On the Monomiality of Nice Error Bases

6 Pages·0.18 MB·

by  Andreas Klappenecker

1
On the Monomiality of Nice Error Bases
Andreas Klappenecker and Martin R¨otteler
Abstract—Unitary error bases generalizethe Pauli matri- entries in the basis matrices are zero. In that sense, the
ces to higher dimensional systems. Twobasicconstructions nice error bases are simpler than one would expect.
ofunitaryerrorbasesareknown: Analgebraicconstruction
We will recall the definition and some properties of nice
by Knill, which yields nice error bases, and a combinato-
rialconstructionbyWerner,whichyieldsshift-and-multiply error bases and shift-and-multiply bases in the next sec-
bases. AnopenproblemposedbySchlingemannandWerner tion. Weintroduceanotionofequivalenceforunitaryerror
relates these two constructions and asks whether each nice
bases in Section III. We show that there exist shift-and-
error basis is equivalent to a shift-and-multiply basis. We
3 solve this problem and show that the answer is negative. multiply bases which are not nice error bases. We then
0 However, we also show that it is always possible to find a go on to prove that there exists an infinite number of nice
0 fairly sparse representation of a nice error basis. error bases, which are not of shift-and-multiply type. We
2
Keywords—Pauli matrices,unitary errorbases, monomial construct an explicit counterexample in dimension 165 in
n representations, Hadamard matrices, Latin squares. Section V.
a
Notations. We denote by C the field of complex num-
J
I. Introduction bers, by Z the ring of integers, by Z the ring of integers
6 d
modulo d. The group of unitary d×d matrices is denoted
1 Unitaryerrorbasesareimportantprimitivesinquantum
by U(d), the general linear group by GL(d,C).
informationtheory. Theyformthebasisofquantumerror-
1
v correcting codes, teleportation, and dense coding schemes. II. Construction of Unitary Error Bases
8 A unitary error basis is by definition an orthonormalbasis
7 ofthevectorspaceofcomplexd×dmatriceswithrespectto AunitaryerrorbasisisasetE ofd2 unitaryd×dmatri-
0 the inner product hA,Bi = 1/dtr(A†B). Such bases have cessuchthattr(E†F)=0foralldistinctE,F ∈E. Theset
1 been studied in numerous works, see for instance [1,2,3, P ofPaulimatricesprovidesthemostwell-knownexample:
0
4,5,6]. However, surprisingly little is known about their
3
1 0 0 1 1 0 0 −i
0 general structure. P = , , , .
(cid:26)(cid:18)0 1(cid:19) (cid:18)1 0(cid:19) (cid:18)0 −1(cid:19) (cid:18)i 0(cid:19)(cid:27)
/ Currently, two fundamentally different constructions of
h
unitary error bases are known: An algebraic construction
p Unitary error bases generalize this example to arbitrary
- duetoKnill[3],whichyieldsniceerrorbases,andacombi-
dimensions. The nonbinary case is more interesting, since
t
n natorialconstructiondue to Werner [7], whichyields shift-
there exist different, non-equivalent, error bases. We re-
a and-multiply bases. The nice error bases of small dimen-
viewinthissectiontheconstructionsofunitaryerrorbases
u sionhave been completely classifiedin [2]. A quick inspec-
q by Knill and by Werner.
tion of this catalogue shows that each nice error basis in
:
v dimensiond≤5isinfactequivalenttoabasisofshift-and-
A. Equivalence
i multiply type. This motivated Schlingemann and Werner
X
to formulate the following problem [8,9]: LetE andE′ be twounitaryerrorbasesinddimensions.
r
We say that E and E are equivalent, in signs E ≡ E , if
a ′ ′
Is every nice error basis equivalent to a basis of
and only if there exist unitary matrices A,B ∈ U(d) and
shift-and-multiply type?
constants c ∈U(1), E ∈E, such that
E
We willgivea precise explanationofthe technicalterms
in the next section. An affirmative answer to this problem E′ ={cEAEB :E ∈E}.
would imply that a nice error basis can be represented by
monomial matrices, which have only one nonzero entry in One readily checks that ≡ is an equivalence relation.
each row and in each column. Lemma 1: Any unitary error basis in dimension 2 is
Do nice error bases really have such a simple structure? equivalent to the Pauli basis.
