Quantum NP - Cont. Classical and Quantum Computation A.Yu Kitaev, A. Shen, M. N. Vyalyi 2002

Transcription

1 Quantum NP - Cont. Classical and Quantum Computation A.Yu Kitaev, A. Shen, M. N. Vyalyi

2 QMA - the quantum analog to MA (and NP). Definition 1 QMA. The complexity class QMA is the class of all languages L for which there exists a quantum polynomial-time verifier V, such that x L y B poly( x ), P rob[v ( x y ) = 1] 2 3 x / L y B poly( x ), P rob[v ( x y ) = 0] 2 3 where B is the Hilbert space of one qubit. Here too it s possible to use amplification. 2

3 Similarly to the class NP, QMA has complete problems. Definition 2 k-local Hamiltonian Problem Input: H 1,..., H r, a set of r Hermitian positive semi-definite matrices on the space of k qubits, B k, with bounded norm H i 1. Each matrix comes with a specification of the k qubits out of n on which it operates. In addition, we are given two real numbers a and b, such that b a > 1/poly(n). Matrix entries, a and b are given with poly(n) many bits. Output: Denote by H i the extension of H i to operate on all n qubits, leaving the other n k qubits unchanged (by tensor product with identity matrices). Is the smallest eigenvalue of H = H 1 + H H r smaller than a or are all eigenvalues larger than b? 3

4 Theorem 1 The k-local Hamiltonians is in QMA for any k = O(log(n)). Proof. If H has an eigenvalue λ smaller than a, Merlin can supply the corresponding eigenvector v in order to convince Arthur. We need to show that there is an efficient procedure which outputs 1 with high probability when λ < a, and 0 when all eigenvalues are larger than b. 4

5 We first show the solution for the simpler case, when all the Hamiltonians H i are projections, H i = α i α i. In this case: λ = v H v = r v H i v = r v α i α i v = r v α i 2 i=1 i=1 i=1 λ r = 1 r r i=1 v α i 2 (1) 5

6 Now, v α i 2 is the probability to get a positive answer when measuring v in the basis α i and the subspace orthogonal to it. Thus we can interpret λ/r as follows: it is the probability to get the answer 1 when picking 1 i r in random and measuring v in the basis α i and the subspace orthogonal to it. 6

7 Based on this, we can construct the following quantum procedure to verify v : 1. Pick an i {1,... r} uniformly at random. 2. Measure v in the basis α i and the subspace orthogonal to it. 3. If the measurement resulted in a projection on α i, output 0; Otherwise, output 1. 7

8 If indeed v corresponds to an eigenvalue λ smaller than a, then the probability for the answer 1 is 1 λ/r 1 a/r. If on the other hand all the eigenvalues are larger than b, then for every v : v H v = r i=1 The probability for 1 in this case is v H i v b. 1 v H v /r 1 b/r. Since we have that b a 1/n g, the probabilities for 1 and 0 are polynomially separated: 1 a/r 1 b/r +1/n g, and can therefor be amplified, thus proving the theorem for simple projections. 8

9 Spectral Decomposition Any normal operator M on a vector space V can be written as: M = i λ i i i Where λ i are the eigenvalues of M, i is an orthonormal basis V, and each i an eigenvector of M with eigenvalue λ i. 9

10 Schmidt Decomposition Let ψ be a pure state of a composite system AB. Then there exist orthonornal states i A for system A and orthonornal states i B for system B, such that ψ = i λ i i A i B where λ i are non-negative real numbers satisfying i λ 2 i = 1 known as the Schmidt co-efficients. 10

11 We now proceed to the general case, where H i are Hermitian positive semi-definite matrices with norm at most 1. Such a matrix can be written in its spectral decomposition: H i = dim(h i ) j=1 h i j αi j αi j 11

12 Our goal is to toss a coin with probability v H i v. We first add to the system a qubit in the state 0. We define the following unitary transformation T i, corresponding to H i, which operates on H i s decomposition states and the extra qubit: T α i j 0 = αi j ( h i j h i j 1 ) 12

