A New Quantum Lower Bound Method, with an Application to a Strong
T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25 www.theoryofcomputing.org
A new quantum lower bound method, with an application to a strong direct product theorem for quantum search Andris Ambainis∗ Received: July 6, 2007; revised: July 24, 2009; published: January 12, 2010.
Abstract: We present a new method for proving lower bounds on quantum query algorithms. The new method is an extension of the adversary method, by analyzing the eigenspace structure of the problem. Using the new method, we prove a strong direct product theorem for quantum search. ˇ This result was previously proved by Klauck, Spalek, and de Wolf (FOCS’04) using the polynomials method. No proof using the adversary method was known before. ACM Classification: F1.2, F2.2 AMS Classification: 81P68, 68Q17 Key words and phrases: quantum computing, quantum algorithms, quantum lower bounds
1
Introduction
Many quantum algorithms (for example, Grover’s algorithm [13] and quantum counting [11]) can be analyzed in the query model where the input is accessed via a black box that answers queries about the values of input bits. There are two main methods for proving lower bounds on quantum query algorithms: the adversary method [3] and the polynomials method [9]. Both of them have been studied in detail. The limits of the adversary method are particularly well understood. The original adversary method [3] has been ∗ Supported by University of Latvia grant ZP01-100, a Marie Curie International Reintegration Grant (IRG), and ESF project
1DP/1.1.1.2.0/09/APIA/VIAA/044. Most of this work was done at the University of Waterloo and supported by NSERC, CIFAR, ARO, MITACS, and an IQC University Professorship.
2010 Andris Ambainis Licensed under a Creative Commons Attribution License
DOI: 10.4086/toc.2010.v006a001
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ˇ generalized in several different ways [4, 18, 8]. Spalek and Szegedy [23] then showed that all the generalizations are equivalent and, for certain problems, cannot improve the best known lower bounds. adversary methods of [4, 18, 8] cannot prove a lower bound For example, it was shown in [23, 24] that the p for a total Boolean function that exceeds O( C0 ( f )C1 ( f )), where C0 ( f ) and C1 ( f ) are the certificate complexities of f on 0-inputs and 1-inputs, resp. This implies that the adversary methods of [4, 18, 8] cannot prove a tight lower bound for element distinctness or improve the best known lower bound for triangle finding. (The complexity of √ element distinctness is Θ(N 2/3 ) [2, 5] but the adversary method cannot prove a bound better than Ω( N). For triangle finding [19], the best known lower bound is Ω(N) and it is known that it cannot be improved using the adversary method. It is, however, possible that the bound is not tight, because the best algorithm uses O(N 1.3 ) queries.) In this paper we describe a new version of the quantum adversary method which may not be subject to those limitations. We then use the new method to prove a strong direct product theorem for the K-fold search problem. In the K-fold search problem, a black box contains x1 , . . . , xN such that |{i : xi = 1}| √ = K and we NK) queries. have to find all the K values of i such that xi = 1. This problem can be solved with O( √ It can be easily shown, using any of the previously known √ methods, that Ω( NK) quantum queries are required. A more difficult problem is to show that Ω( NK) queries are required, even if the algorithm only has to be correct with an exponentially small probability c−K , c > 1. This result is known as the strong direct product theorem for k-fold search. Besides being interesting on its own, the strong direct product theorem is useful for proving time-space tradeoffs for quantum sorting [16] and lower bounds on quantum computations that use advice [1]. ˇ The strong direct product theorem for quantum search was first shown by Klauck, Spalek, and de Wolf [16], using the polynomials method. No proof using the adversary method has been known and, as we show in Section 3, the previously known adversary methods are insufficient to prove a strong direct product theorem for K-fold search. Recent developments opments occurred.
After the author completed the research presented in this paper, several devel-
ˇ 1. Together with Spalek and de Wolf [7], we have used the methods from this paper to prove a direct product theorem for t-threshold functions. This implies time-space tradeoffs for the problem of deciding systems of linear inequalities. This paper and [7] were merged for publication at STOC’06 [6]. We have decided to publish the journal versions separately. The relation between the two papers is as follows. A direct product theorem for k-threshold functions implies a direct product theorem for search. Thus, the result of [7] implies the result in this paper. The proof of our result is, however, substantially simpler than the proof of the more general result in [7]. Because of that, we think that our proof continues to be of interest even though the result in this paper has been generalized by [7]. ˇ 2. Spalek [22] generalized the results of the current paper and [7], obtaining a multiplicative adversary method. This is a general framework for proving lower bounds which includes the results of the current paper and [7] as particular cases. T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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ˇ 3. Høyer, Lee, and Spalek [14] generalized the usual adversary method in a different way, by extending the usual weighted adversary method of [4] to negative weights. Reichardt [21] has shown that this method is optimal: if Adv± ( f ) is the best adversary lower bound that can be proven for a function f , then f can be evaluated using log Adv± ( f ) O Adv ( f ) log log Adv± ( f )
±
queries. Although the negative weight adversary method is guaranteed to provide nearly optimal results, the other lower bound methods also remain interesting because particular lower bounds may follow more easily using different tools.
2
Preliminaries
We consider the following problem. Search for K marked elements, SEARCHK (N): Given a black box containing x1 , . . . , xN ∈ {0, 1} such that xi = 1 for exactly K values i ∈ {1, 2, . . . , N}, find all K values i1 , . . . , iK satisfying xi j = 1. This problem can be viewed as computing an NK -valued function f (x1 , . . . , xN ) in the variables x1 , . . . , xN ∈ {0, 1}, with values of the function being indices for the NK sets S ⊆ [N] of size K, in some canonical ordering of those sets. We study this problem in the quantum query model. (For a survey on the quantum query model, see [12]). In this model, the input bits can be accessed by queries to an oracle X and the complexity of f is the number of quantum queries needed to compute f . A quantum computation with T queries is just a sequence of unitary transformations U0 → O → U1 → O → · · · → UT −1 → O → UT . The U j can be arbitrary unitary transformations that do not depend on the input bits x1 , . . . , xN . The transformations O are query (oracle) transformations which depend on x1 , . . . , xN . To define O, we represent basis states as |i, zi where i consists of dlog(N + 1)e bits and z consists of all other bits. Then, Ox maps |0, zi to itself and |i, zi to (−1)xi |i, zi for i ∈ {1, ..., N} (i. e., we change phase depending on xi , unless i = 0 in which case we do nothing). The computation starts with a state |0i. Then we apply U0 , Ox , . . . , Ox ,UT and measure the final N state. The result of the computation is the string of dlog2 K e rightmost bits of the state obtained by the measurement, which is interpreted as a description of one of the NK subsets S ⊆ {1, . . . , N}, |S| = K.
3
Overview of the adversary method
We describe the adversary method of [3]. Let X be a subset of the set of possible inputs {0, 1}N . We run the algorithm on a superposition of inputs in X. More formally, let HA be the workspace of the algorithm. We consider a bipartite system T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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H = HA ⊗ HI where HI is an “input subspace” spanned by basis vectors |xi corresponding to inputs x ∈ X. Let UT OUT −1 · · ·U0 be the sequence of unitary transformations on HA performed by the algorithm A, where U0 , . . . ,UT denote the transformations that do not depend on the input and O stands for the query transformations. We transform this sequence into a sequence of unitary transformations on H. A unitary transformation Ui on HA corresponds to the transformation Ui0 = Ui ⊗ I on the whole H. The query transformation O corresponds to a transformation O0 that is equal to Ox on the subspace HA ⊗ |xi. We perform the sequence of transformations UT0 O0 UT0 −1 · · ·U00 on the starting state |ψstart i = |0i ⊗ ∑ αx |xi . x∈S
Then, the final state is |ψend i =
∑ αx |ψx i ⊗ |xi ,
x∈X
where |ψx i is the final state of A = UT OUT −1 . . .U0 on input x. This follows from the fact that the restrictions of UT0 , O0 ,UT0 −1 , . . . ,U00 to HA ⊗ |xi are UT , Ox ,UT −1 , . . . ,U0 and these are exactly the transformations of the algorithm A on input x. Let ρend be the reduced density matrix of the HI register of the state |ψend i. The adversary method of [3, 4] works by showing the following two statements. • Let x ∈ X and y ∈ X be such that f (x) 6= f (y) (where f is the function that is being computed). If the algorithm outputs the correct answer with probability 1 − ε on both x and y then p |(ρend )x,y | ≤ 2 ε(1 − ε)|αx ||αy | . • For any algorithm that uses T queries, there exist inputs x, y ∈ S such that f (x) 6= f (y) and p (ρend )x,y > 2 ε(1 − ε)|αx ||αy | . These two statements together imply that any algorithm computing f must use more than T queries. An equivalent approach [15, 4] is to consider the inner productsphψx |ψy i between the final states |ψx i and |ψy i p of the algorithm on inputs x and y. Then, |(ρend )x,y | ≤ 2 ε(1 − ε)|αx ||αy | is equivalent to |hψx |ψy i| ≤ 2 ε(1 − ε). As a result, both of the above statements can be described in terms of inner products hψx |ψy i, without explicitly introducing the register HI . The first statement says that, for the algorithm to succeed on inputs x, y such that f (x) 6= f (y), the states |ψxp i and |ψy i must be sufficiently far apart from one another (so that the inner product |hψx |ψy i| is at most 2 ε(1 − ε)). The second statement says that this is impossible if the algorithm only uses T queries. This approach breaks down if we consider computing a function f with success probability p < 1/2. ( f has to have more than 2 values for this task to be nontrivial.) Then, |ψx i and |ψy i could be the same and the algorithm may still succeed on both inputs, if it outputs x with probability 1/2 and y with probability 1/2. In the case of strong direct product theorems, the situation is even more difficult. Since the algorithm only has to be correct with probability c−K , the algorithm could have almost the same final state on cK different inputs and still “succeed” on every one of them. T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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In this paper, we present a new method that does not suffer from this problem. Our method, described in the next section, uses the idea of augmenting the algorithm with an input register HI , together with two new ingredients: 1. Symmetrization. We symmetrize the algorithm by applying a random permutation π ∈ SN to the input x1 , . . . , xN . 2. Eigenspace analysis. We study the eigenspaces of ρstart , ρend and density matrices describing the state of HI at intermediate steps and use them to bound the progress of the algorithm. The eigenspace analysis is the main new technique. Symmetrization is necessary to simplify the structure of the eigenspaces, to make the eigenspace analysis possible.
4
Our result
Theorem 4.1. There exist ε and c satisfying ε > 0, 0 < c 0, S j = T j ∩ (T j−1 )⊥ , with T j being the space spanned by all states |ψi1 ,...,i j i = q
1 N K− j
∑
|x1 , . . . , xN i .
x1 ,...,xN : x1 +···+xN =K, xi1 =···=xi j =1
Let τ j be the completely mixed state over the subspace S j . Lemma 4.4. There exist pt,0 ≥ 0, . . . , pt,K ≥ 0 such that ρt = ∑Kj=0 pt, j τ j . Proof. By Lemma 4.3, S0 , . . . , SK are the eigenspaces of ρt . Therefore, ρt is a linear combination of the projectors to S0 , . . . , SK . Since τ j is a multiple of the projector to S j , we have K
ρt =
∑ pt, j τ j .
j=0
Since ρt is a density matrix, it must be positive semidefinite. This means that pt,0 ≥ 0, . . . , pt,K ≥ 0. Informally, we can interpret pt, j as the probability that, after t queries, the algorithm HA knows the locations for t out of the K items j with x j = 1. Let qt, j = pt, j + pt, j+1 + · · · + pt,K . The theorem now follows from the following lemmas. Lemma 4.5. p0,0 = 1, p0, j = 0 for j > 0. T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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Proof. In the starting state, HI contains the state |ϕ0 i, independent of HA and HP . Therefore, tracing out HA ⊗ HP leaves the state ρ0 = |ψ0 ihψ0 |. √
Lemma 4.6. For all j ∈ {1, . . . , K} and all t we have qt+1, j+1 ≤ qt, j+1 + 4√NK qt, j . Proof. In Section 5. Lemma 4.7. qt, j ≤
t j
4√K j √ N
.
Proof. By induction on t. The base case, t = 0 follows immediately from p0,0 = 1 and p0,1 = · · · = p0,K = 0 . For the induction step, we have √ √ ! j−1 √ !j √ 4 K t 4 K t 4 K 4 K √ √ + √ qt+1, j ≤ qt, j + √ qt, j−1 ≤ j N N N j−1 N √ !j √ !j t t +1 4 K t 4 K √ √ = , = + j j j−1 N N where the first inequality follows from Lemma 4.6 and the second inequality follows from the inductive assumption. √ Lemma 4.8. If t ≤ 0.03 NK then qt, j < 0.66 j for all j > K/2. Proof. By the well known inequality tj < (et/ j) j for j ≥ 1, we have t qt, j ≤ j
√ !j 4 K √ ≤ N
!j √ 4 Ket √ . Nj
√ Let j > K/2 and t ≤ 0.03 NK. Then √ √ √ 4 Ket 0.12e K NK √ √ ≤ < 0.66 , Nj NK/2 implying the lemma. Lemma 4.9. The success probability of A is at most N p K/2 + 4 qt,K/2+1 . N K
Proof. In Section 6. T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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To complete the proof, given the two Lemmas, we choose a constant c > set ε = 0.03. Then, by Lemma 4.9, the success probability of A is at most N K/2 N K
√ 4 0.66 = 0.90133 . . . and
+4
p 0.66K/2 .
The first term is equal to N K/2 N K
K!(N − K)! K(K − 1) . . . (K/2 + 1) = (K/2)!(N − K/2)! (N − K/2)(N − K/2 − 1) . . . (N − K + 1) K/2 K/2 K N/2 K/2 2 ≤ ≤ = = o(cK ) , N − K/2 3N/4 3
=
√ where the last two inequalities follow from K < N/2. The second term is 4 0.66K/2 = o(cK ).
