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 al-gorithms. 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. This result was previously proved by Klauck, ˇSpalek, 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
generalized in several different ways [4, 18, 8]. ˇSpalek and Szegedy [23] then showed that all the generalizations are equivalent and, for certain problems, cannot improve the best known lower bounds. For example, it was shown in [23,24] that the adversary methods of [4,18,8] cannot prove a lower bound for a total Boolean function that exceedsO(pC0(f)C1(f)), whereC0(f)andC1(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 isΘ(N2/3) [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 usesO(N1.3)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 theK-fold search problem.
In theK-fold search problem, a black box contains x1, . . . ,xN such that|{i:xi=1}|=K and we
have to find all theKvalues ofisuch thatxi=1. This problem can be solved withO(√NK)queries. It can be easily shown, using any of the previously known methods, thatΩ(
√
NK)quantum queries are
required. A more difficult problem is to show thatΩ(
√
NK)queries are required, even if the algorithm only has to be correct with an exponentially small probabilityc−K,c>1. This result is known as the
strong direct product theoremfork-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].
The strong direct product theorem for quantum search was first shown by Klauck, ˇSpalek, and de
Wolf [16], using the polynomials method. No proof using the adversary method has been known and, as
we show inSection 3, the previously known adversary methods are insufficient to prove a strong direct product theorem forK-fold search.
Recent developments After the author completed the research presented in this paper, several devel-opments occurred.
1. Together with ˇSpalek and de Wolf [7], we have used the methods from this paper to prove a direct
product theorem fort-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].
3. Høyer, Lee, and ˇSpalek [14] generalized the usual adversary method in a different way, by extend-ing 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
O
Adv±(f) log 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 forKmarked elements,SEARCHK(N): Given a black box containingx1, . . . ,xN∈ {0,1}such thatxi=1 for exactlyKvaluesi∈ {1,2, . . . ,N}, find allKvaluesi1, . . . ,iKsatisfyingxij =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
setsS⊆[N]of sizeK, 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 oracleXand the complexity of f
is the number of quantum queries needed to compute f. A quantum computation withT queries is just
a sequence of unitary transformations
U0→O→U1→O→ · · · →UT−1→O→UT.
TheUj can be arbitrary unitary transformations that do not depend on the input bitsx1, . . . ,xN. The transformations O are query (oracle) transformations which depend on x1, . . . ,xN. To define O, we represent basis states as|i,ziwhereiconsists ofdlog(N+1)ebits andzconsists of all other bits. Then,
Ox maps|0,zito itself and|i,zito(−1)xi|i,zifori∈ {1, ...,N}(i. e., we change phase depending onxi,
unlessi=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 state. The result of the computation is the string ofdlog2 NKerightmost bits of the state obtained by the measurement, which is interpreted as a description of one of the NKsubsetsS⊆ {1, . . . ,N},|S|=K.
3
Overview of the adversary method
We describe the adversary method of [3].
LetX be a subset of the set of possible inputs{0,1}N. We run the algorithm on a superposition of
H=HA⊗HI whereHI is an “input subspace” spanned by basis vectors|xi corresponding to inputs
x∈X.
LetUTOUT−1· · ·U0be the sequence of unitary transformations onHAperformed by the algorithm
A, whereU0, . . . ,UT denote the transformations that do not depend on the input andO stands for the
query transformations. We transform this sequence into a sequence of unitary transformations onH. A
unitary transformationUi on HA corresponds to the transformationUi0=Ui⊗I on the wholeH. The
query transformationOcorresponds to a transformationO0that is equal toOxon the subspaceHA⊗ |xi.
We perform the sequence of transformationsUT0 O0UT0−1· · ·U00 on the starting state
|ψstarti=|0i ⊗
∑
x∈S αx|xi.
Then, the final state is
|ψendi=
∑
x∈X
αx|ψxi ⊗ |xi,
where|ψxi is the final state of A=UTOUT−1. . .U0 on inputx. This follows from the fact that the restrictions ofUT0,O0,UT0−1, . . . ,U00 toHA⊗ |xiareUT,Ox,UT−1, . . . ,U0and these are exactly the trans-formations of the algorithmAon inputx.
Letρend be the reduced density matrix of theHI register of the state|ψendi. The adversary method of [3,4] works by showing the following two statements.
• Letx∈Xandy∈Xbe 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 bothxandythen
|(ρend)x,y| ≤2
p
ε(1−ε)|αx||αy|.
• For any algorithm that usesT queries, there exist inputsx,y∈Ssuch that f(x)6= f(y)and
(ρend)x,y>2
p
ε(1−ε)|αx||αy|.
These two statements together imply that any algorithm computing f must use more thanT queries.
An equivalent approach [15, 4] is to consider the inner productshψx|ψyi between the final states
|ψxiand|ψyiof the algorithm on inputsxandy. Then,|(ρend)x,y| ≤2
p
ε(1−ε)|αx||αy|is equivalent to
|hψx|ψyi| ≤2
p
ε(1−ε).
As a result, both of the above statements can be described in terms of inner productshψx|ψyi, without
explicitly introducing the registerHI. The first statement says that, for the algorithm to succeed on inputs
x,ysuch that f(x)6=f(y), the states|ψxiand|ψyimust be sufficiently far apart from one another (so that
the inner product|hψx|ψyi|is at most 2
p
ε(1−ε)). The second statement says that this is impossible if
the algorithm only usesT queries.
This approach breaks down if we consider computing a function fwith success probability p<1/2. (f has to have more than 2 values for this task to be nontrivial.) Then, |ψxi and|ψyi could be the
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 registerHI, together with two new ingredients:
1. Symmetrization.We symmetrize the algorithm by applying a random permutationπ∈SN to the inputx1, . . . ,xN.
2. Eigenspace analysis.We study the eigenspaces ofρstart,ρendand density matrices describing the state ofHIat 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<1 such that, for any K≤N/2, solving SEARCHK(N)with probability at least cKrequires(ε−o(1))√NK queries.
4.1 General framework
We first give a high-level overview of the new method, in a form that may be applicable to a variety of problems. After that, inSection 4.2, we will describe how to adapt this general method to theK-fold search problem.
As before, letX ⊆ {0,1}N be a subset of the set of inputs for the function f(x
1, . . . ,xN). LetGbe
the group of all permutationsπon{1, . . . ,N}that fixX:
{(xπ(1), . . . ,xπ(N)):(x1, . . . ,xN)∈X}=X.