The answer is no, as we will show in this correspondence. Proof: Let A = {A1,A2,A3,A4} be an arbitrary
However, we will prove that each nice error basis is equiv- unitary error basis in dimension 2. This basis is equiva-
alent to a unitary error basis where at least half of the lent to a basis of the form {12,diag(1,−1),B3,B4}. The
diagonal elements of B and B are necessarily zero, be-
3 4
TheworkbyA.K.wassupportedinpartbyNSFgrantEIA0218582 cause of the trace orthogonality relations. We may as-
andaTexasA&MTITFgrant. TheworkbyM.R.wassupportedin sume that B and B are of the form B = antidiag(1,a)
partbyECgrantIST-1999-10596(Q-ACTA),byCSEandMITACS. 3 4 3
and B = antidiag(1,−a), where a = exp(iφ) for some
A. Klappenecker is with the Department of Computer Science, 4
Texas A&M University, College Station, TX 77843-3112, USA (e- φ ∈ R, since we are allowed to multiply the matrices with
mail: [e-mail]) scalars. Conjugating the basis elements with the matrix
M. Ro¨tteler is with the Department of Combinatorics and Opti-
diag(1,exp(−iφ/2)) yields the matrices of Pauli basis up
mization, University of Waterloo, Waterloo, Ontario, Canada, N2l
3G1(e-mail: [e-mail]) to scalar multiples, hence A≡P.
2
B.LNeticGe ebreroar bgarsoeusp of order d2 with identity element 1. HL==(j(−ωkiℓ)mk,oℓd∈Zdd),i,wj∈iZthd aωnd=theexpco(2mπpil/edx).HaFdoarmeaxradmmpalet,riixf
d=3, then
A nice error basis in d dimensions is given by a set E =
{ρ(g)∈U(n)|g ∈G} of unitary matrices such that
0 1 2 1 1 1
(i) ρ(1) is the identity matrix, L:= 2 0 1 , H = 1 ω ω2 ,
1 2 0 1 ω2 ω
(ii) trρ(g)=0 for all g ∈G\{1},    
(iii) ρ(g)ρ(h)=ω(g,h)ρ(gh) for all g,h∈G, where ω = exp(2πi/3). According to equation (1), the
basis matrices E and E are respectively given by
where ω(g,h) is a phase factor. Conditions (i) and (iii) 01 12
state that ρ is a projective representation of the group G.
0 1 0 0 0 ω2
Lemma 2 (Knill) A nice error basis is a unitary error
basis. E01 = 0 0 1 , E12 = 1 0 0 .
1 0 0 0 ω 0
Proof: Let E ={ρ(g)∈U(n)|g ∈G} be a nice error    
basis. Notice that ρ(g) = ω(g 1,g) 1ρ(g 1). Assume
† − − − The entries of the middle row of the Latin square L and
that g,h are distinct elements of G, then g 1h 6= 1, hence
− thefirstrowofthecomplexHadamardmatrixH determine
tr(ρ(g) ρ(h)) = ω(g 1,g) 1ω(g 1,h)tr(ρ(g 1h)) = 0 by
† − − − − the matrix E .
property (ii) of a nice error basis. 01
The next example shows that nice error bases exist in D. Abstract Error Groups
arbitrary dimensions:
Let E ={ρ(g):g ∈G} be a nice errorbasis. A groupH
Example 3: Let d ≥ 2 be a integer, ω = exp(2πi/d).
isomorphic to the group generated by the matrices ρ(g) is
Let X denote the cyclic shift X |xi=|x−1moddi, and
d d
let Z denote the diagonalmatrix diag(1,ω,ω2,...,ωd 1). called an abstract error group of E.
d −
Then E := {XiZj|(i,j) ∈ Z ×Z } is a nice error basis. The group H is not necessarily finite. However, if we
d d d d d multiply the representing matrices ρ(g) by scalars c such
This has been shown by an explicit calculation in [10]. g
thatc ρ(g)hasdeterminant1,thentheresultingniceerror
g
basis E = {c ρ(g) : g ∈ G} is equivalent to E, and its
C. Shift-and-Multiply Bases ′ g
abstract error group H is finite.
