\middle\rvert\smash{\psi}\right\rangle,\end{split}$
(168)

where $\chi_{E}$ is the characteristic function, that is $\chi_{E}(x)=1$ if
$x\in E$ and $\chi_{E}(x)=0$ if $x\notin E$. Therefore, by the polarization
identity,
$A^{H}\left\lparen\bigcup_{i}E^{i}\right\rparen=\sum_{i=1}^{\infty}A^{H}(E^{i})$
weakly.

Next, we want that $A^{H}$ is projective. From the work before the lemma, we
have that $A^{H}(\textsf{CS}(H)\otimes\textsf{IB}(H))=\mathbb{I}_{L^{2}(G)}$,
so it is a measurement. To show that $A^{H}$ is projective, note that the
definition directly implies that

$\displaystyle(A^{H}(E)\lvert\psi\rangle)(g)=\sum_{\gamma_{m,n}}d_{\gamma}\chi_{E_{\gamma_{m,n}}}([g])\gamma_{m,n}([g]^{-1}g)\langle\gamma_{m,n},\psi\circ[g]\rangle_{H}.$
(169)

So, as everything is nice and bounded, we can invoke Fubini’s theorem a couple
times to get that

$\displaystyle\begin{split}&\vphantom{\psi
A^{H}(E)A^{H}(F)\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(E)A^{H}(F)}\middle\rvert\smash{\psi}\right\rangle=\int_{G}\overline{(A^{H}(E)\lvert\psi\rangle)(g)}(A^{H}(F)\lvert\psi\rangle)(g)d_{G}g\\\
&=\int_{G}\sum_{\gamma_{m,n},\gamma^{\prime}_{m^{\prime},n^{\prime}}}d_{\gamma}d_{\gamma^{\prime}}\chi_{E_{\gamma_{m,n}}\cap
F_{\gamma^{\prime}_{m^{\prime},n^{\prime}}}}([g])\overline{\gamma_{m,n}}([g]^{-1}g)\gamma^{\prime}_{m^{\prime},n^{\prime}}([g]^{-1}g)\langle\psi\circ[g],\gamma_{m,n}\rangle_{H}\langle\gamma^{\prime}_{m^{\prime},n^{\prime}},\psi\circ[g]\rangle_{H}d_{G}g\\\
&=\sum_{\gamma_{m,n},\gamma^{\prime}_{m^{\prime},n^{\prime}}}d_{\gamma}d_{\gamma^{\prime}}\int_{E_{\gamma_{m,n}}\cap
F_{\gamma^{\prime}_{m^{\prime},n^{\prime}}}}\langle\psi\circ[g],\gamma_{m,n}\rangle_{H}\langle\gamma^{\prime}_{m^{\prime},n^{\prime}},\psi\circ[g]\rangle_{H}\langle\gamma_{m,n},\gamma^{\prime}_{m^{\prime},n^{\prime}}\rangle_{H}d_{G}[g]\\\
&=\sum_{\gamma_{m,n}}d_{\gamma}\int_{E_{\gamma_{m,n}}\cap
F_{\gamma_{m,n}}}\langle\psi\circ[g],\gamma_{m,n}\rangle_{H}\langle\gamma_{m,n},\psi\circ[g]\rangle_{H}d_{G}[g]\\\
&=\vphantom{\psi A^{H}(E\cap
F)\psi}\left\langle\smash{\psi}\middle\rvert\smash{A^{H}(E\cap
F)}\middle\rvert\smash{\psi}\right\rangle.\end{split}$ (170)

Therefore, $A^{H}(E)A^{H}(F)=A^{H}(E\cap F)$. ∎

###  9.2 Compact coset measure game

######  Definition 9.4.

An _compact coset measure monogamy game_ is a tuple $\texttt{G}=\left\lparen
G,\mathcal{S},E\right\rparen$, where $G$ is a compact Hausdorff group,
$\mathcal{S}$ is a finite set of closed subgroups of $G$, and $E\subseteq G$
and is a symmetric neighborhood of identity, _i.e._ $E=E^{-1}$ and $1\in E$.

A _(quantum) strategy_ for a coset measure game G is a tuple
$\texttt{S}=\left\lparen\mathcal{B},\mathcal{C},B,C,\rho\right\rparen$, where
$\mathcal{B}$ and $\mathcal{C}$ are Hilbert spaces,
$B=\\{{B^{H}:\mathscr{B}(\textsf{CS}(H))\rightarrow\mathcal{B}(\mathcal{B})}|{H\in\mathcal{S}}\\}$
and
$C=\\{{C^{H}:\mathscr{P}(\textsf{IB}(H))\rightarrow\mathcal{B}(\mathcal{C})}|{H\in\mathcal{S}}\\}$
are collections of POVM measures, and
$\rho\in\mathcal{D}(L^{2}(G)\otimes\mathcal{B}\otimes\mathcal{C})$.

Let G be a coset measure game and S be a strategy for it. The _winning
probability_ of S is

$\displaystyle\mathfrak{w}_{\texttt{G}}(\texttt{S})=\underset{H\in\mathcal{S}}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\operatorname{Tr}\left[(A^{H}\otimes
B^{H}\otimes C^{H})\lparen E_{H}\rparen\rho\right],$ (171)

where
$E_{H}=\left\\{{([g]_{H},\gamma_{m,n},[eg]_{H},\gamma_{m,n})}\middle|{g\in
G,\gamma_{m,n}\in\textsf{IB}(H),e\in E}\right\\}$ and the expectation with
respect to $H\in\mathcal{S}$ is uniform.

The winning probability of G is
$\mathfrak{w}(\texttt{G})=\sup_{\texttt{S}}\mathfrak{w}_{\texttt{G}}(\texttt{S})$.

######  Theorem 9.5.

Let $\texttt{G}=(G,\mathcal{S},E)$ be a compact coset measure game. Then, for
any complete set of orthogonal permutations
$\pi_{i}:\mathcal{S}\rightarrow\mathcal{S}$, the winning probability is
bounded by

$\displaystyle
w(\texttt{G})\leq\underset{i}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\sup_{\begin{subarray}{c}H\in\mathcal{S}\\\
\gamma\in\textsf{Irr}(H)\\\ g\in G\end{subarray}}\sqrt{d_{\gamma}\mu_{H}(H\cap
gE\pi_{i}(H))}.$ (172)

######  Lemma 9.6.

Let $H,K\leq G$ be closed subgroups and let $E\subseteq G$ be a symmetric open
neighborhood of the identity. Then, for any $\gamma_{m,n}\in\textsf{IB}(H)$
and $q^{\prime}\in\textsf{CS}(K)$,

$\displaystyle\left\|A^{H}\left\lparen\textsf{CS}(H)\times\\{\gamma_{m,n}\\}\right\rparen
A^{K}\left\lparen[Eq^{\prime}]_{K}\times\textsf{IB}(K)\right\rparen\right\|\leq\sup_{g\in
G}\sqrt{d_{\gamma}\mu_{H}(H\cap gEK)}.$ (173)

###### Proof.

Since $\textsf{IB}(K)$ gives rise to an orthonormal basis, for any
$\lvert\varphi\rangle,\lvert\psi\rangle\in L^{2}(G)$,

$\displaystyle\begin{split}\vphantom{\varphi
A^{K}\left\lparen[Eq^{\prime}]_{K}\times\textsf{IB}(K)\right\rparen\psi}\left\langle\smash{\varphi}\middle\rvert\smash{A^{K}\left\lparen[Eq^{\prime}]_{K}\times\textsf{IB}(K)\right\rparen}\middle\rvert\smash{\psi}\right\rangle&=\sum_{\varrho_{i,j}\in\textsf{IB}(K)}d_{\varrho}\int_{[Eq^{\prime}]_{K}}\left\langle\varphi\circ[q],\varrho_{i,j}\right\rangle_{K}\left\langle\varrho_{i,j},\psi\circ[q]\right\rangle_{K}d[q]\\\
&=\int_{[Eq^{\prime}]_{K}}\left\langle\varphi\circ[q],\psi\circ[q]\right\rangle_{K}d[q]=\int_{[Eq^{\prime}]_{K}}\int_{K}\overline{\varphi}([q]k)\psi([q]k)d_{K}kd[q]\\\
&=\int_{Eq^{\prime}K}\overline{\varphi}(q)\psi(q)d_{G}q.\end{split}$ (174)