Additionally we assume for allπ∈Gthat f(xπ(1), . . . ,xπ(N))andπ uniquely determine f(x1, . . . ,xN). We chooseX so thatXconsists of “hard” inputs (inputs on which f is difficult to evaluate with few queries) and the symmetry groupGis as large as possible. For example, in the case ofK-fold search,X
is the set of allx∈ {0,1}N,|x|=KandGconsists of all permutations on{1, . . . ,N}.
LetAbe an algorithm for f(x1, . . . ,xN)and letHAbe the workspace on whichAoperates.
We first “symmetrize”Aby adding an extra registerHSholding a permutationπ∈G. Initially,HS
holds a uniform superposition of all permutationsπ:
1
p
|G|π
∑
∈G|πi.Before each queryO, we insert a transformation|ii|πi 7→ |π−1(i)i|πion the part of the state containing the indexito be queried andHS. After the query, we insert a transformation|ii|πi 7→ |π(i)i|πi. The
effect of the symmetrization is that, on the subspace|si ⊗ |πi, the algorithm is effectively running on
the inputxπ(1), . . . ,xπ(N). At the end of algorithm, we apply a unitary transformation that replaces the
If the original algorithmAsucceeds on every input(x1, . . . ,xN)with probability at leastε, the
sym-metrized algorithm also succeeds with probability at leastε, since its success probability is just the
average of the success probabilities ofA over all(xπ(1), . . . ,xπ(N)), π ∈G. Next, we recast Ainto a
different form, using a register that stores the inputx1, . . . ,xN, as inSection 3.
LetHIbe an|X|-dimensional Hilbert space whose basis states correspond to inputs(x1, . . . ,xN)∈X.
We transform the symmetrized version of A into a sequence of transformations on a Hilbert space
H=HA⊗HS⊗HI. As before, a non-query transformationU on HA⊗HS is replaced byU⊗I on
H. A query is replaced by a transformationOthat is equal toOx1,...,xN⊗I on the subspace consisting
of states of the form |siA,S⊗ |x1. . .xNiI. The starting state of the algorithm on Hilbert space H is
|ϕ0i=|ψstartiA,S⊗ |ψ0iI where|ψstartiA,S is the starting state ofAas an algorithm acting on HA⊗HS
and|ψ0iIis some fixed superposition of states|x1. . .xNi,(x1, . . . ,xN)∈X.
Let|ψtibe the state of the algorithmA, as a sequence of transformations onH, after thetthquery.
Letρt be the mixed state obtained from|ψtiby tracing out theHA⊗HS register and letλ1,λ2, . . .be all the distinct eigenvalues ofρt. LetSi be the subspace ofHI consisting of all eigenvectors with the
eigenvalueλi and letPibe the projector to the subspaceSi. Then,
ρt=
∑
iλiPi.
For a permutationπ∈G, letUπ be the unitary transformation on the registerHIdefined by
Uπ|x1. . .xNi=|xπ(1). . .xπ(N)i.
Then, because of the symmetrization step, ρt is invariant underUπ: ρt =UπρtUπ−1. This means that
every eigenspaceSi is fixed byUπ, for everyπ∈G.
LetVbe the collection of all invariant subspacesS≤HA(subspacesSthat are fixed by allπ∈G). Then, we can always expressρt as
ρt =
∑
S∈VλSPS
where PS is a projector to the subspace S. This means that the state of the algorithm can be fully
described by the vector(λS)S∈V. (For some symmetry groupsG, there could be infinitely many invariant
subspaces. ThenVwill be infinite but only finitely many of theλSwill be nonzero.)
We divideVinto a setVgoodof “good” subspaces and a setVbad of “bad” subspaces so that, if the algorithmAsucceeds, a non-negligible part of the final stateρT must consist ofPS,S∈Vgood.
To prove a lower bound on the number of queries, we show that the initial state ofHIis inVbadand then prove thatT0queries are not enough to move a non-negligible part of the state toS∈Vgood. That is done by bounding the change in(λS)S∈V in one query, as follows. Letρt be the state ofHI before the
(t+1)stquery. We decompose
ρt = N
∑
i=0ρt,i,
withρt,i being the part of the state in which the query register ofAcontainsi. LetG(i) consist of all π∈Swithπ(i) =i. Thenρt,i is fixed byUπ for allπ∈G(i). LetV(i)consist of all subspacesSthat are
fixed by allUπ,π∈G(i). Then
ρt,i=
∑
S∈V(i)We use this decomposition to describe the effect of a query onρt,i. Then we relate subspacesS∈V(i)to
subspacesS∈V. This allows us to bound the change in(λS)S∈V.
Specialization tok-fold search In the case ofk-fold search (described in the next section), there are justK+1 invariant subspaces S0, . . . ,SK with Si intuitively corresponding to the algorithm knowing locations ofiout ofKitems jwithxj=1. The state of the algorithm can then be described by a vector
(λ0, . . . ,λK)of lengthK+1, whereλi describes the probability that the algorithm knows locations ofi
out of theKitems.
The bad subspaces that make upVbad areS0, . . . ,SK/2which correspond toAknowing at mostK/2 of theK locations j with xj =1. The good subspaces that make up Vgood are SK/2+1, . . . ,SK which
correspond toAknowing more thanK/2 of theKlocations jwithxj=1.
4.2 K-fold search
We now proveTheorem 4.1. LetAbe an algorithm for SEARCHK(N)that usesT≤ε
√
NKqueries.
As described in the previous subsection, we first “symmetrize”Aby adding an extra register HS
holding a permutationπ∈SN. Initially,HSholds a uniform superposition of all permutationsπ:
1
√
N!π
∑
∈SN |πi.The result of this step is that, on the subspace consisting of all states|si ⊗ |πi, the algorithm runs on the
inputxπ(1), . . . ,xπ(N). At the end of the algorithm, we apply the transformation
|i1i. . .|iKi|πi 7→ |π−1(i1)i. . .|π−1(iKi|πi
on the part ofHAthat holds the output of the algorithm andHS. This converts the answer for the input
xπ(1), . . . ,xπ(N)into the answer for the actual inputx1, . . . ,xN.
We then add an input registerHI which initially, holds the starting state
|ψ0i= 1
q
N K
x
∑
1,...,xN:
x1+···+xN=K
|x1. . .xNi.
Let|ψtibe the state of the algorithmA, as a sequence of transformations onH=HA⊗HS⊗HI, after
thetthquery. Then,
|ψti=
1
q
N K
N! x1,...,
∑
xN:x1+···+xN=K
∑
π∈SN|ψtxπ(1),...,xπ(N)iA|πiS|x1, . . . ,xNiI
where|ψtxπ(1),...,xπ(N)idenotes the state ofHAaftertsteps, on the inputxπ(1), . . . ,xπ(N).