5
Proof of Lemma 4.6
Let |ψt i be the state before (t + 1)st query. We decompose |ψt i as ∑Ni=0 ai |ψt,i i, where |ψt,i i is the part in which the query register contains |ii. Because of symmetrization, we must have |a1 | = |a2 | = · · · = |aN |. Also, we can choose the relative phases so that a1 , · · · , aN are all positive reals and, thus, a1 = a2 = · · · = aN . Let ρt,i = |ψt,i ihψt,i |. Then N
ρt = ∑ a2i ρt,i .
(5.1)
i=0
Claim 5.1. Let i ∈ {1, . . . , N}. The entry (ρt,i )x,y only depends on xi , yi , and the cardinality of {` : ` 6= i, x` = y` = 1}. Proof. The main idea is similar to Lemma 4.2. This time, we trace out all registers, except for HI and the query register. This gives us a density matrix ρ whose rows and columns are indexed by pairs (x, j) where x is an input and j ∈ {1, . . . , N}. Let x ∈ {0, 1}N , y ∈ {0, 1}N , j ∈ [N], k ∈ [N] and x0 ∈ {0, 1}N , y0 ∈ {0, 1}N , j0 ∈ [N], k0 ∈ [N] be such that there is a permutation π ∈ SN with x0 = π(x), y0 = π(y), i0 = π(i) and j0 = π( j). By an argument similar to the proof of Lemma 4.2, we have ρ(x,i),(y, j) = ρ(x0 ,i0 ),(y0 , j0 ) .
(5.2)
We now observe that ρt,i is the submatrix of ρ consisting of rows and columns indexed by pairs (x, i), with all possible x ∈ {0, 1}N , |{i : xi = 1}| = K. If we have inputs x, y, x0 , y0 with xi = xi0 , yi = y0i and |{` : ` 6= i, x` = y` = 1}| = |{` : ` 6= i, x`0 = y0` = 1}|, then we can construct a permutation π with π(i) = i, π(x) = x0 and π(y) = y0 . Equation (5.2) then implies (ρt,i )x,y = (ρt,i )x0 ,y0 . T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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We now describe the eigenspaces of matrices ρt,i . The proofs of some claims are postponed to Section 7. We define the following subspaces of states. Let i ∈ [N] and j ∈ {0, 1, . . . , K}. We define T ji,0 to be the subspace spanned by all states |ψii,0 i= q 1 ,...,i j
1 N− j−1 K− j
∑
|x1 . . . xN i ,
x:|x|=K xi1 =···=xi j =1,xi =0
with (i1 , . . . , i j ) ranging over all tuples of j distinct elements of [N] − {i}. Similarly, we define T ji,1 to be the subspace spanned by all states |ψii,1 i= q 1 ,...,i j
1 N− j−1 K− j−1
∑
|x1 . . . xN i .
x:|x|=K xi1 =···=xi j =1,xi =1
i,0 i,0 ⊥ i,1 i,1 i,1 ⊥ i,0 i,1 Let Si,0 j = T j ∩ (T j−1 ) and S j = T j ∩ (T j−1 ) . Equivalently, we can define S j and S j as the i, respectively, with i and |ψ˜ ii,1 subspaces spanned by the states |ψ˜ ii,0 1 ,...,i j 1 ,...,i j
|ψ˜ ii,` i = P(T i,` 1 ,...,i j
⊥ j−1 )
i. |ψii,` 1 ,...,i j
i Let Sα,β , j be the subspace spanned by all states
α
i |ψ˜ ii,0 1 ,...,i j ik k|ψ˜ ii,0 1 ,...,i j
+β
|ψ˜ ii,1 i 1 ,...,i j ik k|ψ˜ ii,1 1 ,...,i j
.
(5.3)
where (i1 , . . . , i j ) again ranges over all tuples of j distinct elements of [N] − {i}. i Claim 5.2. Every eigenspace of ρt,i is a direct sum of subspaces Sα,β , j for some α, β , j.
Proof. In Section 7. i i Let τα,β , j be the completely mixed state over Sα,β , j . Similarly to Lemma 4.4, we can write ρt,i as
ρt,i =
i piα,β , j τα,β ,j ,
∑
(5.4)
(α,β , j)∈At,i
where (α, β , j) range over some finite set At,i . (This set is finite because the HI register holding |x1 . . . xN i is finite dimensional and, therefore, decomposes into a direct sum of finitely many eigenspaces.) For every pair (α, β , j) ∈ At,i , we normalize α, β by multiplying them by the same constant so that α 2 +β 2 = 1. Querying xi transforms this state to 0 ρt,i =
∑
i piα,β , j τα,−β ,j ,
(α,β , j)∈At,i
because |ψ˜ ii,` i is a superposition of |xi with xi = ` and, therefore, a query leaves |ψ˜ ii,0 i unchanged 1 ,...,i j 1 ,...,i j 0 = ρ , because, if the query register contains |0i, and flips a phase on |ψ˜ ii,1 i. If i = 0, we have ρt,0 t,0 1 ,...,i j the query maps any state to itself, thus leaving ρt,0 unchanged.
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Claim 5.3. Let α0 =
q
N−K ˜ i,0
N− j ψi1 ,...,i j
and β0 =
q
K− j i,1
˜ N− j ψi1 ,...,i j .
(i) Sαi 0 ,β0 , j ⊆ S j ; (ii) Sβi 0 ,−α0 , j ⊆ S j+1 . Proof. In Section 7. i Corollary 5.4. For any α, β we have Sα,β , j ⊆ S j ⊕ S j+1 . i,1 ˜ ii,0 Proof. We have Sα,β , j ⊆ Si,0 i j ⊕ S j , since Sα,β , j is spanned by linear combinations of states |ψ 1 ,...,i j
˜ ii,1 (which belong to Si,0 i (which belong to Si,1 j ) and states |ψ j ). As shown in the proof of Claim 5.3 1 ,...,i j above, i,1 Si,0 j ⊕ S j ⊆ Sα0 ,β0 , j ⊕ S−β0 ,α0 , j ⊆ S j ⊕ S j+1 . i The next claim quantifies the overlap between Sα,β , j and S j+1 .
Claim 5.5. i Tr PS j+1 τα,β ,j =
|αβ0 − β α0 |2 . α02 + β02
Proof. In Section 7. To be able to use this bound, we also need to bound α0 and β0 . q 4(K− j) Claim 5.6. √ β20 2 ≤ N+3K−4 j. α0 +β0
Proof. In Section 7. We can now complete the proof of Lemma 4.6. By projecting both sides of ρt = ∑i pt,i τi to (T j )⊥ = S j+1 ⊕ · · · ⊕ SK and taking trace, we get K
Tr P(Tj )⊥ ρt =
∑
K
pt, j0 Tr P(T j )⊥ τ j0 =
j0 =0
∑
pt, j0 = qt, j ,
(5.5)
j0 = j+1
with the second equality following because the states τ j0 are uniform mixtures over subspaces S j0 and S0 , . . . , S j are contained in T j while S j+1 , . . . , SK are contained in (T j )⊥ . Because of equations (5.1) and (5.4), this means that N
qt, j+1 = a20 Tr P(Tj )⊥ ρt,0 + ∑ a2i i=1
∑
(α,β , j0 )∈At,i
i piα,β , j0 Tr P(Tj )⊥ τα,β , j0 .
(5.6)
Decomposing the state after the query in a similar way, we get N
0 qt+1, j+1 = a20 Tr P(Tj )⊥ ρt,0 + ∑ a2i i=1
∑
(α,β , j0 )∈At,i
i piα,β , j0 Tr P(Tj )⊥ τα,−β , j0 .
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qt+1, j+1 − qt, j+1 = ∑ a2i i=1
∑
(α,β , j0 )∈At,i
i i piα,β , j0 Tr P(Tj )⊥ (τα,−β , j0 − τα,β , j0 ) .
(5.7)
We now claim that all the terms in this sum with j0 6= j are 0. For j0 < j, Sα,β , j0 ⊆ T j0 +1 ⊆ T j , implying ⊥ i i 0 0 0 that Tr P(Tj )⊥ τα,β , j0 = 0 and, similarly, Tr P(T j )⊥ τα,−β , j0 = 0. For j > j, Sα,β , j0 ⊆ S j ⊕ S j +1 ⊆ (T j ) , implying that i i Tr P(Tj )⊥ τα,−β Tr P(Tj )⊥ τα,β , j0 = 1 , , j0 = 1, and the difference of the two is 0. By removing those terms from (5.7), we get N
qt+1, j+1 − qt, j+1 = ∑ a2i i=1
∑
i i piα,β , j Tr P(Tj )⊥ (τα,−β , j − τα,β , j ) .
(5.8)
(α,β , j)∈At,i
We have i i i i Tr P(Tj )⊥ (τα,−β , j − τα,β , j ) = Tr PS j+1 (τα,−β , j − τα,β , j ) =
|αβ0 + β α0 |2 |αβ0 − β α0 |2 − , α02 + β02 α02 + β02
where the first equality follows from Corollary 5.4, S j ⊆ T j , and S j+1 ⊆ (T j )⊥ , and the second equality follows from Claim 5.5. This is at most s r β0 4(K − j) 4K α0 β0 α0 |αβ α0 β0 | q ≤2 2 = 2q ≤2 ≤2 , 4 2 2 2 N + 3K − 4 j N α0 + β0 α0 + β0 α2 + β 2 α2 + β 2 0
0
0
0
where the first inequality follows from |αβ | ≤
|α|2 + |β |2 1 = 2 2
and the second inequality follows from Claim 5.6 and √ α20 2 ≤ 1. Together with equation (5.7), this α0 +β0 means √ N 4 K qt+1, j+1 − qt, j+1 ≤ √ ∑ a2i (5.9) ∑ pi . N i=1 (α,β , j)∈At,i α,β , j Similarly to equation (5.5) we have pt, j+1 + pt, j = Tr P(S j ⊕S j+1 ) ρt . We can then express the right hand side, similarly to equation (5.6), as a sum of terms p0j0 Tr P(S j ⊕S j+1 ) τ j0 i i and piα,β , j0 Tr P(S j ⊕S j+1 ) τα,β , j0 . Since Sα,β , j ⊆ S j ⊕ S j+1 (by Corollary 5.4), we have i Tr P(S j ⊕S j+1 ) τα,β ,j = 1.
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This means that N
pt, j+1 + pt, j ≥ ∑ a2i i=1
∑
piα,β , j .
(α,β , j)∈At,i
Together with equation (5.9), this implies √ √ 4 K 4 K qt+1, j+1 − qt, j+1 ≤ √ (pt, j + pt, j+1 ) ≤ √ N N
6
K
∑ pt, j
j0 = j
0
√ 4 K = √ qt, j . N
Proof of Lemma 4.9
We start with the case when pT,K/2+1 = · · · = pT,K = 0. Lemma 6.1. If pT,K/2+1 = · · · = pT,K = 0, the success probability of A is at most
N (K/2 ) . (NK )
N Proof. Let |ψi be the final state. The state of register HI lies in TK/2 , which is an K/2 -dimensional N space. Therefore, there is a Schmidt decomposition for |ψi with at most K/2 terms. This means that N the state of HA lies in an K/2 -dimensional subspace of HA ⊗ HS . We express the final state as 1 |ψi = ∑ q |ψx i|xi . N x:|x|=K
K
We can think of |ψx i as a quantum encoding for x and the final measurement as a decoding procedure that takes |ψx i and produces a guess for x. The probability that algorithm A succeeds is then equal to the average success probability of the encoding. We now use Theorem 6.2. [20] For any encoding |ψx i of M classical values by quantum states in d dimensions, the probability of success is at most d/M. N N In our case, M = NK and d = K/2 because the states |ψi all lie in a K/2 -dimensional subspace of HA ⊗ HS . Therefore, Theorem 6.2 √ implies Lemma√6.1. We decompose the state |ψT i as 1 − δ |ψT0 i + δ |ψT00 i where |ψT0 i is in the subspace HA ⊗ TK/2 and |ψT00 i is in HA ⊗ (TK/2 )⊥ . We have K
δ=
∑
pT, j .
j=K/2+1
The success probability of A is the probability that, if we measure both the register of HA containing the result of the computation and HI then we get i1 , . . . , iK and x1 , . . . , xN such that xi1 = · · · = xiK = 1. T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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Consider the probability of getting i1 , . . . , iK and x1 , . . . , xN such that xi1 = · · · = xiK = 1, when meaN / NK . We have suring |ψT0 i (instead of |ψT i). By Lemma 6.1, this probability is at most K/2 kψT − ψT0 k ≤ (1 −
p p √ √ √ 1 − δ 2 )kψT0 k + δ kψT00 k = (1 − 1 − δ 2 ) + δ ≤ 2 δ .
