Self-Testing from Correlation

In the main paper, we have mostly focused on self-testing from the perspective of nonlocal games. A different way to define self-testing is from the perspective of probability distributions, or correlations. In this setting, instead of having a game self-test a strategy \(}\), we have a correlation p that self-tests a strategy \(}\). We will denote the correlation generated by the strategy S as \(p_S\).

We say that the strategy \(S = (\rangle },\\},\\})\) generates the correlation p if for all a, b, s, t

$$p(a,b|s,t) = \otimes B_|\psi }\rangle }.$$

This is the definition of self-testing that is used in [18], though again augmented with the same qualifiers as in Definition 2.7.

Definition A.1

(Self-testing from correlation). Let \(t\subseteq ,\text ,\text \}}\) and let \(\tilde\) be a pure strategy. We say that a correlation \(p_}\) t-self-tests a (reference) strategy \(\tilde\) if for every t-strategy S where \(p_S = p_}\).

In self-testing from correlation, our result for lifting the full-rank or PVM assumption (Theorem 4.5 and 4.3) still hold. However, our results for lifting the purity assumption (Theorem 4.7) does not translate directly into this setting. Here, we show how to translate the proof of Theorem 4.7 for self-testing from correlation, by spotting significant points where they differ. In this appendix, we only consider exact self-testing.

The first one is in the proof of Lemma 4.14, where we have no game operator for correlation. Nevertheless, we can show the following analog of Proposition 4.11:

Proposition A.2

Let \(} = (}\rangle },\_\},\_\})\) be the reference full-rank strategy t-self-tested by correlation \(p_}}\). Then any state \(\rangle }\in \mathcal _} \otimes \mathcal _}\) satisfying

$$ _\otimes \tilde_|\psi }\rangle }=p_}(a,b|s,t)$$

for all a, s, t, b must be that \(\left| \right\rangle \left\langle \right| =\left| }\right\rangle \left\langle }\right| \).

Proof

Let \(S=(\rangle },\_\},\_\})\). By t-self-testing, , which implies

$$\begin (V_A\otimes V_B)(A_\otimes B_)\rangle } = (A_\otimes B_)}\rangle }_\otimes }\rangle }_} }}. \end$$

(40)

By Lemma 4.10, \(\rangle }\) has at least as large Schmidt rank as \(}\rangle }\). The fact that they live on the same space and \(}\rangle }\) has full Schmidt rank implies \(\rangle }\) also has full Schmidt rank. This means that \(}\rangle }\) has Schmidt rank 1, and so is a product state, \(}\rangle }_} }} = }\rangle }_}}\otimes }\rangle }_}}\).

Consider the Schmidt decomposition of \(\rangle } = \sum _i \lambda _i \rangle }\rangle }\). If we now trace out \(\mathcal _}}\otimes \mathcal _}}\) from (40), and sum over a, b we get

$$\begin \sum _i \lambda _i V_A \left| \right\rangle \left\langle \right| V_A^* = \text _}}(\left| }\right\rangle \left\langle }\right| _} }}) \otimes \left| }\right\rangle \left\langle }\right| _}}. \end$$

(41)

By Lemma 5.3 we can conclude that

$$V_A \left| \right\rangle \left\langle \right| V_A^* = \left| \right\rangle \left\langle \right| _}} \otimes \left| }\right\rangle \left\langle }\right| _}}$$

for some state \(\rangle }\in \mathcal _}}\) for all i. This implies that \(V_A \rangle } = \rangle }_}} }\rangle }_}}\) where we have absorbed a potential global phase into \(\rangle }\). Since \(\rangle }\) has full Schmidt rank, \(\,}}(\_i) = \mathcal _A\). This implies for all \(\rangle }\in \mathcal _A\), \( V_A \rangle } = \rangle }}\rangle }_}}\). This directly implies that \((\mathbbm _}}\otimes }|}_}})V_A\) is unitary. A similar argument for \(\mathcal _B\) shows that \((\mathbbm _}}\otimes }|}_}})V_B\) is unitary. So we conclude that exists unitaries \(U_A:=(\mathbbm _}}\otimes }|}_}})V_A\), \(U_B:=(\mathbbm _}}\otimes }|}_}})V_B\) such that for all s, t, a, b

