Self-Testing Graph States

Abstract

We give a construction for a self-test for any connected graph state. In other words, for each connected graph state we give a set of non-local correlations that can only be achieved (quantumly) by that particular graph state and certain local measurements. The number of correlations considered is small, being linear in the number of vertices in the graph. We also prove robustness for the test.

Appendices

Proof of Robustness for Reference Experiment 1

First we note that if $\left\langle \psi \, \right| M\left| \, \psi \right\rangle \ge 1 - \epsilon$ then

$$\begin{aligned} \left| \left| \left| \, \psi \right\rangle - M \left| \, \psi \right\rangle \right| \right| _{2} \le \sqrt{2\epsilon }. \end{aligned}$$

(57)

Next, suppose that we have $\left| \left| \left| \, \psi \right\rangle - M \left| \, \psi \right\rangle \right| \right| _{2} \le \alpha$ and $\left| \left| \left| \, \psi \right\rangle - N \left| \, \psi \right\rangle \right| \right| _{2} \le \beta$. Using the triangle inequality and the fact that $\left| \left| M\right| \right| _{\infty } = 1$ we have

$$\begin{aligned} \left| \left| \left| \, \psi \right\rangle - MN \left| \, \psi \right\rangle \right| \right| _{2} \le \alpha + \beta . \end{aligned}$$

(58)

The remainder of the proof will use these estimates repeatedly, along with the triangle inequality. We need only count the number of operators multiplied together.

First, for Lemma 3 let $c$ be the size of the induced cycle. We multiply $c+1$ operators together. Thus we conclude that for a vertex $u$ in the induced cycle

$$\begin{aligned} \left| \left| X^{\prime }_{u}Z^{\prime }_{u} \left| \, \psi ^{\prime } \right\rangle + Z^{\prime }_{u}X^{\prime }_{u} \left| \, \psi ^{\prime } \right\rangle \right| \right| _{2} \le (c+1)\sqrt{2\epsilon }. \end{aligned}$$

(59)

Next, for Lemma 2 we multiply four operators, then invoke the anti-commuting property on one of the vertices. This gives

$$\begin{aligned} \left| \left| X^{\prime }_{u}Z^{\prime }_{u} \left| \, \psi ^{\prime } \right\rangle + Z^{\prime }_{u}X^{\prime }_{u} \left| \, \psi ^{\prime } \right\rangle \right| \right| _{2} \le 4\sqrt{2\epsilon } + \beta \end{aligned}$$

(60)

where $\beta$ is $\left| \left| X^{\prime }_{v}Z^{\prime }_{v} \left| \, \psi ^{\prime } \right\rangle + Z^{\prime }_{v}X^{\prime }_{v} \left| \, \psi ^{\prime } \right\rangle \right| \right| _{2}$, $v$ being neighbouring vertex. We may apply Lemma 2 along paths from vertices in the induced cycle in $G$. Let $l$ be the length (number of edges) of the longest path. Then for any vertex $u$ we find, at worst,

$$\begin{aligned} \left| \left| X^{\prime }_{u}Z^{\prime }_{u} \left| \, \psi ^{\prime } \right\rangle + Z^{\prime }_{u}X^{\prime }_{u} \left| \, \psi ^{\prime } \right\rangle \right| \right| _{2} \le (4l + c + 1)\sqrt{2 \epsilon }. \end{aligned}$$

(61)

Lastly, for Lemma 5, we multiply $|V^{\prime }|$ operators, and apply the anti-commuting relation $|E^{\prime }|$ times. Thus

$$\begin{aligned} \left| \left| (-1)^{|E^{\prime }|} X^{\prime V^{\prime }} \left| \, \psi ^{\prime } \right\rangle - Z^{\prime N(V^{\prime })} \left| \, \psi ^{\prime } \right\rangle \right| \right| _{2} \le \left( |V^{\prime }| + (4l + c + 1)|E^{\prime }|\right) \sqrt{2\epsilon }. \end{aligned}$$

(62)

We are now ready to analyze the proof of the main theorem for reference experiment 1. To arrive at Eq. 46 we apply the anti-commutation relation. This happens once for each 1 appearing in $x$, for each possible $x$, for a total of $n 2^{n-1}$ times. We may find this by pairing values $x$ and $x \oplus 111\dots 1$. There are $2^{n-1}$ such pairs and each pair contains $n$ 1s all together. We find

$$\begin{aligned} \left| \left| \varPhi (\left| \, \psi ^{\prime } \right\rangle ) - \frac{1}{2^{n}}\sum _{x} \bigotimes _{v \in V} \left( I + Z^{\prime }_{v}\right) X^{\prime x_{v}}_{v} \left| \, \psi ^{\prime } \right\rangle \left| \, x \right\rangle \right| \right| \le n2^{n-1}(4l + c + 1)\sqrt{2\epsilon }. \end{aligned}$$