′
RecallthataLatinsquareoforderdisad×dmatrixsuch
Thus,ifwe consideranice errorbasisuptoequivalence,
thateachelementofthesetZ iscontainedexactlyoncein
d then we may assume without loss of generality that the
eachrowandineachcolumn. AcomplexHadamardmatrix
associated abstract error group is finite.
H oforderdis amatrixinGL(d,C)suchthatH ∈U(1),
ik Example 6: TheabstracterrorgroupH associatedwith
0≤i,k <d, and H H =d1. d
† the nice errorbasis E fromExample 3 is by definition iso-
Let H = (H(i) : 0 ≤ i < d) be a sequence of complex d
morphic to the group generated by X and Z . An ele-
d d
Hadamard matrices, and let L be a Latin square L of or- mentofthe grouphX ,Z iisofthe formωzZyXx,because
der d. Ashift-and-multiply basisE associatedwithL,His d d d d
X Z = ωZ X . Notice that H is isomorphic to the
d d d d d
given by the unitary matrices
unitriangular subgroup of GL(3,Z ) given by
d
E =P diag(H(j) :0≤k <d), i,j ∈Z , (1)
ij j ik d 1 x z
where P denotes the permutation matrix with entries de- Hd ∼= 0 1 y :x,y,z ∈Zd.
j  0 0 1 
fined by Pj(L(j,k),k) = 1, for 0 ≤ k < d, and 0 other-  
wise. In short, E is determined by the ith row of the jth  
ij WeprefertodescribethegroupH abstractlybythesetof
Hadamard matrix H(j), and by the entries of the jth row d
elements(x,y,z)∈Z3 withcompositiongivenby(x,y,z)◦
of the Latin square L; briefly E |ki=H(j)|L(j,k)i. d
If all matrices in H are equalitjo a singliekHadamard ma- (x′,y′,z′)=(x+x′,y+y,z+z′+xy′),wherealloperation
are modulo d.
trixH,thenwerefertothisbasisastheshift-and-multiply
Recall that a finite group H which has an irreducible
basis associated with L,H.
representation of large degree d = (H:Z(H)) is called
Lemma 4 (Werner) A shift-and-multiply basis is a uni-
a group of central type. It has beepn shown in [2] that a
tary error basis.
finite group H is an abstract error group if and only if it
Proof: We have to show that tr(Ei†jEkl) = 0 when is a group of central type with cyclic center. A somewhat
(i,j)6=(k,l). If j 6=l, then the matrix Pj†Pl has a vanish- surprising consequence is that an abstract error group has
ingdiagonal,whencetr(Ei†jEkl)=0foranychoiceofiand to be a solvable group.
k. If j =l and i6=k, then tr(Ei†jEkj) is equal to the inner III. Wicked Error Bases
product of the ith and kth row of the complex Hadamard
matrix H(j), hence tr(Ei†jEkj)=0. A unitary error basis, which is not equivalent to a nice
Example 5: The nice error basis E in Example 3 is a errorbasis,is calledwicked. We show now thatthere exist
d
shift-and-multiply basis. Indeed, choose the Latin square an abundance of wicked shift-and-multiply bases.
3
Theorem 7: Let E be the shift-and-multiply basis asso- induced from a character from the center Z(H), see [11].
α
ciated with L,H , where Since H is an abstract error group, it is in particular a
α
solvable group [2]. It has been shown by Ferguson and
0 1 2 3 1 1 1 1
Isaacs [12] that the multiple of a primitive character of a
L= 3 0 1 2 , H = 1 1 −1 −1 . solvable group can never be induced from an irreducible
2 3 0 1 α 1 −1 eiα −eiα
character of a proper subgroup, contradiction.
   
 1 2 3 0   1 −1 −eiα eiα  It follows that E = {Aρ(t)A 1|t ∈ T} is a nice error
    s −
basis, and half of the entries of Aρ(t)A are zero. By con-
If α∈Q , then E is not equivalent to a nice error basis. †
× α struction, E ≡E ≡E , hence E ≡E as claimed.
