2-Local Hamiltonian with Low Complexity is QCMA
aa r X i v : . [ c s . CC ] S e p -Local Hamiltonian with Low Complexity is QCMAcomplete Ying-hao Chen [email protected]
September 10, 2019
Abstract
We prove that -Local Hamiltonian ( -LH) with Low Complexity problem is QCMA-completeby combining the results from the QMA-completeness[4] of -LH and QCMA-completeness of -LH with Low Complexity[6]. The idea is straightforward. It has been known that -LHis QMA-complete. By putting a low complexity constraint on the input state, we make theproblem QCMA. Finally, we use similar arguments as in [4] to show that all QCMA problemscan be reduced to our proposed problem. Acknowledgements
We thank Professor Scott Aaronson for his advices and help with this work. This work wasa course project of CS 395T Quantum Complexity Theory.
QMA is a quantum version complexity class of NP, where the verifier can be a quantum verifierand the proof is allowed to be quantum proof. QCMA is somewhat between NP and QMA. QCMAcontains MA[2] but contained in QMA[5]. Unfortunately, it is still open whether QCMA is strictlyless powerful than QMA.People tried to study the difference between QCMA and QMA from many kinds of perspec-tives. One way is to study the oracle separation. It has been shown that there exists a quantumcircuit oracle that separates QCMA and QMA[1]. However, we still don’t have any classical oracleseparation between them.We can also study their difference from the perspective of their complete problems. First log( n ) -LH, then -LH, -LH and finally -LH have been proved to be QMA-complete[5, 3, 4]. But we stilldon’t know whether any of them is in QCMA or not, and we don’t have too many QCMA-completeproblems, either.Wocjan et al.[6] show that adding the low complexity constraint to -LH problem makes itQCMA-complete. We simply combine their result with the QMA-completeness of -LH[4] to showthat -LH with low complexity is QCMA-complete.1 .1 QCMA We will think of QCMA, Quantum Classical Merlin-Arthur, as a class of promise problems ratherthan a class of languages. A promise problem L can be divided into 2 disjoint sets L = L yes ∪ L no where instances are promised to be either “Yes” or “No”. If L ∈ QCMA, there exist a quantumpolynomial time verifier V x such that for any instance x ∈ L , x can be verified with the help of abasic state witness | y i only if x ∈ L yes . Formally, let B = C denote the Hilbert space of a qubit. Definition 1 (QCMA) . Fix ǫ = ǫ ( | x | ) s.t. Ω( | x | ) ≤ ǫ ≤ . A promise problem L = L yes ∪ L no isin QCMA if for any x ∈ L , there exists a quantum circuit V x with | V x | elementary quantum gateswhich acts on the Hilbert space H := B ⊗ n x ⊗ B ⊗ m x where there are n x input qubit registers and m x ancilla qubit registers and | V x | , n x , m x ∈ poly ( | x | ) such that x ∈ L yes ⇒ ∃ y ∈ { , } n x , T r ( V x ( | y i h y | ⊗ | i h | ) V † x P ) ≥ − ǫx ∈ L no ⇒ ∀ y ∈ { , } n x , T r ( V x ( | y i h y | ⊗ | i h | ) V † x P ) ≤ ǫ where P is the projection corresponding to the measurement on the first output qubit. T r ( V x ( | y i h y |⊗| i h | ) V † x P ) is the probability for the first output qubit to be state on the measurement. k -LH is a quantum version of the MAX- k -SAT problem. Definition 2 ( k -LH) . Given H = M P i =1 H i where H i is k -local. That is, each H i is a Hamiltonianacting on at most k qubits. It is promised that either1. ∃ | ψ i s.t. h ψ | H | ψ i ≤ a , or2. ∀ | ψ i , h ψ | H | ψ i ≥ b .where < a < b < are constants. The problem is to decide which. Definition 3 (Low Complexity State) . Let L C denote the set of low complexity states. We saythat | ψ i ∈ L C if and only if we can prepare | ψ i by a sequence of elementary quantum gates withsize polynomial in the size of | ψ i . That is, | ψ i = U T U T − · · · U | i for some elementary quantum gates U , . . . , U T where T = poly ( size ( | ψ i )) . In this section, we prove our main result.
