Quantum Merlin-Arthur and proofs without relative phase

Our primary technical contribution is to show a ${\mathsf{QMA}}^{+}$ protocol for a ${\mathsf{NEXP}}$-complete problem. This directly extends the work of Jeronimo and Wu [JW23], who show a ${\mathsf{QMA}}^{+}(2)$ protocol for a ${\mathsf{NEXP}}$-complete problem. As in [JW23], we study constraint satisfaction problems (CSPs) with constant gap. In $(1,\delta)$-GapCSP, either all constraints can be satisfied, or at most a $\delta$ fraction of constraints can be satisfied. These problems are known to be ${\mathsf{NP}}$-hard or ${\mathsf{NEXP}}$-hard (depending on the problem size) using the PCP theorem [AS92, ALM⁺98, Har04].

Before proving ${\mathsf{QMA}}^{+}$ with some constant gap equals ${\mathsf{NEXP}}$, we prove ${\mathsf{QMA}}^{+}_{\log}$ (with some other constant gap) equals ${\mathsf{NP}}$. This choice (also taken by [JW23]) is pedagogical: it allows us to explain the protocol without worrying about input encoding size, since ${\mathsf{QMA}}^{+}$ has a polynomial amount of space and verifier runtime. Here, we consider $(1,\delta)$-GapCSP with polynomially many variables and clauses; the quantum proof must certify that there is a satisfying assignment to all clauses.

The ${\mathsf{QMA}}^{+}_{\log}(2)$ protocol of [JW23] crucially relies on an estimate of sparsity ($\ell_{1}$ norm) of a quantum state without relative phase. The overlap of a $m$-qubit quantum state without relative phase $\ket{\psi}$ with $\ket{+}^{\otimes m}$ is exactly the value $2^{-m/2}\cdot\|\ket{\psi}\|_{1}$. With multiple quantum proofs $\ket{\psi_{1}}\otimes\dots\ket{\psi_{k}}$, one can estimate the sparsity by repeating this “sparsity test” on each $\ket{\psi_{i}}$, and using a swap test to ensure that all $\ket{\psi_{i}}$ are approximately equal. Interestingly, no other part of their protocol requires the no relative phase assumption.³³3Formally, [JW23] studies states with non-negative amplitudes. Recall that the set of these states, up to global phase, are equivalent to states with no relative phase.

In ${\mathsf{QMA}}^{+}_{\log}$, we have a single quantum proof, so we cannot use this test to estimate sparsity. Instead, we design a similar test that directly enforces a rigidity property of the proof.⁴⁴4Note that we use the intuition of rigidity in a more general context, where Arthur’s tests, not a non-local game, enforce states of a certain form. The required form is $\frac{1}{\sqrt{R}}\sum_{j\in[R]}\ket{j}\ket{\vec{v}_{j}}$, where the second register is constant-sized. Using a “sparsity test” over the second register, Arthur ensures that the second register has one $\vec{v}_{j}$ per $j$; but using the complement of a “sparsity test” over the whole proof, Arthur ensures the overall state maximizes $\ell_{1}$ norm. States of the required form are optimal for this combination of tests. We make use of the no relative phase property in Lemmas 6 and 8.

Now we can describe our protocol. For each constraint $j\in R$, Arthur asks for the values $\vec{v}_{j}$ associated with the variables involved in constraint $j$. The protocol either enforces rigidity of the quantum proof, or verifies the constraints of the CSP. Note that we need two kinds of constraint checks: the values $\vec{v}_{j}$ must satisfy constraint $j$, and the value of a variable must be consistent across the constraints it participates in. For states with the rigidity property, checking satisfiability is simple: measure in the computational basis and verify the measured constraint $j,\vec{v}_{j}$. States with the rigidity property will succeed with probability equal to the satisfying fraction of the CSP assignment.

