QUANTUM ERROR-DETECTION AT LOW ENERGIES
2.6 AQEDC at Low Energies: The Excitation Ansatz
2.6.4 Matrix Elements of Local Operators in the Excitation Ansatz Overview of the ProofOverview of the Proof
The norm of the vector (2.6.3) can be bounded as
kฮจ๐ , ๐(๐ผ, ๐ฝ) k2=
= tr
๐ธ๐โ1๐ธ
๐ต(๐)๐ต(๐)๐ธ๐ฟโ๐(|๐ฝih๐ผ| โ |๐ฝih๐ผ|)
โค k๐ธ
๐ต(๐)๐ต(๐)k๐น ยท k๐ธ๐โ1k๐น ยท k๐ธ๐ฟโ๐k๐น
โค k๐ธ
๐ต(๐)๐ต(๐)k๐น .
In the first inequality, we have used (2.5), together with the fact that k |๐ฝih๐ผ| โ |๐ฝih๐ผ| k๐น =1.
In the second inequality, we have used Lemma 2.4.2, along with the fact ๐(๐ธ) =1.
The claim (2.6.3) follows from this.
With a completely analogous proof, we also have k๐ธ๐น(๐ , ๐) k๐น โค ๐ท2k๐นkโ๏ธ
k๐ธ
๐ต(๐)๐ต(๐)k๐น, and k๐ธ๐นk๐น โค ๐ท2k๐นk, which are claims (2.6.3) and (2.6.3).
2.6.4 Matrix Elements of Local Operators in the Excitation Ansatz
for different momenta ๐ โ ๐0. For this purpose, we need to identify the leading order term in the expressionh๐๐|๐น|๐๐i. Higher order terms are again small by the properties of the transfer operator.
To establish these bounds, first observe that an unnormalized excitation ansatz state
|ฮฆ๐(๐ต;๐ด)iis a superposition of the โposition spaceโ states{|ฮฆ๐ , ๐i}๐๐=
1, where each state|ฮฆ๐ , ๐iis given by a simple tensor network with an โinsertionโ of an operator at site ๐0. Correspondingly, we first study matrix elements of the formhฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i. Bounds on these matrix elements are given in Lemma 2.6.5. The idea of the proof of this statement is simple: in the tensor network diagram for the matrix element, subdiagrams associated with powers ๐ธฮ with sufficiently largeฮmay be replaced by the diagram associated with the map |๐iihhโ|, with an error scaling term scaling as๐(๐ฮ/2
2 ). This is due to the Jordan decomposition of the transfer operator. Thanks to the gauge condition (2.6.1), the resulting diagrams then simplify, allowing us to identify the leading order term.
To realize this approach, a key step is to identify suitable subdiagrams corresponding to powers๐ธฮ in the diagram associated withhฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i. These are associated with connected regions of sizeฮwhere the operator๐น acts trivially, and there is no insertion of ๐ต(๐) (respectively๐ต(๐0)), meaning that ๐ and ๐0do not belong to the region. Lemma 2.6.5 provides a careful case-by-case analysis depending on, at the coarsest level of detail, whether or not ๐ and ๐0belong to aฮ-neighborhood of the support of๐น.
Some subleties that arise are the following: to obtain estimates on the leading-order terms for the diagonal matrix elements (see (2) above) as well as related expressions, a bound on the magnitude of the matrix elementhฮฆ๐ , ๐0|๐น|ฮฆ๐ , ๐ionly is not sufficient.
The lowest-order approximating expression to hฮฆ๐ , ๐0|๐น|ฮฆ๐ , ๐iobtained by making the above substitutions of the transfer operators a priori seems to depend on the exact site location ๐. This is awkward because the termhฮฆ๐ , ๐0|๐น|ฮฆ๐ , ๐iappears as a summand (with sum taken over ๐) when computing matrix elements of excitation ansatz states. We argue that in fact, the leading order term of hฮฆ๐ , ๐0|๐น|ฮฆ๐ , ๐i is identical for all values of ๐ not belonging to the support of ๐น. This statement is formalized in Lemma 2.6.6 and allows us to subsequently estimate sums of interest without worry about the explicit dependence on ๐.
Finally, we require a strengthening of the estimates obtained in Lemma 2.6.5 because we are ultimately interested in excitation ansatz states: these are superpositions of
the states |ฮฆ๐ , ๐i, with phases of the form ๐๐ ๐ ๐. Estimating only the magnitude of matrix elements of the form hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i is not sufficient to establish our results. Instead, we need to treat the phases โcoherently,โ which leads to certain cancellations. The corresponding statement is given in Lemma 2.6.7.
The Proof
We will envision the sites {1, . . . , ๐}as points on a ring, i.e., using periodic boundary conditions, and measure the distance between sites ๐ , ๐0by
dist(๐ , ๐0) :=min
๐โZ
|๐โ ๐0+๐ ยท๐|. Forฮโ {0, . . . , ๐} and a subsetF โ {1, . . . , ๐}, let
Bฮ(F )= {๐ โ {1, . . . , ๐} | โ ๐0 โ F such thatdist(๐ , ๐0) โคฮ}
be theฮ-thickeningofF.
We say that ๐0โ {1, . . . , ๐}is aleft neighbor of(or isleft-adjacent to) ๐ โ {1, . . . , ๐} if ๐0 = ๐ โ1 for ๐ > 1, or ๐0 = ๐for ๐ = 1. A connected region R โ {1, . . . , ๐} is said to lie on the left of (or be left-adjacent to) ๐ โ {1, . . . , ๐} if it is of the formR = {๐
1, . . . , ๐๐}, with ๐๐ผ+
1left-adjacent to ๐๐ผ for๐ผ โ {0, . . . , ๐ โ1} with the convention that ๐
0 = ๐๐. Analogous definitions hold for right-adjacency.