That is,
$A^{K}\left\lparen[Eq^{\prime}]_{K}\times\textsf{IB}(K)\right\rparen=\Pi_{Eq^{\prime}K}$,
the projector onto the subspace of states $\lvert\psi\rangle$ such that
$\psi(g)=0$ for almost all $g\notin Eq^{\prime}K$. To get the norm, we
consider
$\|A^{H}\left\lparen\textsf{CS}(H)\times\\{\gamma_{m,n}\\}\right\rparen\Pi_{Eq^{\prime}K}\lvert\psi\rangle\|$
for some unit vector $\lvert\psi\rangle\in L^{2}(G)$ and then take the
supremum. First,

$\displaystyle\begin{split}\|A^{H}\left\lparen\textsf{CS}(H)\times\\{\gamma_{m,n}\\}\right\rparen\Pi_{Eq^{\prime}K}\lvert\psi\rangle\|^{2}&=\vphantom{\psi\Pi_{Eq^{\prime}H}A^{H}\left\lparen\textsf{CS}(H)\times\\{\gamma_{m,n}\\}\right\rparen\Pi_{Eq^{\prime}K}\psi}\left\langle\smash{\psi}\middle\rvert\smash{\Pi_{Eq^{\prime}H}A^{H}\left\lparen\textsf{CS}(H)\times\\{\gamma_{m,n}\\}\right\rparen\Pi_{Eq^{\prime}K}}\middle\rvert\smash{\psi}\right\rangle\\\
&=d_{\gamma}\int_{\textsf{CS}(H)}\left\lvert\left\langle\gamma_{m,n},(\chi_{Eq^{\prime}K}\psi)\circ[g]\right\rangle_{H}\right\rvert^{2}d[g].\end{split}$
(175)

Now, let $p_{[g]}=\int_{H}|\psi([g]h)|^{2}d_{H}h$ – this is finite for almost
every $[g]\in\textsf{CS}(H)$ and the map $[g]\mapsto p_{[g]}$ is in
$L^{1}(\textsf{CS}(H))$ – and then
$\lvert\psi_{[g]}\rangle=\frac{\lvert\psi\rangle}{\sqrt{p_{[g]}}}$. In
particular, $\int_{H}|\psi_{[g]}([g]h)|^{2}d_{H}h=1$ for almost all $[g]$, and
we have

$\displaystyle\begin{split}\|A^{H}\left\lparen\textsf{CS}(H)\times\\{\gamma_{m,n}\\}\right\rparen\Pi_{Eq^{\prime}K}\lvert\psi\rangle\|&=\sqrt{d_{\gamma}\int_{\textsf{CS}(H)}p_{[g]}\left\lvert\left\langle\gamma_{m,n},(\chi_{Eq^{\prime}K}\psi_{[g]})\circ[g]\right\rangle_{H}\right\rvert^{2}d[g]}\\\
&\leq\sqrt{d_{\gamma}}\sup_{[g]\in\textsf{CS}(H)}\left\lvert\left\langle\gamma_{m,n},(\chi_{Eq^{\prime}K}\psi_{[g]})\circ[g]\right\rangle_{H}\right\rvert\\\
&=\sqrt{d_{\gamma}}\sup_{[g]\in\textsf{CS}(H)}\Big{\lvert}\int_{H}\overline{\gamma_{m,n}}(h)\chi_{Eq^{\prime}K}([g]h)\psi_{[g]}([g]h)d_{H}h\Big{\rvert}\\\
&\leq\sqrt{d_{\gamma}}\sup_{[g]\in\textsf{CS}(H)}\int_{H\cap[g]^{-1}Eq^{\prime}K}|\gamma_{m,n}(h)||\psi_{[g]}([g]h)|d_{H}h\\\
&\leq\sqrt{d_{\gamma}}\sup_{[g]\in\textsf{CS}(H)}\int_{H\cap[g]^{-1}Eq^{\prime}K}|\psi_{[g]}([g]h)|d_{H}h.\end{split}$
(176)

As $H\cap[g]^{-1}Eq^{\prime}K$ is a finite measure space
$\|f\|_{1}\leq\sqrt{\mu_{H}(H\cap[g]^{-1}Eq^{\prime}K)}\|f\|_{2}$, giving

$\displaystyle\begin{split}\left\lvert\left\langle\gamma_{m,n},(\chi_{Eq^{\prime}K}\psi)\circ[g]\right\rangle\right\rvert&\leq\sqrt{\mu_{H}(H\cap[g]^{-1}Eq^{\prime}K)}\int_{H\cap[g]^{-1}Eq^{\prime}K}|\psi_{[g]}([g]h)|^{2}d_{H}h\\\
&\leq\sqrt{\mu_{H}(H\cap[g]^{-1}Eq^{\prime}K)}.\end{split}$ (177)

Taking the supremum over $\lvert\psi\rangle$ gives

$\displaystyle\left\|A^{H}\left\lparen\textsf{CS}(H)\times\\{\gamma_{m,n}\\}\right\rparen\Pi_{Eq^{\prime}K}\right\|\leq\sup_{[g]\in\textsf{CS}(H)}\sqrt{d_{\gamma}\mu_{H}(H\cap[g]^{-1}Eq^{\prime}K)}\leq\sup_{g\in
G}\sqrt{d_{\gamma}\mu_{H}(H\cap gEK)}.$ (178)

∎

###### Proof of Theorem 9.5.

We proceed similarly to Theorem 8.7. Let
$\texttt{S}=\left\lparen\mathcal{B},\mathcal{C},B,C,\rho\right\rparen$ be a
strategy for $\texttt{G}=\left\lparen G,\mathcal{S},E,F\right\rparen$, and we
assume that $B$ and $C$ are PVMs by Naimark’s theorem Theorem 8.2. Writing,
for each $H\in\mathcal{S}$, $\Pi^{H}=(A^{H}\otimes B^{H}\otimes
C^{H})(E_{H})$, we use the overlap result Lemma 7.3, getting

$\displaystyle\mathfrak{w}_{\texttt{G}}(\texttt{S})\leq\Big{\|}\underset{S\in\mathcal{S}}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\Pi^{H}\Big{\|}\leq\underset{i}{\text{\raisebox{-3.0pt}{\scalebox{1.5}{$\mathbb{E}$}}}}\sup_{H\in\mathcal{S}}\left\|\Pi^{H}\Pi^{\pi_{i}(H)}\right\|,$
(179)

since $(\Pi^{H})^{2}=\Pi^{H}$. Fixing $H,K\in\mathcal{S}$, it remains to
simplify $\left\|\Pi^{H}\Pi^{K}\right\|$. We have that the set of correct
answers
$E_{H}\subseteq\\{([g]_{H},\gamma_{m,n},[g^{\prime}]_{H},\gamma_{m,n})|g,g^{\prime}\in
G;\gamma_{m,n}\in\textsf{IB}(H);e\in E\\},$ so

$\displaystyle(A^{H}\otimes B^{H}\otimes C^{H})(E_{H})\leq(A^{H}\otimes
C^{H})\left\lparen\\{([g]_{H},\gamma_{m,n},\gamma_{m,n})|g\in
G,\gamma_{m,n}\in\textsf{IB}(H),e\in E\right\rparen\\}\otimes\mathbb{I}_{B};$
(180)

and
$E_{K}\subseteq\\{([eq]_{K},\varrho_{i,j},[q]_{K},\varrho^{\prime}_{i^{\prime},j^{\prime}})|q\in
G;\varrho_{i,j},\varrho^{\prime}_{i^{\prime},j^{\prime}}\in\textsf{IB}(K);e\in
E\\},$ so

$\displaystyle(A^{K}\otimes B^{K}\otimes C^{K})(E_{K})\leq(A^{K}\otimes
B^{K})\left\lparen\\{([eq]_{K},\varrho_{i,j},[q]_{K})|q\in
G,\varrho_{i,j}\in\textsf{IB}(K),e\in E\\}\right\rparen\otimes\mathbb{I}_{C}.$
(181)