Checking consistency is done using a technique called “regularization” from the PCP literature [Din07]; for each constraint $j$, we verify that each variable participating in $j$ has the same value in exactly $d$ other constraints for some constant $d$, in a way that the edges form an expander graph. The expansion property guarantees that cheating on this test is as damaging as cheating on the satisfiability test. Jeronimo and Wu [JW23] use a swap test to implement these new checks, but this requires multiple quantum proofs. We show how to use a Hadamard test (which requires only one quantum proof) to achieve the same result, building on ideas from previous work [BFM22]. Since there exists a $\delta$ such that $(1,\delta)$-GapCSP is ${\mathsf{NP}}$-hard, this completes the proof of ${\mathsf{NP}}\subseteq{\mathsf{QMA}}^{+}_{\log}$ with some constant gap.

When scaling up to ${\mathsf{QMA}}^{+}$, one must be careful of how to succinctly encode the input of a ${\mathsf{NEXP}}$-complete problem. The PCP theorem allows us to choose $(1,\delta)$-GapCSP that is succinct, but we need stronger properties. Following the adjustments taken in [JW23], we choose a PCP system for ${\mathsf{NEXP}}$ that is both doubly explicit and strongly uniform. Doubly explicit means that one can efficiently compute the variables participating in a given constraint and the constraints a given variable participates in; using this, we can implement the consistency checks in polynomial time. Strongly uniform means that the number of constraints a variable participates in is efficiently computable, and one of a fixed number of possibilities; using this, we only need to build a fixed number of expander graphs during regularization. Recent work also shows how to construct exponentially-sized expander graphs in polynomial time [Lub09, Alo21]. Once we are through these input encoding difficulties, our protocol is identical to that for ${\mathsf{NP}}$.

Fundamentally, the no relative phase property allows Arthur to verify a number of constraints exponential in the number of qubits. Attempts to do this for ${\mathsf{QMA}}(2)$ gave too small a promise gap [BT10, Per12, GNN12], too many provers [ABD⁺08, CD10, CF11], or too much space or time [HM13, NZ23]. Jeronimo and Wu [JW23] show that ${\mathsf{QMA}}^{+}(2)$ circumvents this difficulty: using no relative phase and unentanglement, Arthur enforces the sparsity of a quantum proof to solve a ${\mathsf{NEXP}}$-complete problem. At the center of our work is the insight that no relative phase is enough for Arthur to require constant-sized answers to exponentially many questions, solving a ${\mathsf{NEXP}}$-complete problem with a single polynomial-size quantum proof.

Arthur enforces rigidity of a quantum proof using two tests. The first test is the Density test, which maximizes $\ell_{1}$ norm. Here, we measure the state in the Hadamard basis and accept if the outcome is $\ket{+}$.⁷⁷7For simplicity, we denote the uniform superposition over all standard basis states by $\ket{+}$. The dimension is clear from the context. Given $\ket{\psi}$, the success probability of this test is

$\displaystyle D(\ket{\psi})=\left|\bra{+}\ket{\psi}\right|^{2}=\frac{1}{\kappa R}\left|\sum_{j\in[R],x\in\Sigma^{q}}a_{j,x}\right|^{2}=\frac{1}{\kappa R}(\|\ket{\psi}\|_{1})^{2}\,.$

Recall that if $\ket{\psi}$ is a subset state according to subset $S$, its sparsity $\|\ket{\psi}\|_{1}$ is exactly $\sqrt{|S|}$. In completeness, the quantum proof is a subset state with $R$ elements, so this test passes with probability $\frac{1}{\kappa}$.