We now apply Lemma 6.3. [10] For any states |ψi and |ψ 0 i and any measurement, the total variation distance between the probability distributions obtained by applying M to |ψi and to |ψ 0 i is at most 2kψ − ψ 0 k. , . . . , iK and x1 , . . . , xN such that xi1 = · · · = xiK = 1, By Lemma 6.3, the probabilities of getting i1√ √ when measuring |ψT i and |ψT0 i differ by at most 4 δ = 4 qT,K/2+1 . Therefore, the success probability of A is at most N p K/2 + 4 qT,K/2+1 . N K
7
Structure of the eigenspaces of ρt,i
In this section, we prove Claims 5.2, 5.3, 5.5, and 5.6 describing the structure of the eigenspaces of ρt,i . Before giving the proofs, we briefly summarize the results in this section. 1. Let ρt,i be a matrix whose rows and columns are indexed by (x1 , . . . , xN ), xi ∈ {0, 1} with x1 + . . . + xN = K. By Claim 5.2, if ρt,i is symmetric w. r. t. any permutation π ∈ SN that fixes i, then i its eigenspaces are of the form Sα,β , j , which is the subspace spanned by the states of the form α
i |ψ˜ ii,0 1 ,...,i j ik k|ψ˜ ii,0 1 ,...,i j
+β
|ψ˜ ii,1 i 1 ,...,i j ik k|ψ˜ ii,1 1 ,...,i j
i and |ψ˜ ii,1 i are the states defined at the beginning of Section 5. where |ψ˜ ii,0 1 ,...,i j 1 ,...,i j i 2. We then relate the eigenspaces Sα,β , j to the subspaces S0 , . . . , SK defined in Section 4.2. In Claim 5.3 we show that, for certain α0 and β0 , we have
• Sαi 0 ,β0 , j ⊆ S j ; i • S−β ⊆ S j+1 ; 0 ,α0 , j
A corollary of this is that, for any α, β , we have i i i Sα,β , j ⊆ Sα0 ,β0 , j ⊕ S−β0 ,α0 , j ⊆ S j ⊕ S j+1 .
This is shown in Corollary 5.4. i 3. Next, in Claim 5.5, we quantify how close Sα,β , j is to S j and S j+1 . The result is an expression for i i i Tr PS j+1 τα,β , j (where τα,β , j is the maximally mixed state over Sα,β , j ) that involves α, β and α0 , β0 . To use that expression, we also need a bound on the ratio of α0 and β0 (Claim 5.6).
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Proof of Claim 5.2. We rearrange the rows and the columns of ρt,i so that all rows and columns corresponding to |x1 . . . xN i with xi = 0 are before the rows and the columns corresponding to |x1 . . . xN i with xi = 1. We then express ρt,i as A B ρt,i = , C D N−1 N−1 N−1 × square matrix indexed by |x . . . x i with x = 0, D is an × where A is an N−1 1 N i K−1 K−1 K K square matrix indexed by |x1 . . . xN i with xi = 1, and B and C are rectangular matrices with rows (columns) indexed by |x1 . . . xN i with xi = 0 and columns (rows) indexed by |x1 . . . xN i with xi = 1. We claim that ρt,i |ψ˜ ii,0 i = a11 |ψ˜ ii,0 i + a12 |ψ˜ ii,1 i and 1 ,...,i j 1 ,...,i j 1 ,...,i j ρt,i |ψ˜ ii,1 i = a21 |ψ˜ ii,0 i + a22 |ψ˜ ii,1 i, 1 ,...,i j 1 ,...,i j 1 ,...,i j
(7.1)
where a11 , a12 , a21 , a22 are independent of i1 , . . . , i j . To prove that, we first note that A and D are matrices where Axy and Dxy only depend on |{t : xt = yt }|. Therefore Lemma 4.3 applies. This means that Si,0 j and Si,1 j are eigenspaces for A and D, respectively, and i = a11 |ψ˜ ii,0 i and A|ψ˜ ii,0 1 ,...,i j 1 ,...,i j D|ψ˜ ii,1 i = a22 |ψ˜ ii,1 i, 1 ,...,i j 1 ,...,i j i,1 where a11 and a22 are the eigenvalues of A and D for the eigenspaces Si,0 j and S j . It remains to prove that
i = a12 |ψ˜ ii,1 i and B|ψ˜ ii,0 1 ,...,i j 1 ,...,i j
(7.2)
C|ψ˜ ii,1 i = a21 |ψ˜ ii,0 i. 1 ,...,i j 1 ,...,i j
(7.3)
Let M be a rectangular matrix, with entries indexed by x, y, with |x| = |y| = K and xi = 1 and yi = 0. The entries of M are Mxy = 1 if x and y differ in two places, with xi = 1, yi = 0 and xl = 0, yl = 1 for some l 6= i and Mxy = 0 otherwise. We claim M|ψ˜ ii,0 i = c|ψ˜ ii,1 i 1 ,...,i j 1 ,...,i j
(7.4)
for some c that may depend on N, k, and j but not on i1 , . . . , i j . To prove that, we need to prove two things. First, M|ψii,0 i = c|ψii,1 i. 1 ,...,i j 1 ,...,i j T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
(7.5) 16
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This follows by M|ψii,0 i= q 1 ,...,i j
=q
=q
1
∑
M|xi
∑
∑
(N − K) N− j−1
∑
N− j−1 x:xi1 =···=xi j =1, k− j xi =0
1
N− j−1 x:xi1 =···=xi j =1 `:x` =1 K− j xi =0
1
K− j
|x1 . . . x`−1 0x`+1 . . . xi−1 1xi+1 . . . xN i
|yi =
(K − j)(N − K)|ψii,1 i. 1 ,...,i j
p
y:yi1 =···=yi j =1 yi =1
Second, M(T ji,0 ) ⊆ T ji,1 and M(T ji,0 )⊥ ⊆ (T ji,1 )⊥ . The first statement immediately follows from equai and |ψii,1 i, respection (7.5), because the subspaces T ji,0 and T ji,1 are spanned by the states |ψii,0 1 ,...,i j 1 ,...,i j tively. To prove the second statement, let |ψi ∈ (T ji,0 )⊥ and |ψi = ∑x ax |xi. We would like to prove M|ψi ∈ (T ji,1 )⊥ . This is equivalent to hψii,1 |M|ψi = 0 for all i1 , . . . , i j . We have 1 ,...,i j |M|ψi = q hψii,1 1 ,··· ,i j =q
=q
1
∑
N− j−1 y:yi1 =···=yi j =1 K− j−1
1
hy|M|ψi
∑
∑
N− j−1 x:xi1 =···=xi j =1, l:xl =1, K− j−1 l ∈{i / 1 ,...,i j } xi =0
1 N− j−1 K− j−1
(K − j)
∑
ax
ax = 0 .
x:xi1 =···=xi j =1
|, the second equality follows by writing out M. The The first equality follows by writing out hψii,1 1 ,...,i j third equality follows because, for every x with |x| = K and xi1 = · · · = xi j = 1, there are K − j more l ∈ [N] |ψi satisfying xl = 1. The fourth equality follows because ∑x:xi =···=xi j =1 ax is a constant times hψii,0 1 ,...,i j 1
and hψii,0 |ψi = 0, because |ψi ∈ (T ji,0 )⊥ . 1 ,...,i j × N−1 matrix, with (BM)x,y only depending on |{` : x` = y` = 1}|. Furthermore, BM is an N−1 K K i,1 Therefore, S j is an eigenspace of BM and, since |ψ˜ ii,1 i ∈ Si,1 j , we have 1 ,...,i j BM|ψ˜ ii,1 i = λ |ψ˜ ii,1 i 1 ,...,i j 1 ,...,i j for an eigenvalue λ independent of i1 , . . . , i j . Together with equation (7.4), this implies equation (7.2) with a12 = λ /c. Equation (7.3) follows by proving i = c|ψ˜ ii,0 i and CM T |ψ˜ ii,0 i = λ |ψ˜ ii,0 i, M T |ψ˜ ii,1 1 ,...,i j 1 ,...,i j 1 ,...,i j 1 ,...,i j T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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in a similar way. We now diagonalize the matrix 0
M = It has two eigenvectors:
α1 β1
and
α2 β2
a11 a12 a21 a22
.
. Equation (7.1) implies that, for any i1 , . . . , i j ,
α1 |ψ˜ ii,0 i + β1 |ψ˜ ii,1 i 1 ,...,i j 1 ,...,i j is an eigenvector of ρt,i with the same eigenvalue λ . Therefore, Sαi 1 ,β1 , j is an eigenspace of ρt,i . Similarly, Sαi 2 ,β2 , j is an eigenspace of ρt,i . Vectors α1 |ψ˜ ii,0 i + β1 |ψ˜ ii,1 i and α2 |ψ˜ ii,0 i + β2 |ψ˜ ii,1 i together 1 ,...,i j 1 ,...,i j 1 ,...,i j 1 ,...,i j i and |ψ˜ ii,1 i. Since the vectors |ψ˜ ii,` i span Si,` span the same space as vectors |ψ˜ ii,0 j , this means 1 ,...,i j 1 ,...,i j 1 ,...,i j that i,1 Si,0 j ⊕ S j ⊆ Sα1 ,β1 ,i ⊕ Sα2 ,β2 ,i . Therefore, repeating this argument for every j gives a collection of eigenspaces that span the entire state i space for HI . This means that any eigenspace of ρt,i is a direct sum of some of eigenspaces Sα,β , j. Proof of Claim 5.3. For part (i), consider the states |ψi1 ,...,i j i spanning T j . We have s |ψi1 ,...,i j i =
N − k i,0 i+ |ψ N − j i1 ,...,i j
s
K − j i,1 i |ψ N − j i1 ,...,i j
(7.6)
because an (N − K)/(N − j) fraction of the states |x1 . . . xN i with |x| = K and xi1 = · · · = xi j = 1 have i,0 i,1 ⊥ ⊕ T j−1 ) are xi = 0 and the rest have xi = 1. The projection of these states to (T j−1 s
N − K i,0 |ψ˜ i+ N − j i1 ,...,i j
s
K − j i,1 |ψ˜ i N − j i1 ,...,i j
which, by equation (5.3) are exactly the states spanning Sαi 0 ,β0 , j . Furthermore, we claim that i,0 i,1 T j−1 ⊆ T j−1 ⊕ T j−1 ⊆ Tj .
(7.7)
The first containment is true because T j−1 is spanned by the states |ψi1 ,...,i j−1 i which either belong to i,1 i,1 T j−2 ⊆ T j−1 (if one of i1 , . . . , i j−1 is equal to i) or are a linear combination of states |ψii,0 i and 1 ,...,i j−1 i,0 i,1 i which belong to T j−1 and T j−1 . The second containment follows because the states |ψii,1 i |ψii,1 1 ,...,i j−1 1 ,...,i j−1 i,1 spanning T j−1 are the same as the states |ψi,i1 ,...,i j−1 i which belong to T j and the states |ψii,0 i spanning 1 ,...,i j−1 i,0 T j−1 can be expressed as linear combinations of |ψi1 ,...,i j−1 i and |ψi,i1 ,...,i j−1 i which both belong to T j . The first part of (7.7) now implies i,0 i,1 ⊥ Sαi 0 ,β0 , j ⊆ (T j−1 ⊕ T j−1 ) ⊆ (T j−1 )⊥ .
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We also have Sαi 0 ,β0 , j ⊆ T j , because, Sαi 0 ,β0 , j is spanned by the states P(T i,0 ⊕T i,1 )⊥ |ψi1 ,...,i j i = |ψi1 ,...,i j i − PT i,0 ⊕T i,1 |ψi1 ,...,i j i j−1
j−1
j−1
j−1
and |ψi1 ,...,i j i belongs to T j by the definition of T j and PT i,0 ⊕T i,1 |ψi1 ,...,i j i belongs to T j because of the j−1
j−1
second part of (7.7). Therefore, Sαi 0 ,β0 , j ⊆ T j ∩ (T j−1 )⊥ = S j . For the part (ii), we have i,1 i,0 i,1 Sαi 0 ,β0 , j ⊆ Si,0 j ⊕ S j ⊆ T j ⊕ T j ⊆ T j+1 ,
where the first containment is true because Sαi 0 ,β0 , j is spanned by linear combinations of vectors |ψ˜ ii,0 i 1 ,...,i j ˜ ii,1 (which belong to Si,0 i (which belong to Si,1 j ) and vectors |ψ j ) and the last containment is true because 1 ,...,i j of the second part of equation (7.7). Let i |ψ˜ ii,1 i |ψ˜ ii,0 1 ,...,i j 1 ,...,i j − α (7.8) |ψi = β0 0 ik ik k|ψ˜ ii,0 k|ψ˜ ii,1 1 ,...,i j 1 ,...,i j be one of the vectors spanning Sβi 0 ,−α0 , j . To prove that |ψi is in S j+1 = T j+1 − T j , it remains to prove that |ψi is orthogonal to T j . This is equivalent to proving that |ψi is orthogonal to every vector |ψi01 ,...,i0j i spanning T j . i,0 i,1 ⊥ Case 1 {i01 , . . . , i0j } = {i1 , . . . , i j }: Since |ψi belongs to (T j−1 ⊕ T j−1 ) , it suffices to prove that |ψi i,0 i,1 ⊥ ⊕ T j−1 ) which, by the discussion after the equais orthogonal to the projection of |ψi1 ,...,i j i to (T j−1 tion (7.6), is equal to i |ψ˜ ii,1 i |ψ˜ ii,0 1 ,...,i j 1 ,...,i j + β . (7.9) α0 0 i,0 i,1 k|ψ˜ i1 ,...,i j ik k|ψ˜ i1 ,...,i j ik
¿From equations (7.8) and (7.9) we see that the inner product of the two states is α0 β0 − β0 α0 = 0. Case 2 {i01 , . . . , i0j } 6= {i1 , . . . , i j } but one of i01 , . . . , i0j is equal to i: For simplicity, assume i = i0j . Then i,1 i |ψi01 ,...,i0j i is the same as |ψii,1 i which belongs to T j−1 . By the definition of Sα,β 0 ,...,i0 , j , the vector |ψi 1
j−1
i,0 i,1 ⊥ belongs to (T j−1 ⊕ T j−1 ) and is therefore orthogonal to |ψii,1 i. 0 ,...,i0 1
j−1
Case 3 {i01 , . . . , i0j } 6= {i1 , . . . , i j } and none of i01 , . . . , i0j is equal to i: One of i01 , . . . , i0j must be not in {i1 , . . . , i j }. For simplicity, assume it is i0j . We have |ψi01 ,...,i0j−1 i =
∑
i0 ∈{i / 01 ,...,i0j−1 }
|ψi01 ,...,i0j−1 ,i0 i .
i,0 i,1 Also, hψi01 ,...,i0j−1 |ψi = 0, because |ψi01 ,...,i0j−1 i is in T j−1 ⊕ T j−1 . As proved in the previous case,
hψi01 ,...,i0j−1 ,i |ψi = 0 . T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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We therefore have
∑
hψi01 ,...,i0j−1 ,i0 |ψi = 0 .