13 We now show that the measurement of the extra qubit output 1 with probability 1 v H i v. The Schmidt decomposition of v is: v = j v j α i j βi j Applying the transformation T i yields the state: T v 0 = j v j α i j βi j ( h i j h i j 1 ) 13

14 The probability to measure 1 is the squared norm of the following vector: which is just: But, we have that j j v j 1 h i j αi j βi j (1 h i j ) v j 2 = 1 j h i j v j 2 v H i v = j h i j v j 2 14

15 so the probability to get 1 is exactly 1 v H i v We thus get the following procedure for verifying v : 1. Pick an i {1,... r} uniformly at random. 2. Add the extra qubit. 3. Apply T i and measure the extra qubit. 15

16 The result is 1 with probability: 1 r i (1 v H i v ) = 1 1 r v H v which, just as in the simple projections case, gives the probabilistic gap between the two possibilities. 16

17 Theorem 2 The 5-Local Hamiltonians is QMA-hard Proof idea. The reduction idea is similar to that in the Cook-Levin theorem. However, the quantum counterpart is not as simple... 17

18 Cook-Levin Theorem - Overview We have a language in Np that is polynomially verifiable, and we want to construct a reduction to 3-SAT. Any configuration is polynomially long, and there are at most a polynomial number of configuration. Thus, a complete computation is polynomially long. If we have the details of the verifier, we can construct a formula corresponding to a legal computation from the starting configuration to an accepting configuration. This is possible since the TM works locally. 18

19 The Quantum analog idea Here we also want the computation as a witness. Given a language L QMA, there exists a quantum circuit that for x L, accepts x y for some witness y with high probability, and for x / L, rejects x y for any witness y with high probability. Our guess for a witness would be a computation: x y, U 1 x y, U 2 U 1 x y,..., U T U 2 U 1 x y Our goal is to construct a sequence of Hamiltonians that will verify the witness locally. 19

20 Alas, this is not possible locally. For example, suppose you want to verify that in a certain step only the identity matrix was applied. It is possible to construct two states which agree on all the density matrices of 5 qubits, but are completely different due to their overall correlations or entanglement. 20

21 The solution: use entanglement. For example, consider the state: 0 a + 1 b 2 The reduced density matrix of the first qubit tells us the angle between the states a and b. 21

22 So the main idea of the reduction is to require a superposition of the computation history: w = 1 T + 1 T t=0 U T U 1 x, y and to show how to verify it locally. by the way, the idea of moving to a time-independent superposition is attributed to Feynman. 22

23 The Reduction. Let L QMA. Then there exists a classical polynomial TM that for a given input x with length n, produces a (uniform) quantum circuit Q with two-qubit gates U = U 1,..., U T, which acts on N qubits, where the first m qubits initially contain the witness w, the rest are initialized to 0, and T, N = poly(n). 23

24 The Reduction Cont. The following holds for Q: If x L, there exists a witness w for which the output qubit is 1 with high probability. If x / L, for any witness w the output qubit is 1 with low probability. As usual, these probabilities can be assumed to be exponentially close to 1 or 0. 24

25 The Reduction Cont. Given Q, we will construct a sequence of matrices that will be used as an input to the local Hamiltonians problem. In the beginning, these matrices will not be local, but will operate on 2 qubits out of the input bits, and on an extra (T + 1)- dimensional Hilbert space that serves as a step counter. Later, we shall see how to transform the matrices to 5-local matrices. 25

26 The Reduction Cont. The Hamiltonian is consisted of three main terms H = H in + H out + H prop. We use the following notation: An operator subscript i means that it operates on the ith qubit. α denotes the projection operator that projects on the subspace spanned by α. The subscript C denotes the subspace of the counter. 26

27 The Reduction Cont. H in verifies that the initial state is legal, by projecting to time 0, and verifying that the ith bit is 0 for any m + 1 i N: H in = N j=m+1 Π 1 j 0 0 C 27

28 The Reduction Cont. H out verifies that the output is 1 at time T, by projecting to time T, and verifying that the first qubit (which WLOG is the answer qubit) is 1. H out = Π 0 1 T T C 28