Proof: Suppose there exist A,B ∈ U(4), and scalars ′ s s
c such that the set {c AU B : i,j = 1,...,4} is a nice V. Nonmonomial Abstract Error Groups
ij ij ij
errorbasis. Withoutlossofgenerality,wemayassumethat
Inthissection,wewillfinallyanswerthe questionraised
the group G generated by the matrices c AU B is finite.
ij ij
by Schlingemann and Werner:
Notice that the unitary error basis E contains the ma-
α
trices M = diag(1,−1,eiα,−eiα), and M = 1 . Con- Theorem 9: There exist nice error bases which are not
1 2 4
equivalent to bases of shift-and-multiply type.
sequently, (c1AM1B)(c2AM2B)−1 = c1c−21AM1M2†A† = We will prove this result with the help of abstract error
c1c−21AM1A−1 is an element of the group G. Since G is groups. The following result will play a key role in our
finite, it follows that c1c−21AM1A−1 and hence also that proof:
ctr1ice−2s1oMft1hiissmofatfirnixitethoisrdimerp.liLesoothkaintgc1act−21thaendincd1ivc−2id1ueaiαl aenre- cenTthraeolrteympe1w0i(tDhacdyec,liIcsacaecnst)erT, whehriechexhiasstaangornomuopnoHmioafl
roots of unity. It follows that eiα would have to be a root
irreducible character χ of degree χ(1)= (H:Z(H)).
of unity as well, contradicting the assumption α∈Q .
×
Proof of Theorem 9: We actuallypshow a stronger
IV. Sparsity of Nice Error Bases statement, namely that there are nice error bases which
are not equivalent to monomial bases.
Amatrixissaidtobemonomialifandonlyifitcontains
By Theorem 10, there exists an abstract error group H
exactly onenonzeroentryin eachrowandineachcolumn.
thathasanon-monomialirreducibleunitaryrepresentation
IfaniceerrorbasisE isalsoashift-and-multiplybasis,then
ρ of degree d = (H:Z(H)). Denote by E a nice error
allmatricesM ∈E aremonomial. Thisdoesnothavetobe
basis associated wpith ρ, that is,
the case. However, our next result shows that a nice error
basis is always equivalent to a fairly sparse error basis:
E ={ρ(t)|t∈T}
Theorem 8: Let E be a nice error basis. There exists an
equivalentnice errorbasis E ≡E suchthatat leasthalfof
s where T is a transversal of H modulo Z(H), with 1 ∈ T.
the entries of each matrix M ∈E are zero.
s Since ρ is nonmonomial, it is impossible to find a base
Proof: ThereexistsaniceerrorbasisE ≡E suchthat
′ changeA such that Aρ(t)A is monomialfor allt∈T. We
†
the abstract error group H of E is finite. The group gen-
′ shownextthatthispropertyisevenpreservedwithrespect
eratedby the matrices of E is an irreducible matrix group
′ to the equivalence ≡.
isomorphic to H. In other words, H has an irreducible
Seekingacontradiction,wesupposethatthereexistuni-
unitary representation ρ such that
tary matrices A,B and scalars c such that c Aρ(t)B is
t t
a monomial unitary error basis. Since the identity ma-
E ={ρ(t)|t∈T},
′ trix 1 = ρ(1) is part of the nice error basis, we can
d
where T is a transversalof H modulo Z(H). conclude that the matrix C = c1AB is monomial. But
RecallthatacharacterχofH issaidtobeinducedfrom ctAρ(t)B = ctAρ(t)(A†A)B = ct/c1Aρ(t)A†C shows that
an irreducible character ψ of a proper subgroup K of H if the resulting equivalent error basis is nonmonomial. In-
χ is of the form deed,amongthematricesAρ(t)A† isatleastonenonmono-
mialmatrixU. MultiplyingU withthemonomialmatrixC
χ(x)= 1 ψ(hxh 1). and the scalarprefactor ct/c1 cannot result in a monomial
−
|K| matrix, leading to a contradiction.
hX∈H
hxh−1∈K Remark. We haveshownin the proofofTheorem8 that
Let χ be the irreducible character of H corresponding to an irreducible character χ of large degree of an abstract
therepresentationρ. Ifχisinducedbyanirreduciblechar- error group H is always induced from an irreducible char-
acter of a proper subgroup of H, then there exists a base acterψofapropersubgroup. TheessenceofTheorem10is
change A ∈ U(d) such that at least half of the entries of thatingeneralwecannotchooseψtobealinearcharacter.