Letρt be the mixed state obtained from|ψtiby tracing out theHAregister. We claim that the states ρt have a special form, due to our symmetrization step.
Lemma 4.2. The entries(ρt)x,y are the same for all x= (x1, . . . ,xN), y= (y1, . . . ,yN) with the same
Proof. Sinceρt is independent of the way how theHA⊗HSis traced out, we first measureHS (in the
|πi basis) and then measureHA (arbitrarily). When measuringHS, everyπ is obtained with an equal
probability. Letρt,πbe the reduced density matrix ofHI, conditioned on the measurement ofHSgiving π. Then
ρt=
∑
π1
N!ρt,π.
The entry(ρt,π)x,y is the same as the entry (ρt,id)π−1(x),π−1(y) because the symmetrization by π maps
π−1(x),π−1(y)tox,y. Letx,y,x0,y0be such that|{i:xi=yi=1}|=|{i:xi0=y
0
i=1}|. Then there is a
permutationτ∈SNsuch thatτ(x) =x0,τ(y) =y0. Therefore, (ρt)x,y=
1
N!π
∑
∈SN
(ρt,π)x,y=
1
N!π
∑
∈SN
(ρt,id)π−1(x),π−1(y)
= 1
N!π
∑
∈SN
(ρt,id)π−1τ(x),π−1τ(y)= (ρt)τ(x),τ(y)= (ρt)x0,y0.
This means that(ρt)x,yonly depends on|{`:x`=y`=1}|.
Any NK× N
K
matrix with this property shares the same eigenspaces. Namely, we have
Lemma 4.3(Knuth [17]). Let A be an NK× NK
matrix whose rows and columns are indexed by 0-1 sequences x1, . . . ,xNwith x1+· · ·+xN=K. Assume that A is such that Ax,yonly depends on the number of bits{i:xi=yi=1}. Then, the eigenspaces of A are S0,S1, . . . ,SKwhere T0=S0consists of multiples
of|ψ0iand, for j>0, Sj=Tj∩(Tj−1)⊥, with Tjbeing the space spanned by all states
|ψi1,...,iji=
1
q
N K−j
∑
x1,...,xN:
x1+···+xN=K,
xi1=···=xi j=1
|x1, . . . ,xNi.
Letτjbe the completely mixed state over the subspaceSj.
Lemma 4.4. There exist pt,0≥0, . . . ,pt,K≥0such thatρt =∑Kj=0pt,jτj.
Proof. ByLemma 4.3,S0, . . . ,SK are the eigenspaces ofρt. Therefore,ρt is a linear combination of the
projectors toS0, . . . ,SK. Sinceτjis a multiple of the projector toSj, we have
ρt = K
∑
j=0pt,jτj.
Sinceρt is a density matrix, it must be positive semidefinite. This means thatpt,0≥0, . . . ,pt,K≥0.
Informally, we can interpret pt,j as the probability that, aftertqueries, the algorithmHA knows the
locations fortout of theK items jwithxj =1. Letqt,j =pt,j+pt,j+1+· · ·+pt,K. The theorem now
follows from the following lemmas.
Proof. In the starting state, HI contains the state|ϕ0i, independent ofHA andHP. Therefore, tracing outHA⊗HPleaves the stateρ0=|ψ0ihψ0|.
Lemma 4.6. For all j∈ {1, . . . ,K}and all t we have qt+1,j+1≤qt,j+1+4
√
K
√
Nqt,j. Proof. InSection 5.
Lemma 4.7. qt,j≤ tj
4 √ K √ N j .
Proof. By induction ont. The base case,t=0 follows immediately fromp0,0=1 and
p0,1=· · ·=p0,K=0.
For the induction step, we have
qt+1,j≤qt,j+
4√K
√
N qt,j−1≤
t j
4√K
√ N !j +4 √ K √ N t j−1
4√K
√
N
!j−1
= t j + t j−1
4√K
√
N
!j
=
t+1
j
4√K
√
N
!j
,
where the first inequality follows fromLemma 4.6and the second inequality follows from the inductive
assumption.
Lemma 4.8. If t≤0.03√NK then qt,j<0.66jfor all j>K/2.
Proof. By the well known inequality tj
<(et/j)jfor j≥1, we have
qt,j≤
t j
4√K
√
N
!j
≤ 4 √
Ket
√
N j
!j
.
Let j>K/2 andt≤0.03√NK. Then
4√Ket
√
N j ≤
0.12e√K√NK
√
NK/2 <0.66,
implying the lemma.
Lemma 4.9. The success probability ofAis at most
N K/2 N K +4 p
qt,K/2+1.
To complete the proof, given the two Lemmas, we choose a constantc>√40.66=0.90133. . . and setε=0.03. Then, byLemma 4.9, the success probability ofAis at most
N K/2
N K
+4
p
0.66K/2.
The first term is equal to
N K/2
N K
=
K!(N−K)! (K/2)!(N−K/2)! =
K(K−1). . .(K/2+1)
(N−K/2)(N−K/2−1). . .(N−K+1)
≤
K N−K/2
K/2
≤
N/2 3N/4
K/2
=
2 3
K/2
=o(cK),
where the last two inequalities follow fromK<N/2. The second term is 4
√
0.66K/2=o(cK).
5
Proof of
Lemma 4.6
Let|ψtibe the state before(t+1)stquery. We decompose|ψtias∑iN=0ai|ψt,ii, where|ψt,iiis 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 thata1,· · ·,aN are all positive reals and, thus,a1=a2=· · ·=
aN. Letρt,i=|ψt,iihψt,i|. Then
ρt= N
∑
i=0a2iρt,i. (5.1)
Claim 5.1. Leti∈ {1, . . . ,N}. The entry(ρt,i)x,y only depends onxi,yi, and the cardinality of{`:`6= i,x`=y`=1}.
Proof. The main idea is similar toLemma 4.2.
This time, we trace out all registers, except forHI and the query register. This gives us a density matrixρwhose rows and columns are indexed by pairs(x,j)wherexis an input and j∈ {1, . . . ,N}. Let x∈ {0,1}N,y∈ {0,1}N, j∈[N],k∈[N]andx0∈ {0,1}N,y0∈ {0,1}N, j0∈[N], k0∈[N]be such that
there is a permutationπ∈SNwithx0=π(x),y0=π(y),i0=π(i)and j0=π(j). By an argument similar
to the proof ofLemma 4.2, we have
ρ(x,i),(y,j)=ρ(x0,i0),(y0,j0). (5.2)
We now observe thatρt,i is the submatrix ofρ consisting of rows and columns indexed by pairs(x,i),
with all possiblex∈ {0,1}N,|{i:x
i=1}|=K.