For an operator ๐น acting on (Cp)โ๐, let supp(๐น) โ {1, . . . , ๐} denote its support, i.e., the sites of the system that the operator acts on non-trivially. We say that๐น is ๐-local if |supp(๐น) | = ๐. Let us assume that supp(๐น) decomposes into ๐ disjoint connected components
supp(๐น) =
๐ โ1
ร
๐ผ=0
F๐ผ . (2.56)
We may, without loss of generality, assume that this gives a partition of{1, . . . , ๐} into disjoint connected sets
{1, . . . , ๐}=A0โช F0โช A1โช F1โช ยท ยท ยท โช A๐ โ1โช F๐ โ1
whereA๐ผis left-adjacent toF๐ผfor๐ผโ {0, . . . , ๐ โ1},A๐ผ+1is right-adjacent toF๐ผ
for๐ผ โ {0, . . . , ๐ โ2}, andA0 is right-adjacent toF๐ โ1. We may then decompose the operator๐น as
๐น =โ๏ธ
๐ ๐ โ1
ร
๐ผ=0
(๐ผA
๐ผ โ๐น๐,๐ผ),
๐น =
๐น(๐(๐
1))=๐น(๐(๐
4)) =๐น(0) = ,
๐น(๐(๐
2)) =๐น(1) = ,
๐น(๐(๐
3)) =๐น(๐ โ1) = .
Figure 2.6: Example for๐นand sites๐
1, ๐
2, ๐
3, ๐
4 โ {1, . . . , ๐}with๐(๐
1) =๐(๐
4) =7, ๐(๐
2) =19, and๐(๐
3) =35.
where we write๐น as a sum of decomposable tensor operators (indexed by๐), with each๐น๐,๐ผ being an operator acting on the componentF๐ผ.
Let us define a function๐ :{1, . . . , ๐}\supp(๐น) โ {0, . . . , ๐ โ1}which associates to every site ๐ โ supp(F ) the unique index ๐(๐) for the component A๐(๐) of the complement ofsupp(๐น)such that ๐ โ A๐(๐).
It is also convenient to introduce the following operators {๐น(๐)}๐ โ1
๐=0. The operator ๐น(๐)is obtained by removing the identity factor on the sitesA๐ of๐น, and cyclically permuting the remaining components in such a way that F๐ ends up on the sites {1, . . . ,|F๐|}. More precisely, we define๐น(๐) โ B ( (Cp)โ(๐โ|A๐|))by
๐น(๐) =โ๏ธ
๐
๐น๐,๐ โ
๐+๐ โ1
ร
๐ผ=๐+1
๐ผ
โ|A๐ผ(mod๐ )|
Cp โ ๐น๐,๐ผ (
mod๐ )
!
, (2.57)
for๐ โ {0, . . . , ๐ โ1}. We note that ๐ โฆโ ๐น(๐(๐)) associates a permuted operator to each site ๐ not belonging to the support of ๐น. Let us also define๐(๐) to be the index of the site which gets cyclically shifted to the first site when defining ๐น๐(๐). An example is shown diagrammatically in Figure 2.6.
For two excitation ansatz states |ฮฆ๐iand |ฮฆ๐0i, and an operator ๐น on (Cp)โ๐, we
may write the corresponding matrix element as hฮฆ๐0|๐น|ฮฆ๐i =
๐
โ๏ธ
๐ , ๐0=1
๐๐(๐ ๐โ๐
0๐0)hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i, (2.58)
where|ฮฆ๐ , ๐iare the โposition spaceโ states introduced in Equation (2.6.1). We are interested in bounding the magnitude of this quantity.
We begin by bounding the individual terms in the sum (2.6.4).
Lemma 2.6.5. Let ๐ , ๐0 โ {1, . . . , ๐} and let ๐, ๐0be arbitrary non-zero momenta.
Consider the states|ฮฆ๐ , ๐iand|ฮฆ๐0, ๐0idefined by(2.6.1). Letฮ = ฮ(๐)and๐ =๐(๐) be monotonically increasing functions of๐. Suppose further that we have
10ฮ๐ < ๐ .
Assume ๐น is a ๐-local operator of unit norm on (Cp)โ๐ whose support has ๐ connected components as in(2.6.4). Then we have the following:
(i) There is some fixed๐ โ [๐]such that for all ๐ , ๐0โ Bฮ(supp(๐น)), we have hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i= hhโ|๐ธ
๐ผโฮโ๐น๐(๐)โ๐ผโฮ(๐ , ๐,ห ๐ห0, ๐0) |๐ii +๐(๐ฮ
2), where ๐ห= ๐ โ๐(๐) +ฮ+1(mod๐)and ๐ห0= ๐0โ๐(๐) +ฮ+1 (mod๐).
Furthermore,
hฮฆ๐0, ๐0|ฮฆ๐ , ๐i = ฮ๐ , ๐0๐๐ ๐0+๐(๐ฮ
2). (2.59)
(ii) If ๐ , ๐0โBฮ(supp(๐น)), then (a) |hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i|=๐(๐ฮ/2
2 )if ๐ โ ๐0.
(b) hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i=hhโ|๐ธ๐น(๐(๐))|๐ii ยท๐๐ ๐0+๐(๐ฮ/2
2 ).
Here the operator๐น(๐(๐))is defined by Equation(2.6.4). (iii) If ๐ โ Bฮ(supp(๐น)) and ๐0โBฮ(supp(๐น)), then
(a) |hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i|=๐(๐ฮ/2
2 )if ๐0โB2ฮ(supp(๐น)).
(b) There exists some fixed๐ โ [๐] such that, for all ๐ โ Bฮ(supp(๐น)) and ๐0 โ B2ฮ(supp(๐น))\Bฮ(supp(๐น)), we have
hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i= hhโ|๐ธ๐น(๐(๐))(๐ , ๐,ห ๐ห0, ๐0,2ฮ) |๐ii +๐(๐2ฮ
2 ), where ๐ห= ๐โ๐(๐) +2ฮ+1(mod๐) and ๐ห0= ๐0โ๐(๐) +2ฮ+1 (mod๐).
(iv) If ๐0โ Bฮ(supp(๐น))and ๐ โBฮ(supp(๐น)), then (a) |hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i|=๐(๐ฮ/2
2 )if ๐ โB2ฮ(supp(๐น)).