(7.10)
i0 ∈{i / 01 ,...,i0j−1 ,i}
/ {i01 , . . . , i0j−1 , i}. Therefore, By symmetry, the inner product hψi01 ,...,i0j−1 ,i0 |ψi is the same for every i0 ∈ / {i01 , . . . , i0j−1 , i}. equation (7.10) means hψi01 ,...,i0j−1 ,i0 |ψi = 0 for every i0 ∈ i i Proof of Claim 5.5. τα,β , j is a mixture of states |ψi from the subspace Sα,β , j . We prove the claim by showing that, for any of those states |ψi, the squared norm of its projection to S j+1 is equal to the right i hand side of Claim 5.5. Since |ψi ∈ Sα,β , j we can write it as
|ψi =
∑
i1 ,...,i j
i + β |ψ˜ ii,1 i) ai1 ,...,i j (α|ψ˜ ii,0 1 ,...,i j 1 ,...,i j
for some ai1 ,...,i j . Let |ψ + i =
∑
i − α0 |ψ˜ ii,1 i) and ai1 ,...,i j (β0 |ψ˜ ii,0 1 ,...,i j 1 ,...,i j
∑
i + β0 |ψ˜ ii,1 i) . ai1 ,...,i j (α0 |ψ˜ ii,0 1 ,...,i j 1 ,...,i j
i1 ,...,i j −
|ψ i =
i1 ,...,i j
Then, |ψi is a linear combination of |ψ + i which belongs to Sβi 0 ,−α0 , j ⊂ S j+1 (by Claim 5.3) and |ψ − i which belongs to Sαi 0 ,β0 , j ⊆ S j . Moreover, all three states are linear combinations of |ψ 0 i, |ψ 1 i defined by i. |ψ ` i = ∑ ai1 ,...,i j |ψ˜ ii,` 1 ,...,i j i1 ,...,i j
We have |ψi = α|ψ 0 i + β |ψ 1 i ,
|ψ + i = β0 |ψ 0 i − α0 |ψ 1 i and |ψ − i = α0 |ψ 0 i + β0 |ψ 1 i .
Since |ψ + i and |ψ − i belong to subspaces S j+1 and S j which are orthogonal, we must have hψ + |ψ − i = 0. This means α0 β0 kψ 0 k2 − β0 α0 kψ 1 k2 = 0 . By dividing the equation by α0 β0 , we get kψ 0 k2 = kψ 1 k2 and kψ 0 k = kψ 1 k. Since kψk = 1, this means that 1 kψ 0 k = kψ 1 k = p = 1. 2 α +β2 Since |ψi lies in the subspace spanned by |ψ + i which belongs to S j+1 and |ψ − i which belongs to S j , the norm of the projection of |ψi to S j+1 is equal to |hψ|ψ + i|/kψ + k. By expressing |ψi and |ψ + i in terms of |ψ 0 i and |ψ 1 i, we get |hψ|ψ + i| αβ0 kψ 0 k2 − α0 β kψ 1 k2 |αβ0 − α0 β | = q = q , kψ + k β02 kψ 0 k2 + α02 kψ 0 k2 α02 + β02 proving the claim. T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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Proof of Claim 5.6. We will prove kψ˜ ii,0 k ≥ 12 kψ˜ ii,1 k, because that means 1 ,...,i j 1 ,...,i j √ √ √ √ N − K i,0 N −K 1 N − K K − j i,1 √ α0 = √ kψ˜ k≥ √ kψ˜ k= √ β0 2 K − j N − j i1 ,...,i j N − j i1 ,...,i j 2 K− j and β0 q
α02 + β02
≤q
p
4(K − j) =q . =√ N−K N + 3K − 4 j 1 + 4(K− j) 1
β0 N−K 2 2 4(K− j) β0 + β0
To prove kψ˜ ii,0 k ≥ kψ˜ ii,1 k, we calculate the vector 1 ,...,i j 1 ,...,i j |ψ˜ ii,0 i = P(T i,0 )⊥ |ψii,0 i. 1 ,...,i j 1 ,...,i j j−1
i,0 i and the subspace T j−1 Both the vector |ψii,0 are fixed by 1 ,...,i j
Uπ |xi = |xπ(1) . . . xπ(N) i i is fixed by for any permutation π that fixes i and maps {i1 , . . . , i j } to itself. This means that |ψ˜ ii,0 1 ,...,i j i only depends on any such Uπ as well. Therefore, the amplitude of |xi, |x| = K, xi = 0 in |ψ˜ ii,0 1 ,...,i j i is of the form |{i1 , . . . , i j } ∩ {t : xt = 1}|. This means |ψ˜ ii,0 1 ,...,i j j
|ψ0 i =
∑
∑ αm
m=0
|xi .
x:|x|=K,xi =0 |{i1 ,...,i j }∩{t:xt =1}|=m
To simplify the following calculations, we multiply α0 , . . . , α j by the same constant so that αj = q
1 N− j−1 K− j
.
Then, |ψ˜ ii,0 i remains a multiple of |ψ0 i but may no longer be equal to |ψ0 i. 1 ,...,i j The coefficients α0 , . . . , α j−1 should be such that the state is orthogonal to T j−1 and, in particular, i for ` ∈ {0, . . . , j − 1}. By writing out hψ0 |ψii,0 i = 0, we get orthogonal to states |ψii,0 1 ,...,i` 1 ,...,i` j
∑ αm m=`
N − j−1 K −m
j−` = 0. m−`
(7.11)
i is a uniform superposition of all |xi, |x| = K, xi = 0, xi1 = · · · = To show that, we first note that |ψii,0 1 ,...,i` xil = 1. If we want to choose x subject to those constraints and also satisfying |{i1 , . . . , i j }∩{t : xt = 1}| = m, we have to set xt = 1 for m − ` different t ∈ {i`+1 , . . . , i j } and for K − m different t ∈ / {i, i1 , . . . , i j }. j−` N− j−1 This can be done in m−` and K−m different ways, respectively. T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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By solving the system of equations (7.11), we get that the only solution is N− j−1 j−m K− j (−1) αj . N− j−1 K−m
αm =
(7.12)
Let |ψ00 i = |ψ0 i/kψ0 k be the normalized version of |ψ0 i. Then |ψ˜ ii,0 i = hψ00 |ψii,0 i|ψ00 i and 1 ,...,i j 1 ,...,i j kψ˜ ii,0 k = hψ00 |ψii,0 i= 1 ,...,i j 1 ,...,i j
hψ0 |ψii,0 i 1 ,...,i j kψ0 k
.
(7.13)
First, we have hψ0 |ψii,0 i = 1, 1 ,...,i j j−1 because |ψii,0 i consists of N− basis states |xi, xi = 0, xi1 = · · · = xi j = 1, each of which has K− j 1 ,...,i j q N− j−1 in both |ψ0 i and |ψii,0 i. Second, amplitude 1/ K− j 1 ,...,i j 2 j N− j−1 j j N − j − 1 K− j kψ0 k2 = ∑ αm2 = ∑ α 2j N− j−1 m K − m m m=0 m=0 K−m j j N− j−1 j j (K − m)!(N − K + m − j − 1)! K− j = ∑ = ∑ N− j−1 (K − j)!(N − K − 1)! m=0 m m=0 m K−m j j (K − m) · · · (K − j + 1) = ∑ m=0 m (N − K − 1) · · · (N − K + m − j) j
(7.14)
j−1 with the first equality following because there are mj N− K−m vectors x such that |x| = K, xi = 0, xt = 1 for m different t ∈ {i1 , . . . , i j } and K − m different t ∈ / {i, i1 , . . . , i j }, the second equality following from q N− j−1 equation (7.12) and the third equality following from our choice α j = 1/ K− j . We can similarly calculate kψ˜ ii,1 k. We omit the details and just state the result. The counterpart 1 ,...,i j of equation (7.13) is kψ˜ ii,1 k= 1 ,...,i j
hψ1 |ψii,1 i 1 ,...,i j kψ1 k
,
with |ψ1 i being the counterpart of |ψ0 i: j
|ψ1 i =
∑ αm
m=0
∑
|xi ,
x:|x|=K,xi =1 |{i1 ,...,i j }∩{`:x` =1}|=m
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q with α0 = 1/
N− j−1 K− j−1
. Similarly as before, we get hψ1 |ψii,1 i = 1 and 1 ,...,i j N− j−1 j K− j−1 kψ1 k2 = ∑ N− j−1 m m=0 K−m−1 j j (K − m − 1) . . . (K − j) = ∑ . m=0 m (N − K) . . . (N − K + m − j + 1) j
(7.15)
Each term in (7.14) is (K − m)(N − K + m − j) (K − j)(N − K) times the corresponding term in equation (7.15). We have K K −m N −K +m− j ≤ ·2 = 4, K− j N −K K/2 because j ≤ K/2 and N − K + m − j ≤ N − K (because of m ≤ j). Therefore, kψ0 k2 ≤ 4kψ1 k2 which implies 1 1 1 k= kψ˜ ii,0 ≥√ = kψ˜ ii,1 ,...,i k . 1 ,...,i j kψ0 k 4kψ1 k 2 1 j ˇ Acknowledgment. I would like to thank Frederic Magniez, Robert Spalek, Ronald de Wolf, and several anonymous referees for very helpful comments on a draft of this paper.
References [1] S. A ARONSON: Limitations of quantum advice and one-way communication. Theory of Computing, 1:1–28, 2005. Conference version in CCC’2004. [ToC:v001/a001, arXiv:quant-ph/0402095]. 2 [2] S. A ARONSON AND Y. S HI: Quantum lower bounds for the collision and the element distinctness problems. J. ACM, 51(4):595–605, 2004. [JACM:1008731.1008735]. 2 [3] A. A MBAINIS: Quantum lower bounds by quantum arguments. J. Comput. System Sci., 64(4):750–767, 2002. Conference version in STOC’00. [JCSS:10.1006/jcss.2002.1826, STOC:335305.335394, arXiv:quant-ph/0002066]. 1, 3, 4 [4] A. A MBAINIS: Polynomial degree vs. quantum query complexity. J. Comput. System Sci., 72(2):220–238, 2006. Conference version in FOCS’03. [JCSS:10.1016/j.jcss.2005.06.006, FOCS:10.1109/SFCS.2003.1238197, arXiv:quant-ph/0305028]. 2, 3, 4 [5] A. A MBAINIS: Quantum walk algorithm for element distinctness. SIAM J. Comput., 37(1):210–239, 2007. Conference version in FOCS’04. [SICOMP:10.1137/S0097539705447311, arXiv:quant-ph/0311001]. 2 T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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[6] A. A MBAINIS , R. Sˇ PALEK , AND R. DE W OLF: A new quantum lower bound method, with applications to direct product theorems and time-space tradeoffs. In Proc. 38th STOC, pp. 618–633. ACM Press, 2006. [STOC:10.1145/1132516.1132604]. 2 [7] A. A MBAINIS , R. Sˇ PALEK , AND R. DE W OLF: Quantum direct product theorems for symmetric functions and time-space tradeoffs. Algorithmica, 55(3):422–461, 2009. 2 [8] H. BARNUM , M. S AKS , AND M. S ZEGEDY: Quantum decision trees and semidefinite programming. In Proc. 18th IEEE Conf. on Comput. Complex. (CCC’03), pp. 179–193. IEEE Press, 2003. [CCC:10.1109/CCC.2003.1214419]. 2 [9] R. B EALS , H. B UHRMAN , R. C LEVE , M. M OSCA , AND R. DE W OLF: Quantum lower bounds by polynomials. J. ACM, 48(4):778–797, 2001. Conference version in FOCS’98. [JACM:502090.502097, FOCS:10.1109/SFCS.1998.743485, arXiv:quant-ph/9802049]. 1 [10] E. B ERNSTEIN AND U. VAZIRANI: Quantum complexity theory. SIAM J. Comput., 26(5):1411– 1473, 1997. [SICOMP:10.1137/S0097539796300921]. 15 [11] G. B RASSARD , P. H ØYER , AND A. TAPP: Quantum counting. In Proc. 25th Intern. Conf. Automata, Languages and Programming (ICALP’98), volume 1443 of LNCS, pp. 820–831. Springer, 1998. [ICALP:ap2mrf08d8gcqppe, arXiv:quant-ph/9805082]. 1 [12] H. B UHRMAN AND R. DE W OLF: Complexity measures and decision tree complexity: A survey. Theoret. Comput. Sci., 288:21–43, 2002. [TCS:10.1016/S0304-3975(01)00144-X]. 3 [13] L. G ROVER: A fast quantum mechanical algorithm for database search. In Proc. 28th STOC, pp. 212–219. ACM Press, 1996. [STOC:237814.237866, arXiv:quant-ph/9605043]. 1 [14] P. H ØYER , T. L EE , AND R. Sˇ PALEK: Negative weights make adversaries stronger. In Proc. 39th STOC, pp. 526–535. ACM Press, 2007. [STOC:10.1145/1250790.1250867]. 3 [15] P. H OYER , J. N EERBEK , AND Y. S HI: Quantum lower bounds of ordered searching, sorting and element distinctness. Algorithmica, 34:429–448, 2002. Conference version at ICALP’01. [Algorithmica:25gl9elr5rxr3q6a, arXiv:quant-ph/0102078]. 4 [16] H. K LAUCK , R. Sˇ PALEK , AND R. DE W OLF: Quantum and classical strong direct product theorems and optimal time-space tradeoffs. SIAM J. Comput., 36(5):1472–1493, 2007. Conference version at FOCS’04. [SICOMP:10.1137/05063235X, FOCS:10.1109/FOCS.2004.52, arXiv:quantph/0402123]. 2 [17] D. K NUTH: Combinatorial matrices. In Selected Papers on Discrete Mathematics, CSLI Lecture Notes, no. 106. University of Chicago Press, 2003. 8 [18] S. L APLANTE AND F. M AGNIEZ: Lower bounds for randomized and quantum query complexity using Kolmogorov arguments. SIAM J. Comput., 38(1):46–62, 2008. Conference version at CCC’04. [SICOMP:10.1137/050639090, CCC:10.1109/CCC.2004.1313852, arXiv:quantph/0311189]. 2 T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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[19] F. M AGNIEZ , M. S ANTHA , AND M. S ZEGEDY: Quantum algorithms for the triangle problem. SIAM J. Comput., 37(2):413–424, 2007. Conference version at SODA’05. [SICOMP:10.1137/050643684, arXiv:quant-ph/0310134]. 2 [20] A. NAYAK: Optimal lower bounds for quantum automata and random access codes. In Proc. 40th FOCS, pp. 369–377. IEEE Press, 1999. [FOCS:10.1109/SFFCS.1999.814608, arXiv:quantph/9904093]. 14 [21] B. R EICHARDT: Span programs and quantum query complexity: The general adversary bound is nearly tight for every Boolean functionnote. Quant-ph, arXiv:0904.2759, 2009. [arXiv:0904.2759]. 3 [22] R. Sˇ PALEK: The multiplicative quantum adversary. In Proc. 23rd IEEE Conf. on Comput. Complex., pp. 237–248. IEEE Press, 2008. [CCC:10.1109/CCC.2008.9]. 2 [23] R. Sˇ PALEK AND M. S ZEGEDY: All quantum adversary methods are equivalent. Theory of Computing, 2(1):1–18, 2006. Conference version at ICALP’05. [ToC:v002/a001, arXiv:quantph/0409116]. 2 [24] S. Z HANG: On the power of Ambainis’s lower bounds. 3):241–256, 2005. Conference version in ICALP’04. ICALP:gm2ff6wpc0q39v3x, arXiv:quant-ph/0311060]. 2
Theoret. Comput. Sci., 339(2– [TCS:10.1016/j.tcs.2005.01.019,
AUTHOR Andris Ambainis professor University of Latvia Raina bulv. 19, Riga, LV-1586, Latvia andris ambainis lu lv http://home.lanet.lv/∼ambainis ABOUT THE AUTHOR A NDRIS A MBAINIS received his Ph. D. from the University of California, Berkeley in 2001, supervised by Umesh Vazirani. After that, he was a postdoc at the Institute for Advanced Study, Princeton and at the University of California, Berkeley, and a faculty member at the University of Waterloo, Canada. In 2007, Andris returned to his native Latvia and became Professor at the University of Latvia. His research interests include many areas of quantum computing and quantum information theory (quantum algorithms, quantum complexity theory, quantum cryptography, pseudorandom quantum states, etc.), as well as the classical theory of computation.