If we have inputsx,y,x0,y0withxi=x0i,yi=y0iand|{`:`6=i,x`=y`=1}|=|{`:`6=i,x0`=y0`=1}|, then we can construct a permutationπwithπ(i) =i,π(x) =x0andπ(y) =y0. Equation (5.2) then implies
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. Leti∈[N]and j∈ {0,1, . . . ,K}. We defineTji,0to be the subspace spanned by all states
|ψii,0 1,...,iji=
1
q
N−j−1
K−j
∑
x:|x|=K xi1=···=xi j=1,xi=0
|x1. . .xNi,
with(i1, . . . ,ij)ranging over all tuples of jdistinct elements of[N]− {i}. Similarly, we defineTji,1to be
the subspace spanned by all states
|ψii,1 1,...,iji=
1
q
N−j−1
K−j−1
∑
x:|x|=K xi1=···=xi j=1,xi=1
|x1. . .xNi.
Let Sij,0=Tji,0∩(Tji−,01)⊥ and Sij,1 =Tji,1∩(Tji−,11)⊥. Equivalently, we can define Sij,0 and Sij,1 as the subspaces spanned by the states|ψ˜ii,0
1,...,ijiand|ψ˜
i,1
i1,...,iji, respectively, with |ψ˜ii,`
1,...,iji=P(Tji−1,` )⊥|ψ i,`
i1,...,iji.
LetSiα,β,jbe the subspace spanned by all states
α
|ψ˜ii,0
1,...,iji k|ψ˜ii,0
1,...,ijik
+β
|ψ˜ii,1
1,...,iji k|ψ˜ii,1
1,...,ijik
. (5.3)
where(i1, . . . ,ij)again ranges over all tuples of jdistinct elements of[N]− {i}.
Claim 5.2. Every eigenspace ofρt,iis a direct sum of subspacesSiα,β,jfor someα,β, j. Proof. InSection 7.
Letταi,β,jbe the completely mixed state overSiα,β,j. Similarly toLemma 4.4, we can writeρt,i as
ρt,i=
∑
(α,β,j)∈At,i
piα,β,jταi,β,j, (5.4)
where(α,β,j)range over some finite setAt,i. (This set is finite because theHIregister holding|x1. . .xNi is finite dimensional and, therefore, decomposes into a direct sum of finitely many eigenspaces.) For every pair(α,β,j)∈At,i, we normalizeα,βby multiplying them by the same constant so thatα2+β2=
1. Queryingxitransforms this state to
ρt0,i=
∑
(α,β,j)∈At,i
piα,β,jταi,−β,j,
because|ψ˜ii,`
1,...,ijiis a superposition of|xiwithxi=`and, therefore, a query leaves|ψ˜
i,0
i1,...,ijiunchanged
and flips a phase on|ψ˜i1i,1,...,iji. Ifi=0, we haveρt0,0=ρt,0, because, if the query register contains|0i,
Claim 5.3. Letα0=
q
N−K N−j
ψ˜
i,0
i1,...,ij
andβ0=
q
K−j N−j
ψ˜
i,1
i1,...,ij
.
(i) Si
α0,β0,j⊆Sj;
(ii) Si
β0,−α0,j⊆Sj+1. Proof. InSection 7.
Corollary 5.4. For anyα,β we haveSiα,β,j⊆Sj⊕Sj+1.
Proof. We have Sα,β,j ⊆Sij,0⊕Sij,1, since Sα,β,j is spanned by linear combinations of states|ψ˜ii,0
1,...,iji
(which belong toSij,0) and states|ψ˜ii,1
1,...,iji(which belong toS
i,1
j ). As shown in the proof ofClaim 5.3
above,
Sij,0⊕Sij,1⊆Sα0,β0,j⊕S−β0,α0,j⊆Sj⊕Sj+1.
The next claim quantifies the overlap betweenSi
α,β,j andSj+1. Claim 5.5.
TrPSj+1τ i α,β,j=
|α β0−β α0|2
α02+β02 .
Proof. InSection 7.
To be able to use this bound, we also need to boundα0andβ0.
Claim 5.6. √β0 α02+β02
≤qN4+(K3K−−j)4j.
Proof. InSection 7.
We can now complete the proof ofLemma 4.6. By projecting both sides ofρt =∑ipt,iτi to(Tj)⊥= Sj+1⊕ · · · ⊕SK and taking trace, we get
TrP(Tj)⊥ρt= K
∑
j0=0pt,j0TrP(
Tj)⊥τj0= K
∑
j0=j+1pt,j0 =qt,j, (5.5)
with the second equality following because the statesτj0 are uniform mixtures over subspacesSj0 and S0, . . . ,Sj are contained inTj whileSj+1, . . . ,SK are contained in(Tj)⊥. Because of equations (5.1) and (5.4), this means that
qt,j+1=a20TrP(Tj)⊥ρt,0+
N
∑
i=1a2i
∑
(α,β,j0)∈A
t,i
piα,β,j0TrP(T
j)⊥τ
i
α,β,j0. (5.6)
Decomposing the state after the query in a similar way, we get
qt+1,j+1=a02TrP(Tj)⊥ρ
0
t,0+
N
∑
i=1a2i
∑
(α,β,j0)∈A
t,i
piα,β,j0TrP(T
j)⊥τ
By substracting the two sums and usingρt0,0=ρt,0, we get
qt+1,j+1−qt,j+1=
N
∑
i=1a2i
∑
(α,β,j0)∈A
t,i
piα,β,j0TrP(T
j)⊥(τ
i
α,−β,j0−ταi,β,j0). (5.7)
We now claim that all the terms in this sum with j06= jare 0. For j0< j,Sα,β,j0 ⊆Tj0+1⊆Tj, implying
that TrP(Tj)⊥ταi,β,j0 =0 and, similarly, TrP(Tj)⊥ταi,−β,j0 =0. For j0> j, Sα,β,j0 ⊆Sj0⊕Sj0+1⊆(Tj)⊥,
implying that
TrP(Tj)⊥ταi,β,j0=1, TrP(T
j)⊥τ
i
α,−β,j0=1,
and the difference of the two is 0. By removing those terms from (5.7), we get
qt+1,j+1−qt,j+1=
N
∑
i=1a2i
∑
(α,β,j)∈At,i
piα,β,jTrP(T
j)⊥(τ
i
α,−β,j−τ i
α,β,j). (5.8)
We have
TrP(T
j)⊥(τ
i
α,−β,j−τ i
α,β,j) =TrPSj+1(τ i
α,−β,j−τ i α,β,j) =
|α β0+β α0|2
α02+β02
−|α β0−β α0|
2
α02+β02 ,
where the first equality follows fromCorollary 5.4,Sj⊆Tj, andSj+1⊆(Tj)⊥, and the second equality follows fromClaim 5.5. This is at most
4|α β α0β0|
α02+β02
≤2 α0β0
α02+β02 =2 α0
q
α02+β02 β0
q
α02+β02
≤2
s
4(K−j)
N+3K−4j ≤2
r
4K N ,
where the first inequality follows from
|α β| ≤|α|
2+|
β|2
2 =
1 2
and the second inequality follows fromClaim 5.6and √α0
α02+β02 ≤1. Together with equation (5.7), this
means
qt+1,j+1−qt,j+1≤ 4√K
√
N N
∑
i=1a2i
∑
(α,β,j)∈At,i
piα,β,j. (5.9)
Similarly to equation (5.5) we have
pt,j+1+pt,j=TrP(Sj⊕Sj+1)ρt.