Using the definition of the tensor product operator measure,
$\Pi_{H}\leq\int_{\textsf{IB}(H)}A^{H}(\textsf{CS}(H)\times\\{\gamma_{m,n}\\})\otimes\mathbb{I}_{B}\otimes
dC^{H}(\gamma_{m,n})$ and
$\Pi_{K}\leq\int_{\textsf{CS}(K)}A^{K}([Eq]_{K}\times\textsf{IB}(K))\otimes
dB^{K}([q]_{K})\otimes\mathbb{I}_{C}$, so

$\displaystyle\begin{split}\|\Pi^{H}\Pi^{K}\|&\leq\Big{\|}\int_{\textsf{CS}(K)\times\textsf{IB}(H)}A^{H}(\textsf{CS}(H)\times\\{\gamma_{m,n}\\})A^{K}([Eq]_{K}\times\textsf{IB}(K))\otimes
d(B^{K}\otimes C^{H})([q]_{k},\gamma_{m,n})\Big{\|}\\\
&\leq\sup_{\begin{subarray}{c}[q]_{K}\in\textsf{CS}(K)\\\
\gamma_{m,n}\in\textsf{IB}(H)\end{subarray}}\left\|A^{H}(\textsf{CS}(H)\times\\{\gamma_{m,n}\\})A^{K}([Eq]_{K}\times\textsf{IB}(K))\right\|,\end{split}$
(182)

using the result of Lemma 8.8. Then, using Lemma 9.6, we get

$\displaystyle\|\Pi^{H}\Pi^{K}\|\leq\sup_{\begin{subarray}{c}g\in G\\\
\gamma\in\textsf{Irr}(H)\end{subarray}}\sqrt{d_{\gamma}\mu_{H}(H\cap gEK)},$
(183)

which gives the result. ∎

##  Appendix A Integration by an operator-valued measure

In this section, we fix a measurable space $(X,\mathscr{S})$, a Hilbert space
$\mathcal{H}$, a separable Hilbert space $\mathcal{K}$, and
$P:\mathscr{S}\rightarrow\mathcal{B}(\mathcal{H})$ a POVM measure. We
generally follow the standard notation of [Axl20].

######  Definition A.1.

Let $f:X\rightarrow\mathbb{C}$ be a measurable function. If

$\displaystyle\sup\left\\{{\int|f|d\vphantom{vPv}\left\langle\smash{v}\middle\rvert\smash{P}\middle\rvert\smash{v}\right\rangle}\middle|{\lvert
v\rangle\in\mathcal{H},\|\lvert v\rangle\|\leq 1}\right\\}<\infty,$ (184)

we say $f$ is operator integrable and define the integral $\int
fdP\in\mathcal{B}(\mathcal{H})$ as the operator such that

$\displaystyle\vphantom{u\int
fdPv}\left\langle\smash{u}\middle\rvert\smash{\int
fdP}\middle\rvert\smash{v}\right\rangle=\int
fd\vphantom{uPv}\left\langle\smash{u}\middle\rvert\smash{P}\middle\rvert\smash{v}\right\rangle$
(185)

for all $\lvert u\rangle,\lvert v\rangle\in\mathcal{H}$.

As the complex measure
$\vphantom{uPv}\left\langle\smash{u}\middle\rvert\smash{P}\middle\rvert\smash{v}\right\rangle$
is linear in $\lvert v\rangle$ and antilinear in $\lvert u\rangle$, $\int fdP$
is in fact a linear operator; and, by the polarization identity, $\|\int
fdP\|<\infty$.

######  Lemma A.2.

Let $f=\sum_{i}c_{i}\chi_{A_{i}}$ be a measurable simple function. Then, $\int
fdP=\sum_{i}c_{i}P(A_{i})$.

###### Proof.

Let $\lvert u\rangle,\lvert v\rangle\in\mathcal{H}$. Then,

$\displaystyle\vphantom{u\int
fdPv}\left\langle\smash{u}\middle\rvert\smash{\int
fdP}\middle\rvert\smash{v}\right\rangle=\sum_{i}c_{i}\int\chi_{A_{i}}d\vphantom{uPv}\left\langle\smash{u}\middle\rvert\smash{P}\middle\rvert\smash{v}\right\rangle=\vphantom{u\sum_{i}c_{i}P(A_{i})v}\left\langle\smash{u}\middle\rvert\smash{\sum_{i}c_{i}P(A_{i})}\middle\rvert\smash{v}\right\rangle$
(186)

∎

We can, as for a usual measure, approximate the operator integral of a
function by the integrals of simple functions that approximate it. Let
$(f_{n})$ be a sequence of simple functions that converges pointwise to $f$.
Then, as for each $\lvert u\rangle,\lvert v\rangle\in\mathcal{H}$, $\int
f_{n}d\vphantom{uPv}\left\langle\smash{u}\middle\rvert\smash{P}\middle\rvert\smash{v}\right\rangle\rightarrow\int
fd\vphantom{uPv}\left\langle\smash{u}\middle\rvert\smash{P}\middle\rvert\smash{v}\right\rangle$,
the sequence $\left\lparen\int f_{n}dP\right\rparen$ in
$\mathcal{B}(\mathcal{H})$ converges _weakly_ to $\int fdP$.

We would like to able to integrate not just scalar functions, but operator-
valued functions by a POVM measure. The following proceeds similarly to
Pettis’ integral [Yos95].

######  Definition A.3.

We say that $F:X\rightarrow\mathcal{B}(\mathcal{K})$ is _weakly measurable_
if, for every $\lvert u\rangle,\lvert v\rangle\in\mathcal{K}$,
$\vphantom{uFv}\left\langle\smash{u}\middle\rvert\smash{F}\middle\rvert\smash{v}\right\rangle$
is measurable.

######  Lemma A.4.

Let $\Delta\geq 0$. The open ball

$\displaystyle
B_{\Delta}:=\left\\{{A\in\mathcal{B}(\mathcal{K})}\middle|{\|A\|<\Delta}\right\\}$
(187)

is separable in the strong operator topology.

In particular, $\mathcal{B}(\mathcal{K})=\bigcup_{N=1}^{\infty}B_{N}$ is
separable.

###### Proof.

Let $D$ be a countable dense subset of $\mathcal{K}$ and $\left\\{{\lvert
n\rangle}\middle|{n\in\mathbb{N}}\right\\}$ be an orthonormal basis. Set

$\displaystyle
D_{\Delta}=\\{{\vphantom{v_{1}1}\left\lvert\smash{v_{1}}\middle\rangle\\!\middle\langle\smash{1}\right\rvert+\ldots+\vphantom{v_{n}n}\left\lvert\smash{v_{n}}\middle\rangle\\!\middle\langle\smash{n}\right\rvert}|{n\in\mathbb{N};\lvert
v_{1}\rangle,\ldots,\lvert v_{n}\rangle\in D}\\}\cap B_{\Delta}.$ (188)

$D_{\Delta}$ is countable as $D$ is countable, and I claim $D_{\Delta}$ is
dense in $B_{\Delta}$. Let $A\in B_{\Delta}$. By density of $D$, for each
$n\in\mathbb{N}$ and $i\leq n$, there exists $\lvert v^{n}_{i}\rangle\in D$
such that $\|\lvert v^{n}_{i}\rangle-A\lvert
i\rangle\|<\min\left\\{{\frac{1}{n^{2}},\frac{\Delta-\|A\|}{n}}\right\\}$. Set
$A_{n}=\sum_{i=1}^{n}\vphantom{v^{n}_{i}i}\left\lvert\smash{v^{n}_{i}}\middle\rangle\\!\middle\langle\smash{i}\right\rvert$,
so we have