The second test is the Validity test, which minimizes $\ell_{1}$ norm only on the second register. Here, we measure the color register in the Hadamard basis, and reject if the outcome is $\ket{+}$. Given $\ket{\psi}$, the success probability of this test is

$\displaystyle V(\ket{\psi})=1-\bra{+}\Tr_{R}(\ket{\psi}\bra{\psi})\ket{+}=1-\frac{1}{\kappa}\sum_{j\in[R]}\left|\sum_{x\in\Sigma^{q}}a_{j,x}\right|^{2}\,,$

where $\Tr_{R}$ is partial trace over the constraint register. In completeness, recall that the proof has the form $\frac{1}{\sqrt{R}}\sum_{j\in[R]}\ket{j}\ket{\vec{v}_{j}}$, so the success probability is

$\displaystyle 1-\bra{+}\left(\frac{1}{R}\sum_{j\in[R]}\outerproduct{\vec{v}_{j}}{\vec{v}_{j}}\right)\ket{+}=1-\frac{1}{\kappa}\,.$

In fact, no quantum state without relative phase can pass the Validity test with a higher probability:

It turns out that is impossible to score high on both the Validity test and the Density test. We use this to enforce the rigidity property of $\ket{\psi}$.

Why does this help with rigidity? Suppose Arthur inputs a quantum proof (without relative phase) $\ket{\psi}$ and runs Density test with probability $p_{1}$ and Validity test with probability $p_{2}$. Suppose also that $p_{2}>p_{1}$. Then the expected success probability is $p_{1}D(\ket{\psi})+p_{2}V(\ket{\psi})\leq p_{1}+(p_{2}-p_{1})V(\ket{\psi})\leq p_{1}+(p_{2}-p_{1})(1-\frac{1}{\kappa})$. Note that this upper bound is achieved in completeness, and for any state of the form $\frac{1}{\sqrt{R}}\sum_{j\in R}\ket{j}\ket{\vec{v}_{j}}$. We show that quantum proofs must have this form to reach the upper bound.

One requirement to get close to the upper bound is near-optimal success probability on Validity test. We prove that any quantum proof that has this property must be close to a state that assigns one color to each constraint.

Another requirement to get close to the upper bound is near-optimal success probability on Density test, up to Lemma 7. Consider any quantum proof that passes Validity test with probability close to $1-\frac{1}{\kappa}$ and Density test with probability close to $\frac{1}{\kappa}$; we show it must be close to a state of the form $\frac{1}{\sqrt{R}}\sum_{j\in R}\ket{j}\ket{\vec{v}_{j}}$. Now we can prove the soundness of the rigidity test by relying on the following fact:

Intuitively, Lemma 10 allows us to tune the probability of each test in the ${\mathsf{NP}}$ protocol. As we explain in the analysis (Section 3.3), if the probabilities of running Validity test and Density test are much higher than that for constraint tests, then if $d_{1}$ or $d_{2}$ is large, these two tests catch a “deceptive” quantum proof in soundness. This allows constraint tests to focus on the case of small $d_{1}$ and $d_{2}$; i.e. nearly rigid quantum proofs.

We analyze the constraint tests on rigid quantum proofs, i.e. states of the form $\ket{\psi}=\frac{1}{\sqrt{R}}\sum_{j\in[R}\ket{j}\ket{\vec{v}_{j}}$. The verifier needs to check two properties:

1. (i)

   (satisfiability) For all $j\in[R]$, the assignment $\vec{v}_{j}$ satisfies $\mathcal{C}_{j}$.

2. (ii)

   (consistency) Each variable is assigned the same value when participating in different constraints.

One may ask why we even need to check for consistency. Couldn’t we ask for the assignment of each variable $a:[N]\to\Sigma$, for example as the quantum proof $\frac{1}{\sqrt{N}}\sum_{i\in[N]}\ket{i}\ket{a(i)}$? The problem with this is checking satisfiability becomes difficult, since the assigned values are given in superposition.⁸⁸8There is a way around this limitation for CSP systems consisting of unique game constraints, where each (binary) constraint involving variables $i_{1},i_{2}$ accepts exactly one $a(i_{2})$ for each $a(i_{1})$. See [JW23, Section 6] for more discussion.

Instead, with a state $\frac{1}{\sqrt{R}}\sum_{j\in[R}\ket{j}\ket{\vec{v}_{j}}$, satisfiability is easy to verify: measure the first register (observing some $\ket{j}\ket{\vec{v}_{j}}$), and compute $\mathcal{C}_{j}(\vec{v}_{j})$. Let $u_{s}$ be the number of unsatisfied constraints. The outcome is $1$ with probability $1-\frac{u_{s}}{R}$.