We can then express the right hand side, similarly to equation (5.6), as a sum of termsp0j0TrP(Sj⊕Sj+1)τj0
andpi
α,β,j0TrP(Sj⊕Sj+1)τ i
α,β,j0. SinceS i
α,β,j⊆Sj⊕Sj+1(byCorollary 5.4), we have
This means that
pt,j+1+pt,j≥ N
∑
i=1a2i
∑
(α,β,j)∈At,i
piα,β,j.
Together with equation (5.9), this implies
qt+1,j+1−qt,j+1≤ 4√K
√
N (pt,j+pt,j+1)≤
4√K
√
N K
∑
j0=jpt,j0=4
√
K
√
N qt,j.
6
Proof of
Lemma 4.9
We start with the case whenpT,K/2+1=· · ·=pT,K=0.
Lemma 6.1. If pT,K/2+1=· · ·=pT,K=0, the success probability ofAis at most
(N K/2)
(N K)
.
Proof. Let|ψi be the final state. The state of registerHI lies inTK/2, which is an KN/2
-dimensional
space. Therefore, there is a Schmidt decomposition for|ψiwith at most KN/2terms. This means that the state ofHAlies in an KN/2-dimensional subspace ofHA⊗HS.
We express the final state as
|ψi=
∑
x:|x|=K1
q
N K
|ψxi|xi.
We can think of|ψxias a quantum encoding forx and the final measurement as a decoding procedure
that takes|ψxiand produces a guess forx. The probability that algorithmAsucceeds is then equal to
the average success probability of the encoding. We now use
Theorem 6.2. [20] For any encoding|ψxiof M classical values by quantum states in d dimensions, the probability of success is at most d/M.
In our case,M= NK andd= KN/2because the states|ψiall lie in a KN/2-dimensional subspace
ofHA⊗HS. Therefore,Theorem 6.2impliesLemma 6.1.
We decompose the state|ψTias
√
1−δ|ψT0i+
√
δ|ψT00i where|ψT0i is in the subspaceHA⊗TK/2 and|ψT00iis inHA⊗(TK/2)⊥. We have
δ= K
∑
j=K/2+1pT,j.
The success probability ofAis the probability that, if we measure both the register of HA containing
Consider the probability of gettingi1, . . . ,iKandx1, . . . ,xN such thatxi1 =· · ·=xiK =1, when
mea-suring|ψT0i(instead of|ψTi). ByLemma 6.1, this probability is at most KN/2
/ NK
. We have
kψT−ψT0k ≤(1−
p
1−δ2)kψT0k+
√
δkψT00k= (1−
p
1−δ2) +
√
δ ≤2
√
δ.
We now apply
Lemma 6.3. [10] For any states|ψiand|ψ0iand any measurement, the total variation distance between the probability distributions obtained by applying M to|ψiand to|ψ0iis at most2kψ−ψ0k.
By Lemma 6.3, the probabilities of getting i1, . . . ,iK andx1, . . . ,xN such that xi1 =· · ·=xiK =1,
when measuring|ψTiand|ψT0idiffer by at most 4
√
δ=4√qT,K/2+1. Therefore, the success probability ofAis at most
N K/2
N K
+4
p
qT,K/2+1.
7
Structure of the eigenspaces of
ρ
t,iIn this section, we prove Claims5.2,5.3,5.5, and5.6describing the structure of the eigenspaces ofρt,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} withx1+
. . .+xN=K. ByClaim 5.2, ifρt,i is symmetric w. r. t. any permutationπ∈SN that fixesi, then
its eigenspaces are of the formSi
α,β,j, which is the subspace spanned by the states of the form
α
|ψ˜i1i,0,...,iji
k|ψ˜i1i,0,...,ijik +β
|ψ˜i1i,1,...,iji
k|ψ˜i1i,1,...,ijik
where|ψ˜ii,0
1,...,ijiand|ψ˜
i,1
i1,...,ijiare the states defined at the beginning ofSection 5.
2. We then relate the eigenspaces Sαi,β,j to the subspaces S0, . . . ,SK defined in Section 4.2. In
Claim 5.3we show that, for certainα0andβ0, we have
• Siα0,β0,j⊆Sj;
• Si−β0,α0,j⊆Sj+1;
A corollary of this is that, for anyα,β, we have
Siα,β,j⊆Siα0,β0,j⊕Si−β0,α0,j⊆Sj⊕Sj+1. This is shown inCorollary 5.4.
3. Next, inClaim 5.5, we quantify how closeSiα,β,j is toSj andSj+1. The result is an expression for TrPSj+1τ
i
α,β,j(whereτ i
α,β,jis the maximally mixed state overS i
α,β,j) that involvesα,β andα0,β0.