(b) There exists some fixed๐ โ [๐] such that, for all ๐0โ Bฮ(supp(๐น))) and ๐ โ B2ฮ(supp(๐น))\Bฮ(supp(๐น)), we have
hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i= hhโ|๐ธ๐น(๐(๐))(๐ , ๐,ห ๐ห0, ๐0,2ฮ) |๐ii +๐(๐2ฮ
2 ), where ๐ห= ๐โ๐(๐) +2ฮ+1(mod๐) and ๐ห0= ๐0โ๐(๐) +2ฮ+1 (mod๐). Proof. For the proof of (i), suppose that ๐ , ๐0 โ Bฮ(supp(๐น)). Pick any site ๐ โB2ฮ(supp(๐น)). We note that such a site always exists since
|B2ฮ(supp(๐น)) | โค5ฮ|supp(๐น) | =5ฮ๐ <10ฮ๐ < ๐ by assumption. Let us define the shifted indices
ห
๐ = ๐ โ๐(๐) +ฮ+1(mod๐), and ๐ห0= ๐0โ๐(๐) +ฮ+1 (mod๐). Then we may write
hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i=tr(๐ธ๐น(๐ , ๐, ๐0, ๐0))
=tr
๐ธ๐ ๐ธ
๐ผโฮโ๐น๐(๐)โ๐ผโฮ(๐ , ๐,ห ๐ห0, ๐0)
(2.60) where ๐ โฅ 2ฮ. This is because by the choice of ๐, there are at least 2ฮsites not belonging to supp(๐น) both on the left and the right of ๐. Each of these 4ฮsites contributes a factor๐ธ =๐ธ๐ผ (i.e., a single transfer operator) to the expression within the trace. The term ๐ธ
๐ผโฮโ๐น๐(๐)โ๐ผโฮ(๐ , ๐,ห ๐ห0, ๐0) incorporates ฮ of the associated transfer operators๐ธ =๐ธ๐ผ on the left- and right of๐, respectively, such that at least 2ฮfactors of๐ธ remain. By the cyclicity of the trace, these can be consolidated into a single term๐ธ๐ with ๐ โฅ 2ฮ. The operator ๐ผโฮ โ ๐น๐(๐) โ ๐ผโฮ (i.e., the additional ๐ผโฮfactors) in the term๐ธ
๐ผโฮโ๐น๐(๐)โ๐ผโฮ(๐ , ๐,ห ๐ห0, ๐0)is used to ensure that ๐and ๐0are correctly โretainedโ when going from the first to the second line in (2.6.4). Inserting the Jordan decomposition๐ธ =|๐iihhโ| โ๐ธห, we obtain
hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i=hhโ|๐ธ
๐ผโฮโ๐น๐(๐)โ๐ผโฮ(๐ , ๐,ห ๐ห0, ๐0) |๐ii +tr
ห ๐ธ๐ ๐ธ
๐ผโฮโ๐น๐(๐)โ๐ผโฮ(๐ , ๐,ห ๐ห0, ๐0) . (2.61)
By Lemma 2.4.2(ii) and Lemma 2.6.3, we have the bound
tr
ห ๐ธ๐ ๐ธ
๐ผโฮโ๐น๐(๐)โ๐ผโฮ(๐ , ๐,ห ๐ห0, ๐0)
โค k๐ธห๐ k๐น ยท k๐ธ
๐ผโฮโ๐น๐(๐)โ๐ผโฮ(๐ , ๐,ห ๐ห0, ๐0) k๐น
โค ๐๐ /2
2 ยท๐ท2k๐นk ยทโ๏ธ
k๐ธ
๐ต(๐0)๐ต(๐0)k๐นk๐ธ
๐ต(๐0)๐ต(๐)k๐น
=๐(๐ฮ
2), where we have used the fact that๐
๐ /2 2 โค ๐ฮ
2 in the last line. We have also absorbed the dependence on the constants ๐ท, k๐นk, and
โ๏ธk๐ธ
๐ต(๐0)๐ต(๐0)k๐นk๐ธ
๐ต(๐0)๐ต(๐)k๐น into the big-O notation. Inserting this into (2.6.4) gives the first claim of (i).
Now consider the inner product hฮฆ๐0, ๐0|ฮฆ๐ , ๐i = tr(๐ธ(๐ , ๐, ๐0, ๐0)), which corre- sponds to the case where ๐น is the identity. By the cyclicity of the trace, this can be written as hฮฆ๐0, ๐0|ฮฆ๐ , ๐i = tr(๐ธ๐ ๐ธ(๐ , ๐,ห ๐ห0, ๐0)) for some ๐ โฅ 2ฮ and suitably defined ห๐ ,๐ห0. Repeating the same argument as above and using the fact that
hhโ|๐ธ(๐ , ๐,ห ๐ห0, ๐0) |๐ii = ฮ๐ ,ห๐ห0๐๐ ๐0 = ฮ๐ , ๐0๐๐ ๐0
by definition of ๐ธ(๐ , ๐,ห ๐ห0, ๐0), Equation (2.4.1) (i.e., the fact that |โii and |๐ii are left, respectively right, eigenvectors of๐ธ), and the gauge identities (2.6.1) of๐ธ๐ต(๐) and๐ธ
๐ต(๐), we obtain the claim (i).
Now consider claim (ii). Suppose that ๐ , ๐0 โ Bฮ(supp(๐น)). We consider the following two cases:
(iia) If ๐ โ ๐0, then there is a connected region of at least ฮ sites not belonging tosupp(๐น)to either the left of ๐0and not containing ๐, or the left of ๐ and not containing ๐0. Without loss of generality, we assume the former is the case.