the matrices Aρ(h)A 1 are zero. In the next section, we want to construct an explicit
−
Seekingacontradiction,weassumethatχisnotinduced example of a nice error basis that is not equivalent to a
by an irreducible characterof some proper subgroupof H. shift-and-multiply basis. We will need an explicit example
In other words, we assume that χ is a primitive character. of a group H satisfying the assumptions of Theorem 10
Notice that the degree of χ is large, χ(1) = (H:Z(H)). for that purpose. Theorem 10 was independently proved
It follows that the multiple χ(1)χ of the cpharacter χ is by Everett Dade and by Martin Isaacs, but unfortunately
4
their results remained unpublished. Our exposition is a Let p,q, and r be three distinct odd primes such that r
variation of Dade’s approach, which we include here with divides p+1 and q+1. Define
the kind permission of Professor Dade.
G=(H ×H )⋊ H ,
p q ϕ r
A. Semidirect Products
where the action ϕ is chosen such that ϕ(g) acts trivially
Let N,H be finite groups, and let ϕ be a group ho- on H ×H for all g ∈Z(H ).
p q r
momorphism from H to Aut(N). Recall that the (outer) Lemma 12: The group G is of central type with cyclic
semidirect product G=N ⋊ϕH is a group defined on the center. The center Z(G) is of order pqr.
set N ×H, with composition given by (n1,h1)(n2,h2) = Proof: The three Sylow subgroups Hp, Hq, and Hr
(n1ϕ(h1)(n2),h1h2). If the center of H acts trivially on of the group G are of central type. Since the center of
N, then the center of the semidirect product is given by H acts trivially on H ×H , we have Z(G) = Z(H )×
r p q p
Z(G)=Z(N)×Z(H). A detaileddiscussion ofsemidirect Z(H )× Z(H ). In particular, H ∩ Z(G) = Z(H ) for
q r ℓ ℓ
products can be found for instance in [13]. ℓ∈{p,q,r}. It follows from Theorem 2 of [16] that G is a
group of central type.
B. Automorphisms of the Heisenberg group H
p The center of a Heisenberg group H is given by the
d
Letpbeanoddprime. TheHeisenberggroupHpdefined cyclic subgroup Z(Hd) = {(0,0,z)|z ∈ Zd}. The center
inExample6hasp3elements. Recallthatthespeciallinear of G is thus a direct product of cyclic groups of coprime
group SL(2,Fp) is a matrix groupof order (p+1)p(p−1), orders, hence Z(G) is a cyclic group of order pqr.
which is generated by the matrices
We want to choose ϕ such that G does not contain a
0 −1 1 0 subgroupof index pqr. If this is the case, then a character
α= and β = .
(cid:18) 1 0 (cid:19) (cid:18) 1 1 (cid:19) χ ∈ Irr(G) of degree χ(1)= (G:Z(G)) =pqr cannot be
monomial, because a monompial character is induced from
The special linear group acts as an automorphisms group
a linear character of a subgroup of index pqr.
on the Heisenberg group H . Indeed, define
p Lemma 13: If G has a subgroup H of index (G:H) =
α(x,y,z) = (−y,x,z−xy), pqr, then there exists a conjugate subgroup K = Hg such
β(x,y,z) = (x,x+y,z+ (p+1)x2). that |K∩Hp|=p2, |K∩Hq|=q2, and |K∩Hr|=r2,
2 Proof: The Sylow subgroups H and H of G are
p q
Itisstraightforwardtocheckthatα,β ∈Aut(Hp). Wewill normal, hence respectively contain the subgroups of order
construct the abstract error group by semidirect products p2 and of order q2 of Hg for any g ∈ G. The subgroup of
of Heisenberg groups. We will need some more detailed order r2 of H is contained in some Sylow subgroup of G.
knowledge about these automorphisms to tailor this con- Theclaimfollows,sincether-SylowsubgroupsinGareall
struction to our needs. conjugate by Sylow’s theorem.