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A new quantum lower bound method, with an application to a strong direct product theorem for quantum search Andris Ambainis∗ Received: July 6, 2007; revised: July 24, 2009; published: January 12, 2010.
Abstract: We present a new method for proving lower bounds on quantum query algorithms. The new method is an extension of the adversary method, by analyzing the eigenspace structure of the problem. Using the new method, we prove a strong direct product theorem for quantum search. ˇ This result was previously proved by Klauck, Spalek, and de Wolf (FOCS’04) using the polynomials method. No proof using the adversary method was known before. ACM Classification: F1.2, F2.2 AMS Classification: 81P68, 68Q17 Key words and phrases: quantum computing, quantum algorithms, quantum lower bounds
1
Introduction
Many quantum algorithms (for example, Grover’s algorithm [13] and quantum counting [11]) can be analyzed in the query model where the input is accessed via a black box that answers queries about the values of input bits. There are two main methods for proving lower bounds on quantum query algorithms: the adversary method [3] and the polynomials method [9]. Both of them have been studied in detail. The limits of the adversary method are particularly well understood. The original adversary method [3] has been ∗ Supported by University of Latvia grant ZP01-100, a Marie Curie International Reintegration Grant (IRG), and ESF project
1DP/1.1.1.2.0/09/APIA/VIAA/044. Most of this work was done at the University of Waterloo and supported by NSERC, CIFAR, ARO, MITACS, and an IQC University Professorship.
2010 Andris Ambainis Licensed under a Creative Commons Attribution License
DOI: 10.4086/toc.2010.v006a001
A NDRIS A MBAINIS
ˇ generalized in several different ways [4, 18, 8]. Spalek and Szegedy [23] then showed that all the generalizations are equivalent and, for certain problems, cannot improve the best known lower bounds. adversary methods of [4, 18, 8] cannot prove a lower bound For example, it was shown in [23, 24] that the p for a total Boolean function that exceeds O( C0 ( f )C1 ( f )), where C0 ( f ) and C1 ( f ) are the certificate complexities of f on 0-inputs and 1-inputs, resp. This implies that the adversary methods of [4, 18, 8] cannot prove a tight lower bound for element distinctness or improve the best known lower bound for triangle finding. (The complexity of √ element distinctness is Θ(N 2/3 ) [2, 5] but the adversary method cannot prove a bound better than Ω( N). For triangle finding [19], the best known lower bound is Ω(N) and it is known that it cannot be improved using the adversary method. It is, however, possible that the bound is not tight, because the best algorithm uses O(N 1.3 ) queries.) In this paper we describe a new version of the quantum adversary method which may not be subject to those limitations. We then use the new method to prove a strong direct product theorem for the K-fold search problem. In the K-fold search problem, a black box contains x1 , . . . , xN such that |{i : xi = 1}| √ = K and we NK) queries. have to find all the K values of i such that xi = 1. This problem can be solved with O( √ It can be easily shown, using any of the previously known √ methods, that Ω( NK) quantum queries are required. A more difficult problem is to show that Ω( NK) queries are required, even if the algorithm only has to be correct with an exponentially small probability c−K , c > 1. This result is known as the strong direct product theorem for k-fold search. Besides being interesting on its own, the strong direct product theorem is useful for proving time-space tradeoffs for quantum sorting [16] and lower bounds on quantum computations that use advice [1]. ˇ The strong direct product theorem for quantum search was first shown by Klauck, Spalek, and de Wolf [16], using the polynomials method. No proof using the adversary method has been known and, as we show in Section 3, the previously known adversary methods are insufficient to prove a strong direct product theorem for K-fold search. Recent developments opments occurred.
After the author completed the research presented in this paper, several devel-
ˇ 1. Together with Spalek and de Wolf [7], we have used the methods from this paper to prove a direct product theorem for t-threshold functions. This implies time-space tradeoffs for the problem of deciding systems of linear inequalities. This paper and [7] were merged for publication at STOC’06 [6]. We have decided to publish the journal versions separately. The relation between the two papers is as follows. A direct product theorem for k-threshold functions implies a direct product theorem for search. Thus, the result of [7] implies the result in this paper. The proof of our result is, however, substantially simpler than the proof of the more general result in [7]. Because of that, we think that our proof continues to be of interest even though the result in this paper has been generalized by [7]. ˇ 2. Spalek [22] generalized the results of the current paper and [7], obtaining a multiplicative adversary method. This is a general framework for proving lower bounds which includes the results of the current paper and [7] as particular cases. T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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ˇ 3. Høyer, Lee, and Spalek [14] generalized the usual adversary method in a different way, by extending the usual weighted adversary method of [4] to negative weights. Reichardt [21] has shown that this method is optimal: if Adv± ( f ) is the best adversary lower bound that can be proven for a function f , then f can be evaluated using log Adv± ( f ) O Adv ( f ) log log Adv± ( f )
±
queries. Although the negative weight adversary method is guaranteed to provide nearly optimal results, the other lower bound methods also remain interesting because particular lower bounds may follow more easily using different tools.
2
Preliminaries
We consider the following problem. Search for K marked elements, SEARCHK (N): Given a black box containing x1 , . . . , xN ∈ {0, 1} such that xi = 1 for exactly K values i ∈ {1, 2, . . . , N}, find all K values i1 , . . . , iK satisfying xi j = 1. This problem can be viewed as computing an NK -valued function f (x1 , . . . , xN ) in the variables x1 , . . . , xN ∈ {0, 1}, with values of the function being indices for the NK sets S ⊆ [N] of size K, in some canonical ordering of those sets. We study this problem in the quantum query model. (For a survey on the quantum query model, see [12]). In this model, the input bits can be accessed by queries to an oracle X and the complexity of f is the number of quantum queries needed to compute f . A quantum computation with T queries is just a sequence of unitary transformations U0 → O → U1 → O → · · · → UT −1 → O → UT . The U j can be arbitrary unitary transformations that do not depend on the input bits x1 , . . . , xN . The transformations O are query (oracle) transformations which depend on x1 , . . . , xN . To define O, we represent basis states as |i, zi where i consists of dlog(N + 1)e bits and z consists of all other bits. Then, Ox maps |0, zi to itself and |i, zi to (−1)xi |i, zi for i ∈ {1, ..., N} (i. e., we change phase depending on xi , unless i = 0 in which case we do nothing). The computation starts with a state |0i. Then we apply U0 , Ox , . . . , Ox ,UT and measure the final N state. The result of the computation is the string of dlog2 K e rightmost bits of the state obtained by the measurement, which is interpreted as a description of one of the NK subsets S ⊆ {1, . . . , N}, |S| = K.
3
Overview of the adversary method
We describe the adversary method of [3]. Let X be a subset of the set of possible inputs {0, 1}N . We run the algorithm on a superposition of inputs in X. More formally, let HA be the workspace of the algorithm. We consider a bipartite system T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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H = HA ⊗ HI where HI is an “input subspace” spanned by basis vectors |xi corresponding to inputs x ∈ X. Let UT OUT −1 · · ·U0 be the sequence of unitary transformations on HA performed by the algorithm A, where U0 , . . . ,UT denote the transformations that do not depend on the input and O stands for the query transformations. We transform this sequence into a sequence of unitary transformations on H. A unitary transformation Ui on HA corresponds to the transformation Ui0 = Ui ⊗ I on the whole H. The query transformation O corresponds to a transformation O0 that is equal to Ox on the subspace HA ⊗ |xi. We perform the sequence of transformations UT0 O0 UT0 −1 · · ·U00 on the starting state |ψstart i = |0i ⊗ ∑ αx |xi . x∈S
Then, the final state is |ψend i =
∑ αx |ψx i ⊗ |xi ,
x∈X
where |ψx i is the final state of A = UT OUT −1 . . .U0 on input x. This follows from the fact that the restrictions of UT0 , O0 ,UT0 −1 , . . . ,U00 to HA ⊗ |xi are UT , Ox ,UT −1 , . . . ,U0 and these are exactly the transformations of the algorithm A on input x. Let ρend be the reduced density matrix of the HI register of the state |ψend i. The adversary method of [3, 4] works by showing the following two statements. • Let x ∈ X and y ∈ X be such that f (x) 6= f (y) (where f is the function that is being computed). If the algorithm outputs the correct answer with probability 1 − ε on both x and y then p |(ρend )x,y | ≤ 2 ε(1 − ε)|αx ||αy | . • For any algorithm that uses T queries, there exist inputs x, y ∈ S such that f (x) 6= f (y) and p (ρend )x,y > 2 ε(1 − ε)|αx ||αy | . These two statements together imply that any algorithm computing f must use more than T queries. An equivalent approach [15, 4] is to consider the inner productsphψx |ψy i between the final states |ψx i and |ψy i p of the algorithm on inputs x and y. Then, |(ρend )x,y | ≤ 2 ε(1 − ε)|αx ||αy | is equivalent to |hψx |ψy i| ≤ 2 ε(1 − ε). As a result, both of the above statements can be described in terms of inner products hψx |ψy i, without explicitly introducing the register HI . The first statement says that, for the algorithm to succeed on inputs x, y such that f (x) 6= f (y), the states |ψxp i and |ψy i must be sufficiently far apart from one another (so that the inner product |hψx |ψy i| is at most 2 ε(1 − ε)). The second statement says that this is impossible if the algorithm only uses T queries. This approach breaks down if we consider computing a function f with success probability p < 1/2. ( f has to have more than 2 values for this task to be nontrivial.) Then, |ψx i and |ψy i could be the same and the algorithm may still succeed on both inputs, if it outputs x with probability 1/2 and y with probability 1/2. In the case of strong direct product theorems, the situation is even more difficult. Since the algorithm only has to be correct with probability c−K , the algorithm could have almost the same final state on cK different inputs and still “succeed” on every one of them. T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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In this paper, we present a new method that does not suffer from this problem. Our method, described in the next section, uses the idea of augmenting the algorithm with an input register HI , together with two new ingredients: 1. Symmetrization. We symmetrize the algorithm by applying a random permutation π ∈ SN to the input x1 , . . . , xN . 2. Eigenspace analysis. We study the eigenspaces of ρstart , ρend and density matrices describing the state of HI at intermediate steps and use them to bound the progress of the algorithm. The eigenspace analysis is the main new technique. Symmetrization is necessary to simplify the structure of the eigenspaces, to make the eigenspace analysis possible.