By the cyclicity of the trace, we may also assume without loss of generality that ๐0= ฮ+1, ๐ > ๐0, and that๐นis supported on the sites{2ฮ+2, . . . , ๐}. Let
ห
๐นdenote the restriction of๐นto the sites{ฮ+2, . . . , ๐}, and let ห๐ := ๐โ (ฮ+1). Then we may write
hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i=tr
๐ธฮ๐ธ
๐ต(๐0)๐ธ
ห
๐น(๐ , ๐ห ) . Substituting the Jordan decomposition๐ธฮ =|๐iihhโ| โ๐ธหฮ, we have
hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i= hhโ|๐ธ
๐ต(๐0)๐ธ
ห
๐น(๐ , ๐ห ) |๐ii +tr
ห ๐ธฮ๐ธ
๐ต(๐0)๐ธ
ห ๐น(๐ , ๐ห )
.
Since we assume that๐ โ 0, the gauge condition (2.6.1) states thathhโ|๐ธ
๐ต(๐) =0, hence the first term vanishes and it follows that
|hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i|= tr
ห ๐ธฮ๐ธ
๐ต(๐0)๐ธ
ห
๐น(๐ , ๐ห )
โค k๐ธหฮk๐น ยท k๐ธ
๐ต(๐0)k๐น ยท k๐ธ
ห
๐น(๐ , ๐ห ) k๐น
โค๐ฮ/2
2 k๐ธ
๐ต(๐0)k๐น ยท๐ท2k๐นหkโ๏ธ
k๐ธ
๐ต(๐)๐ต(๐)k๐น
=๐(๐ฮ/2
2 ) ,
as claimed in (iia). In the last line, we have again absorbed the constants into the big-๐-expression. This proves part (iia) of Claim (ii).
(iib) If ๐ = ๐0, then there are at leastฮsites to the left and right of ๐ which do not belong tosupp(๐น). Therefore we may write
hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i=tr
๐ธ๐ ๐ธ
๐ต(๐0)๐ต(๐)๐ธ๐ก๐ธ๐น(๐(๐))
,
where๐ and๐ก are integers greater thanฮ, representing the sites surrounding ๐ which are not in the support of๐น.
Applying the Jordan decomposition ๐ธฮ = |๐iihhโ| โ ๐ธหฮ twice (for ๐ธ๐ and ๐ธ๐ก) then gives four terms
hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i= hhโ|๐ธ
๐ต(๐0)๐ต(๐)|๐iihhโ|๐ธ๐น(๐(๐))|๐ii +tr
|๐iihhโ|๐ธ
๐ต(๐0)๐ต(๐)๐ธห๐ ๐ธ๐น(๐(๐))
+tr
ห ๐ธ๐ก๐ธ
๐ต(๐0)๐ต(๐)|๐iihhโ|๐ธ๐น(๐(๐))
+tr
ห ๐ธ๐ก๐ธ
๐ต(๐0)๐ต(๐)๐ธห๐ ๐ธ๐น(๐(๐))
.
Since ๐ and ๐ก are both larger than ฮ, by the same arguments from before, it is clear that the last three terms can each be bounded by๐(๐ฮ/2
2 ). The claim follows sincehhโ|๐ธ
๐ต(๐0)๐ต(๐)|๐ii=๐๐ ๐0.
Next, we give the proof of claim (iii). Let us consider the situation where ๐ โ Bฮ(supp(๐น)) and ๐0 โBฮ(supp(๐น)). The proof of the other setting is analogous.
We consider two cases:
(iiia) Suppose ๐0โB2ฮ(supp(๐น)). Let us define the shifted index ห๐ = ๐โ๐(๐0) +ฮ+ 1(mod๐). Then we may write
hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i =tr
๐ธ๐ ๐ธ
๐ต(๐0)๐ธ๐ก๐ธ
๐ผโฮโ๐น(๐(๐0))โ๐ผโฮ(๐ , ๐ห ) ,
where๐ and๐กare integers larger thanฮ, representing the number of sites adjacent to ๐0 on the left and right which are not in Bฮ(supp(๐น)). We use the Jordan decomposition๐ธ =|๐iihhโ| โ๐ธห on๐ธ๐ to get
tr
๐ธ๐ ๐ธ
๐ต(๐0)๐ธ๐ก๐ธ
๐ผโฮโ๐น(๐(๐0))โ๐ผโฮ(๐ , ๐ห )
=hhโ|๐ธ
๐ต(๐0)๐ธ๐ก๐ธ
๐ผโฮโ๐น(๐(๐0))โ๐ผโฮ(๐ , ๐ห ) |๐ii +tr
ห ๐ธ๐ ๐ธ
๐ต(๐0)๐ธ๐ก๐ธ
๐ผโฮโ๐น(๐(๐0))โ๐ผโฮ(๐ , ๐ห )
=tr
ห ๐ธ๐ ๐ธ
๐ต(๐0)๐ธ๐ก๐ธ
๐ผโฮโ๐น(๐(๐0))โ๐ผโฮ(๐ , ๐ห ) ,
where the first term vanishes due to the gauge condition (2.6.1). From Lemma 2.4.2(ii) we have k๐ธ๐กk๐น โค 1, and repeating the same arguments as before, we get the bound
|hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i|= tr
ห ๐ธ๐ ๐ธ
๐ต(๐0)๐ธ๐ก๐ธ
๐ผโฮโ๐น(๐(๐0))โ๐ผโฮ(๐ , ๐ห )
โค k๐ธห๐ k๐น ยท k๐ธ
๐ต(๐0)k๐น ยท k๐ธ๐กk๐นยท ๐ท2k๐นk ยท k๐ธ
๐ต(๐)๐ต(๐)k๐น
โค๐
๐ /2 2 k๐ธ
๐ต(๐0)k๐น ยท ๐ท2k๐นkโ๏ธ
k๐ธ
๐ต(๐)๐ต(๐)k๐น . Since๐ โฅ ฮ, we conclude that
|hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i|=๐
๐ฮ/2
2
.
(iiib) Suppose now that ๐0 โ B2ฮ(supp(๐น)). Then by repeating the argument for case (i), withฮreplaced by 2ฮ, we obtain
hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i= hhโ|๐ธ
๐ผโ2ฮโ๐น(๐(๐))โ๐ผโ2ฮ(๐ , ๐,ห ๐ห0, ๐0) |๐ii +๐(๐2ฮ
2 ),
where we now have๐ โB4ฮ(F ). Again, the existence of such a๐is guaranteed by the condition 10ฮ๐ < ๐.