But this form of quantum proof gives Merlin a new way to “deceive”: for a given variable, send different values depending on the constraint! We prevent this by checking for consistency, similarly to the pre-processing step of [Din07] sometimes called regularization. As in [JW23, Section 7], we add “consistency constraints” to the CSP system $\mathcal{C}$ as follows:⁹⁹9Note that since $N$ and $R$ are polynomially-sized, this process is efficient.

• •

  For each variable $i\in[N]$, let $V_{i}$ represent the constraints that $i$ participates in.

• •

  Fix a constant $d$. For each $i\in[N]$, draw a $d$-regular graph with vertices $V_{i}$ that is expanding.¹⁰¹⁰10For technical reasons of 11, we require that the Cheeger constant is at least 2.

• •

  Each edge $(j_{1},j_{2})$ of each expander $V_{i}$ represents a “consistency constraint”, where we assert that the value of variable $i$ sent with constraint $j_{1}$ equals that sent with constraint $j_{2}$.

Using expander graphs allows us to prevent this kind of “deceptive” Merlin: either the proof fails many of the original constraints, or it fails many “consistency constraints”. Let $u_{e}$ be the number of unsatisfied “consistency constraints” out of $Rq\cdot\frac{d}{2}$:

How do we check these “consistency constraints”? Over the next few paragraphs, we construct a unitary related to permutations on the constraint graph. In completeness, the quantum proof is an eigenvector of this unitary, but in soundness, all rigid quantum proofs are detectably far (i.e. using a Hadamard test) from an eigenvector. We study the graph $\widetilde{G}$ with $R\cdot q$ vertices, where each vertex $(j,i)$ corresponds to a clause $j$ and a variable $i$ that participates in clause $j$. Let $\widetilde{G}$ be the union of all consistency edges created during regularization, i.e. $(j_{1},j_{2})$ for variable $i$ becomes the edge $((j_{1},i),(j_{2},i))$. Note that $\widetilde{G}$ contains a copy of each expander graph, so it is $d$-regular.

We now choose $d$ permutations. It is a classical fact that the adjacency matrix of any $d$-regular graph can always be decomposed to $d$ permutations. Let $\pi_{1},\ldots,\pi_{d}$ be the decomposition of $\widetilde{G}$; recall that these are permutations on $V(\widetilde{G})$ where $|V(\widetilde{G})|=R\cdot q$. For each $k\in[d]$, we identify $\pi_{k}$ with a permutation on $[R]\times[N]$, where any $(j,i)\in[R]\times[N]$ that is not a vertex of $\widetilde{G}$ (i.e. variable $i$ does not participate in constraint $j$) is mapped to itself.¹¹¹¹11These permutations (and their inverses) are all efficient because $N$ and $R$ are polynomially-sized. Note that this map always preserves the variable $i\in[N]$; without loss of generality, we also identify $\pi_{k}$ with its restriction $[R]\times[N]\to[R]$. From here on, we use this last definition of $\pi_{k}$, which maps constraint $j_{1}$ that variable $i$ participates in to another constraint $j_{2}$ that variable $i$ participates in, and identity otherwise.

Now consider a rigid quantum proof, i.e. of the form $\ket{\psi}=\frac{1}{\sqrt{R}}\sum_{j\in[R]}\ket{j}\ket{\vec{v}_{j}}$. Since there are a polynomial number of variables and constraints, we can efficiently transform $\ket{\psi}$ to $\ket{\phi^{\prime}}$, where

$\displaystyle\ket{\phi^{\prime}}:=\frac{1}{\sqrt{q\cdot R}}\sum_{j\in[R]}\sum_{i\in\mathcal{C}_{j}}\ket{j}\ket{\vec{v}_{j}}\ket{i}\ket{v_{j}(i)}\,.$

Here, $i\in\mathcal{C}_{j}$ are the variables participating in $\mathcal{C}_{j}$, and $v_{j}(i)$ is the value of this variable according to $\vec{v}_{j}$.