(63)

For Eq. 47 we use Lemma 5, once for each possible value of $x$. Again, the estimate depends on the number of 1s in $x$, summed over all possible $x$s. As well, it depends on the number of edges in the induced subgraph. An edge $(u,v)$ will be counted only when $x_{u} = x_{v} = 1$. This occurs for $1/4$ of all $x$s. Summed over all possible $x$s and edges, then, the number of times edges are counted is $2^{n-2}|E|$. This gives our final estimate:

$$\begin{aligned} \left| \left| \varPhi (\left| \, \psi ^{\prime } \right\rangle ) - \left( \frac{1}{2^{n}}\bigotimes _{v \in V} \left( I + Z^{\prime }_{v}\right) \left| \, \psi ^{\prime } \right\rangle \right) \sum _{x} (-1)^{e(x)}\left| \, x \right\rangle \right| \right| _{2} \end{aligned}$$

(64)

$$\begin{aligned} \le \left( n 2^{n-1}(4l + c + 1) + n2^{n-1} + (4l + c + 1)2^{n-2} |E|\right) \sqrt{2\epsilon } \end{aligned}$$

(65)

$$\begin{aligned} = 2^{n-2}\left( (4l + c + 1)\left( 2n + |E| \right) + 2n\right) \sqrt{2\epsilon } \end{aligned}$$

(66)

where $e(x)$ is the number of edges in the induced subgraph on the set $V_{x} = \{v \in V | x_{v} = 1\}$.

Note that when calculating $\varPhi \left( X^{\prime }_{u} \left| \, \psi ^{\prime } \right\rangle \right)$ etc. we did not use any more estimations, we simply rearrange when Lemma 5 is applied. Thus the same robustness applies. For $\varPhi \left( Z^{\prime }_{v} \left| \, \psi ^{\prime } \right\rangle \right)$ we use $(I + (-1)^{x_{v}} Z^{\prime }_{v}) Z^{\prime }_{v} = -(I + (-1)^{x_{v}} Z^{\prime }_{v})$, which does not involve an estimation, so again the same robustness applies.

As a last estimation, we note that $l$ and $c$ cannot be larger than $n$, and $|E| \le n^{2}$. We may thus set $\delta = (5n^{2} + 11n + 4)n2^{n-2}\sqrt{2\epsilon }$.

Note that we may make much better estimates if some properties of the graph are known. For example, if every vertex lies in a triangle and the max degree is $6$, as in the case of a lattice of triangles, we may instead set $\delta = 17n2^{n-1} \sqrt{2\epsilon }$.

Proof of Robustness for Reference Experiment 2

Much of the same analysis may be used for experiment 2. Indeed, since the only difference in the proofs for the non-robust results is how the anti-commuting property is proved, we may simply replace the estimation for Lemma 3 with that of Lemma 4.

We begin, then, with $\epsilon$-simulation and prove a robust version of Lemma 4. First we wish to estimate $\alpha = \left| \left| D^{\prime }_{u} \left| \, \psi \right\rangle - \frac{X^{\prime }_{u} + Z^{\prime }_{u}}{\sqrt{2}}\left| \, \psi \right\rangle \right| \right| _{2}$. Using techniques from the previous section, we have

$$\begin{aligned} \left| \left| X^{\prime }_{u} \left| \, \psi ^{\prime } \right\rangle - Z^{\prime N_{u}} \left| \, \psi \right\rangle \right| \right| _{2}&\le 2 \sqrt{\epsilon } \end{aligned}$$

(67)

$$\begin{aligned} \left| \left| Z^{\prime }_{u} \left| \, \psi ^{\prime } \right\rangle - X^{\prime }_{v} Z^{\prime N_{v} \setminus u}\left| \, \psi ^{\prime } \right\rangle \right| \right| _{2}&\le 2 \sqrt{\epsilon }. \end{aligned}$$

(68)

These along with the triangle inequality give an upper bound for $\alpha$ of

$$\begin{aligned} 2\sqrt{2\epsilon } + \left| \left| D^{\prime }_{u} \left| \, \psi \right\rangle - \frac{Z^{\prime N_{u}} + X^{\prime }_{v} Z^{\prime N_{v} \setminus u}}{\sqrt{2}}\left| \, \psi \right\rangle \right| \right| _{2} \end{aligned}$$

(69)