We note that (iv) follows immediately from (iii) by interchanging the roles of (๐ , ๐)and (๐0, ๐0). Note that we can write
hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i= hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i=hฮฆ๐ , ๐|๐นโ |ฮฆ๐0, ๐0i.
The last expression within the parentheses is precisely what we had calculated in (iii), so this implies the following:
(iva) If ๐ โB2ฮ(supp(๐น)), then hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i
=
hฮฆ๐ , ๐|๐นโ |ฮฆ๐0, ๐0i
=๐(๐ฮ/2
2 ),
where we note that the exact same bound holds for๐นand๐นโ sincek๐นk = k๐นโ k. (ivb) If ๐ โ B2ฮ(supp(๐น)), then
hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i=hฮฆ๐ , ๐|๐นโ |ฮฆ๐0, ๐0i
=hhโ|๐ธ
๐ผโ2ฮโ๐นโ (๐(๐))โ๐ผโ2ฮ(๐ห0, ๐0, ๐ , ๐ห ) |๐ii +๐(๐2ฮ
2 )
=hhโ|๐ธ
๐ผโ2ฮโ๐นโ (๐(๐))โ๐ผโ2ฮ(๐ห0, ๐0, ๐ , ๐ห ) |๐ii +๐(๐2ฮ
2 )
=hhโ|๐ธ
๐ผโ2ฮโ๐น(๐(๐))โ๐ผโ2ฮ(๐ , ๐,ห ๐ห0, ๐0) |๐ii +๐(๐2ฮ
2 ). This proves the claim.8
Note that in the statement (iib), the dependence on๐in the expressionhhโ|๐ธ๐น(๐(๐))|๐ii can be eliminated as follows:
Lemma 2.6.6. Suppose ๐
1, ๐
2โBฮ(supp(๐น)). Then
|hhโ|๐ธ๐น(๐(๐
1))|๐ii โ hhโ|๐ธ๐น(๐(๐
2))|๐ii| =๐(๐ฮ
2). (2.62)
In particular, for any fixed ๐
0โBฮ(supp(๐น)) we have hฮฆ๐ , ๐0|๐น|ฮฆ๐ , ๐i =hhโ|๐ธ๐น(๐(๐
0))|๐ii ยท๐๐ ๐0 +๐(๐ฮ/2
2 ), (2.63)
for all ๐ โBฮ(supp(๐น)).
8To clarify how the termhhโ|๐ธ๐ผโฮโ๐นโ (๐(๐)) โ๐ผโฮ(๐ห0, ๐0,๐ , ๐ห ) |๐iiis complex conjugated, first write
hhโ|๐ธ๐ผโฮโ๐นโ (๐(๐)) โ๐ผโฮ(๐ห0, ๐0,๐ , ๐ห ) |๐ii=hฮฆ๐ฟ
ห
๐0, ๐0|๐ผโ๐ผโ2ฮโ๐นโ
๐(๐)โ๐ผโ2ฮโ๐ผ|ฮฆ๐ฟ
ห ๐ , ๐i, where|ฮฆ๐ฟ
ห
๐ , ๐iare the states defined by (2.6.3), for some appropriate length๐ฟ. Then we can proceed to conjugate the matrix element, giving us
hฮฆ๐ฟ
ห
๐0, ๐0|๐ผโ๐ผโ2ฮโ๐นโ
๐(๐)โ๐ผโ2ฮโ๐ผ|ฮฆ๐ฟ
ห
๐ , ๐i=hฮฆ๐ฟ
ห
๐ , ๐|๐ผโ๐ผ2ฮโ๐น๐(๐)โ๐ผ2ฮโ๐ผ|ฮฆ๐ฟ
ห ๐0, ๐0i
=hhโ|๐ธ๐ผโ2ฮโ๐น(๐(๐)) โ๐ผโ2ฮ(๐ , ๐,ห ๐ห0, ๐0) |๐ii.
Proof. The claim (2.6.6) follows immediately from (2.6.6) and claim (iib) of Lemma 2.6.5 since |๐๐ ๐0|=๐(1).
If๐(๐
1) =๐(๐
2), there is nothing to prove. Suppose๐(๐
1) โ ๐(๐
2). Without loss of generality, assume that๐(๐
1) =0 and๐(๐
2) =๐. Then we may write ๐น(๐(๐
1))=โ๏ธ
๐
๐น๐,
0โ๐ผโ๐1 โ๐น๐,
1โ๐ผโ๐2ยท ยท ยท โ๐ผโ๐๐ โ1 โ๐น๐,๐ โ
1, and ๐น(๐(๐
2))=โ๏ธ
๐
๐น๐,๐ โ๐ผโ๐๐+1 โ ๐น๐,๐+
1โ ๐ผโ๐๐+2ยท ยท ยท โ ๐ผโ๐๐ โ ๐น๐,๐ โ
1โ ๐ผโ๐0
โ๐น๐,
0โ๐ผโ๐1 โ๐น๐,
1โ๐ผโ๐2 โ ยท ยท ยท โ๐น๐,๐โ
1, where๐๐ผ =|A๐ผ|for๐ผ โ {0, . . . , ๐ }. Defining the operators
ห
๐น๐ =๐น๐,๐ โ ๐ผโ๐๐+1 โ๐น๐,๐+
1โ ๐ผโ๐๐+2ยท ยท ยท โ ๐ผโ๐๐ โ1 โ๐น๐,๐ โ
1, ห
๐บ๐ =๐น๐,
0โ ๐ผโ๐1 โ ๐น๐,
1โ ๐ผโ๐2 โ ยท ยท ยท โ ๐น๐,๐โ
1, we have
๐น(๐(๐
1)) =โ๏ธ
๐
ห
๐บ๐ โ๐ผโ๐๐ โ๐นห๐, and ๐น(๐(๐
2))=โ๏ธ
๐
ห
๐น๐โ ๐ผโ๐0 โ๐บห๐.