$$U_A\otimes U_B(A_\otimes B_)\rangle } = (A_\otimes B_)}\rangle }.$$

Since \(}\rangle }\) has full Schmidt rank, \(\,}}_}}(\rangle }) = \mathcal _}}\). This implies that for all \(\rangle }\in \mathcal _}}\), \(U_A }_ U_A^*\rangle } = }_\rangle }\) for all s, a. This directly gives that \(U_A }_ U_A^* = }_\), and so \([}_,U_A] = 0\) for all s, a. By Lemma 4.10, the state \(}\rangle }\) has minimum Schmidt rank across all states that can give rise to \(p_}}\) using some local measurements. Hence, strategy \(}\) has the minimum (local) dimension among all strategies that give rise to p. It now follows that the matrix algebra generated by all the \(\tilde_\) is irreducible and thus \(\langle }_ \rangle _ = \mathbb _\), where \(d=\dim (}_}})\). The only matrix that commutes with all \(d\times d\) matrices is \(U_A = c\mathbbm \). After applying a similar argument to Bob, we obtain that \((U_A\otimes U_B)\rangle } = }\rangle }\), where both \(U_A\) and \(U_B\) are proportional to identity. This establishes the desired statement. \(\square \)

The second major difference is in the proof of Lemma 4.14, where we decomposed \(X\rangle }_\) in the eigenspace of \(}\). In Definition 2.7, we require the strategies to be optimal, which gives us an extremality conditions on the game value. Here, in the correlation case, we would instead have the correlation be extreme in the set of quantum correlations.

Lemma A.3

Let \(}\) be a pure full-rank strategy self-tested by a extreme correlation \(p_}}\). Then for any mixed state \(\rho \in \mathcal (\mathcal _}}\otimes \mathcal _}})\) satisfying

$$ \text [\rho }_\otimes }_]=p_}}(a,b|s,t), \forall a,b,s,t $$

must be so that \(\rho =\left| }\right\rangle \left\langle }\right| \).

Proof

Consider the spectral decomposition of \(\rho \)

$$ \rho =\sum _i\lambda _i\left| \right\rangle \left\langle \right| ,\rangle }\in \mathcal _}}\otimes \mathcal _}}. $$

Then

$$ p_}}(a,b|s,t)=\text [\rho }_\otimes }_]=\sum _i \lambda _i}_\otimes }_|\phi _i}\rangle }. $$

From the extremality of \(p_}}\), each \(\rangle }\) must have the same correlation using the measurements \(}_\otimes }_\). By Proposition A.2, \(\left| \right\rangle \left\langle \right| =\left| }\right\rangle \left\langle }\right| \). So \(\rho =\left| }\right\rangle \left\langle }\right| \). \(\square \)

Now we can state the primary result of this appendix. We are not going to fully prove this, since the proof is essentially the same as Theorem 4.7. We are instead going to state the places where they differ.

Theorem A.4

Let \(t\in \\}\), and correlation p pure t self-tests \(\tilde\), where \(}\rangle }\) has full Schmidt rank. If \(p_}\) is extreme in the set of quantum correlations, then p mixed t self-tests \(}\).

Sketch

The proof is very similar to the one in Theorem 4.7. Following the proof of Lemma 4.12, we can show that there exists local isometry X such that \((X\rangle },\}_\otimes \mathbbm _}\},\}_\otimes \mathbbm _}\otimes \mathbbm _)\) is a local dilation of the purification of a physical strategy. Let

$$\begin \rho _}}}':= \text _\hatP}[X\left| \right\rangle \left\langle \right| X^*] = \sum _^ \rho _}}}'-1} p_i\left| \right\rangle \left\langle \right| _}}}. \end$$

Observe that by the extremality of p, each of the \(\rangle }_}}}\) must have the same correlation using the same measurements \(}_\otimes }_\). By Lemma A.3, \(X\rangle }=}\rangle }}\rangle }\) for some auxiliary state \(}\rangle }\in \mathcal _\hatP}\).

This proves that \(p_}}\) mixed t-self-tests \(}\). \(\square \)