4
Our result
Theorem 4.1. There exist ε and c satisfying ε > 0, 0 < c 0, S j = T j ∩ (T j−1 )⊥ , with T j being the space spanned by all states |ψi1 ,...,i j i = q
1 N K− j
∑
|x1 , . . . , xN i .
x1 ,...,xN : x1 +···+xN =K, xi1 =···=xi j =1
Let τ j be the completely mixed state over the subspace S j . Lemma 4.4. There exist pt,0 ≥ 0, . . . , pt,K ≥ 0 such that ρt = ∑Kj=0 pt, j τ j . Proof. By Lemma 4.3, S0 , . . . , SK are the eigenspaces of ρt . Therefore, ρt is a linear combination of the projectors to S0 , . . . , SK . Since τ j is a multiple of the projector to S j , we have K
ρt =
∑ pt, j τ j .
j=0
Since ρt is a density matrix, it must be positive semidefinite. This means that pt,0 ≥ 0, . . . , pt,K ≥ 0. Informally, we can interpret pt, j as the probability that, after t queries, the algorithm HA knows the locations for t out of the K items j with x j = 1. Let qt, j = pt, j + pt, j+1 + · · · + pt,K . The theorem now follows from the following lemmas. Lemma 4.5. p0,0 = 1, p0, j = 0 for j > 0. T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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Proof. In the starting state, HI contains the state |ϕ0 i, independent of HA and HP . Therefore, tracing out HA ⊗ HP leaves the state ρ0 = |ψ0 ihψ0 |. √
Lemma 4.6. For all j ∈ {1, . . . , K} and all t we have qt+1, j+1 ≤ qt, j+1 + 4√NK qt, j . Proof. In Section 5. Lemma 4.7. qt, j ≤
t j
4√K j √ N
.
Proof. By induction on t. The base case, t = 0 follows immediately from p0,0 = 1 and p0,1 = · · · = p0,K = 0 . For the induction step, we have √ √ ! j−1 √ !j √ 4 K t 4 K t 4 K 4 K √ √ + √ qt+1, j ≤ qt, j + √ qt, j−1 ≤ j N N N j−1 N √ !j √ !j t t +1 4 K t 4 K √ √ = , = + j j j−1 N N where the first inequality follows from Lemma 4.6 and the second inequality follows from the inductive assumption. √ Lemma 4.8. If t ≤ 0.03 NK then qt, j < 0.66 j for all j > K/2. Proof. By the well known inequality tj < (et/ j) j for j ≥ 1, we have t qt, j ≤ j
√ !j 4 K √ ≤ N
!j √ 4 Ket √ . Nj
√ Let j > K/2 and t ≤ 0.03 NK. Then √ √ √ 4 Ket 0.12e K NK √ √ ≤ < 0.66 , Nj NK/2 implying the lemma. Lemma 4.9. The success probability of A is at most N p K/2 + 4 qt,K/2+1 . N K
Proof. In Section 6. T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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To complete the proof, given the two Lemmas, we choose a constant c > set ε = 0.03. Then, by Lemma 4.9, the success probability of A is at most N K/2 N K
√ 4 0.66 = 0.90133 . . . and
+4
p 0.66K/2 .
The first term is equal to N K/2 N K
K!(N − K)! K(K − 1) . . . (K/2 + 1) = (K/2)!(N − K/2)! (N − K/2)(N − K/2 − 1) . . . (N − K + 1) K/2 K/2 K N/2 K/2 2 ≤ ≤ = = o(cK ) , N − K/2 3N/4 3
=
√ where the last two inequalities follow from K < N/2. The second term is 4 0.66K/2 = o(cK ).
5
Proof of Lemma 4.6
Let |ψt i be the state before (t + 1)st query. We decompose |ψt i as ∑Ni=0 ai |ψt,i i, where |ψt,i i is the part in which the query register contains |ii. Because of symmetrization, we must have |a1 | = |a2 | = · · · = |aN |. Also, we can choose the relative phases so that a1 , · · · , aN are all positive reals and, thus, a1 = a2 = · · · = aN . Let ρt,i = |ψt,i ihψt,i |. Then N
ρt = ∑ a2i ρt,i .
(5.1)
i=0
Claim 5.1. Let i ∈ {1, . . . , N}. The entry (ρt,i )x,y only depends on xi , yi , and the cardinality of {` : ` 6= i, x` = y` = 1}. Proof. The main idea is similar to Lemma 4.2. This time, we trace out all registers, except for HI and the query register. This gives us a density matrix ρ whose rows and columns are indexed by pairs (x, j) where x is an input and j ∈ {1, . . . , N}. Let x ∈ {0, 1}N , y ∈ {0, 1}N , j ∈ [N], k ∈ [N] and x0 ∈ {0, 1}N , y0 ∈ {0, 1}N , j0 ∈ [N], k0 ∈ [N] be such that there is a permutation π ∈ SN with x0 = π(x), y0 = π(y), i0 = π(i) and j0 = π( j). By an argument similar to the proof of Lemma 4.2, we have ρ(x,i),(y, j) = ρ(x0 ,i0 ),(y0 , j0 ) .
(5.2)
We now observe that ρt,i is the submatrix of ρ consisting of rows and columns indexed by pairs (x, i), with all possible x ∈ {0, 1}N , |{i : xi = 1}| = K. If we have inputs x, y, x0 , y0 with xi = xi0 , yi = y0i and |{` : ` 6= i, x` = y` = 1}| = |{` : ` 6= i, x`0 = y0` = 1}|, then we can construct a permutation π with π(i) = i, π(x) = x0 and π(y) = y0 . Equation (5.2) then implies (ρt,i )x,y = (ρt,i )x0 ,y0 . T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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We now describe the eigenspaces of matrices ρt,i . The proofs of some claims are postponed to Section 7. We define the following subspaces of states. Let i ∈ [N] and j ∈ {0, 1, . . . , K}. We define T ji,0 to be the subspace spanned by all states |ψii,0 i= q 1 ,...,i j
1 N− j−1 K− j
∑
|x1 . . . xN i ,
x:|x|=K xi1 =···=xi j =1,xi =0
with (i1 , . . . , i j ) ranging over all tuples of j distinct elements of [N] − {i}. Similarly, we define T ji,1 to be the subspace spanned by all states |ψii,1 i= q 1 ,...,i j
1 N− j−1 K− j−1
∑
|x1 . . . xN i .
x:|x|=K xi1 =···=xi j =1,xi =1
i,0 i,0 ⊥ i,1 i,1 i,1 ⊥ i,0 i,1 Let Si,0 j = T j ∩ (T j−1 ) and S j = T j ∩ (T j−1 ) . Equivalently, we can define S j and S j as the i, respectively, with i and |ψ˜ ii,1 subspaces spanned by the states |ψ˜ ii,0 1 ,...,i j 1 ,...,i j
|ψ˜ ii,` i = P(T i,` 1 ,...,i j
⊥ j−1 )
i. |ψii,` 1 ,...,i j
i Let Sα,β , j be the subspace spanned by all states
α
i |ψ˜ ii,0 1 ,...,i j ik k|ψ˜ ii,0 1 ,...,i j
+β
|ψ˜ ii,1 i 1 ,...,i j ik k|ψ˜ ii,1 1 ,...,i j
.
(5.3)
where (i1 , . . . , i j ) again ranges over all tuples of j distinct elements of [N] − {i}. i Claim 5.2. Every eigenspace of ρt,i is a direct sum of subspaces Sα,β , j for some α, β , j.
Proof. In Section 7. i i Let τα,β , j be the completely mixed state over Sα,β , j . Similarly to Lemma 4.4, we can write ρt,i as
ρt,i =
i piα,β , j τα,β ,j ,
∑
(5.4)
(α,β , j)∈At,i
where (α, β , j) range over some finite set At,i . (This set is finite because the HI register holding |x1 . . . xN i is finite dimensional and, therefore, decomposes into a direct sum of finitely many eigenspaces.) For every pair (α, β , j) ∈ At,i , we normalize α, β by multiplying them by the same constant so that α 2 +β 2 = 1. Querying xi transforms this state to 0 ρt,i =
∑
i piα,β , j τα,−β ,j ,
(α,β , j)∈At,i
because |ψ˜ ii,` i is a superposition of |xi with xi = ` and, therefore, a query leaves |ψ˜ ii,0 i unchanged 1 ,...,i j 1 ,...,i j 0 = ρ , because, if the query register contains |0i, and flips a phase on |ψ˜ ii,1 i. If i = 0, we have ρt,0 t,0 1 ,...,i j the query maps any state to itself, thus leaving ρt,0 unchanged.
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Claim 5.3. Let α0 =
q
N−K ˜ i,0
N− j ψi1 ,...,i j
and β0 =
q
K− j i,1
˜ N− j ψi1 ,...,i j .
(i) Sαi 0 ,β0 , j ⊆ S j ; (ii) Sβi 0 ,−α0 , j ⊆ S j+1 . Proof. In Section 7. i Corollary 5.4. For any α, β we have Sα,β , j ⊆ S j ⊕ S j+1 . i,1 ˜ ii,0 Proof. We have Sα,β , j ⊆ Si,0 i j ⊕ S j , since Sα,β , j is spanned by linear combinations of states |ψ 1 ,...,i j
˜ ii,1 (which belong to Si,0 i (which belong to Si,1 j ) and states |ψ j ). As shown in the proof of Claim 5.3 1 ,...,i j above, i,1 Si,0 j ⊕ S j ⊆ Sα0 ,β0 , j ⊕ S−β0 ,α0 , j ⊆ S j ⊕ S j+1 . i The next claim quantifies the overlap between Sα,β , j and S j+1 .
Claim 5.5. i Tr PS j+1 τα,β ,j =
|αβ0 − β α0 |2 . α02 + β02
Proof. In Section 7. To be able to use this bound, we also need to bound α0 and β0 . q 4(K− j) Claim 5.6. √ β20 2 ≤ N+3K−4 j. α0 +β0
Proof. In Section 7. We can now complete the proof of Lemma 4.6. By projecting both sides of ρt = ∑i pt,i τi to (T j )⊥ = S j+1 ⊕ · · · ⊕ SK and taking trace, we get K
Tr P(Tj )⊥ ρt =
∑
K
pt, j0 Tr P(T j )⊥ τ j0 =
j0 =0
∑
pt, j0 = qt, j ,
(5.5)
j0 = j+1
with the second equality following because the states τ j0 are uniform mixtures over subspaces S j0 and S0 , . . . , S j are contained in T j while S j+1 , . . . , SK are contained in (T j )⊥ . Because of equations (5.1) and (5.4), this means that N
qt, j+1 = a20 Tr P(Tj )⊥ ρt,0 + ∑ a2i i=1
∑
(α,β , j0 )∈At,i
i piα,β , j0 Tr P(Tj )⊥ τα,β , j0 .
(5.6)
Decomposing the state after the query in a similar way, we get N
0 qt+1, j+1 = a20 Tr P(Tj )⊥ ρt,0 + ∑ a2i i=1
∑
(α,β , j0 )∈At,i
i piα,β , j0 Tr P(Tj )⊥ τα,−β , j0 .
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qt+1, j+1 − qt, j+1 = ∑ a2i i=1
∑
(α,β , j0 )∈At,i
i i piα,β , j0 Tr P(Tj )⊥ (τα,−β , j0 − τα,β , j0 ) .
(5.7)
We now claim that all the terms in this sum with j0 6= j are 0. For j0 < j, Sα,β , j0 ⊆ T j0 +1 ⊆ T j , implying ⊥ i i 0 0 0 that Tr P(Tj )⊥ τα,β , j0 = 0 and, similarly, Tr P(T j )⊥ τα,−β , j0 = 0. For j > j, Sα,β , j0 ⊆ S j ⊕ S j +1 ⊆ (T j ) , implying that i i Tr P(Tj )⊥ τα,−β Tr P(Tj )⊥ τα,β , j0 = 1 , , j0 = 1, and the difference of the two is 0. By removing those terms from (5.7), we get N
qt+1, j+1 − qt, j+1 = ∑ a2i i=1
∑
i i piα,β , j Tr P(Tj )⊥ (τα,−β , j − τα,β , j ) .
(5.8)
(α,β , j)∈At,i
We have i i i i Tr P(Tj )⊥ (τα,−β , j − τα,β , j ) = Tr PS j+1 (τα,−β , j − τα,β , j ) =
|αβ0 + β α0 |2 |αβ0 − β α0 |2 − , α02 + β02 α02 + β02
where the first equality follows from Corollary 5.4, S j ⊆ T j , and S j+1 ⊆ (T j )⊥ , and the second equality follows from Claim 5.5. This is at most s r β0 4(K − j) 4K α0 β0 α0 |αβ α0 β0 | q ≤2 2 = 2q ≤2 ≤2 , 4 2 2 2 N + 3K − 4 j N α0 + β0 α0 + β0 α2 + β 2 α2 + β 2 0
0
0
0
where the first inequality follows from |αβ | ≤
|α|2 + |β |2 1 = 2 2
and the second inequality follows from Claim 5.6 and √ α20 2 ≤ 1. Together with equation (5.7), this α0 +β0 means √ N 4 K qt+1, j+1 − qt, j+1 ≤ √ ∑ a2i (5.9) ∑ pi . N i=1 (α,β , j)∈At,i α,β , j Similarly to equation (5.5) we have pt, j+1 + pt, j = Tr P(S j ⊕S j+1 ) ρt . We can then express the right hand side, similarly to equation (5.6), as a sum of terms p0j0 Tr P(S j ⊕S j+1 ) τ j0 i i and piα,β , j0 Tr P(S j ⊕S j+1 ) τα,β , j0 . Since Sα,β , j ⊆ S j ⊕ S j+1 (by Corollary 5.4), we have i Tr P(S j ⊕S j+1 ) τα,β ,j = 1.
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This means that N
pt, j+1 + pt, j ≥ ∑ a2i i=1
∑
piα,β , j .