29 The Reduction Cont. H prop verifies that the computation follows the given circuit. It is a sum of T Hamiltonians, where the ith Hamiltonian verifies that the propagation from time i 1 to time i is done according to U i : where H prop = T j=1 H j, H j = 1 2 (I j j + I j 1 j 1 U j j j 1 U j j 1 j ). 29

30 Change of Basis. We change the basis according to the following operator: W = T j=0 U j U 1 j j. Under the basis change, a vector ψ is represented by a vector ψ such that ψ = W ψ, and the Hamiltonian is transformed by H = W HW. 30

31 Change of Basis cont. Recall that: H in = N j=m+1 Π 1 j H out = Π 0 1 T T C W = T j=0 The term H in is not changed: The term H out is changed to: U j U 1 j j H in = W H in W = H in H out = W H out W = (U Π 0 1 U) T T 31

32 Change of Basis cont. Recall that: H prop = T H j j=1 H j = 1 2 (I j j + I j 1 j 1 U j j j 1 U j j 1 j ) W = T j=0 U j U 1 j j We examine how the four terms of H j in H prop are changed. 32

33 Change of Basis cont. The term I j j transforms to: W (I j j )W = (U j U 1 j j ) j j (U j U 1 j j ) = (U j U 1 ) (U j U 1 ) ( j j ) j j ( j j ) = I j j and similarly the term I j 1 j 1 becomes: W (I j 1 j 1 )W = I j 1 j 1. 33

34 Change of Basis cont. The term U j j j 1 transforms to: W (I j j 1 )W = (U j U 1 j j ) U j j j 1 (U j 1 U 1 j 1 j 1 ) = (U j U 1 ) U j (U j 1 U 1 ) ( j j ) j j 1 ( j 1 j 1 ) = I j j 1 and similarly the term U j j 1 j becomes: W (I j 1 j )W = I j 1 j 34

35 Change of Basis cont. We get: H j = W H j W and then, = I 1 ( j 1 j 1 + j j j 1 j j j 1 ) 2 = I E j, H prop = W H prop W = I E, 35

36 where E = T j=1 E j =

37 Completness The completeness proof shows that if an input is in the given QMA language, then the reduction creates an instance which is in the Local Hamiltonians language. If U on an input ξ accepts (i.e. yields 1 in the first qubit) with probability 1 ε, then by definition P (0) = ξ, 0 U Π 0 1 U ξ, 0 ε. 36

38 Completness cont. To show that H has a small eigenvalue, it is sufficient to find a vector w, such that w H w is small enough. For the counter space we choose the vector and set w = ξ, 0 ψ. ψ = 1 T + 1 T j=0 j, 37

39 Completness cont. We first note that since all the auxiliary qubits are 0 in w, we immediately have w H in w = 0. Also, for any 1 j T, we have 2E j ψ = ( j 1 + j j 1 j ) = 0, thus E ψ = 0, and we get w H prop w = ξ, 0 I ξ, 0 E ψ = 0. 38

40 Completness cont. To estimate the last term, we note that w H out w = w ((U Π 0 1 U) T T ) w = ξ, 0 U Π 0 1 U ξ, 0 ψ T T ψ 1 = P (0) T + 1 ε T

41 Completness cont. Thus we showed that w H w ε T + 1, So H has an eigenvalue with the same upper bound. 40

42 Soundness We assume now that for any vector ξ, ξ, 0 U Π 0 1 U ξ, 0 1 ε. Let A 1 = H in + H out and A 2 = H prop, such that H = A 1 + A 2. Denote by K 1 and K 2 the null subspaces of A 1 and A 2 respectively. Note that A 1 and A 2 are nonnegative, and that K 1 K 2 = {0}. 41

43 Soundness Cont. Definition 3 The angle θ(k 1, K 2 ) between subspaces K 1 and K 2 with trivial intersection is given by cosθ(k 1, K 2 ) = max k 1 K 1, k 2 K 2 k 1 k 2, 0 < θ(k 1, K 2 ) < π 2 42