Expanding the second term, we get

$$\begin{aligned} \sqrt{1 + \left| \left| \frac{Z^{\prime N_{u}} +X^{\prime }_{v} Z^{\prime N_{v} \setminus u}}{\sqrt{2}} \left| \, \psi ^{\prime } \right\rangle \right| \right| _{2}^{2} - \sqrt{2}\left( \left\langle \psi ^{\prime } \, \right| D^{\prime }_{u}Z^{\prime N_{u}} \left| \, \psi ^{\prime } \right\rangle + \left\langle \psi ^{\prime } \, \right| D^{\prime }_{u} X^{\prime }_{v}Z^{\prime N_{v} \setminus u} \left| \, \psi ^{\prime } \right\rangle \right) }. \end{aligned}$$

(70)

Since $\left| \left| Z^{\prime }_{u} \left| \, \psi ^{\prime } \right\rangle - X^{\prime }_{v} Z^{\prime N_{v} \setminus u}\left| \, \psi ^{\prime } \right\rangle \right| \right| _{2} \le 2 \sqrt{\epsilon }$ and $\left| \left| Z^{\prime N_{u}}\left| \, \psi ^{\prime } \right\rangle \right| \right| _{2} = 1$ we find

$$\begin{aligned} \left| \left\langle \psi ^{\prime } \, \right| Z^{\prime N_{u}}Z^{\prime }_{u} \left| \, \psi ^{\prime } \right\rangle - \left\langle \psi ^{\prime } \, \right| Z^{\prime N_{u}}X^{\prime }_{v} Z^{\prime N_{v} \setminus u}\left| \, \psi ^{\prime } \right\rangle \right| \le 2 \sqrt{\epsilon }. \end{aligned}$$

(71)

By hypothesis, $\left| \left\langle \psi ^{\prime } \, \right| Z^{\prime N_{u}}Z^{\prime }_{u} \left| \, \psi ^{\prime } \right\rangle \right| \le \epsilon$, so $\left| \left\langle \psi ^{\prime } \, \right| Z^{\prime N_{u}}X^{\prime }_{v} Z^{\prime N_{v} \setminus u}\left| \, \psi ^{\prime } \right\rangle \right| \le 2 \sqrt{\epsilon } + \epsilon$.

Meanwhile $\beta ^{2} = \left| \left| \frac{Z^{\prime N_{u}} +X^{\prime }_{v} Z^{\prime N_{v} \setminus u}}{\sqrt{2}} \left| \, \psi ^{\prime } \right\rangle \right| \right| _{2}^{2} = 1 + \text {Re} \left\langle \psi ^{\prime } \, \right| Z^{\prime N_{u}} X^{\prime }_{v} Z^{\prime N_{v} \setminus u} \left| \, \psi ^{\prime } \right\rangle$, so $|1- \beta ^{2}| \le 2\sqrt{\epsilon } + \epsilon$.

Finally, by hypothesis $\left| \left\langle \psi ^{\prime } \, \right| D^{\prime }_{u}Z^{\prime N_{u}} \left| \, \psi ^{\prime } \right\rangle + \left\langle \psi ^{\prime } \, \right| D^{\prime }_{u} X^{\prime }_{v}Z^{\prime N_{v} \setminus u} \left| \, \psi ^{\prime } \right\rangle - \sqrt{2}\right| \le 2\epsilon$. Combining these facts we find $\alpha \le 2\sqrt{2 \epsilon } + \sqrt{2 \sqrt{\epsilon } + (1 + 2\sqrt{2}) \epsilon }$.

Now we wish to estimate

$$\begin{aligned} \left| \left| (D_{u}^{\prime })^{2}\left| \, \psi ^{\prime } \right\rangle - \frac{\left( X^{\prime }_{u} + Z^{\prime }_{u}\right) ^{2}}{2} \left| \, \psi ^{\prime } \right\rangle \right| \right| _{2} \end{aligned}$$

(72)

By the fact $\left| \left| D^{\prime }_{u}\right| \right| _{\infty } = 1$ we have $\left| \left| (D_{u}^{\prime })^{2}\left| \, \psi ^{\prime } \right\rangle - D^{\prime }_{u}\frac{X^{\prime }_{u} +Z^{\prime }_{u}}{\sqrt{2}} \left| \, \psi ^{\prime } \right\rangle \right| \right| _{2} \le \alpha$. Similarly, since $\left| \left| X^{\prime }_{u} + Z^{\prime }_{u}\right| \right| _{\infty } \le 2$ we find $\left| \left| D_{u}^{\prime }\frac{X^{\prime }_{u} Z^{\prime }_{u}}{\sqrt{2}}\left| \, \psi ^{\prime } \right\rangle - \frac{\left( X^{\prime }_{u} + Z^{\prime }_{u}\right) ^{2}}{2} \left| \, \psi ^{\prime } \right\rangle \right| \right| _{2} \le \sqrt{2} \alpha$. Using these facts, the triangle inequality, and $(D^{\prime }_{u})^{2} = I$, we obtain