(We give an example for the operator ๐น, ๐น(๐(๐
1)), and ๐น(๐(๐
2)) in Figure 2.7.) Therefore we can write
hhโ|๐ธ๐น(๐(๐
1))|๐ii=โ๏ธ
๐
hhโ|๐ธ
ห ๐บ๐๐ธ๐๐๐ธ
ห ๐น๐|๐ii, hhโ|๐ธ๐น(๐(๐
2))|๐ii=โ๏ธ
๐
hhโ|๐ธ
ห ๐น๐
๐ธ๐0๐ธ
ห ๐บ๐|๐ii. Inserting the Jordan decomposition๐ธ = |๐iihhโ| โ๐ธห gives
hhโ|๐ธ๐น(๐(๐
1))|๐ii=โ๏ธ
๐
hhโ|๐ธ
ห
๐บ๐|๐iihhโ|๐ธ
ห
๐น๐|๐ii + hhโ|๐ธ
ห ๐บ๐๐ธห๐๐๐ธ
ห ๐น๐|๐ii
, hhโ|๐ธ๐น(๐(๐
2))|๐ii=โ๏ธ
๐
hhโ|๐ธ
ห
๐น๐|๐iihhโ|๐ธ
ห
๐บ๐|๐ii + hhโ|๐ธ
ห ๐น๐๐ธห๐0๐ธ
ห ๐บ๐|๐ii
. Taking the difference, the first terms of the sums cancel, and we are left with hhโ|๐ธ๐น(๐(๐
1))|๐ii โ hhโ|๐ธ๐น(๐(๐
2))|๐ii =
โ๏ธ
๐
hhโ|๐ธ
ห ๐บ๐๐ธห๐๐๐ธ
ห
๐น๐|๐ii โโ๏ธ
๐
hhโ|๐ธ
ห ๐น๐๐ธห๐0๐ธ
ห ๐บ๐|๐ii
โค
โ๏ธ
๐
hhโ|๐ธ
ห ๐บ๐๐ธห๐๐๐ธ
ห ๐น๐|๐ii
+
โ๏ธ
๐
hhโ|๐ธ
ห ๐น๐๐ธห๐0๐ธ
ห ๐บ๐|๐ii
. (2.64)
๐น = ,
๐น(๐(๐
1)) = ,
๐น(๐(๐
2)) = .
Figure 2.7: Example for the operator ๐น and the corresponding ๐น(๐(๐
1)) and ๐น(๐(๐
2)).
We can bound the first term
ร
๐hhโ|๐ธ
ห ๐บ๐๐ธห๐๐๐ธ
ห ๐น๐|๐ii
as follows. First, we write
โ๏ธ
๐
hhโ|๐ธ
ห ๐บ๐
ห ๐ธ๐๐๐ธ
ห ๐น๐
|๐ii
=tr ๐ธห๐๐
โ๏ธ
๐
๐ธ๐นห๐
|๐iihhโ|๐ธ
ห ๐บ๐
!
โค k๐ธห๐๐k๐น
โ๏ธ
๐
๐ธ๐นห๐|๐iihhโ|๐ธ
ห ๐บ๐
๐น
โค๐ฮ
2
โ๏ธ
๐
๐ธ๐นห๐|๐iihhโ|๐ธ
ห ๐บ๐
๐น
, where the last inequality comes from the fact that ๐
2 โBฮ(supp(๐น)) and ๐
2 โ A๐
implies that ๐๐ โฅ 2ฮ, so Lemma 2.4.2(ii) gives k๐ธห๐๐k๐น โค ๐ฮ
2. Proceeding as we did in the proof of Lemma 2.6.3, we can write the latter Frobenius norm as
โ๏ธ
๐
๐ธ๐นห๐|๐iihhโ|๐ธ
ห ๐บ๐
2
๐น
=
๐ท
โ๏ธ
๐ผ1,๐ผ
2, ๐ฝ
1, ๐ฝ
2=1
h๐ผ
1|h๐ผ
2| โ๏ธ
๐
๐ธ๐นห๐|๐iihhโ|๐ธ
ห ๐บ๐
!
|๐ฝ
1i|๐ฝ
2i
2
.
The individual terms in the sum can be depicted diagrammatically as h๐ผ
1|h๐ผ
2| โ๏ธ
๐
๐ธ๐นห๐|๐iihhโ|๐ธ
ห ๐บ๐
!
|๐ฝ
1i|๐ฝ
2i = .
Defining the vectors
|ฮจ(๐ผ, ๐ฝ)i = ,
we can then write h๐ผ
1|h๐ผ
2| โ๏ธ
๐
๐ธ๐นห๐|๐iihhโ|๐ธ
ห ๐บ๐
!
|๐ฝ
1i|๐ฝ
2i =hฮจ(๐ผ
1, ๐ฝ
1) | โ๏ธ
๐
ห
๐น๐ โ ๐ผ๐ท โ ๐ผ๐ท โ๐บห๐
!
|ฮจ(๐ผ
2, ๐ฝ
2)i.
Applying the Cauchy-Schwarz inequality, we get
h๐ผ
1|h๐ผ
2| โ๏ธ
๐
๐ธ๐นห๐
|๐iihhโ|๐ธ
ห ๐บ๐
!
|๐ฝ
1i|๐ฝ
2i
2
โค kฮจ(๐ผ
1, ๐ฝ
1) k2ยท kฮจ(๐ผ
2, ๐ฝ
2) k2ยท
โ๏ธ
๐
ห
๐น๐ โ๐ผ๐ท โ๐ผ๐ท โ๐บห๐
2
.