Proof ofClaim 5.2. We rearrange the rows and the columns ofρt,iso that all rows and columns
corre-sponding to|x1. . .xNiwithxi=0 are before the rows and the columns corresponding to|x1. . .xNiwith
xi=1. We then expressρt,ias
ρt,i=
A B
C D
,
whereAis an NK−1× NK−1
square matrix indexed by|x1. . .xNi withxi=0, Dis an NK−−11
× NK−−11
square matrix indexed by |x1. . .xNi with xi =1, and B and C are rectangular matrices with rows
(columns) indexed by|x1. . .xNiwithxi=0 and columns (rows) indexed by|x1. . .xNiwithxi=1. We claim that
ρt,i|ψ˜ i,0
i1,...,iji=a11|ψ˜
i,0
i1,...,iji+a12|ψ˜
i,1
i1,...,iji and
ρt,i|ψ˜ i,1
i1,...,iji=a21|ψ˜
i,0
i1,...,iji+a22|ψ˜
i,1
i1,...,iji, (7.1)
wherea11,a12,a21,a22are independent ofi1, . . . ,ij. To prove that, we first note thatAandDare matrices
whereAxyandDxy only depend on|{t:xt =yt}|. ThereforeLemma 4.3 applies. This means thatSij,0
andSij,1are eigenspaces forAandD, respectively, and
A|ψ˜ii,0
1,...,iji=a11|ψ˜
i,0
i1,...,iji and
D|ψ˜ii,1
1,...,iji=a22|ψ˜
i,1
i1,...,iji,
wherea11 anda22 are the eigenvalues ofAandDfor the eigenspacesSij,0 andS i,1
j . It remains to prove
that
B|ψ˜i1i,0,...,i
ji=a12|ψ˜
i,1
i1,...,iji and (7.2)
C|ψ˜ii,1
1,...,iji=a21|ψ˜
i,0
i1,...,iji. (7.3)
LetMbe a rectangular matrix, with entries indexed byx,y, with|x|=|y|=Kandxi=1 andyi=0. The entries ofMareMxy=1 ifxandydiffer in two places, withxi=1,yi=0 andxl =0,yl=1 for some
l6=iandMxy=0 otherwise. We claim
M|ψ˜i1i,0,...,iji=c|ψ˜i1i,1,...,iji (7.4)
for somec that may depend onN,k, and j but not oni1, . . . ,ij. To prove that, we need to prove two things. First,
M|ψii,0
1,...,iji=c|ψ
i,1
This follows by
M|ψii,0 1,...,iji=
1
q
N−j−1
k−j
∑
x:xi1=···=xi j=1,
xi=0
M|xi
=q 1
N−j−1
K−j
∑
x:xi1=···=xi j=1
xi=0
∑
`:x`=1
|x1. . .x`−10x`+1. . .xi−11xi+1. . .xNi
=q 1
N−j−1
K−j
(N−K)
∑
y:yi1=···=yi j=1
yi=1
|yi=p(K−j)(N−K)|ψi1i,1,...,iji.
Second,M(Tji,0)⊆Tji,1 andM(Tji,0)⊥⊆(Tji,1)⊥. The first statement immediately follows from equa-tion (7.5), because the subspacesTji,0andTji,1 are spanned by the states|ψii,0
1,...,ijiand|ψ
i,1
i1,...,iji,
respec-tively. To prove the second statement, let|ψi ∈(Tji,0)⊥ and|ψi=∑xax|xi. We would like to prove M|ψi ∈(Tji,1)⊥. This is equivalent tohψii,1
1,...,ij|M|ψi=0 for alli1, . . . ,ij. We have
hψii,1
1,···,ij|M|ψi=
1
q
N−j−1
K−j−1
∑
y:yi1=···=yi j=1
hy|M|ψi
=q 1
N−j−1
K−j−1
∑
x:xi1=···=xi j=1,
xi=0
∑
l:xl=1,l∈{/ i1,...,ij}
ax
=q 1
N−j−1
K−j−1
(
K−j)
∑
x:xi1=···=xi j=1
ax=0.
The first equality follows by writing outhψii,1
1,...,ij|, the second equality follows by writing out M. The
third equality follows because, for everyxwith|x|=Kandxi1=· · ·=xij=1, there areK−jmorel∈[N] satisfyingxl =1. The fourth equality follows because∑x:xi1=···=xi j=1axis a constant timeshψ
i,0
i1,...,ij|ψi
andhψii,0
1,...,ij|ψi=0, because|ψi ∈(T
i,0
j )⊥.
Furthermore,BMis an NK−1× NK−1
matrix, with(BM)x,y only depending on|{`:x`=y`=1}|. Therefore,Sij,1is an eigenspace ofBMand, since|ψ˜ii,1
1,...,iji ∈S
i,1
j , we have
BM|ψ˜ii,1
1,...,iji=λ|ψ˜
i,1
i1,...,iji
for an eigenvalueλ independent ofi1, . . . ,ij. Together with equation (7.4), this implies equation (7.2)
witha12=λ/c.
Equation (7.3) follows by proving
MT|ψ˜ii,1
1,...,iji=c|ψ˜
i,0
i1,...,iji and CM
T|˜ ψii,0
1,...,iji=λ|ψ˜
i,0
in a similar way.
We now diagonalize the matrix
M0=
a11 a12 a21 a22
.
It has two eigenvectors:
α1
β1
and
α2
β2
. Equation (7.1) implies that, for anyi1, . . . ,ij,
α1|ψ˜ii,0
1,...,iji+β1|ψ˜
i,1
i1,...,iji
is an eigenvector ofρt,iwith the same eigenvalueλ. Therefore,Sα1i ,β1,jis an eigenspace ofρt,i. Similarly, Si
α2,β2,jis an eigenspace ofρt,i. Vectorsα1|ψ˜ i,0
i1,...,iji+β1|ψ˜
i,1
i1,...,ijiandα2|ψ˜
i,0
i1,...,iji+β2|ψ˜
i,1
i1,...,ijitogether
span the same space as vectors|ψ˜ii,0
1,...,ijiand|ψ˜
i,1
i1,...,iji. Since the vectors|ψ˜
i,`
i1,...,ijispanS
i,`
j , this means
that
Sij,0⊕Sij,1⊆Sα1,β1,i⊕Sα2,β2,i.
Therefore, repeating this argument for every jgives a collection of eigenspaces that span the entire state space forHI. This means that any eigenspace ofρt,iis a direct sum of some of eigenspacesSiα,β,j.