Recall that a matrix M is said to act irreducibly on a Theorem 14: Letp,q,r be distinctoddprimessuchthat
vector space V if and only if {0} and V are the only M- r divides p+1 and q+1. It is possible to choose ϕ such
invariant subspaces of V. that G=(H ×H )⋊ H is a group of central type that
p q ϕ r
Lemma 11: Let r,p be odd prime numbers such that does not contain a subgroup of index pqr.
r|(p+1). ThegroupSL(2,Fp)containsmatricesoforderr, Proof: First,wedefinetheactionϕofHr onHp×Hq.
and all such matrices act irreducibly on Fp×Fp. Recall that the Heisenberg group Hr is generated by the
Proof: It is known that the group SL(2,Fp) has a two elements a = (1,0,0) and b = (0,1,0). Let A,B ∈
cyclic subgroup of order p+1, hence contains matrices of SL(2,F ) be matrices of order r. The element a acts with
p
order r, see [14, p. 42]. Let g ∈ SL(2,Fp) be an element AonHp,andtriviallyonHq. Similarly,the elementb acts
of order r. The subgroup R = hgi of order r acts on the trivially on H , and with B on H .
p p
vector space V = Fp×Fp. The orbit length |Rv| ∈ {1,r} Notice that c = (0,0,1) = aba−1b−1 generates the cen-
for all v ∈V. If we denote by U the centralizer of R, then ter of H . It follows that c acts trivially on H × H ,
r p q
rdivides|V|−|U|=p2−pk,wherek ∈{0,1}. Sincerdoes hence Z(H ) acts trivially on H × H . An immediate
r p q
not divide p−1 or p, it follows that k has to be 0. Thus, consequence is that the center of G is given by Z(G) =
0 ∈ V is the only fixed point. It follows that {0} and V Z(H )×Z(H )×Z(H ).
p q r
are the only R-invariant subspaces of V. Indeed, suppose TheSylowsubgroupsofGareisomorphictotheHeisen-
that V′ is an R-invariant subspace with p elements, then berg groups Hp, Hq, and Hr, whence all Sylow subgroups
the number of elements in V′−{0} must be a multiple of of G are of central type. Moreover, the construction of G
r, contradicting r ∤(p−1). ensures that the intersection of a Sylow subgroup P of G
Remark: The preceding lemma also follows from Theo- with the center Z(G) gives Z(P). It follows from Theo-
rem 3.5 in Hering [15]. rem 2 of [16] that G is a group of central type.
Seeking a contradiction, we suppose that G has a sub-
C. A nonmonomial nice error group
group of index pqr. Lemma 13 shows that G must then
We take now the Heisenberg groups as a starting point have a subgroup K such that the intersection of K with
toconstructanabstracterrorgroupGofaniceerrorbasis, H , H , and H contains p2, q2, and r2 elements, respec-
p q r
which is not equivalent to a shift-and-multiply basis. tively.
5
Let X = ha,ci and Y = hb,ci; both are subgroups C. A Nonmonomial Error Basis in Dimension 165
of order r2 of the Heisenberg group H . The subgroup
r Weneedanelementoforder3ofSL(2,F )tospecifythe
K =H ∩K cannotcoincidewithbothX andY. Suppose p
r r action of H on H ×H . We can choose for instance the
that K 6=X. Since H =hK ,Xi,the groupK mustact 3 5 11
r r r r element γ =βαβα in SL(2,F ), i.e.,
irreduciblyonH /Z(H ). ThesubgroupK =K∩H can- p
q q q q
not exist, because K would have to normalizeK /Z(K ),
r q q 1 0 0 −1 1 0 0 −1
which is impossible. Similarly, if K 6= Y, then the sub- γ = .
r (cid:18) 1 1 (cid:19)(cid:18) 1 0 (cid:19)(cid:18) 1 1 (cid:19)(cid:18) 1 0 (cid:19)
group K = K ∩H cannot exist. Therefore, the group
p p
K cannot exist. This proves that G does not contain a One easily verfies that γ3 =1.
subgroup of index pqr.