The norm of the vector|ฮจ(๐ผ, ๐ฝ)iis given by
kฮจ(๐ผ, ๐ฝ) k2= = = h๐ผ|๐|๐ผih๐ฝ|โ|๐ฝi,
where in the second equality, we have used the fixed-point equations (2.4.1). There- fore we have
โ๏ธ
๐
๐ธ๐นห๐|๐iihhโ|๐ธ
ห ๐บ๐
2
๐น
โค
โ๏ธ
๐
ห
๐น๐ โ๐ผ๐ท โ๐ผ๐ท โ๐บห๐
2 ๐ท
โ๏ธ
๐ผ1,๐ผ
2, ๐ฝ
1, ๐ฝ
2=1
h๐ผ
1|๐|๐ผ
1ih๐ผ
2|๐|๐ผ
2ih๐ฝ
1|โ|๐ฝ
1ih๐ฝ
2|โ|๐ฝ
2i
=
โ๏ธ
๐
ห
๐น๐ โ๐ผ๐ท โ๐ผ๐ท โ๐บห๐
2
ยท |tr(๐)tr(โ) |2= ๐ท2
โ๏ธ
๐
ห
๐น๐โ ๐ผ๐ท โ ๐ผ๐ท โ๐บห๐
2
,
where the last equality follows from the fact that we gauge-fix the left and right fixed-points such that๐ =๐ผ
C๐ท and tr(โ) =1. Finally, we note that since the operator
norm is multiplicative over tensor products, i.e., k๐ดโ๐ตk =k๐ดk ยท k๐ตk, we have
โ๏ธ
๐
ห
๐น๐โ ๐ผ๐ท โ ๐ผ๐ท โ๐บห๐
=
โ๏ธ
๐
ห ๐น๐ โ๐บห๐
=k๐นk. Therefore, we have
โ๏ธ
๐
hhโ|๐ธ
ห ๐บ๐
ห ๐ธ๐๐๐ธ
ห ๐น๐|๐ii
โค ๐ทk๐นk๐ฮ
2. The term involving๐
0in (2.6.4) can be bounded identically, and so hhโ|๐ธ
๐น(๐(๐
1))|๐ii โ hhโ|๐ธ
๐น(๐(๐
2))|๐ii
โค2๐ทk๐นk๐ฮ
2 , which proves (2.6.6).
We also need a different version of statement (i), as well as statements (iiib) and (ivb) derived from it.
Lemma 2.6.7. Forฮฉ โ [๐]2, let us define ๐๐ ๐0(ฮฉ) = โ๏ธ
(๐ , ๐0)โฮฉ
๐๐(๐ ๐โ๐
0๐0)hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i.
Let us write F :=supp(๐น) andA๐ = [๐]\A for the complement of a subsetA โ [๐]. Then:
๐๐ ๐0(Bฮ(F ) ร Bฮ(F ))
โค |Bฮ(F ) | ยท k๐นkโ
๐๐๐๐0+๐ โ
๐ ๐ฮ/2
2
, (2.65)
๐๐ ๐0(Bฮ(F ) ร Bฮ(F )๐)
โค |B2ฮ(F ) | ยท k๐นkโ
๐๐๐๐0 +๐
๐2๐ฮ/2
2
, (2.66)
๐๐ ๐0(Bฮ(F )๐ร Bฮ(F ))
โค |B2ฮ(F ) | ยท k๐นkโ
๐๐๐๐0 +๐
๐2๐ฮ/2
2
. (2.67) Finally, we have the following: There exists some fixed ๐
0 โ [๐]such that for ๐= ๐0, we have
๐๐ ๐(Bฮ(F )๐ร Bฮ(F )๐) =|Bฮ(F )๐| ยท hhโ|๐ธ๐น
๐(๐
0)|๐ii๐๐+๐
๐2๐ฮ/2
2
.(2.68) For ๐ โ ๐0, we have
๐๐ ๐0(Bฮ(F )๐ร Bฮ(F )๐)
โค |Bฮ(F ) | ยท k๐นkโ
๐๐๐๐0+๐
๐2๐ฮ/2
2
. (2.69) We observe that the first expression on the right-hand side of the above bound scales linearly with the support size of F instead of the support size of F๐, as may be naively expected. For (2.6.7), this is due to a cancellation of phases, see (2.6.4) below.
Proof. For the proof of (2.6.7), let us first define the vectors
|ฮจ(๐)i = โ๏ธ
๐โBฮ(F )
๐๐ ๐ ๐|ฮฆ๐ , ๐i. Then we can write
|๐๐ ๐0(Bฮ(F ) ร Bฮ(F )) | = |hฮจ(๐0) |๐น|ฮจ(๐)i|
โค k๐นk ยท kฮจ(๐) k ยท kฮจ(๐0) k , (2.70) where the last inequality follows by Cauchy-Schwarz along with the definition of the operator normk๐นk. The vector norm is given by
kฮจ(๐) k2 = โ๏ธ
๐ , ๐0โBฮ(F )
๐๐ ๐(๐โ๐
0)hฮฆ๐0, ๐|ฮฆ๐ , ๐i,
and together with Equation (i), we get
kฮจ(๐) k2= |Bฮ(F ) | ยท๐๐+๐(๐ฮ/2
2 ) . Taking the square root and inserting into Equation (2.6.4), we get
|๐๐ ๐0(Bฮ(F ) ร Bฮ(F )) | =k๐นk
โ๏ธ
|Bฮ(F ) | ยท๐๐+๐(๐ฮ/2
2 ) โ๏ธ
|Bฮ(F ) | ยท๐๐0 +๐(๐ฮ/2
2 )
=|Bฮ(F ) | ยท k๐นkโ
๐๐๐๐0 +๐
โ๏ธ|Bฮ(F ) | ยท๐ฮ/2
2
. Using the bound
Bฮ(F )
โค 5๐ฮ< ๐gives (2.6.7).
Next, let us look at (2.6.7). We have ๐๐ ๐0(Bฮ(F ) ร Bฮ(F )๐) = โ๏ธ
๐โBฮ(F )
โ๏ธ
๐0โBฮ(F )๐
๐๐(๐ ๐โ๐
0๐0)hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i= ฮฃ1+ฮฃ2,
where we define
ฮฃ1 := โ๏ธ
๐โBฮ(F )
โ๏ธ
๐0โB2ฮ(F )\Bฮ(F )
๐๐(๐ ๐โ๐
0๐0)hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i,
and
ฮฃ2:= โ๏ธ
๐โBฮ(F )
โ๏ธ
๐0โB2ฮ(F )๐
๐๐(๐ ๐โ๐
0๐0)hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i.