$\displaystyle\begin{split}\|A_{n}\|\leq\Big{\|}A\sum_{i=1}^{n}\vphantom{i}\left\lvert\smash{i}\middle\rangle\\!\middle\langle\smash{i}\right\rvert\Big{\|}+\sum_{i=1}^{n}\|\lvert
v^{n}_{i}\rangle-A\lvert i\rangle\|<\|A\|+(\Delta-\|A\|)=\Delta,\end{split}$
(189)

giving $A_{n}\in D_{\Delta}$. I claim now that the sequence $(A_{n})$
converges strongly to $A$. Let $\varepsilon>0$ and $\lvert
v\rangle\in\mathcal{K}$. There exists $N\in\mathbb{N}$ such that both
$\sum_{i=N+1}^{\infty}\left\lvert\vphantom{iv}\left\langle\smash{i}\middle|\smash{v}\right\rangle\right\rvert^{2}<\left\lparen\frac{\varepsilon}{2\Delta}\right\rparen^{2}$
and $\frac{\|\lvert v\rangle\|}{N}<\frac{\varepsilon}{2}$. Then, for $n\geq
N$,

$\displaystyle\begin{split}\|(A-A_{n})\lvert
v\rangle\|&\leq\Big{\|}\sum_{i=1}^{n}(A\lvert i\rangle-\lvert
v^{n}_{i}\rangle)\vphantom{iv}\left\langle\smash{i}\middle|\smash{v}\right\rangle\Big{\|}+\Big{\|}A\sum_{i=n+1}^{\infty}\vphantom{iv}\left\langle\smash{i}\middle|\smash{v}\right\rangle\lvert
i\rangle\Big{\|}\\\ &\leq\sum_{i=1}^{n}\|A\lvert i\rangle-\lvert
v^{n}_{i}\rangle\|\|v\|+\|A\|\sqrt{\sum_{i=n+1}^{\infty}\left\lvert\vphantom{iv}\left\langle\smash{i}\middle|\smash{v}\right\rangle\right\rvert^{2}}\\\
&<\frac{\|v\|}{n}+\Delta\frac{\varepsilon}{2\Delta}<\varepsilon.\end{split}$
(190)

∎

######  Lemma A.5.

Let $F:X\rightarrow\mathcal{B}(\mathcal{K})$. Then, there exists a sequence of
simple functions $(F_{n})$ that converges strongly pointwise to $F$.
Additionally, if $F$ is weakly measurable, then each $F_{n}$ can be chosen to
be weakly measurable; and if $F$ is bounded, then for all $\delta>0$,
$(F_{n})$ can be chosen so that $\|F_{n}(x)\|<\sup_{x}\|F(x)\|+\delta$.

###### Proof.

Let $\delta>0$. Then, if $F$ is bounded, we can corestrict it to
$U=B_{\sup_{x}\|F(X)\|+\delta}$; and if $F$ is not bounded, we do not change
its codomain, $U=\mathcal{B}(\mathcal{K})$. Since $\mathcal{K}$ is separable,
there exists a countable dense sequence $(\lvert
v_{n}\rangle)_{n\in\mathbb{N}}$ in $\mathcal{K}$. For $n\in\mathbb{N}$ and
$M\in\mathcal{B}(\mathcal{K})$, define the cylinder sets

$\displaystyle B_{n}(M)=\left\\{{A\in U}\middle|{\|A\|<\|M\|+1;\|(M-A)\lvert
v_{i}\rangle\|<\frac{1}{n}\text{ for }i=1,\ldots,n}\right\\}.$ (191)

By separability of $U$, there exists a countable collection
$M^{n}_{1},M^{n}_{2},\ldots\in U$ such that
$U\subseteq\bigcup_{i=1}^{\infty}B_{n}(M^{n}_{i})$. Next, define the sets
$E^{n}_{m}$ as

$\displaystyle E^{n}_{m}=F^{-1}\left\lparen
B_{n}(M^{n}_{m})\right\rparen\backslash\bigcup_{i=1}^{m-1}F^{-1}\left\lparen
B_{n}(M^{n}_{i})\right\rparen,$ (192)

which are in $\mathscr{S}$ if $F$ is weakly measurable, and set
$F^{n}_{m}=\sum_{i=1}^{m}M^{n}_{m}\chi_{E^{n}_{m}}$. Finally, define
$F_{n}:X\rightarrow U$ as $F_{n}(x)=F^{i}_{n}(x)$, where $i\leq n$ is the
largest such that $F^{i}_{n}(x)\neq 0$, and $F_{n}(x)=0$ if no such $i$
exists. Then, $F_{n}$ is weakly measurable if $F$ is, takes at most $n^{2}$
values, so is simple, and $F_{n}(X)\subseteq U$.

It remains to show that $(F_{n})$ converges pointwise strongly to $F$. Let
$\varepsilon>0$, $x\in X$, and $\lvert v\rangle\in\mathcal{K}$. Then, there
exists $N\in\mathbb{N}$ such that $\frac{1}{N}<\frac{\varepsilon}{2}$ and
$\|\lvert v_{N}\rangle-\lvert v\rangle\|<\frac{\varepsilon}{2(2\|F(x)\|+1)}$
As $F(x)\in U$, there exists $i$ such that $F(x)\in B_{n}(M^{N}_{i})$. Let
$n\geq\max\\{i,N\\}$. Then, there exists $m\geq N$ and $j\leq n$ such that
$F_{n}(x)=M^{m}_{j}$. We must have $F(x)\in B_{m}(M^{m}_{j})$, so

$\displaystyle\begin{split}\|(F(x)-F_{n}(x))\lvert
v\rangle\|&\leq\|(F(x)-M^{m}_{j})\lvert
v_{N}\rangle\|+\|F(x)-M^{m}_{j}\|\|\lvert v\rangle-\lvert v_{N}\rangle\|\\\
&<\frac{1}{m}+(2\|F(x)\|+1)\frac{\varepsilon}{2(2\|F(x)\|+1)}\\\
&<\varepsilon.\end{split}$ (193)

∎

######  Definition A.6.

Let $F:X\rightarrow\mathcal{B}(\mathcal{K})$. If the set

$\displaystyle\left\\{{\sum_{i,j=1}^{m}\sqrt{p_{i}p_{j}}\int\vphantom{k_{i}Fk_{j}}\left\langle\smash{k_{i}}\middle\rvert\smash{F}\middle\rvert\smash{k_{j}}\right\rangle
d\vphantom{h_{i}Ph_{j}}\left\langle\smash{h_{i}}\middle\rvert\smash{P}\middle\rvert\smash{h_{j}}\right\rangle}\middle|{m\in\mathbb{N};\sum_{i=1}^{m}p_{i}\leq
1;\\{\lvert h_{i}\rangle\\}\subseteq\mathcal{H},\\{\lvert
k_{i}\rangle\\}\subseteq\mathcal{K}\text{ orthonormal}}\right\\}$ (194)

is bounded in $\mathbb{C}$, we say $F$ is _weakly operator integrable_. In
that case, we take $\int F\otimes
dP\in\mathcal{B}(\mathcal{K}\otimes\mathcal{H})$ to be the operator such that

$\displaystyle\langle k\rvert\otimes\langle h\rvert\int F\otimes dP\lvert
k^{\prime}\rangle\otimes\lvert
h^{\prime}\rangle=\int\vphantom{kFk^{\prime}}\left\langle\smash{k}\middle\rvert\smash{F}\middle\rvert\smash{k^{\prime}}\right\rangle
d\vphantom{hPh^{\prime}}\left\langle\smash{h}\middle\rvert\smash{P}\middle\rvert\smash{h^{\prime}}\right\rangle,$
(195)

for any $\lvert h\rangle,\lvert h^{\prime}\rangle\in\mathcal{H}$ and $\lvert
k\rangle,\lvert k^{\prime}\rangle\in\mathcal{K}$.

As for integration of scalar-valued functions. this is well-defined as a
bounded linear operator via the polarization identity and linearity of
integrals with respect to complex measures.