The action of γ on the matrix representation ρ (H ) of
5 5
VI. An Explicit Counterexample the Heisenberg group H5 is realized by conjugation with
the matrix R = D Z3·F ·D Z3·F ∈ U(5). Similarly,
We take now a closer look at the examples given in the 5 5 5 5 5 5 5
the action of γ on the matrix version of H is given by
11
previous section. Specifically, it is our goal is to make the
conjugationwithR =D Z3 ·F ·D Z3 ·F ∈U(11).
constructionexplicit for the smallestpossible choice of pa- 11 11 11 11 11 11 11
Recall that H is generatedby the two elements (1,0,0)
3
rameters. This will give us a concrete example of a nice
and (0,1,0). According to the construction of Section V,
error basis in dimension 165=3·5·11 that is not equiva-
the action of H is chosen such that the generator (1,0,0)
3
lent to a shift-and-multiply basis.
acts with γ on H and trivially on H , and the generator
5 11
(0,1,0) acts trivially on H and by γ on H . Explicitly,
A. A Representation of H 5 11
p
we obtain the matrix group:
Letpbeanoddprime. TheHeisenberggroupH hasan
p
irreducible representationρp:Hp →U(p), which associates G=h13⊗X5⊗111, 13⊗Z5⊗111, 13⊗15⊗X11,
to an element (x,y,z)∈Hp the matrix 13⊗15⊗Z11, X3⊗R5⊗111, Z3⊗15⊗R11i.
ρ ((x,y,z))=ωzZyXx.
p p p The group G is an inner semidirect product of the form
H ⋉(H ×H ). Indeed, the subgroup generated by the
Hereω denotestheprimitiverootofunityω =exp(2πi/p), 3 5 11
first four generators is isomorphic to N = H ×H since
and X and Z denote the generalized Pauli matrices, as 5 11
p p
theirreduciblerepresentationsofadirectproductaregiven
defined in Example 3.
bytensorproductsoftheirreduciblerepresentationsofthe
We will derive a faithful irreducible matrix representa-
factors. WehaveN(cid:1)GbecauseofthechoiceofR andR .
tionofdegree165ofthegroupG=(H ×H )⋊ H bya 5 11
5 11 ϕ 3
The complement H of N is given by the group generated
suitable compositionofthe representationsρ ,ρ ,andρ .
3 5 11
bythe toremaininggenerators. Obviouslythe intersection
B. Automorphisms of H H ∩N is trivial and G = HN by definition. This shows
p
that G=H ⋉N.
Recall that a group G is said to be an inner semidirect
We obtain a non-monomial nice error basis by choosing
product of the two subgroups H and N if and only if N is
a transversal of Z(G) in G. This nice error basis is in
a normal subgroup of G such that HN =G and H ∩N =
particular not equivalent to a shift-and-multiply basis.
{1 }.
G
The matrix group representing G = (H ×H )⋊ H
5 11 ϕ 3 VII. Conclusions
isaninnersemidirectproduct. This meansthatthe action
of the automorphism is realized by a conjugation with a We have studied unitary error bases for higher-
matrix. It suffices to find matrices which realize the ac- dimensional systems. We have shown that all such bases
tion of the generators α and β of SL(2,F ). Recall that areequivalentindimension2. Thischangesdramaticallyin
p
α(1,0,0)=(0,1,0) and α(0,1,0)=(−1,0,0). This means higherdimensions. Wehaveshownthatniceerrorbasesare
we need to find a matrix A∈U(p) such that in general not equivalent to shift-and-multiply bases, and
viceversa. ThissolvesanopenproblemposedbySchlinge-
XpA =Zp and ZpA =Xp−1. mann and Werner.
Similarly, the action of the automorphism β is determined Acknowledgments
by β(1,0,0) = (1,1,(p + 1)/2) and β(0,1,0) = (0,1,0).
WethankEverettDadeandMartyIsaacsfordiscussions
Hence we need to find a matrix B ∈U(p) such that
onnon-monomialrepresentationsofgroupsofcentraltype.
XB =ω(p+1)/2Z X and ZB =Z . Theconstructionofthenonmonomialcharacterofextremal
p p p p p
degreegiveninSectionVisduetoEverettDade. Wethank
We can choose A to be the discrete Fourier transform Thomas Beth for his encouragement and support.
F = 1 (ωkℓ) , with ω = exp(2πi/p). Notice
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