Proof ofClaim 5.3. For part (i), consider the states|ψi1,...,ijispanningTj. We have
|ψi1,...,iji=
s
N−k N−j|ψ
i,0
i1,...,iji+
s
K−j N−j|ψ
i,1
i1,...,iji (7.6)
because an(N−K)/(N−j)fraction of the states |x1. . .xNiwith|x|=K andxi1 =· · ·=xij =1 have
xi=0 and the rest havexi=1. The projection of these states to(T i,0
j−1⊕T
i,1
j−1)
⊥are
s
N−K N−j|ψ˜
i,0
i1,...,iji+
s
K−j N−j|ψ˜
i,1
i1,...,iji
which, by equation (5.3) are exactly the states spanningSiα0,β0,j. Furthermore, we claim that
Tj−1⊆Tji−,01⊕T
i,1
j−1⊆Tj. (7.7)
The first containment is true becauseTj−1 is spanned by the states |ψi1,...,ij−1iwhich either belong to Tji−,12 ⊆Tji−,11 (if one of i1, . . . ,ij−1 is equal to i) or are a linear combination of states |ψi1i,0,...,ij−1i and
|ψi1i,1,...,ij−1iwhich belong toTji−,01andTji−,11. The second containment follows because the states|ψi1i,1,...,ij−1i
spanningTji−,11are the same as the states|ψi,i1,...,ij−1iwhich belong toTjand the states|ψ i,0
i1,...,ij−1ispanning Tji−,01can be expressed as linear combinations of|ψi1,...,ij−1iand|ψi,i1,...,ij−1iwhich both belong toTj.
The first part of (7.7) now implies
We also haveSi
α0,β0,j⊆Tj, because,S i
α0,β0,jis spanned by the states
P(Ti,0
j−1⊕T
i,1
j−1)⊥
|ψi1,...,iji=|ψi1,...,iji −PTji−1,0⊕Tji−1,1|ψi1,...,iji
and|ψi1,...,iji belongs toTj by the definition ofTj andPTji−1,0⊕Tji−1,1|ψi1,...,iji belongs toTj because of the
second part of (7.7). Therefore,Si
α0,β0,j⊆Tj∩(Tj−1)
⊥=S
j.
For the part (ii), we have
Siα0,β0,j⊆Sij,0⊕Sij,1⊆Tji,0⊕Tji,1⊆Tj+1,
where the first containment is true becauseSi
α0,β0,jis spanned by linear combinations of vectors|ψ˜ i,0
i1,...,iji
(which belong toSij,0) and vectors|ψ˜ii,1
1,...,iji(which belong toS
i,1
j ) and the last containment is true because
of the second part of equation (7.7). Let
|ψi=β0
|ψ˜ii,0
1,...,iji k|ψ˜ii,0
1,...,ijik −α0
|ψ˜ii,1
1,...,iji k|ψ˜ii,1
1,...,ijik
(7.8)
be one of the vectors spanningSi
β0,−α0,j. To prove that|ψiis inSj+1=Tj+1−Tj, it remains to prove
that|ψiis orthogonal toTj. This is equivalent to proving that|ψiis orthogonal to every vector|ψi01,...,i0ji
spanningTj.
Case 1 {i01, . . . ,i0j}={i1, . . . ,ij}: Since|ψi belongs to (T
i,0
j−1⊕T
i,1
j−1)⊥, it suffices to prove that|ψi is orthogonal to the projection of|ψi1,...,iji to(T
i,0
j−1⊕T
i,1
j−1)
⊥ which, by the discussion after the
equa-tion (7.6), is equal to
α0
|ψ˜ii,0
1,...,iji k|ψ˜ii,0
1,...,ijik
+β0
|ψ˜ii,1
1,...,iji k|ψ˜ii,1
1,...,ijik
. (7.9)
¿From equations (7.8) and (7.9) we see that the inner product of the two states isα0β0−β0α0=0.
Case 2 {i01, . . . ,i0j} 6={i1, . . . ,ij}but one ofi01, . . . ,i0j is equal toi: For simplicity, assumei=i0j. Then
|ψi0
1,...,i0ji is the same as|ψ
i,1
i01,...,i0j−1i which belongs toT i,1
j−1. By the definition of S
i
α,β,j, the vector|ψi
belongs to(Tji−,01⊕Tji−,11)⊥and is therefore orthogonal to|ψii0,1 1,...,i
0
j−1i.
Case 3 {i01, . . . ,i0j} 6={i1, . . . ,ij}and none of i01, . . . ,i0j is equal to i: One of i
0
1, . . . ,i
0
j must be not in
{i1, . . . ,ij}. For simplicity, assume it isi0j. We have
|ψi01,...,i0j−1i=
∑
i0∈{/ i01,...,i 0
j−1}
|ψi01,...,i0j−1,i0i.
Also,hψi01,...,i0j−1|ψi=0, because|ψi01,...,i0j−1iis inT i,0
j−1⊕T
i,1
j−1. As proved in the previous case,
We therefore have
∑
i0∈{/ i01,...,i 0
j−1,i}
hψi01,...,i0j−1,i0|ψi=0. (7.10)
By symmetry, the inner producthψi01,...,i0j−1,i0|ψi is the same for every i0∈ {/ i01, . . . ,i0j−1,i}. Therefore,
equation (7.10) meanshψi01,...,i0j−1,i0|ψi=0 for everyi0∈ {/ i01, . . . ,i0j−1,i}.
Proof ofClaim 5.5. ταi,β,j is a mixture of states|ψi from the subspaceSiα,β,j. We prove the claim by
showing that, for any of those states|ψi, the squared norm of its projection toSj+1is equal to the right hand side ofClaim 5.5. Since|ψi ∈Siα,β,jwe can write it as
|ψi=
∑
i1,...,ijai1,...,ij(α|ψ˜
i,0
i1,...,iji+β|ψ˜
i,1
i1,...,iji)
for someai1,...,ij. Let
|ψ+i=
∑
i1,...,ijai1,...,ij(β0|ψ˜
i,0
i1,...,iji −α0|ψ˜
i,1
i1,...,iji) and
|ψ−i=
∑
i1,...,ijai1,...,ij(α0|ψ˜
i,0
i1,...,iji+β0|ψ˜
i,1
i1,...,iji).
Then,|ψi is a linear combination of|ψ+iwhich belongs toSiβ0,−α0,j⊂Sj+1(byClaim 5.3) and|ψ−i
which belongs toSiα0,β0,j⊆Sj. Moreover, all three states are linear combinations of|ψ0i,|ψ1idefined
by
|ψ`i=
∑
i1,...,ijai1,...,ij|ψ˜
i,`
i1,...,iji.
We have
|ψi=α|ψ0i+β|ψ1i, |ψ+i=β0|ψ0i −α0|ψ1i and |ψ−i=α0|ψ0i+β0|ψ1i.