We now would like to construct a unitary on $\ket{\phi^{\prime}}$ that maps $\ket{j}\ket{i}\ket{value}$ to $\ket{j^{\prime}}\ket{i}\ket{value}$ for some other constraint $j^{\prime}$ that $i$ participates in. In completeness, this unitary would leave the state unchanged. Notice that from the perspective of such a unitary, the second register containing $\ket{\vec{v}_{j}}$ is “junk”. Fortunately, we can measure out the second register in the Hadamard basis, and reject if the outcome is not $\ket{+}$. All rigid states will observe outcome $\ket{+}$ with probability $\frac{1}{\kappa}$; one can see this by writing the second register in the Hadamard basis.

Suppose the observed outcome is $\ket{+}$; let us call the postselected state $\ket{\phi}$, where

$\displaystyle\ket{\phi}:=\frac{1}{\sqrt{q\cdot R}}\sum_{j\in R}\sum_{i\in\mathcal{C}_{j}}\ket{j}\ket{i}\ket{v_{j}(i)}\,.$

For each $k\in[d]$, we now implement the in-place transformation $\Pi_{k}$ according to $\pi_{k}:[R]\times[N]\to[R]$, where

$\displaystyle\Pi_{k}:\ket{j}\ket{i}\ket{v_{j}(i)}\rightarrow\ket{\pi_{k}(j,i)}\ket{i}\ket{v_{j}(i)}\,.$

Recall that the map $(j,i)\mapsto(\pi_{k}(j,i),i)$ is a permutation. Since we have access both to this permutation and its inverse, we can implement $\Pi_{k}$.

Note that in a satisfiable instance, $\Pi_{k}\ket{\phi}=\ket{\phi}$. By contrast, if $v_{j}(i)\neq v_{j^{\prime}}(i)$, $\ket{j^{\prime}}\ket{i}\ket{v_{j}(i)}$ is orthogonal to $\ket{\phi}$. Hence, $\bra{\phi}\Pi_{k}\ket{\phi}$ is the fraction of satisfied “consistency constraints” observed by $\pi_{k}$. We use the Hadamard test to measure this value, in a similar way to the Spectral test in [BFM22]. Note that unlike the swap test, the Hadamard test only uses one copy of a quantum state.

We now can describe the constraint tests together:

1. (i)

   With probability $\frac{1}{qd\kappa+1}$, check satisfiability. This succeeds with probability $1-\frac{u_{s}}{R}$.

2. (ii)

   With probability $\frac{qd\kappa}{qd\kappa+1}$, generate $\ket{\phi^{\prime}}$, and measure the second register in the Hadamard basis. If the output state is not $\ket{+}$, reject. Otherwise, choose a random $k\in[d]$, and perform a Hadamard test with $\Pi_{k}$. This succeeds with probability $\frac{1}{\kappa}(\frac{1}{2}+\frac{1}{2}\mathbb{E}_{k}[\real\bra{\phi}\Pi_{k}\ket{\phi}])=\frac{1}{\kappa}(1-\frac{u_{e}}{qdR})$.¹²¹²12Note that in our protocol, $\bra{\phi}\Pi_{k}\ket{\phi}$ is always real because $\ket{\phi}$ and $\Pi_{k}$ have real values.