(α,β , j)∈At,i
Together with equation (5.9), this implies √ √ 4 K 4 K qt+1, j+1 − qt, j+1 ≤ √ (pt, j + pt, j+1 ) ≤ √ N N
6
K
∑ pt, j
j0 = j
0
√ 4 K = √ qt, j . N
Proof of Lemma 4.9
We start with the case when pT,K/2+1 = · · · = pT,K = 0. Lemma 6.1. If pT,K/2+1 = · · · = pT,K = 0, the success probability of A is at most
N (K/2 ) . (NK )
N Proof. Let |ψi be the final state. The state of register HI lies in TK/2 , which is an K/2 -dimensional N space. Therefore, there is a Schmidt decomposition for |ψi with at most K/2 terms. This means that N the state of HA lies in an K/2 -dimensional subspace of HA ⊗ HS . We express the final state as 1 |ψi = ∑ q |ψx i|xi . N x:|x|=K
K
We can think of |ψx i as a quantum encoding for x and the final measurement as a decoding procedure that takes |ψx i and produces a guess for x. The probability that algorithm A succeeds is then equal to the average success probability of the encoding. We now use Theorem 6.2. [20] For any encoding |ψx i of M classical values by quantum states in d dimensions, the probability of success is at most d/M. N N In our case, M = NK and d = K/2 because the states |ψi all lie in a K/2 -dimensional subspace of HA ⊗ HS . Therefore, Theorem 6.2 √ implies Lemma√6.1. We decompose the state |ψT i as 1 − δ |ψT0 i + δ |ψT00 i where |ψT0 i is in the subspace HA ⊗ TK/2 and |ψT00 i is in HA ⊗ (TK/2 )⊥ . We have K
δ=
∑
pT, j .
j=K/2+1
The success probability of A is the probability that, if we measure both the register of HA containing the result of the computation and HI then we get i1 , . . . , iK and x1 , . . . , xN such that xi1 = · · · = xiK = 1. T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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Consider the probability of getting i1 , . . . , iK and x1 , . . . , xN such that xi1 = · · · = xiK = 1, when meaN / NK . We have suring |ψT0 i (instead of |ψT i). By Lemma 6.1, this probability is at most K/2 kψT − ψT0 k ≤ (1 −
p p √ √ √ 1 − δ 2 )kψT0 k + δ kψT00 k = (1 − 1 − δ 2 ) + δ ≤ 2 δ .
We now apply Lemma 6.3. [10] For any states |ψi and |ψ 0 i and any measurement, the total variation distance between the probability distributions obtained by applying M to |ψi and to |ψ 0 i is at most 2kψ − ψ 0 k. , . . . , iK and x1 , . . . , xN such that xi1 = · · · = xiK = 1, By Lemma 6.3, the probabilities of getting i1√ √ when measuring |ψT i and |ψT0 i differ by at most 4 δ = 4 qT,K/2+1 . Therefore, the success probability of A is at most N p K/2 + 4 qT,K/2+1 . N K
7
Structure of the eigenspaces of ρt,i
In this section, we prove Claims 5.2, 5.3, 5.5, and 5.6 describing the structure of the eigenspaces of ρt,i . Before giving the proofs, we briefly summarize the results in this section. 1. Let ρt,i be a matrix whose rows and columns are indexed by (x1 , . . . , xN ), xi ∈ {0, 1} with x1 + . . . + xN = K. By Claim 5.2, if ρt,i is symmetric w. r. t. any permutation π ∈ SN that fixes i, then i its eigenspaces are of the form Sα,β , j , which is the subspace spanned by the states of the form α
i |ψ˜ ii,0 1 ,...,i j ik k|ψ˜ ii,0 1 ,...,i j
+β
|ψ˜ ii,1 i 1 ,...,i j ik k|ψ˜ ii,1 1 ,...,i j
i and |ψ˜ ii,1 i are the states defined at the beginning of Section 5. where |ψ˜ ii,0 1 ,...,i j 1 ,...,i j i 2. We then relate the eigenspaces Sα,β , j to the subspaces S0 , . . . , SK defined in Section 4.2. In Claim 5.3 we show that, for certain α0 and β0 , we have
• Sαi 0 ,β0 , j ⊆ S j ; i • S−β ⊆ S j+1 ; 0 ,α0 , j
A corollary of this is that, for any α, β , we have i i i Sα,β , j ⊆ Sα0 ,β0 , j ⊕ S−β0 ,α0 , j ⊆ S j ⊕ S j+1 .
This is shown in Corollary 5.4. i 3. Next, in Claim 5.5, we quantify how close Sα,β , j is to S j and S j+1 . The result is an expression for i i i Tr PS j+1 τα,β , j (where τα,β , j is the maximally mixed state over Sα,β , j ) that involves α, β and α0 , β0 . To use that expression, we also need a bound on the ratio of α0 and β0 (Claim 5.6).
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Proof of Claim 5.2. We rearrange the rows and the columns of ρt,i so that all rows and columns corresponding to |x1 . . . xN i with xi = 0 are before the rows and the columns corresponding to |x1 . . . xN i with xi = 1. We then express ρt,i as A B ρt,i = , C D N−1 N−1 N−1 × square matrix indexed by |x . . . x i with x = 0, D is an × where A is an N−1 1 N i K−1 K−1 K K square matrix indexed by |x1 . . . xN i with xi = 1, and B and C are rectangular matrices with rows (columns) indexed by |x1 . . . xN i with xi = 0 and columns (rows) indexed by |x1 . . . xN i with xi = 1. We claim that ρt,i |ψ˜ ii,0 i = a11 |ψ˜ ii,0 i + a12 |ψ˜ ii,1 i and 1 ,...,i j 1 ,...,i j 1 ,...,i j ρt,i |ψ˜ ii,1 i = a21 |ψ˜ ii,0 i + a22 |ψ˜ ii,1 i, 1 ,...,i j 1 ,...,i j 1 ,...,i j
(7.1)
where a11 , a12 , a21 , a22 are independent of i1 , . . . , i j . To prove that, we first note that A and D are matrices where Axy and Dxy only depend on |{t : xt = yt }|. Therefore Lemma 4.3 applies. This means that Si,0 j and Si,1 j are eigenspaces for A and D, respectively, and i = a11 |ψ˜ ii,0 i and A|ψ˜ ii,0 1 ,...,i j 1 ,...,i j D|ψ˜ ii,1 i = a22 |ψ˜ ii,1 i, 1 ,...,i j 1 ,...,i j i,1 where a11 and a22 are the eigenvalues of A and D for the eigenspaces Si,0 j and S j . It remains to prove that
i = a12 |ψ˜ ii,1 i and B|ψ˜ ii,0 1 ,...,i j 1 ,...,i j
(7.2)
C|ψ˜ ii,1 i = a21 |ψ˜ ii,0 i. 1 ,...,i j 1 ,...,i j
(7.3)
Let M be a rectangular matrix, with entries indexed by x, y, with |x| = |y| = K and xi = 1 and yi = 0. The entries of M are Mxy = 1 if x and y differ in two places, with xi = 1, yi = 0 and xl = 0, yl = 1 for some l 6= i and Mxy = 0 otherwise. We claim M|ψ˜ ii,0 i = c|ψ˜ ii,1 i 1 ,...,i j 1 ,...,i j
(7.4)
for some c that may depend on N, k, and j but not on i1 , . . . , i j . To prove that, we need to prove two things. First, M|ψii,0 i = c|ψii,1 i. 1 ,...,i j 1 ,...,i j T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
(7.5) 16
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This follows by M|ψii,0 i= q 1 ,...,i j
=q
=q
1
∑
M|xi
∑
∑
(N − K) N− j−1
∑
N− j−1 x:xi1 =···=xi j =1, k− j xi =0
1
N− j−1 x:xi1 =···=xi j =1 `:x` =1 K− j xi =0
1
K− j
|x1 . . . x`−1 0x`+1 . . . xi−1 1xi+1 . . . xN i
|yi =
(K − j)(N − K)|ψii,1 i. 1 ,...,i j
p
y:yi1 =···=yi j =1 yi =1
Second, M(T ji,0 ) ⊆ T ji,1 and M(T ji,0 )⊥ ⊆ (T ji,1 )⊥ . The first statement immediately follows from equai and |ψii,1 i, respection (7.5), because the subspaces T ji,0 and T ji,1 are spanned by the states |ψii,0 1 ,...,i j 1 ,...,i j tively. To prove the second statement, let |ψi ∈ (T ji,0 )⊥ and |ψi = ∑x ax |xi. We would like to prove M|ψi ∈ (T ji,1 )⊥ . This is equivalent to hψii,1 |M|ψi = 0 for all i1 , . . . , i j . We have 1 ,...,i j |M|ψi = q hψii,1 1 ,··· ,i j =q
=q
1
∑
N− j−1 y:yi1 =···=yi j =1 K− j−1
1
hy|M|ψi
∑
∑
N− j−1 x:xi1 =···=xi j =1, l:xl =1, K− j−1 l ∈{i / 1 ,...,i j } xi =0
1 N− j−1 K− j−1
(K − j)
∑
ax
ax = 0 .
x:xi1 =···=xi j =1
|, the second equality follows by writing out M. The The first equality follows by writing out hψii,1 1 ,...,i j third equality follows because, for every x with |x| = K and xi1 = · · · = xi j = 1, there are K − j more l ∈ [N] |ψi satisfying xl = 1. The fourth equality follows because ∑x:xi =···=xi j =1 ax is a constant times hψii,0 1 ,...,i j 1
and hψii,0 |ψi = 0, because |ψi ∈ (T ji,0 )⊥ . 1 ,...,i j × N−1 matrix, with (BM)x,y only depending on |{` : x` = y` = 1}|. Furthermore, BM is an N−1 K K i,1 Therefore, S j is an eigenspace of BM and, since |ψ˜ ii,1 i ∈ Si,1 j , we have 1 ,...,i j BM|ψ˜ ii,1 i = λ |ψ˜ ii,1 i 1 ,...,i j 1 ,...,i j for an eigenvalue λ independent of i1 , . . . , i j . Together with equation (7.4), this implies equation (7.2) with a12 = λ /c. Equation (7.3) follows by proving i = c|ψ˜ ii,0 i and CM T |ψ˜ ii,0 i = λ |ψ˜ ii,0 i, M T |ψ˜ ii,1 1 ,...,i j 1 ,...,i j 1 ,...,i j 1 ,...,i j T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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in a similar way. We now diagonalize the matrix 0
M = It has two eigenvectors:
α1 β1
and
α2 β2
a11 a12 a21 a22
.
. Equation (7.1) implies that, for any i1 , . . . , i j ,
α1 |ψ˜ ii,0 i + β1 |ψ˜ ii,1 i 1 ,...,i j 1 ,...,i j is an eigenvector of ρt,i with the same eigenvalue λ . Therefore, Sαi 1 ,β1 , j is an eigenspace of ρt,i . Similarly, Sαi 2 ,β2 , j is an eigenspace of ρt,i . Vectors α1 |ψ˜ ii,0 i + β1 |ψ˜ ii,1 i and α2 |ψ˜ ii,0 i + β2 |ψ˜ ii,1 i together 1 ,...,i j 1 ,...,i j 1 ,...,i j 1 ,...,i j i and |ψ˜ ii,1 i. Since the vectors |ψ˜ ii,` i span Si,` span the same space as vectors |ψ˜ ii,0 j , this means 1 ,...,i j 1 ,...,i j 1 ,...,i j that i,1 Si,0 j ⊕ S j ⊆ Sα1 ,β1 ,i ⊕ Sα2 ,β2 ,i . Therefore, repeating this argument for every j gives a collection of eigenspaces that span the entire state i space for HI . This means that any eigenspace of ρt,i is a direct sum of some of eigenspaces Sα,β , j. Proof of Claim 5.3. For part (i), consider the states |ψi1 ,...,i j i spanning T j . We have s |ψi1 ,...,i j i =
N − k i,0 i+ |ψ N − j i1 ,...,i j
s
K − j i,1 i |ψ N − j i1 ,...,i j
(7.6)
because an (N − K)/(N − j) fraction of the states |x1 . . . xN i with |x| = K and xi1 = · · · = xi j = 1 have i,0 i,1 ⊥ ⊕ T j−1 ) are xi = 0 and the rest have xi = 1. The projection of these states to (T j−1 s
N − K i,0 |ψ˜ i+ N − j i1 ,...,i j
s
K − j i,1 |ψ˜ i N − j i1 ,...,i j
which, by equation (5.3) are exactly the states spanning Sαi 0 ,β0 , j . Furthermore, we claim that i,0 i,1 T j−1 ⊆ T j−1 ⊕ T j−1 ⊆ Tj .
(7.7)
The first containment is true because T j−1 is spanned by the states |ψi1 ,...,i j−1 i which either belong to i,1 i,1 T j−2 ⊆ T j−1 (if one of i1 , . . . , i j−1 is equal to i) or are a linear combination of states |ψii,0 i and 1 ,...,i j−1 i,0 i,1 i which belong to T j−1 and T j−1 . The second containment follows because the states |ψii,1 i |ψii,1 1 ,...,i j−1 1 ,...,i j−1 i,1 spanning T j−1 are the same as the states |ψi,i1 ,...,i j−1 i which belong to T j and the states |ψii,0 i spanning 1 ,...,i j−1 i,0 T j−1 can be expressed as linear combinations of |ψi1 ,...,i j−1 i and |ψi,i1 ,...,i j−1 i which both belong to T j . The first part of (7.7) now implies i,0 i,1 ⊥ Sαi 0 ,β0 , j ⊆ (T j−1 ⊕ T j−1 ) ⊆ (T j−1 )⊥ .