Since|ψ+iand|ψ−ibelong to subspacesSj+1andSjwhich are orthogonal, we must havehψ+|ψ−i=0.
This means
α0β0kψ0k2−β0α0kψ1k2=0.
By dividing the equation byα0β0, we getkψ0k2=kψ1k2andkψ0k=kψ1k. Sincekψk=1, this means that
kψ0k=kψ1k= p 1
α2+β2
=1.
Since|ψilies in the subspace spanned by|ψ+iwhich belongs toSj+1and|ψ−iwhich belongs to Sj, the norm of the projection of|ψitoSj+1is equal to|hψ|ψ+i|/kψ+k. By expressing|ψiand|ψ+i in terms of|ψ0iand|ψ1i, we get
|hψ|ψ+i|
kψ+k =
α β0kψ0k2−α0βkψ1k2
q
β02kψ0k2+α02kψ0k2
=|α βq0−α0β|
α02+β02
,
Proof ofClaim 5.6. We will provekψ˜ii,0
1,...,ijk ≥
1 2kψ˜
i,1
i1,...,ijk, because that means
α0=
√
N−K
√
N−jkψ˜ i,0
i1,...,ijk ≥
1 2
√
N−K
√
K−j
√
K−j
√
N−jkψ˜ i,1
i1,...,ijk= √
N−K
2√K−jβ0
and
β0
q
α02+β02
≤q β0
N−K
4(K−j)β
2 0+β02
=q 1
1+4N(K−−Kj) =
p
4(K−j)
√
N+3K−4j.
To provekψ˜ii,0
1,...,ijk ≥ kψ˜
i,1
i1,...,ijk, we calculate the vector
|ψ˜ii,0
1,...,iji=P(Tji−1,0)⊥|ψ i,0
i1,...,iji.
Both the vector|ψii,0
1,...,ijiand the subspaceT
i,0
j−1are fixed by
Uπ|xi=|xπ(1). . .xπ(N)i
for any permutationπ that fixes iand maps {i1, . . . ,ij}to itself. This means that|ψ˜ii,0
1,...,ijiis fixed by
any suchUπ as well. Therefore, the amplitude of |xi, |x|=K, xi =0 in |ψ˜ i,0
i1,...,iji only depends on |{i1, . . . ,ij} ∩ {t:xt=1}|. This means|ψ˜ii1,0,...,ijiis of the form
|ψ0i=
j
∑
m=0αm
∑
x:|x|=K,xi=0
|{i1,...,ij}∩{t:xt=1}|=m |xi.
To simplify the following calculations, we multiplyα0, . . . ,αjby the same constant so that
αj=
1
q
N−j−1
K−j
.
Then,|ψ˜i1i,0,...,ijiremains a multiple of|ψ0ibut may no longer be equal to|ψ0i.
The coefficientsα0, . . . ,αj−1 should be such that the state is orthogonal toTj−1and, in particular, orthogonal to states|ψi1i,0,...,i`ifor`∈ {0, . . . ,j−1}. By writing outhψ0|ψi1i,0,...,i`i=0, we get
j
∑
m=`αm
N−j−1
K−m
j−`
m−`
=0. (7.11)
To show that, we first note that|ψii,0
1,...,i`iis a uniform superposition of all|xi,|x|=K,xi=0,xi1 =· · ·= xil=1. If we want to choosexsubject to those constraints and also satisfying|{i1, . . . ,ij}∩{t:xt=1}|= m, we have to set xt =1 form−`differentt∈ {i`+1, . . . ,ij}and forK−mdifferentt∈ {/ i,i1, . . . ,ij}.
By solving the system of equations (7.11), we get that the only solution is
αm= (−1)j−m N−j−1
K−j
N−j−1
K−m
αj. (7.12)
Let|ψ00i=|ψ0i/kψ0kbe the normalized version of|ψ0i. Then
|ψ˜ii,0
1,...,iji=hψ
0
0|ψ
i,0
i1,...,iji|ψ
0
0i and
kψ˜ii,0
1,...,ijk=hψ
0
0|ψ
i,0
i1,...,iji=
hψ0|ψii1,0,...,iji kψ0k
. (7.13)
First, we have
hψ0|ψii,0
1,...,iji=1,
because|ψii,0
1,...,iji consists of
N−j−1
K−j
basis states |xi, xi =0, xi1 =· · ·=xij =1, each of which has
amplitude 1/q NK−−j−j1
in both|ψ0iand|ψi1i,0,...,iji. Second,
kψ0k2=
j
∑
m=0
j m
N−j−1
K−m
αm2 = j
∑
m=0
j m
N−j−1
K−j
2
N−j−1
K−m
α 2 j = j
∑
m=0
j m
N−j−1
K−j
N−j−1
K−m
=
j
∑
m=0
j m
(K−m)!(N−K+m−j−1)! (K−j)!(N−K−1)!
=
j
∑
m=0
j m
(K−m)· · ·(K−j+1)
(N−K−1)· · ·(N−K+m−j) (7.14)
with the first equality following because there are mj NK−−j−m1vectorsxsuch that|x|=K,xi=0,xt=1 formdifferentt∈ {i1, . . . ,ij}andK−mdifferentt∈ {/ i,i1, . . . ,ij}, the second equality following from
equation (7.12) and the third equality following from our choiceαj=1/
q
N−j−1
K−j
.
We can similarly calculatekψ˜i1i,1,...,ijk. We omit the details and just state the result. The counterpart of equation (7.13) is
kψ˜i1i,1,...,ijk=
hψ1|ψii,1 1,...,iji kψ1k
,
with|ψ1ibeing the counterpart of|ψ0i:
|ψ1i=
j
∑
m=0αm
∑
x:|x|=K,xi=1
withα0=1/
q
N−j−1
K−j−1
. Similarly as before, we gethψ1|ψ
i,1
i1,...,iji=1 and
kψ1k2=
j
∑
m=0
j m
N−j−1
K−j−1
N−j−1
K−m−1
=
j
∑
m=0
j m
(K−m−1). . .(K−j)
(N−K). . .(N−K+m−j+1). (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−m K−j
N−K+m−j
N−K ≤
K
K/2·2=4,
because j≤K/2 andN−K+m−j≤N−K(because ofm≤ j). Therefore, kψ0k2≤4kψ1k2which implies
kψ˜ii,0
1,...,ijk=
1
kψ0k
≥√ 1
4kψ1k = 1
2kψ˜
i,1
i1,...,ijk.
Acknowledgment. I would like to thank Frederic Magniez, Robert ˇSpalek, 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