######  Lemma A.7.

Let $F=\sum_{l}M_{l}\chi_{E_{l}}$ be a simple function. Then, $F$ is weakly
operator integrable and

$\displaystyle\int F\otimes dP=\sum_{l}M_{l}\otimes P(E_{l}).$ (196)

###### Proof.

Let $\\{{\lvert h_{i}\rangle}|{i=1,\ldots,m}\\}\subseteq\mathcal{H}$ and
$\\{{\lvert k_{i}\rangle}|{i=1,\ldots,m}\\}\subseteq\mathcal{K}$ be
orthonormal, and $\sum_{i=1}^{m}p_{i}\leq 1$. Write $\lvert
v\rangle=\sum_{i=1}^{m}\sqrt{p_{i}}\lvert k_{i}\rangle\otimes\lvert
h_{i}\rangle\in\mathcal{K}\otimes\mathcal{H}$. Then, for each $i,j$,
$\sum_{l}\vphantom{k_{i}Fk_{j}}\left\langle\smash{k_{i}}\middle\rvert\smash{F}\middle\rvert\smash{k_{j}}\right\rangle\chi_{E_{l}}$
is simple and weakly operator integrable, so

$\displaystyle\begin{split}\left\lvert\sum_{i,j=1}^{m}\sqrt{p_{i}p_{j}}\int\vphantom{k_{i}Fk_{j}}\left\langle\smash{k_{i}}\middle\rvert\smash{F}\middle\rvert\smash{k_{j}}\right\rangle
d\vphantom{h_{i}Ph_{j}}\left\langle\smash{h_{i}}\middle\rvert\smash{P}\middle\rvert\smash{h_{j}}\right\rangle\right\rvert&=\left\lvert\sum_{i,j=1}^{m}\sqrt{p_{i}p_{j}}\sum_{l}\vphantom{k_{i}M_{l}k_{j}}\left\langle\smash{k_{i}}\middle\rvert\smash{M_{l}}\middle\rvert\smash{k_{j}}\right\rangle\vphantom{h_{i}P(E_{l})h_{j}}\left\langle\smash{h_{i}}\middle\rvert\smash{P(E_{l})}\middle\rvert\smash{h_{j}}\right\rangle\right\rvert\\\
&=\left\lvert\vphantom{v\sum_{l}M_{l}\otimes
P(E_{l})v}\left\langle\smash{v}\middle\rvert\smash{\sum_{l}M_{l}\otimes
P(E_{l})}\middle\rvert\smash{v}\right\rangle\right\rvert\\\
&\leq\max_{l}\left\|M_{l}\right\|<\infty.\end{split}$ (197)

Thus, $F$ is weakly operator integrable and its integral is
$\sum_{l}M_{l}\otimes P(E_{l})$. ∎

######  Lemma A.8.

Suppose $F:X\rightarrow\mathcal{B}(\mathcal{K})$ is bounded and weakly
measurable. Then, $F$ is weakly operator integrable.

###### Proof.

Fix $m\in\mathbb{N}$, $\left\\{{\lvert
h_{i}\rangle}\middle|{i=1,\ldots,m}\right\\}\subseteq\mathcal{H}$ and
$\left\\{{\lvert
k_{i}\rangle}\middle|{i=1,\ldots,m}\right\\}\subseteq\mathcal{K}$ orthonormal,
and $p_{i}\geq 0$ such that $\sum_{i=1}^{m}p_{i}\leq 1$; set
$S=\left\lvert\sum_{i,j=1}^{m}\sqrt{p_{i}p_{j}}\int\vphantom{k_{i}Fk_{j}}\left\langle\smash{k_{i}}\middle\rvert\smash{F}\middle\rvert\smash{k_{j}}\right\rangle
d\vphantom{h_{i}Ph_{j}}\left\langle\smash{h_{i}}\middle\rvert\smash{P}\middle\rvert\smash{h_{j}}\right\rangle\right\rvert$.
To show that $F$ is weakly operator integrable, it suffices to show that we
can upper bound $S$ by a constant independent of $m$, or the $\lvert
h_{i}\rangle$, $\lvert k_{i}\rangle$, and $p_{i}$. Using Lemma A.5, let
$\left\lparen F_{n}=\sum_{l}M_{l}^{n}\chi_{E^{n}_{l}}\right\rparen$ be a
sequence of simple functions that converges strongly pointwise to $F$ and
$\|F_{n}(x)\|\leq\sup_{y}\|F(y)\|+1$ for all $x\in X$. In particular,
$\vphantom{k_{i}F_{n}k_{j}}\left\langle\smash{k_{i}}\middle\rvert\smash{F_{n}}\middle\rvert\smash{k_{j}}\right\rangle\rightarrow\vphantom{k_{i}Fk_{j}}\left\langle\smash{k_{i}}\middle\rvert\smash{F}\middle\rvert\smash{k_{j}}\right\rangle$
pointwise for all $i,j$, so there exists $N\in\mathbb{N}$ such that for all
$n\geq N$,

$\displaystyle\Big{\lvert}\int\vphantom{k_{i}F_{n}-Fk_{j}}\left\langle\smash{k_{i}}\middle\rvert\smash{F_{n}-F}\middle\rvert\smash{k_{j}}\right\rangle
d\vphantom{h_{i}Ph_{j}}\left\langle\smash{h_{i}}\middle\rvert\smash{P}\middle\rvert\smash{h_{j}}\right\rangle\Big{\rvert}\leq\sqrt{p_{i}p_{j}}.$
(198)

As such, writing $\lvert v\rangle=\sum_{i=1}^{n}\sqrt{p_{i}}\lvert
k_{i}\rangle\otimes\lvert h_{i}\rangle$, we get that

$\displaystyle\begin{split}S&\leq\Big{\lvert}\sum_{i,j=1}^{m}\sqrt{p_{i}p_{j}}\int\vphantom{k_{i}F_{n}k_{j}}\left\langle\smash{k_{i}}\middle\rvert\smash{F_{n}}\middle\rvert\smash{k_{j}}\right\rangle
d\vphantom{h_{i}Ph_{j}}\left\langle\smash{h_{i}}\middle\rvert\smash{P}\middle\rvert\smash{h_{j}}\right\rangle\Big{\rvert}+\sum_{i,j}^{m}p_{i}p_{j}=\Big{\lvert}\vphantom{v\sum_{l}M^{n}_{l}\otimes
P(E^{n}_{l})v}\left\langle\smash{v}\middle\rvert\smash{\sum_{l}M^{n}_{l}\otimes
P(E^{n}_{l})}\middle\rvert\smash{v}\right\rangle\Big{\rvert}+1\\\
&\leq\sup_{l}\|M^{n}_{l}\|+1=\sup_{x\in X}\|F_{n}(x)\|+1,\end{split}$ (199)

and thus $S\leq\sup_{x}\|F(x)\|+2$, giving the result. ∎

######  Theorem A.9.

Let $F:X\rightarrow\mathcal{B}(\mathcal{K})$ be norm bounded and weakly
measurable. For any sequence of weakly operator integrable simple functions
$(F_{n})$ that converges to $F$ strongly pointwise and $\sup_{n}\left\|\int
F_{n}\otimes dP\right\|<\infty$, we have that $\int F_{n}\otimes
dP\rightarrow\int F\otimes dP$ weakly.

Note that Lemma A.5 shows that such a sequence of simple functions exists.

###### Proof.

Write $F_{n}=\sum_{l}M^{n}_{l}\chi_{E^{n}_{l}}$ and $W=\max\left\\{{\|\int
F\otimes dP\|,\sup_{n}}\right\\}<\infty$. Fix $\lvert
v\rangle\in\mathcal{K}\otimes\mathcal{H}$ and $\varepsilon>0$. Via the Schmidt
decomposition, we may write $\lvert v\rangle=\sum_{i}\sqrt{p_{i}}\lvert
k_{i}\rangle\otimes\lvert h_{i}\rangle$. As the norm of $\lvert v\rangle$ is
finite, there exists $I\in\mathbb{N}$ such that
$\sum_{i=I+1}^{\infty}p_{i}<\min\\{\frac{\varepsilon}{4(2\|\lvert
v\rangle\|+1)W},1\\}$; let $\lvert
v^{\prime}\rangle=\sum_{i=I+1}^{\infty}\sqrt{p_{i}}\lvert
k_{i}\rangle\otimes\lvert h_{i}\rangle$. Next, as
$\vphantom{k_{i}F_{n}k_{j}}\left\langle\smash{k_{i}}\middle\rvert\smash{F_{n}}\middle\rvert\smash{k_{j}}\right\rangle\rightarrow\vphantom{k_{i}Fk_{j}}\left\langle\smash{k_{i}}\middle\rvert\smash{F}\middle\rvert\smash{k_{j}}\right\rangle$
pointwise, there exists $N\in\mathbb{N}$ such that for all $n\geq N$,
$\left\lvert\int\vphantom{k_{i}F_{n}-Fk_{j}}\left\langle\smash{k_{i}}\middle\rvert\smash{F_{n}-F}\middle\rvert\smash{k_{j}}\right\rangle
d\vphantom{h_{i}Ph_{j}}\left\langle\smash{h_{i}}\middle\rvert\smash{P}\middle\rvert\smash{h_{j}}\right\rangle\right\rvert\leq\frac{\varepsilon}{2I\|\lvert
v\rangle\|^{2}}$ for all $i,j\leq I$. Then, for any $n\geq N$,