$$\begin{aligned} 2\left| \left| \left| \, \psi ^{\prime } \right\rangle - \frac{\left( X^{\prime }_{u} + Z^{\prime }_{u}\right) ^{2}}{2} \left| \, \psi ^{\prime } \right\rangle \right| \right| _{2} = \left| \left| X^{\prime }_{u} Z^{\prime }_{u}\left| \, \psi ^{\prime } \right\rangle + Z^{\prime }_{u} X^{\prime }_{u} \left| \, \psi ^{\prime } \right\rangle \right| \right| _{2} \end{aligned}$$

$$\begin{aligned} \le 2(1 + \sqrt{2})\left( 2\sqrt{2 \epsilon } + \sqrt{2 \sqrt{\epsilon } + (1 + 2\sqrt{2}) \epsilon }\right) \le 26 \epsilon ^{\frac{1}{4}} \end{aligned}$$

(73)

with the last inequality valid for $\epsilon \le 1$.

Using this estimate, and working through the estimations as in the previous section, we find that we may set

$$\begin{aligned} \delta = 2^{n-2}\left( (2n + |E|)(26 \epsilon ^{\frac{1}{4}} + 4l \sqrt{2 \epsilon }) + 2n \sqrt{2 \epsilon } \right) . \end{aligned}$$

(74)

For a simpler expression, we may use $l \le n$ and $|E| \le n^{2}$, obtaining

$$\begin{aligned} \delta = n 2^{n-2}\left( (n + 2)26 \epsilon ^{\frac{1}{4}} + (4n^{2} + 10n) \sqrt{2 \epsilon } \right) \end{aligned}$$

(75)

Again, we may find a better estimate with more information about the graph. For cluster states, which have a square lattice graph, we have $|E| \le 4n$. We may also perform $D_{u}$ measurements on all vertices and set $l=0$. In this case we may set $\delta =n 2^{n-2}\left( 156 \epsilon ^{\frac{1}{4}} + 2 \sqrt{2 \epsilon }\right)$.

Classical Hidden Variable Model for Bipartite Graph States with $X$ and $Z$ Measurements

Let $G$ be a bipartite graph and $\left| \, \psi \right\rangle$ the corresponding graph state. We give a local hidden variable model that is consistent will all measurements which are tensor products of $X$ and $Z$ on this state.

We construct a local hidden variable model by randomly choosing a value $\pm 1$ for $Z^{\prime }_{v}$ for each $v$ in the graph. We then set $X^{\prime }_{v}$ to be

$$\begin{aligned} X^{\prime }_{v} = \prod _{u \in N_{v}} Z^{\prime }_{u}. \end{aligned}$$

(76)

Now we show that this is consistent with all possible tensor product $X$ and $Z$ measurements on $\left| \, \psi \right\rangle$. Let $M = X^{S}Z^{T}$, $S \cap T = \emptyset$ be such a measurement. First, suppose that $\pm M$ can be written as a product of stabilizers of $\left| \, \psi \right\rangle$. Using Lemma 1 we have

$$\begin{aligned} M = X^{S} Z^{N(S)} = (-1)^{|E(S)|} \prod _{x \in S} S_{v}. \end{aligned}$$

(77)

Note that, by assumption, $M$ has only $X$ and $Z$ factors, so each $v \in S$ must have an even number of neighbours in $S$. Then the induced subgraph on $S$ is Eulerian and we can partition the edges of the subgraph into cycles with no common edges (see Diestel [Die10] for a proof). Suppose that $|E(S)|$ is odd. Then there must be at least one odd cycle in this partition and then $S$ has an odd cycle and so does $G$. Since $G$ is bipartite this must not be the case and in fact $|E(S)|$ is even. Hence $M = \prod _{x \in S} S_{v}$ and $\left\langle \psi \, \right| M\left| \, \psi \right\rangle = 1$. By construction $M^{\prime } = X^{\prime S} Z^{\prime N(S)} = \prod _{v \in S} X^{\prime }_{v} Z^{\prime N_{v}} = 1$ and the expected value of $M^{\prime }$ matches that of $M$.

Now suppose that $M$ is not a product of stabilizers of $\left| \, \psi \right\rangle$. Then $M$ must anti-commute with at least one stabilizer and hence $\left\langle \psi \, \right| M\left| \, \psi \right\rangle =0$. Meanwhile, by construction

$$\begin{aligned} M^{\prime } = X^{\prime S} Z^{\prime T} = Z^{\prime N(S)}Z^{\prime T.}. \end{aligned}$$

(78)

If $N(S) = T$ then $M$ is in fact a product of stabilizers. This is not the case, so there is at least one $Z^{\prime }_{v}$ in the above equation which is not cancelled. Since all the $Z^{\prime }_{v}$s are chosen randomly, the product of the $Z^{\prime }_{v}$s not cancelled will also be uniformly random. Thus the expected value of $M^{\prime }$ is 0.