The overall success probability of the constraint tests is

$\displaystyle\frac{1}{qd\kappa+1}(1-\frac{u_{s}}{R})+\frac{qd\kappa}{qd\kappa+1}\left(\frac{1}{\kappa}\cdot(1-\frac{u_{e}}{qdR})\right)=\frac{qd+1}{qd\kappa+1}-\frac{u_{e}+u_{s}}{R\cdot(qd\kappa+1)}\,.$

We now show a constant gap between completeness and soundness. In completeness, $u_{e}=u_{s}=0$, so $\ket{\psi}$ passes the constraint tests with probability $C^{\textnormal{YES}}:=\frac{qd+1}{qd\kappa+1}$. In soundness, recall that $\mathop{\bf val\/}(\mathcal{C})\leq\delta$, so by 11, any rigid quantum proof passes the constraint tests with probability at most $C^{\textnormal{YES}}-\frac{1-\delta}{qd\kappa+1}$. We now apply Lemma 10: any quantum proof that passes Density test and Validity test with probabilities too similar to that in completeness must pass the constraint tests with probability less than $C^{\textnormal{YES}}$.

In the protocol, Arthur applies Density test, Validity test, or constraint tests with probability $p_{1},p_{2},p_{3}$, respectively, where $p_{3}=1-p_{1}-p_{2}$.

We start by analyzing the success probability of the protocol in completeness. Here, $\mathop{\bf val\/}(\mathcal{C})=1$, and the quantum proof $\ket{\psi}=\frac{1}{R}\sum_{j}\ket{j}\ket{\vec{v}_{j}}$ is such that $\vec{v}_{j}$ is a satisfying assignment to the variables that participate in $\mathcal{C}_{j}$. The success probability for each test is exactly $\frac{1}{\kappa}$, $1-\frac{1}{\kappa}$, and $C^{\textnormal{YES}}$, respectively. So the success probability of the protocol in completeness is $P_{\textnormal{YES}}=\frac{p_{1}}{\kappa}+p_{2}(1-\frac{1}{\kappa})+p_{3}C^{\textnormal{YES}}$.

We now choose the probabilities $p_{1},p_{2},p_{3}$. Choose $\lambda:=\frac{1-\delta}{qd\kappa+1}$.

1. 1.

   We first set a distance threshold $\varepsilon:=\frac{\lambda}{\Gamma}$ for a large enough constant $\Gamma(\kappa,q,d,\delta)$ satisfying

   $\displaystyle\left(\kappa\varepsilon+(\kappa+1)\sqrt{\kappa\cdot\varepsilon}\right)^{1/2}\leq\frac{\lambda}{2}\,.$

2. 2.

   Let $Z:=\frac{1}{2}+1+\frac{\varepsilon}{4(1-C^{\textnormal{YES}})}$. Then let

   $\displaystyle p_{1}$

   $\displaystyle=\frac{1}{2}\cdot\frac{1}{Z}$

   $\displaystyle p_{2}$

   $\displaystyle=\frac{1}{Z}$

   $\displaystyle p_{3}$

   $\displaystyle=\frac{\varepsilon}{4(1-C^{\textnormal{YES}})}\cdot\frac{1}{Z}\,.$

Now we study soundness, i.e. when $\mathop{\bf val\/}(\mathcal{C})\leq\delta$. We again denote the quantum proof as $\ket{\psi}$. We divide up the analysis into a few parts:

1. 1.

   A quantum proof that is “too sparse” (i.e. $D(\ket{\psi})=\frac{1}{\kappa}-d$ for any $d\geq\varepsilon$) is detected by Density test.

   	$\displaystyle P_{\textnormal{NO}}$	$\displaystyle=p_{1}(\frac{1}{\kappa}-d)+p_{2}V(\ket{\psi})+p_{3}C(\ket{\psi})$
   		$\displaystyle\leq P_{\textnormal{YES}}-p_{1}d+p_{3}(1-C^{\textnormal{YES}})$
   		$\displaystyle\leq P_{\textnormal{YES}}-p_{1}\varepsilon+p_{3}(1-C^{\textnormal{YES}})$
   		$\displaystyle=P_{\textnormal{YES}}-\frac{\varepsilon}{2Z}+\frac{\varepsilon}{4Z}=P_{\textnormal{YES}}-\frac{\varepsilon}{4Z}\,.$

2. 2.

   A quantum proof that is “too dense” (i.e. $D(\ket{\psi})=\frac{1}{\kappa}+d$ for any $d\geq\varepsilon$) is detected by Validity test.