$\displaystyle\begin{split}\Big{\lvert}\vphantom{v\int(F_{n}-F)\otimes
dPv}\left\langle\smash{v}\middle\rvert\smash{\int(F_{n}-F)\otimes
dP}\middle\rvert\smash{v}\right\rangle\Big{\rvert}&\leq\Big{\lvert}\sum_{i,j=1}^{I}\sqrt{p_{i}p_{j}}\int\vphantom{k_{i}F_{n}-Fk_{j}}\left\langle\smash{k_{i}}\middle\rvert\smash{F_{n}-F}\middle\rvert\smash{k_{j}}\right\rangle
d\vphantom{h_{i}Ph_{j}}\left\langle\smash{h_{i}}\middle\rvert\smash{P}\middle\rvert\smash{h_{j}}\right\rangle\Big{\rvert}\\\
&\qquad+(2\left\|\lvert v\rangle\right\|+1)\|\lvert
v^{\prime}\rangle\|\Big{\|}\int(F_{n}-F)\otimes dP\Big{\|}\\\
&\leq\sum_{i,j=1}^{I}\sqrt{p_{i}p_{j}}\Big{\lvert}\int\vphantom{k_{i}F_{n}-Fk_{j}}\left\langle\smash{k_{i}}\middle\rvert\smash{F_{n}-F}\middle\rvert\smash{k_{j}}\right\rangle
d\vphantom{h_{i}Ph_{j}}\left\langle\smash{h_{i}}\middle\rvert\smash{P}\middle\rvert\smash{h_{j}}\right\rangle\Big{\rvert}+(2\left\|\lvert
v\rangle\right\|+1)\|\lvert v^{\prime}\rangle\|2W\\\
&<\Big{\lparen}\sum_{i=1}^{I}\sqrt{p_{i}}\Big{\rparen}^{2}\frac{\varepsilon}{2I\|\lvert
v\rangle\|^{2}}+\frac{\varepsilon}{2}\leq\varepsilon.\end{split}$ (200)

∎

######  Definition A.10.

Let $(X,\mathscr{S},P)$ and $(Y,\mathscr{T},Q)$ be POVM operator measure
spaces. Then, the _tensor product measure_ on the measurable space $(X\times
Y,\mathscr{S}\otimes\mathscr{T})$ is

$\displaystyle(P\otimes Q)(E)=\int_{Y}P((E)^{y})\otimes dQ(y),$ (201)

where $(E)^{y}=\left\\{{x\in X}\middle|{(x,y)\in E}\right\\}$ is the cross
section of $E$ at $Y$.

This definition is sensible since the map $y\mapsto P((E)^{y})$ is weakly
measurable and bounded in norm by $1$. This definition allows us to
immediately generalise Fubini’s theorem, as the matrix elements must all
satify it: for any weakly operator integrable $F:X\times
Y\rightarrow\mathcal{B}(\mathcal{K})$,

$\displaystyle\int F\otimes d(P\otimes
Q)=\int_{Y}\Big{\lparen}\int_{X}F(x,y)\otimes dP(x)\Big{\rparen}\otimes
dQ(y)=\int_{X}\int_{Y}F(x,y)\otimes dP(x)\otimes dQ(y).$ (202)

##  Appendix B Damping operators and maximally-entangled states

In this section, let $\mathcal{H}$ be any Hilbert space. Write
$\mathcal{T}_{1}(\mathcal{H})$ for the set of trace-class operator and for
$T\in\mathcal{T}_{1}(\mathcal{H})$ the trace norm
$\|T\|_{1}=\operatorname{Tr}(\sqrt{A^{\dagger}A})$; and write
$\mathcal{T}_{2}(\mathcal{H})$ for the Hilbert-Schmidt operators and for
$T\in\mathcal{T}_{2}(\mathcal{H})$ the Hilbert-Schmidt norm
$\|T\|_{2}=\sqrt{\operatorname{Tr}(A^{\dagger}A)}$.

######  Definition B.1.

A _complex conjugate_ on $\mathcal{H}$ is an antilinear involutive isometry
$c:\mathcal{H}\rightarrow\mathcal{H}$, _i.e._
$c(\alpha\lvert\psi\rangle+\beta\lvert\phi\rangle)=\overline{\alpha}c\lvert\psi\rangle+\overline{\beta}c\lvert\phi\rangle$,
$c^{2}=1$, and
$\left\|c\lvert\psi\rangle\right\|=\left\|\lvert\psi\rangle\right\|$ for all
$\alpha,\beta\in\mathbb{C}$ and
$\lvert\psi\rangle,\lvert\phi\rangle\in\mathcal{H}$.

We quickly work out some properties of a complex conjugate, and use that to
show that any complex conjugate as defined above may be expressed as a complex
conjugate with respect to some basis — the map that acts by taking the
conjugates of the coefficients in the expansion with respect to a fixed
orthonormal basis. For any $\lvert\psi\rangle\in\mathcal{H}$,
$\lvert\psi\rangle=\frac{1}{2}(\lvert\psi\rangle+c\lvert\psi\rangle)+\frac{1}{2}(\lvert\psi\rangle-c\lvert\psi\rangle)$
so we may decompose $\mathcal{H}=\mathcal{H}_{+}\oplus\mathcal{H}_{-}$ where
$H_{\pm}=\left\\{{\lvert\psi\rangle\in
H}\middle|{c\lvert\psi\rangle=\pm\lvert\psi\rangle}\right\\}$. As $c$ is
$\mathbb{R}$-linear, $\mathcal{H}_{+}$ and $\mathcal{H}_{-}$ are
$\mathbb{R}$-vector spaces, and $\mathcal{H}_{-}=i\mathcal{H}_{+}$. For
$\lvert\psi\rangle,\lvert\phi\rangle\in\mathcal{H}$

$\displaystyle\vphantom{c\psi
c\phi}\left\langle\smash{c\psi}\middle|\smash{c\phi}\right\rangle=\frac{1}{4}\sum_{k=0}^{3}i^{k}\|c\lvert\psi\rangle+i^{k}c\lvert\phi\rangle\|^{2}=\frac{1}{4}\sum_{k=0}^{3}i^{k}\|\lvert\psi\rangle+(-i)^{k}\lvert\phi\rangle\|^{2}=\overline{\vphantom{\psi\phi}\left\langle\smash{\psi}\middle|\smash{\phi}\right\rangle},$
(203)

which implies that
$\vphantom{c\psi\phi}\left\langle\smash{c\psi}\middle|\smash{\phi}\right\rangle=\vphantom{c\psi
c^{2}\phi}\left\langle\smash{c\psi}\middle|\smash{c^{2}\phi}\right\rangle=\overline{\vphantom{\psi
c\phi}\left\langle\smash{\psi}\middle|\smash{c\phi}\right\rangle}$. Thus, for
$\lvert\psi\rangle,\lvert\phi\rangle\in\mathcal{H}_{+}$,
$\vphantom{\psi\phi}\left\langle\smash{\psi}\middle|\smash{\phi}\right\rangle=\overline{\vphantom{\psi\phi}\left\langle\smash{\psi}\middle|\smash{\phi}\right\rangle}$,
so $\mathcal{H}_{+}$ is a real inner product space. So, there exists an
orthonormal basis $\left\\{{\lvert
n\rangle}\middle|{n\in\Gamma}\right\\}\subseteq\mathcal{H}_{+}$; this basis is
countable iff $\mathcal{H}_{+}$ is separable, which is iff $\mathcal{H}$ is.
Using the decomposition, $\mathcal{H}$ is the closure of
$\operatorname{span}_{\mathbb{C}}\left\\{{\lvert
n\rangle}\middle|{n\in\Gamma}\right\\}$. Thus, we may expand any
$\lvert\psi\rangle\in\mathcal{H}$ as
$\lvert\psi\rangle=\sum_{n\in\Gamma}\psi_{n}\lvert n\rangle$, so

$\displaystyle c\lvert\psi\rangle=\sum_{n\in\Gamma}\overline{\psi}_{n}c\lvert
n\rangle=\sum_{n\in\Gamma}\overline{\psi}_{n}\lvert n\rangle,$ (204)

giving that any complex conjugate may be expressed as the complex conjugate
with respect to a basis.