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We also have Sαi 0 ,β0 , j ⊆ T j , because, Sαi 0 ,β0 , j is spanned by the states P(T i,0 ⊕T i,1 )⊥ |ψi1 ,...,i j i = |ψi1 ,...,i j i − PT i,0 ⊕T i,1 |ψi1 ,...,i j i j−1
j−1
j−1
j−1
and |ψi1 ,...,i j i belongs to T j by the definition of T j and PT i,0 ⊕T i,1 |ψi1 ,...,i j i belongs to T j because of the j−1
j−1
second part of (7.7). Therefore, Sαi 0 ,β0 , j ⊆ T j ∩ (T j−1 )⊥ = S j . For the part (ii), we have i,1 i,0 i,1 Sαi 0 ,β0 , j ⊆ Si,0 j ⊕ S j ⊆ T j ⊕ T j ⊆ T j+1 ,
where the first containment is true because Sαi 0 ,β0 , j is spanned by linear combinations of vectors |ψ˜ ii,0 i 1 ,...,i j ˜ ii,1 (which belong to Si,0 i (which belong to Si,1 j ) and vectors |ψ j ) and the last containment is true because 1 ,...,i j of the second part of equation (7.7). Let i |ψ˜ ii,1 i |ψ˜ ii,0 1 ,...,i j 1 ,...,i j − α (7.8) |ψi = β0 0 ik ik k|ψ˜ ii,0 k|ψ˜ ii,1 1 ,...,i j 1 ,...,i j be one of the vectors spanning Sβi 0 ,−α0 , j . To prove that |ψi is in S j+1 = T j+1 − T j , it remains to prove that |ψi is orthogonal to T j . This is equivalent to proving that |ψi is orthogonal to every vector |ψi01 ,...,i0j i spanning T j . i,0 i,1 ⊥ Case 1 {i01 , . . . , i0j } = {i1 , . . . , i j }: Since |ψi belongs to (T j−1 ⊕ T j−1 ) , it suffices to prove that |ψi i,0 i,1 ⊥ ⊕ T j−1 ) which, by the discussion after the equais orthogonal to the projection of |ψi1 ,...,i j i to (T j−1 tion (7.6), is equal to i |ψ˜ ii,1 i |ψ˜ ii,0 1 ,...,i j 1 ,...,i j + β . (7.9) α0 0 i,0 i,1 k|ψ˜ i1 ,...,i j ik k|ψ˜ i1 ,...,i j ik
¿From equations (7.8) and (7.9) we see that the inner product of the two states is α0 β0 − β0 α0 = 0. Case 2 {i01 , . . . , i0j } 6= {i1 , . . . , i j } but one of i01 , . . . , i0j is equal to i: For simplicity, assume i = i0j . Then i,1 i |ψi01 ,...,i0j i is the same as |ψii,1 i which belongs to T j−1 . By the definition of Sα,β 0 ,...,i0 , j , the vector |ψi 1
j−1
i,0 i,1 ⊥ belongs to (T j−1 ⊕ T j−1 ) and is therefore orthogonal to |ψii,1 i. 0 ,...,i0 1
j−1
Case 3 {i01 , . . . , i0j } 6= {i1 , . . . , i j } and none of i01 , . . . , i0j is equal to i: One of i01 , . . . , i0j must be not in {i1 , . . . , i j }. For simplicity, assume it is i0j . We have |ψi01 ,...,i0j−1 i =
∑
i0 ∈{i / 01 ,...,i0j−1 }
|ψi01 ,...,i0j−1 ,i0 i .
i,0 i,1 Also, hψi01 ,...,i0j−1 |ψi = 0, because |ψi01 ,...,i0j−1 i is in T j−1 ⊕ T j−1 . As proved in the previous case,
hψi01 ,...,i0j−1 ,i |ψi = 0 . T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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We therefore have
∑
hψi01 ,...,i0j−1 ,i0 |ψi = 0 .
(7.10)
i0 ∈{i / 01 ,...,i0j−1 ,i}
/ {i01 , . . . , i0j−1 , i}. Therefore, By symmetry, the inner product hψi01 ,...,i0j−1 ,i0 |ψi is the same for every i0 ∈ / {i01 , . . . , i0j−1 , i}. equation (7.10) means hψi01 ,...,i0j−1 ,i0 |ψi = 0 for every i0 ∈ i i Proof of Claim 5.5. τα,β , j is a mixture of states |ψi from the subspace Sα,β , j . We prove the claim by showing that, for any of those states |ψi, the squared norm of its projection to S j+1 is equal to the right i hand side of Claim 5.5. Since |ψi ∈ Sα,β , j we can write it as
|ψi =
∑
i1 ,...,i j
i + β |ψ˜ ii,1 i) ai1 ,...,i j (α|ψ˜ ii,0 1 ,...,i j 1 ,...,i j
for some ai1 ,...,i j . Let |ψ + i =
∑
i − α0 |ψ˜ ii,1 i) and ai1 ,...,i j (β0 |ψ˜ ii,0 1 ,...,i j 1 ,...,i j
∑
i + β0 |ψ˜ ii,1 i) . ai1 ,...,i j (α0 |ψ˜ ii,0 1 ,...,i j 1 ,...,i j
i1 ,...,i j −
|ψ i =
i1 ,...,i j
Then, |ψi is a linear combination of |ψ + i which belongs to Sβi 0 ,−α0 , j ⊂ S j+1 (by Claim 5.3) and |ψ − i which belongs to Sαi 0 ,β0 , j ⊆ S j . Moreover, all three states are linear combinations of |ψ 0 i, |ψ 1 i defined by i. |ψ ` i = ∑ ai1 ,...,i j |ψ˜ ii,` 1 ,...,i j i1 ,...,i j
We have |ψi = α|ψ 0 i + β |ψ 1 i ,
|ψ + i = β0 |ψ 0 i − α0 |ψ 1 i and |ψ − i = α0 |ψ 0 i + β0 |ψ 1 i .
Since |ψ + i and |ψ − i belong to subspaces S j+1 and S j which are orthogonal, we must have hψ + |ψ − i = 0. This means α0 β0 kψ 0 k2 − β0 α0 kψ 1 k2 = 0 . By dividing the equation by α0 β0 , we get kψ 0 k2 = kψ 1 k2 and kψ 0 k = kψ 1 k. Since kψk = 1, this means that 1 kψ 0 k = kψ 1 k = p = 1. 2 α +β2 Since |ψi lies in the subspace spanned by |ψ + i which belongs to S j+1 and |ψ − i which belongs to S j , the norm of the projection of |ψi to S j+1 is equal to |hψ|ψ + i|/kψ + k. By expressing |ψi and |ψ + i in terms of |ψ 0 i and |ψ 1 i, we get |hψ|ψ + i| αβ0 kψ 0 k2 − α0 β kψ 1 k2 |αβ0 − α0 β | = q = q , kψ + k β02 kψ 0 k2 + α02 kψ 0 k2 α02 + β02 proving the claim. T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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Proof of Claim 5.6. We will prove kψ˜ ii,0 k ≥ 12 kψ˜ ii,1 k, because that means 1 ,...,i j 1 ,...,i j √ √ √ √ N − K i,0 N −K 1 N − K K − j i,1 √ α0 = √ kψ˜ k≥ √ kψ˜ k= √ β0 2 K − j N − j i1 ,...,i j N − j i1 ,...,i j 2 K− j and β0 q
α02 + β02
≤q
p
4(K − j) =q . =√ N−K N + 3K − 4 j 1 + 4(K− j) 1
β0 N−K 2 2 4(K− j) β0 + β0
To prove kψ˜ ii,0 k ≥ kψ˜ ii,1 k, we calculate the vector 1 ,...,i j 1 ,...,i j |ψ˜ ii,0 i = P(T i,0 )⊥ |ψii,0 i. 1 ,...,i j 1 ,...,i j j−1
i,0 i and the subspace T j−1 Both the vector |ψii,0 are fixed by 1 ,...,i j
Uπ |xi = |xπ(1) . . . xπ(N) i i is fixed by for any permutation π that fixes i and maps {i1 , . . . , i j } to itself. This means that |ψ˜ ii,0 1 ,...,i j i only depends on any such Uπ as well. Therefore, the amplitude of |xi, |x| = K, xi = 0 in |ψ˜ ii,0 1 ,...,i j i is of the form |{i1 , . . . , i j } ∩ {t : xt = 1}|. This means |ψ˜ ii,0 1 ,...,i j j
|ψ0 i =
∑
∑ αm
m=0
|xi .
x:|x|=K,xi =0 |{i1 ,...,i j }∩{t:xt =1}|=m
To simplify the following calculations, we multiply α0 , . . . , α j by the same constant so that αj = q
1 N− j−1 K− j
.
Then, |ψ˜ ii,0 i remains a multiple of |ψ0 i but may no longer be equal to |ψ0 i. 1 ,...,i j The coefficients α0 , . . . , α j−1 should be such that the state is orthogonal to T j−1 and, in particular, i for ` ∈ {0, . . . , j − 1}. By writing out hψ0 |ψii,0 i = 0, we get orthogonal to states |ψii,0 1 ,...,i` 1 ,...,i` j
∑ αm m=`
N − j−1 K −m
j−` = 0. m−`
(7.11)
i is a uniform superposition of all |xi, |x| = K, xi = 0, xi1 = · · · = To show that, we first note that |ψii,0 1 ,...,i` xil = 1. If we want to choose x subject to those constraints and also satisfying |{i1 , . . . , i j }∩{t : xt = 1}| = m, we have to set xt = 1 for m − ` different t ∈ {i`+1 , . . . , i j } and for K − m different t ∈ / {i, i1 , . . . , i j }. j−` N− j−1 This can be done in m−` and K−m different ways, respectively. T HEORY OF C OMPUTING, Volume 6 (2010), pp. 1–25
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By solving the system of equations (7.11), we get that the only solution is N− j−1 j−m K− j (−1) αj . N− j−1 K−m
αm =
(7.12)
Let |ψ00 i = |ψ0 i/kψ0 k be the normalized version of |ψ0 i. Then |ψ˜ ii,0 i = hψ00 |ψii,0 i|ψ00 i and 1 ,...,i j 1 ,...,i j kψ˜ ii,0 k = hψ00 |ψii,0 i= 1 ,...,i j 1 ,...,i j
hψ0 |ψii,0 i 1 ,...,i j kψ0 k
.
(7.13)
First, we have hψ0 |ψii,0 i = 1, 1 ,...,i j j−1 because |ψii,0 i consists of N− basis states |xi, xi = 0, xi1 = · · · = xi j = 1, each of which has K− j 1 ,...,i j q N− j−1 in both |ψ0 i and |ψii,0 i. Second, amplitude 1/ K− j 1 ,...,i j 2 j N− j−1 j j N − j − 1 K− j kψ0 k2 = ∑ αm2 = ∑ α 2j N− j−1 m K − m m m=0 m=0 K−m j j N− j−1 j j (K − m)!(N − K + m − j − 1)! K− j = ∑ = ∑ N− j−1 (K − j)!(N − K − 1)! m=0 m m=0 m K−m j j (K − m) · · · (K − j + 1) = ∑ m=0 m (N − K − 1) · · · (N − K + m − j) j
(7.14)
j−1 with the first equality following because there are mj N− K−m vectors x such that |x| = K, xi = 0, xt = 1 for m different t ∈ {i1 , . . . , i j } and K − m different t ∈ / {i, i1 , . . . , i j }, the second equality following from q N− j−1 equation (7.12) and the third equality following from our choice α j = 1/ K− j . We can similarly calculate kψ˜ ii,1 k. We omit the details and just state the result. The counterpart 1 ,...,i j of equation (7.13) is kψ˜ ii,1 k= 1 ,...,i j
hψ1 |ψii,1 i 1 ,...,i j kψ1 k
,
with |ψ1 i being the counterpart of |ψ0 i: j
|ψ1 i =
∑ αm
m=0
∑
|xi ,
x:|x|=K,xi =1 |{i1 ,...,i j }∩{`:x` =1}|=m
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q with α0 = 1/
N− j−1 K− j−1
. Similarly as before, we get hψ1 |ψii,1 i = 1 and 1 ,...,i j N− j−1 j K− j−1 kψ1 k2 = ∑ N− j−1 m m=0 K−m−1 j j (K − m − 1) . . . (K − j) = ∑ . m=0 m (N − K) . . . (N − K + m − j + 1) j
(7.15)
Each term in (7.14) is (K − m)(N − K + m − j) (K − j)(N − K) times the corresponding term in equation (7.15). We have K K −m N −K +m− j ≤ ·2 = 4, K− j N −K K/2 because j ≤ K/2 and N − K + m − j ≤ N − K (because of m ≤ j). Therefore, kψ0 k2 ≤ 4kψ1 k2 which implies 1 1 1 k= kψ˜ ii,0 ≥√ = kψ˜ ii,1 ,...,i k . 1 ,...,i j kψ0 k 4kψ1 k 2 1 j ˇ Acknowledgment. I would like to thank Frederic Magniez, Robert Spalek, Ronald de Wolf, and several anonymous referees for very helpful comments on a draft of this paper.
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AUTHOR Andris Ambainis professor University of Latvia Raina bulv. 19, Riga, LV-1586, Latvia andris ambainis lu lv http://home.lanet.lv/∼ambainis ABOUT THE AUTHOR A NDRIS A MBAINIS received his Ph. D. from the University of California, Berkeley in 2001, supervised by Umesh Vazirani. After that, he was a postdoc at the Institute for Advanced Study, Princeton and at the University of California, Berkeley, and a faculty member at the University of Waterloo, Canada. In 2007, Andris returned to his native Latvia and became Professor at the University of Latvia. His research interests include many areas of quantum computing and quantum information theory (quantum algorithms, quantum complexity theory, quantum cryptography, pseudorandom quantum states, etc.), as well as the classical theory of computation.
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