The norm of the second sum can be bounded using Lemma 2.6.5(iiia), giving us
|ฮฃ2| โค โ๏ธ
๐โBฮ(F )
โ๏ธ
๐0โB2ฮ(F )๐
|hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i|
โค |Bฮ(F ) | ยท |B2ฮ(F )๐| ยท๐(๐ฮ/2
2 )
=๐(๐2๐ฮ/2
2 ) , (2.71)
where we again use the trivial bound
Bฮ(F ) ,
B2ฮ(F )๐
โค ๐in the last line. Using Lemma 2.6.5 (iiib), we can express the first sum, with some fixed๐ โ [๐], as
ฮฃ1= โ๏ธ
๐โBฮ(F )
โ๏ธ
๐0โB2ฮ(F )\Bฮ(F )
๐๐(๐ ๐โ๐
0๐0)hhโ|๐ธ๐น(๐(๐))(๐ , ๐,ห ๐ห0, ๐0) |๐ii + |Bฮ(F ) | ยท |B2ฮ(F )\Bฮ(F ) | ยท๐(๐ฮ/2
2 )
= โ๏ธ
๐โBฮ(F )
โ๏ธ
๐0โB2ฮ(F )\Bฮ(F )
๐๐(๐ ๐โ๐
0๐0)hhโ|๐ธ๐น(๐(๐))(๐ , ๐,ห ๐ห0, ๐0) |๐ii +๐
๐2๐ฮ/2
2
, where the indices ห๐ and ห๐0are defined as in Lemma 2.6.5. To bound the remaining sum, let us introduce the states
|ฮจ1(๐)i := โ๏ธ
๐โBฮ(F )
๐๐ ๐ ๐|ฮฆ๐ฟ
ห
๐ , ๐i, and
|ฮจ2(๐0)i := โ๏ธ
๐0โB2ฮ(F )\Bฮ(F )
๐๐ ๐
0๐0
|ฮฆ๐ฟ
ห ๐0, ๐0i,
where we set๐ฟ =|supp(๐น(๐(๐))) |. Here,|ฮฆ๐ฟ๐ , ๐iare as defined in (2.6.3). Then we can write
ฮฃ1=hฮจ2(๐0) |๐น(๐(๐)) |ฮจ1(๐)i +๐
๐2๐ฮ/2
2
.
By the Cauchy-Schwarz inequality and the orthogonality relations (2.6.4), we have
|hฮจ2(๐0) |๐น(๐(๐)) |ฮจ1(๐)i| โค k๐นk ยท kฮจ1(๐) k ยท kฮจ2(๐0) k
= k๐นk
โ๏ธ
๐๐๐๐0|Bฮ(F ) | ยท |B2ฮ(F )\Bฮ(F ) | , where we bound the states|ฮจ1,2(๐)iin exactly the same way as we did in the proof of (2.6.7). Using the fact that|Bฮ(F ) |,|B2ฮ(F )\Bฮ(F ) | โค |B2ฮ(F ) |, we conclude that
|ฮฃ1| โค |B2ฮ(F ) | ยท k๐นkโ
๐๐๐๐0+๐
๐2๐ฮ/2
2
.
Combining this with (2.6.4) gives the claim (2.6.7). The proof of (2.6.7) is analo- gous, using Lemma 2.6.5(iv).
Finally, consider (2.6.7) and (2.6.7). We have ๐๐ ๐0(Bฮ(F )๐ร Bฮ(F )๐) = โ๏ธ
๐โBฮ(F )๐
๐๐ ๐(๐โ๐
0)hฮฆ๐ , ๐0|๐น|ฮฆ๐ , ๐i
| {z }
=:ฮ1
+ โ๏ธ
๐ , ๐0โBฮ(F )๐ ๐โ ๐0
๐๐(๐ ๐โ๐
0๐0)hฮฆ๐0, ๐0|๐น|ฮฆ๐ , ๐i
| {z }
=:ฮ2
.
Using Lemma 2.6.5(iia), we have
|ฮ2| โค k๐นk ยท๐(๐2๐ฮ/2
2 ). (2.72)
On the other hand, by Lemma 2.6.5(iib), or more precisely its refinement in the form of Equation (2.6.6) from Lemma 2.6.6, we have
ฮ1=ยฉ
ยญ
ยซ
โ๏ธ
๐โBฮ(F )๐
๐๐ ๐(๐โ๐
0)ยช
ยฎ
ยฌ
hhโ|๐ธ๐น(๐(๐
0))|๐ii๐๐ ๐0 +๐(๐๐ฮ/2
2 ) for some fixed ๐
0 โ Bฮ(F )๐. For ๐0 = ๐, the sum above is given trivially by ร
๐โBฮ(F )๐1=
Bฮ(F )๐
. For๐ โ ๐0, we haveร
๐โ[๐]๐๐ ๐(๐โ๐
0) =0, and hence
โ๏ธ
๐โBฮ(F )๐
๐๐ ๐(๐โ๐
0)
=
โ๏ธ
๐โBฮ(F )
๐๐ ๐(๐โ๐
0)
โค |Bฮ(F ) |. (2.73)
Therefore, for ๐ =๐0, we have ฮ1=
Bฮ(F )๐
hhโ|๐ธ๐น(๐(๐
0))|๐ii๐๐+๐(๐๐ฮ/2
2 ) , and for๐ โ ๐0, we have
|ฮ1| โค
Bฮ(F )
hhโ|๐ธ๐น(๐(๐
0))|๐ii๐๐ ๐0+๐(๐๐ฮ/2
2 )
โค
Bฮ(F )
ยท k๐นk๐๐ ๐0+๐(๐๐ฮ/2
2 ). Note that we also have๐๐ ๐0 โค โ
๐๐๐๐0by the Cauchy-Schwarz inequality. Combining these results with (2.6.4) proves claims (2.6.7) and (2.6.7).