######  Definition B.2.

  * •

A _damping sequence_ is a sequence $(\Delta_{n})$ in
$\mathcal{T}_{2}(\mathcal{H})$ such that $\|\Delta_{n}\|\leq 1$, and
$\Delta_{n}\lvert\psi\rangle\rightarrow\lvert\psi\rangle$ and
$\Delta_{n}^{\dagger}\lvert\psi\rangle\rightarrow\lvert\psi\rangle$ for all
$\lvert\psi\rangle\in\mathcal{H}$.

  * •

The _pure tensor norm_ on $\mathcal{H}\otimes\mathcal{H}$ is
$\|\lvert\Psi\rangle\|_{\otimes}=\sup\big{\\{}{\left\lvert(\langle\psi\rvert\otimes\langle\phi\rvert)\lvert\Psi\rangle\right\rvert}\big{|}{\lvert\psi\rangle,\lvert\phi\rangle\in\mathcal{H};\|\lvert\psi\rangle\|,\|\lvert\phi\rangle\|\leq
1}\big{\\}}$.

  * •

Let $c:\mathcal{H}\rightarrow\mathcal{H}$ be a complex conjugate. An
_approximate maximally entangled state_ is a sequence
$(\lvert\Psi_{n}\rangle)$ in $\mathcal{H}\otimes\mathcal{H}$ such that
$\|\lvert\Psi_{n}\rangle\|=1$, and
$\frac{1}{\|\lvert\Psi_{n}\rangle\|_{\otimes}}(\langle\psi\rvert\otimes\mathbb{I})\lvert\Psi_{n}\rangle\rightarrow
c\lvert\psi\rangle$ and
$\frac{1}{\|\lvert\Psi_{n}\rangle\|_{\otimes}}(\mathbb{I}\otimes\langle\psi\rvert)\lvert\Psi_{n}\rangle\rightarrow
c\lvert\psi\rangle$ for all $\lvert\psi\rangle\in\mathcal{H}$.

It is direct to see that every damping sequence generates an approximate
maximally entangled state, and vice versa.

######  Lemma B.3.

  * •

Let $(\Delta_{n})$ be a damping sequence. Writing the singular-value
decompositions
$\Delta_{n}=\sum_{i=1}^{\infty}s_{n,i}\vphantom{\phi_{n,i}\chi_{n,i}}\left\lvert\smash{\phi_{n,i}}\middle\rangle\\!\middle\langle\smash{\chi_{n,i}}\right\rvert$,
the sequence $(\lvert\Psi_{n}\rangle)$ defined as
$\lvert\Psi_{n}\rangle=\frac{1}{\|\Delta_{n}\|_{2}}\sum_{i=1}^{\infty}s_{n,i}\lvert\phi_{n,i}\rangle\otimes
c\lvert\chi_{n,i}\rangle$ is an approximate maximally entangled state.

  * •

Let $(\lvert\Psi_{n}\rangle)$ be an approximate maximally entangled state.
Writing the Schmidt decomposition
$\lvert\Psi_{n}\rangle=\sum_{i=1}^{\infty}\sqrt{p_{n,i}}\lvert\phi_{n,i}\rangle\otimes\lvert\chi_{n,i}\rangle$,
the sequence $(\Delta_{n})$ defined as
$\Delta_{n}=\frac{1}{\|\lvert\Psi_{n}\rangle\|_{\otimes}}\sum_{i=1}^{\infty}\sqrt{p_{n,i}}\vphantom{\phi_{n,i}c\chi_{n,i}}\left\lvert\smash{\phi_{n,i}}\middle\rangle\\!\middle\langle\smash{c\chi_{n,i}}\right\rvert$
is a damping sequence.

###### Proof.

For the first point, note first that
$\vphantom{\Psi_{n}}\left\langle\smash{\Psi_{n}}\middle|\smash{\Psi_{n}}\right\rangle=\frac{1}{\operatorname{Tr}(\Delta_{n}^{\dagger}\Delta_{n})}\sum_{i=1}^{\infty}s_{n,i}^{2}=1$.
Next, for any $\lvert\psi\rangle\in\mathcal{H}$,

$\displaystyle\left\lparen\langle\psi\rvert\otimes\mathbb{I}\right\rparen\lvert\Psi_{n}\rangle=\frac{1}{\|\Delta_{n}\|_{2}}\sum_{i=1}^{\infty}s_{n,i}\vphantom{\psi\phi_{n,i}}\left\langle\smash{\psi}\middle|\smash{\phi_{n,i}}\right\rangle
c\lvert\chi_{n,i}\rangle=\frac{1}{\|\Delta_{n}\|_{2}}c\sum_{i=1}^{\infty}s_{n,i}\lvert\chi_{n,i}\rangle\\!\\!\vphantom{\phi_{n,i}\psi}\left\langle\smash{\phi_{n,i}}\middle|\smash{\psi}\right\rangle=c\frac{\Delta_{n}^{\dagger}}{\|\Delta_{n}\|_{2}}\lvert\psi\rangle.$
(205)

Since
$\left\|\lvert\Psi_{n}\rangle\right\|_{\otimes}=\frac{\|\Delta_{n}\|}{\|\Delta_{n}\|_{2}}$
and $\|\Delta_{n}\|\rightarrow 1$, we have that
$\frac{1}{\|\lvert\Psi_{n}\rangle\|_{\otimes}}\left\lparen\langle\psi\rvert\otimes\mathbb{I}\right\rparen\lvert\Psi_{n}\rangle=c\frac{\Delta_{n}^{\dagger}}{\|\Delta_{n}\|}\lvert\psi\rangle\rightarrow
c\lvert\psi\rangle$. In the same way,

$\displaystyle\left\lparen\mathbb{I}\otimes\langle\psi\rvert\right\rparen\lvert\Psi_{n}\rangle=\frac{1}{\|\Delta_{n}\|_{2}}\sum_{i=1}^{\infty}s_{n,i}\vphantom{\psi
c\chi_{n,i}}\left\langle\smash{\psi}\middle|\smash{c\chi_{n,i}}\right\rangle\lvert\phi_{n,i}\rangle=\frac{1}{\|\Delta_{n}\|_{2}}\sum_{i=1}^{\infty}s_{n,i}\lvert\phi_{n,i}\rangle\vphantom{\chi_{n,i}c\psi}\left\langle\smash{\chi_{n,i}}\middle|\smash{c\psi}\right\rangle=\frac{\Delta_{n}}{\|\Delta_{n}\|_{2}}c\lvert\psi\rangle,$
(206)

and hence
$\frac{1}{\|\lvert\Psi_{n}\rangle\|_{\otimes}}\left\lparen\langle\psi\rvert\otimes\mathbb{I}\right\rparen\lvert\Psi_{n}\rangle=\frac{\Delta_{n}}{\|\Delta_{n}\|}c\lvert\psi\rangle\rightarrow\lvert\psi\rangle$.