Theorem 1 ( -LH with Low Complexity is QCMA-complete) . Given any -local Hamiltonian H = M P i =1 H i , and promised that either . There exists a low energy and low complexity state | y ′ i ∈ L C s.t. h y ′ | H | y ′ i ≤ ǫ
2. or for any low complexity states | y ′ i ∈ L C , h y ′ | H | y ′ i ≥ − ǫ The -Local Hamiltonian with Low Complexity ( -LHLC) problem is to decide which. -LHLC isQCMA-complete.Proof. (Contained in QCMA) First note that the restriction to low complexity states makesevery k -LHLC problem a QCMA problem. If the Hamiltonian H has a low complexity low energystate | y ′ i , we can use a classical proof to encode how to prepare such a state. It can be verified inquantum polynomial time.The only thing we need to check is that after applying the low complexity constraint, all QCMAproblems can still be reduced to the -LHLC problem. (completeness of k -LHLC) We start from k -LHLC problem with k = O (log n ) , where n = | x | .Given a QCMA problem L , by definition, for each instance x ∈ L , there exists a quantum circuits V x = U T U T − · · · U such that1. x ∈ L yes ⇒ ∃ | y i ∈ { , } n x such that P [ get | i on the first qubit of V x ( | y i | i ) after measurement ] ≤ ǫ x ∈ L no ⇒ ∀ | y i ∈ { , } n x , P [ get | i on the first qubit of V x ( | y i | i ) after measurement ] ≥ − ǫ For this instance, we can construct a k -local Hamiltonian[4] by Kempe’s construction for ( n x + m x +log T ) -qubits low energy low complexity states. H = J in H in + H out + J prop T X t =1 H prop,t H in = n x + m x X i = n x +1 | ih | i ⊗ | ih | H out = ( T + 1) | ih | ⊗ | T ih T | H prop,t = 12 ( I ⊗ | t ih t | + I ⊗ | t − ih t − | − U t ⊗ | t i h t − | − U † t ⊗ | t − i h t | ) | y ′ i = 1 √ T + 1 T X t =0 U t U t − · · · U ( | y i | i ) ⊗ | t i ( n x + m x ) computational qubits and the second part is log( T + 1) clock qubits.These hamiltonians are O (log( n )) -local because they acts on at most T + 1) qubits at atime.Clearly, if x ∈ L yes , then | y ′ i ∈ L C is low complexity because we only need polynomial numberof quantum gates to prepare the basic state ( | y i | i ) and it takes at most poly ( T ) quantum gatesoperating on them to get | y ′ i . Moreover, h y ′ | H | y ′ i = h y ′ | H out | y ′ i = ( V x ( | y i | i )) † | ih | V x ( | y i | i ) ≤ ǫ (soundness of k -LHLC) For the soundness, we need the projection lemma in [4],
Lemma 1 (Projection Lemma) . (Please refer to the proof in [4]) Given two hamiltonians H , H .Let S be the zero eigen space of H and the eigenvectors in S ⊥ has eigenvalue at least J > || H || .Then, λ ( H | S ) − || H || J − || H || ≤ λ ( H + H ) ≤ λ ( H | S ) where λ ( · ) denote the smallest eigenvalue and λ ( H | S ) is the smallest eigenvalue of H correspond-ing to all eigenvectors orthogonal to S ⊥ . Moreover, we can choose J large enough so that JJ > || H || + 8 || H || and hence λ ( H | S ) − ≤ λ ( H + J H ) Let S in denote the zero eigenspace of H in . We can see that the space is actually a space for validinputs. That is, a n x qubits states follows by m x ancilla qubits. With projection lemma, we canlower bound λ ( H ) by λ (( H out + J prop T X t =1 H prop,t ) | S in ) − ≤ λ ( H ) In other words, we can simply rule out other invalid input states by choosing J in large enough. Wecan regard H in and H prop,t as constraints that force the state to do exactly what we want. Forexample, valid input and states going through U T , . . . , U . If any of them violated, it would causelarge energy to H .Similar to H in , we can also choose J prop large enough so that λ ( H out | S in ∩S prop ) − ≤ λ (( H out + J prop T X t =1 H prop,t ) | S in ) and hence λ ( H out | S in ∩S prop ) − ≤ λ ( H ) By the definition of QCMA, if x ∈ L no , then λ ( H out | S in ∩S prop ) ≥ − ǫ λ ( H ) ≥ − ǫ Note that we should write λ ( H out | L C ∩S in ∩S prop ) − ≤ λ ( H | L C ) . For simplicity, we ignore thenotation for low complexity constraint L C while writing λ ( · ) . Therefore, k -LHLC with k = O (log n ) is QCMA-complete. (From k -LHLC to -LHLC) Note that if we use unary representation to keep the clock qubits, k can be reduce to . We can replace | t i h t − | by | i h | t − and hence the bottleneck wouldbe the term U t ⊗ | i h | t − , and U † t | i h | t − They operate on at most qubits. for computational qubits in U t and for clock qubits.It has also been shown that, actually, we just need qubit to keep the clock if we can somehowensure the clock to be always valid. i.e. always happens before . It turns out that we can simplyadd more clock constraints to H to ensure this. Let H = J in H in + H out + J prop T X t =1 H prop,t + J clock H clock where H clock = P ≤ i 48= 12 − ǫ The elimination of J prop H prop is not exactly the same as with other hamiltonians, but the resultsare similar. For more details, please refer to [4]. The only difference between our H and theirs is that6hroughout the whole argument, we restrict the input state to be low complexity L C . Even while werefer to the smallest eigenvalue of some hamiltonian λ ( H ) , we only refer to those corresponding tolow complexity eigenvectors. Most of the results inherit directly from the original -LH construction. References [1] S. Aaronson and G. Kuperberg. Quantum versus classical proofs and advice. In Twenty-SecondAnnual IEEE Conference on Computational Complexity (CCC’07) , pages 115–128. IEEE, 2007.[2] L. Babai and S. Moran. Arthur-merlin games: a randomized proof system, and a hierarchy ofcomplexity classes. Journal of Computer and System Sciences , 36(2):254–276, 1988.[3] J. Kempe and O. Regev. 3-local hamiltonian is qma-complete. arXiv preprint quant-ph/0302079 ,2003.[4] J. Kempe, A. Kitaev, and O. Regev. The complexity of the local hamiltonian problem. SIAMJournal on Computing , 35(5):1070–1097, 2006.[5] A. Y. Kitaev, A. Shen, M. N. Vyalyi, and M. N. Vyalyi. Classical and quantum computation .Number 47. American Mathematical Soc., 2002.[6] P. Wocjan, D. Janzing, and T. Beth. Two qcma-complete problems. arXiv preprint quant-ph/0305090arXiv preprint quant-ph/0305090