44 Soundness Cont. Lemma 1 Let A 1 and A 2 be nonnegative operators, and K 1 and K 2 their null subspaces, where K 1 K 2 = {0}. Suppose that no nonzero eigenvalue of A 1 or A 2 is smaller than τ.then A 1 + A 2 τ 2sin 2 θ 2 where θ = θ(k 1, K 2 ) is the angle between K 1 and K 2. Remark. The notation A 0 where A is an operator, means that the operator is nonnegative, and thus all its eigenvalues are greater than or equal to 0. Similarly, the notation A a where A is an operator and a is a number, means A ai 0, thus A s eigenvalues are greater than or equal to a. 43

45 Proof of the Lemma. It holds that A 1 τ(i Π K1 ) and A 2 τ(i Π K2 ), so it is sufficient to prove that (I Π K2 ) + (I Π K2 ) 2sin 2 θ 2, which is equivalent to Π K2 + Π K2 1 + cosθ. 44

46 Proof of the Lemma. Cont. Let ξ be an eigenvector of the operator Π K2 + Π K2 corresponding to an eigenvalue λ > 0. Let Π K1 ξ = u 1 η 1 and Π K2 ξ = u 2 η 2, where η 1 K 1 and η 2 K 2 are unit vectors, and u 1 and u 2 are nonnegative real numbers. Then u 1 η 1 + u 2 η 2 = λ ξ, and λ 2 = (u 1 η 1 +u 2 η 2 )(u 1 η 1 +u 2 η 2 ) = u 2 1 +u2 2 +2u 1u 2 Re η 1 η2. 45

47 Proof of the Lemma. Cont. Also, λ = ξ (Π K2 + Π K2 ) ξ = u u2 2. Using some simple algebraic operations, we get: where x = Re η 1 η 2. Thus (1 + x)λ λ 2 = x(u 1 ± u 2 ) 2 0, λ 1 + x 1 + cosθ. 46

48 Soundness Cont. It is left to bound the nonzero eigenvectors of A 1 and A 2 and the angle between the subspaces. First note that A 1 is a sum of commuting projections, so all its eigenvalues are nonnegative integers. Therefore A 1 K

49 Soundness Cont. We note now that the eigenvectors and eigenvalues of E are T ψ m = α m cos(q m (j )) j, λ m = 1 cos(q m ), j=0 where q m = πm/(t + 1), m = 0,..., T. It follows that A 2 K 2 π 1 cos( T + 1 ) c T 2. 48

50 Soundness Cont. Lemma 2 Let ε be the maximum probability of acceptance of the initial circuit. Then for the angle θ between K 1 and K 2 it holds that sin 2 θ 1 ε T

51 Soundness Cont. It is left to derive the desired inequality H c(1 ε)t 3. 50

52 Realization of the Counter. Using log(t ) qubits to implement the counter will cause the Hamiltonians to be log-local. Instead, we will use an unary representation for the counter, using T qubits: j 1,..., 1, 0,..., 0. } {{ } j } {{ } T j 51

53 Realization of the Counter Cont. The counter projections are changed accordingly: j j j 1 j j j where the 3-qubit projections act on the qubits j-1, j and j+1 of the counter. Remark. For times 0 and T we check only 2 qubits. 52

54 Realization of the Counter Cont. Denote the replaced Hamiltonian by H ext. Our final Hamiltonian will contain another term: H final = H ext + H stab where H stab = I B N T 1 j=1 Π 0 j Π 1 j+1 H stab does not change the upper bound on the minimal eigenvalue for the witness in case of acceptance, since the counter states in the witness are proper. 53

55 Realization of the Counter Cont. For the rejection case, we examine the eigenvalues of H final = H ext + H stab. Let L = B N C T +1, the space which the original Hamiltonian acts on. On L we have H ext c(1 ε)l 3 and H stab = 0. On L we have H ext 0 and H stab 1. In both cases H ext + H stab c(1 ε)t 3. and H final is 5-local. This completes the proof of the theorem. 54

56 FIN 55