A Noncommutative Nullstellensatz for Perfect Two-Answer Quantum Nonlocal Games

By the Hahn-Banach theorem (Barvinok, 2002, Theorem III.1.7) and Archimedeanity of ${\mathrm{SOS}}_{\mathcal{A}}$, there exists a functional $f:\mathcal{A}\rightarrow{\mathbb{C}}$ which strictly separate $-1$ and ${\mathrm{SOS}}_{\mathcal{A}}+\mathcal{L}(\mathcal{N})+\mathcal{L}(\mathcal{N})^{*}$, i.e

We list the properties of $f$ as follows:

• •

  $f(\mathcal{L}(\mathcal{N}))=\{0\}\text{~{}and~{}}f({\mathrm{SOS}}_{\mathcal{A}})\subseteq{\mathbb{R}}_{\geqslant 0}.$

• •

  $f(h^{*})=f(h)^{*}$ for every $h\in\mathcal{A}$.

Now, the GNS construction yields the desired *-representation $\sigma$ and a cyclic vector $\psi$. Define the sesquilinear form on $\mathcal{A}$

and

By Cauchy-Schwarz inequality, $M$ is a left ideal of $\mathcal{A}$. Form the quotient space $\widetilde{\mathcal{H}}:=\mathcal{A}/M$, and equip it with the inner product $\langle\cdot\mid\cdot\rangle$. We can complete $\widetilde{\mathcal{H}}$ to the Hilbert space $\mathcal{H}$.

It is worth mentioning that we can assume $\mathcal{H}$ to be a separable Hilbert space, as this assumption holds because $\mathcal{A}$ has only a finite number of generators, allowing us to generate a countable dense subset of $\mathcal{A}$ using these generators with rational coefficients. By applying this to the quotient space, we establish the separability of $\mathcal{H}$.

Define the quotient map

the cyclic vector

and the left regular representation

By Archimedeanity of ${\mathrm{SOS}}_{\mathcal{A}}$, it is easy to verify that $\sigma(\alpha)$ is bounded for every $\alpha\in\mathcal{A}$, and thus $\sigma$ is a $*-$representation. Finally, the result

follows from

∎

We construct the one-dimensional representation $\rho$ as follows. Since

for every fixed pair $(x,y)$, we know that there exist $(x,y,a,b)\in X\times Y\times A\times B$ such that

Let

we have

since

and thus

for any $(x,y,a,b)\in\Lambda$, where $\Lambda$ is the index of the invalid determining set $\mathcal{N}$ (3.2), see Remark 1.

Using the generators $A_{x}$ and $B_{y}$ defined in (2.2), we can rewrite:

Since $\mathcal{H}$ is separable, we can choose an orthogonal basis of $\mathcal{H}$ named

where $\psi_{1}=\psi$. Define

Unlike the finite-dimensional case considered in the proof of (Cleve et al., 2004, Theorem 4), here we need to show that for every $x\in X$, $y\in Y$, $k(x)$ and $l(y)$ are well-defined.

As $\psi\neq 0$ and $\sigma(A_{x})^{2}=1$, there must exist a $j\in{\mathbb{N}}$ such that $\psi_{j}^{*}\sigma(A_{x})\psi\neq 0$ (otherwise, $\sigma(A_{x})\psi=0$, which contradicts the assumption that $\psi\neq 0$ and $\sigma(A_{x})^{2}=1$). Similarly, we can also prove that $l(y)$ is well-defined.

Let

We have the following claim:

We will provide the proof of Claim 3.4 after completing the proof of Proposition 3.3.

We construct the one-dimensional $*-$representation $\rho$ as follows. For every $x\in X$, define

and for every $y\in Y$, define

Then, by linearity and homogeneity, we extend $\rho$ to the entire game algebra $\mathcal{A}$.

Since $\rho(e_{a}^{x})$ and $\rho(f_{b}^{y})$ are either $0$ or $1$, they are naturally commutative. It is straightforward to check that $\rho$ satisfies:

and

for all $e_{a}^{x}$, $f_{b}^{y}\in\mathcal{A}$, i.e., $\rho(e_{a}^{x})$ and $\rho(f_{b}^{y})$ satisfy the same relations as $e_{a}^{x}$ and $f_{b}^{y}$ in $\mathcal{A}$. Therefore, $\rho$ is indeed a $*-$representation of $\mathcal{A}$.

Since

By Claim 3.4, we have

Since the value of $\rho(e_{a}^{x}f_{b}^{y})$ can only be $1$ or $0$, as

the condition (3.8) implies that for every $(x,y,a,b)\in\Lambda$, i.e., for every $e_{a}^{x}f_{b}^{y}\in\mathcal{N}$,

holds, which completes the proof. ∎

We set $a=u(x)$ and $b=v(y)$ in equations (3.5), and compute

Notice that $\sigma(A_{x})$ and $\sigma(B_{y})$ are commutative self-adjoint operators, so $\psi^{*}\sigma(A_{x})\psi,~{}\psi^{*}\sigma(B_{y})\psi$ and $\psi^{*}\sigma(A_{x}B_{y})\psi$ are all real numbers.

If $\psi^{*}\sigma(A_{x})\psi\neq 0$, and since $\psi_{1}=\psi$, we conclude that $k(x)=1$. Moreover, according to (3.6), if $\psi^{*}\sigma(A_{x})\psi>0$, we have

if $\psi^{*}\sigma(A_{x})\psi<0$, we have

Therefore, the value below is always positive:

Similarly, if $\psi^{*}\sigma(B_{y})\psi\neq 0$, we have

Therefore, if either $\psi^{*}\sigma(A_{x})\psi$ or $\psi^{*}\sigma(B_{y})\psi$ is nonzero, we have

Since $\dfrac{1}{4}+\dfrac{1}{4}(-1)^{u(x)+v(y)}\psi^{*}\sigma(A_{x}B_{y})\psi\geqslant 0$, we have

Hence, we only need to consider the case

Since we are working with infinite-dimensional separable Hilbert spaces, we modify the argument in (Cleve et al., 2004, Theorem 4) by incorporating Cauchy-Schwarz inequality and Parseval’s identity for proving that

Conversely, suppose

holds. By Cauchy-Schwarz’s inequality, we know that

Since $\psi$ is a unit vector and the eigenvalues of $\sigma(A_{x}),\sigma(B_{y})$ can only be $\pm 1$, we know

Applying the equality condition of the Cauchy-Schwarz inequality, and the assumption (3.10), we obtain

By Parseval’s identity, we have

and

Then equation (3.11) gives

which implies that

holds for every $j\in\{1,2,\ldots\ldots\}$. However, equation (3.12) must fail to hold for $j=\min\{k(x),l(y)\}$. It is clear that equation (3.12) fails when $k(x)\neq l(y)$. Assume $k(x)=l(y)=j$, we find that the arguments of $\arg\left((-1)^{u(x)}\psi_{j}^{*}\sigma(A_{x})\psi\right)$ and $\arg\left((-1)^{v(y)}\psi_{j}^{*}\sigma(B_{y})\psi\right)$ both lie in the range $\left[0,\pi\right)$, which contradicts equation (3.12) once again!

Therefore, when $\psi^{*}\sigma(A_{x})\psi=\psi^{*}\sigma(B_{y})\psi=0$, we have shown that

That is,

which always holds, thereby proving the claim. ∎

($\Longleftarrow$) This direction is straightforward. Suppose, for the sake of contradiction, that this direction does not hold, i.e.,

and there exists a $*-$representation $\rho$ such that

then we have

which is a contradiction!

($\Longrightarrow$) This follows from Proposition 3.2 and Proposition 3.3. ∎