For the second point, we have that
$\|\Delta_{n}\|=\frac{\|\lvert\Psi_{n}\rangle\|_{\otimes}}{\|\lvert\Psi_{n}\rangle\|_{\otimes}}=1$
and also
$\|\Delta_{n}\|_{2}^{2}=\frac{1}{\|\lvert\Psi_{n}\rangle\|_{\otimes}^{2}}\sum_{i=1}^{\infty}p_{n,i}=\frac{1}{\|\lvert\Psi_{n}\rangle\|_{\otimes}^{2}}<\infty$,
so it is well-defined. Finally, for any $\lvert\psi\rangle\in\mathcal{H}$,

$\displaystyle\Delta_{n}\lvert\psi\rangle=\frac{1}{\|\lvert\Psi_{n}\rangle\|_{\otimes}}\sum_{i=1}^{\infty}\sqrt{p_{n,i}}\lvert\phi_{n,i}\rangle\\!\\!\vphantom{c\psi\chi_{n,i}}\left\langle\smash{c\psi}\middle|\smash{\chi_{n,i}}\right\rangle=\frac{1}{\|\lvert\Psi_{n}\rangle\|_{\otimes}}(\mathbb{I}\otimes\langle
c\psi\rvert)\lvert\Psi_{n}\rangle\rightarrow
c^{2}\lvert\psi\rangle=\lvert\psi\rangle$ (207)

and

$\displaystyle\Delta_{n}^{\dagger}\lvert\psi\rangle=\frac{1}{\|\lvert\Psi_{n}\rangle\|_{\otimes}}c\sum_{i=1}^{\infty}\sqrt{p_{n,i}}\lvert\chi_{n,i}\rangle\\!\\!\vphantom{\psi\phi_{n,i}}\left\langle\smash{\psi}\middle|\smash{\phi_{n,i}}\right\rangle=c\frac{1}{\|\lvert\Psi_{n}\rangle\|_{\otimes}}(\langle\psi\rvert\otimes\mathbb{I})\lvert\Psi_{n}\rangle\rightarrow
c^{2}\lvert\psi\rangle=\lvert\psi\rangle.$ (208)

∎

Finally, we use damping sequences to construct states from operator-valued
measures.

######  Definition B.4.

Let $(X,\mathscr{S},\mu)$ be a measure space and
$P:\mathscr{S}\rightarrow\mathcal{B}(\mathcal{H})$ be an operator-valued
measure. A damping sequence $(\Delta_{n})$ in $\mathcal{B}(\mathcal{H})$
_$\mu$ -damps $P$_ if $\|\Delta_{n}^{\dagger}P\Delta_{n}\|_{1}\ll\mu$, _i.e._
if $\mu(E)=0$ then $\Delta_{n}^{\dagger}P(E)\Delta_{n}=0$.

Note first that this is well-defined. As
$\Delta_{n}\in\mathcal{T}_{2}(\mathcal{H})$,
$P(E)\Delta_{n}\in\mathcal{T}_{2}(\mathcal{H})$, and therefore
$\Delta_{n}^{\dagger}P(E)\Delta_{n}\in\mathcal{T}_{1}(\mathcal{H})$, so the
trace norm is defined on this operator. Also,
$(\Delta_{n}^{\dagger}P(E)\Delta_{n})$ converges weakly to $P(E)$, so the
sequence can be seen to approximate the operator-valued measure. This is
because, for any $\lvert\phi\rangle,\lvert\psi\rangle\in\mathcal{H}$,

$\displaystyle\begin{split}\big{\lvert}\vphantom{\psi\Delta_{n}^{\dagger}P(E)\Delta_{n}-P(E)\phi}\left\langle\smash{\psi}\middle\rvert\smash{\Delta_{n}^{\dagger}P(E)\Delta_{n}-P(E)}\middle\rvert\smash{\phi}\right\rangle\big{\rvert}&\leq\big{\lvert}\vphantom{\psi\Delta_{n}^{\dagger}P(E)\Delta_{n}-\Delta_{n}^{\dagger}P(E)\phi}\left\langle\smash{\psi}\middle\rvert\smash{\Delta_{n}^{\dagger}P(E)\Delta_{n}-\Delta_{n}^{\dagger}P(E)}\middle\rvert\smash{\phi}\right\rangle\big{\rvert}+\big{\lvert}\vphantom{\psi\Delta_{n}^{\dagger}P(E)-P(E)\phi}\left\langle\smash{\psi}\middle\rvert\smash{\Delta_{n}^{\dagger}P(E)-P(E)}\middle\rvert\smash{\phi}\right\rangle\big{\rvert}\\\
&\leq\big{\|}P(E)^{\dagger}\Delta_{n}\lvert\psi\rangle\big{\|}\big{\|}(\Delta_{n}-\mathbb{I})\lvert\phi\rangle\big{\|}+\big{\|}\lvert\psi\rangle\big{\|}\big{\|}(\Delta_{n}^{\dagger}-\mathbb{I})P(E)\lvert\phi\rangle\big{\|}\\\
&\rightarrow 0.\end{split}$ (209)

The definition of $\mu$-damping is meant to be a way to formalise the idea of
damping unnormalizable states. That is, for Casimir operator $C$, we have a
candidate damping sequence $(e^{-\frac{C}{2n^{2}}})$, where $\frac{1}{n}$
represents the damping strength, which we refer to as _Casimir damping_. In
order to extract a state-valued function
$X\rightarrow\mathcal{D}(\mathcal{H})$ from the damped measure
$\Delta_{n}^{\dagger}P\Delta_{n}$, we may make use of the Radon-Nikodym
theorem.

######  Lemma B.5.

Let $P:\mathscr{S}\rightarrow\mathcal{B}(\mathcal{H})$ be a POVM measure, and
let $(\Delta_{n})$ be a sequence in $\mathcal{B}(\mathcal{H})$ that
$\mu$-damps $P$, for $\mu:\mathscr{S}\rightarrow[0,\infty]$ a $\sigma$-finite
measure. Then, for each $n$, there exists a measurable function
$\rho_{n}:X\rightarrow\mathcal{D}(\mathcal{H})$ and an integrable function
$\pi_{n}:X\rightarrow[0,\infty)$ such that, for all $E\in\mathscr{S}$,

$\displaystyle\Delta_{n}^{\dagger}P(E)\Delta_{n}=\int_{E}\rho_{n}\pi_{n}d\mu$
(210)

###### Proof.

As
$\|\Delta_{n}^{\dagger}P\Delta_{n}\|=\operatorname{Tr}(\Delta_{n}^{\dagger}P\Delta_{n})\ll\mu$,
this implies that, in particular,
$\vphantom{\psi\Delta_{n}^{\dagger}P\Delta_{n}\phi}\left\langle\smash{\psi}\middle\rvert\smash{\Delta_{n}^{\dagger}P\Delta_{n}}\middle\rvert\smash{\phi}\right\rangle\ll\mu$
for all $\lvert\psi\rangle,\lvert\phi\rangle\in\mathcal{H}$. Thus, by the
Radon-Nikodym theorem, there exist integrable functions
$f_{\psi,\phi}:X\rightarrow\mathbb{C}$ and $g:X\rightarrow[0,\infty)$ such
that
$\vphantom{\psi\Delta_{n}^{\dagger}P(E)\Delta_{n}\phi}\left\langle\smash{\psi}\middle\rvert\smash{\Delta_{n}^{\dagger}P(E)\Delta_{n}}\middle\rvert\smash{\phi}\right\rangle=\int
f_{\psi,\phi}d\mu$ and
$\operatorname{Tr}(\Delta_{n}^{\dagger}P\Delta_{n})=\int gd\mu$. Then, it is
direct to see that, for almost every $x\in X$,
$(\lvert\psi\rangle,\lvert\phi\rangle)\mapsto f_{\psi,\phi}(x)$ is linear in
$\lvert\phi\rangle$ and antilinear in $\lvert\psi\rangle$. Then, there exists
a function $F:X\rightarrow\mathcal{L}(\mathcal{H})$ such that $\vphantom{\psi
F(x)\phi}\left\langle\smash{\psi}\middle\rvert\smash{F(x)}\middle\rvert\smash{\phi}\right\rangle=f_{\psi,\phi}(x)$
almost everywhere. Next, as $\|F(x)\|_{1}=g(x)=\operatorname{Tr}(F(x))$ almost
everywhere, we may assume that $F$ is bounded, trace class, and positive.
Finally, we take $\rho_{n}(x)=\frac{F(x)}{\operatorname{Tr}(F(x))}$ and
$\pi_{n}(x)=\operatorname{Tr}(F(x))$. ∎

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