   	$\displaystyle P_{\textnormal{NO}}$	$\displaystyle=p_{1}(\frac{1}{\kappa}+d)+p_{2}V(\ket{\psi})+p_{3}C(\ket{\psi})$
   		$\displaystyle\leq p_{1}(\frac{1}{\kappa}+d)+p_{2}(1-\frac{1}{\kappa}-d)+p_{3}$
   		$\displaystyle=P_{\textnormal{YES}}-(p_{2}-p_{1})d+p_{3}(1-C^{\textnormal{YES}})$
   		$\displaystyle\leq P_{\textnormal{YES}}-(p_{2}-p_{1})\varepsilon+p_{3}(1-C^{\textnormal{YES}})$
   		$\displaystyle=P_{\textnormal{YES}}-\frac{\varepsilon}{2Z}+\frac{\varepsilon}{4Z}=P_{\textnormal{YES}}-\frac{\varepsilon}{4Z}\,,$

   where the first inequality follows from Lemma 7.

3. 3.

   A quantum proof that is “the right density” (i.e. $D(\ket{\psi})=\frac{1}{\kappa}+d_{1}$ for $|d_{1}|\leq\varepsilon$) but far from “valid” ($V(\ket{\psi})=1-\frac{1}{\kappa}-d_{2}$ for $d_{2}\geq\varepsilon$) is detected by Validity test when $p_{2}>p_{1}$.

   	$\displaystyle P_{\textnormal{NO}}$	$\displaystyle\leq p_{1}(\frac{1}{\kappa}+|d_{1}|)+p_{2}(1-\frac{1}{\kappa}-d_{2})+p_{3}$
   		$\displaystyle\leq P_{\textnormal{YES}}+p_{1}|d_{1}|-p_{2}d_{2}+p_{3}(1-C^{\textnormal{YES}})$
   		$\displaystyle\leq P_{\textnormal{YES}}-(p_{2}-p_{1})\varepsilon+p_{3}(1-C^{\textnormal{YES}})$
   		$\displaystyle=P_{\textnormal{YES}}-\frac{\varepsilon}{4Z}\,.$

4. 4.

   Lastly, a quantum proof that is nearly rigid (i.e. $D(\ket{\psi})=\frac{1}{\kappa}+d_{1}$ and $V(\ket{\psi})=1-\frac{1}{\kappa}-d_{2}$ for any $|d_{1}|,d_{2}\leq\varepsilon$) is detected by the constraint tests.

   	$\displaystyle P_{\textnormal{NO}}$	$\displaystyle=p_{1}(\frac{1}{\kappa}+d_{1})+p_{2}(1-\frac{1}{\kappa}-d_{2})+p_{3}C(\ket{\psi})$
   		$\displaystyle\leq P_{\textnormal{YES}}+p_{1}d_{1}-p_{2}d_{2}+p_{3}\left(-\frac{1-\delta}{qd\kappa+1}+\left(\kappa d_{1}+(\kappa+1)\sqrt{\kappa\cdot d_{2}}\right)^{1/2}\right)$
   		$\displaystyle\leq P_{\textnormal{YES}}+p_{1}d_{1}-p_{2}d_{2}-p_{3}\frac{\lambda}{2}\,.$

   The first inequality follows from Corollary 13, and the second inequality holds by our choice of $\varepsilon$. Note that $d_{2}\geq 0$ by Lemma 6. By Lemma 7, if $d_{1}\geq 0$, then $d_{1}\leq d_{2}$; otherwise $d_{1}\leq d_{2}$ trivially. So then

   $\displaystyle P_{\textnormal{NO}}$

   $\displaystyle\leq P_{\textnormal{YES}}-(p_{2}-p_{1})d_{2}-p_{3}\frac{\lambda}{2}\leq P_{\textnormal{YES}}-\frac{\varepsilon\lambda}{8(1-C^{\textnormal{YES}})Z}\,.$

Combining these cases proves the following result: