Monomial MUBs. September 27, 2005

Transcription

1 Monoia MUBs Seteber 7, 005 Abstract We rove that axia sets of couniting onoia unitaries are equivaent to a grou under utiication. We aso show that hadaard that underies this set of unitaries is equivaent to a hadaard with the sae grou structure. MUBs, Hadaards, and Notation Suose there are n MUBs in diension d: B i = { i, i,..., i d } for i (, n) such that: j i = δ i, δ j, + ( δ i, ) d We ca B 0 the standard basis. A set of MUBs then defines a set of Hadaard atrices: H(, ) = j j j The MUB condition is: H (, )H(, ) = j j j j = j j jj = j j j j j j j = = H(, ) With H(, ) being a hadaard atrix for a. Since H(, ) = I, then H (, ) = H(, ) and we can write: H(, )H(, ) = H(, ) We use H(, ) q,r q H(, ) r

2 Monoia Unitaries We now that if we have sets of d traceess couniting unitary atrices, such that a (d ) airs are orthogona, then the eigenvectors of the sets are MUBs. Suose that in addition to the above contraints, we aso have the roerty that a of the unitary atrices are onoia atrices. Consider such a onoia untiary: U λ(α) = j λ(α) j j j The index α (, d) and we define U λ() = I, thus λ() i =. Orthogonaity between the U λ(α) and U λ(β) is equivaent to orthogonaity of λ(α) and λ(β), with the usua vector inner roduct. Moiaity of U eans that: Uλ(α) {0 () = e iφ = () () We can rewrite equation in ters of the hadaard atrices: Uλ(α) = j λ(α) j j j = j λ(α) j H(, ) j,h(, ) j, () Proosition.. Every set of orthogona onoia unitaries ay be noraized such that: Uλ(α) = with U λ(α) aso being a set of orthogona onoia unitaries. Equivaenty, we ay aways choose λ(α) =. Proof. Given soe set of orthogona unitaries U λ(α), we can define U λ(α) = λ (α) U λ(α). Ceary, U λ(α) =. Mutiication by a constant does not change the orthogonaity nor the onoiaity, thus the set U λ(α) is a set of orthogona onoias with the desired roerty. Now we define two casses of diagona unitaries: ζ(α) λ(α) Z(, n) d H(, n), See that ζ(α) is just a diagonaization of U λ (α), and Z(, n) is th coun of H(, n), ut on the diagona of a atrix, and renooraized so that the atrix is unitary. Lea.. Z(, ) Z (, ) = c(, )ζ(α(, )) for c(, ) a function fro Z d Z d C and α(, ) a function fro Z d Z d Z d Proof. Note that due to the orthonoraity of the set of d atrices U λ(α), the set ζ(α) fors an orthonora basis for diagona atrices. Aso note that due to the unitarity of a hadaard atrices, the set Z(, n) for (, d), fors an orthonora basis for diagona atrices. Rewriting equation in ters of the above diagonas we have: Uα = d T r ( ζ(α)z (, ) Z(, ) )

3 We use a atrix inner roduct of: U, V d T r(u V ) Thus: U α = Z(, ) Z (, ), ζ(α) We can exand Z(, ) Z (, ) in ters of ζ(α) since ζ(α) is a basis. Z(, ) Z (, ) = β c β (, )ζ(β) Z(, ) Z (, ), ζ(α) = c α(, ) Since Z(, ) Z (, ) is unitary, we now that α c α(, ) =. Since U α is onoia, we now that c α (, ) is 0 or. Thus, c α (, ) = 0 for a α α(, ). Hence: Z(, ) Z (, ) = c(, )ζ(α(, )) Lea.3. α(, ) is a atin square. Proof. This iies that α(, ) α(, ) if and iewise for. We can rove this by direct coutation. See that if we assue that α(, ) = α(, ) then we can show that =. The sae aroach wors for. Z(, ) Z (, ) c (, ) = ζ(α(, )) = ζ(α(, )) = Z(, ) Z (, ) c (, ) Z (, ) Z(, ) = c(, )c (, ) Z(, ), Z(, ) = c(, )c (, ) δ, = = Lea.4. c(, )c(, n) = c(, n) and ζ(α) for a grou of order d under utiication. Proof. Due to Lea., we now that: Z(, ) = c(, n)ζ(α(, n))z(, ) n Z(, ) = c(, n)ζ(α(, n))z(, ) n Z(, ) Z (, ) = c(, )ζ(α(, )) c(, n)ζ(α(, n))z(, ) n = c(, )c(, n)ζ(α(, ))ζ(α(, n))z(, ) n c(, n)ζ(α(, n)) = c(, )c(, n)ζ(α(, ))ζ(α(, n)) Using the fact that ζ(α) = we have that c(, )c(, n) = c(, n), roving this first art of the Lea. Aying this resut we obtain: ζ(α(, n)) = ζ(α(, ))ζ(α(, n)) Thus the atrices ζ(α) are cosed under utiication. Note that since Z(, ) Z (, ) = I = c(, )ζ(α(, )) we have that c(, ) = and ζ(α(, )) = ζ() = I. Finay, any set of d invertibe atrices cosed under utiication that contains I is grou of order d. 3

4 Coroary.5. Any set of couniting unitary onoias has an equivaent onoia reresentation which fors a grou of order d under utiication. Proof. ζ(α) is a diagonaization of U λ(α). By Lea.4, ζ(α) fors a grou under utiication, but since if ζ(α)ζ(β) = ζ(γ) then U λ(α) U λ(β) = U λ(γ), thus we rove the resut. Lea.6. H(, ) is equivaent to a hadaard whose couns are the diagonas of ζ(α), and thus for a grou. We ca such Hadaards grou-ie. Proof. Two hadaard atrices are equivaent if H a = M H b M where M and M are onoia unitaries. Note that Z(, ) Z (, ) c (, ) = ζ(α(, )). Since α(, ) is a atin square, α(, ) is just a eruation of the ordering of ζ(α). See that if we define ξ() = n c(n, ) n n, then the couns of H (, ) = Z (, ) H(, )ξ() are exacty the diagonas of the set ζ(α). This oeration is identica to noraizing the first row and the first coun of H(, ) to be. Thus, the noraized hadaard ust have couns which for a grou under coonent-wise utiication. Theore.7. There exist sets of d couting traceess onoia orthogona unitaries if and ony if there exist utuay unbiased Hadaards each of which is equivaent to a grou-ie Hadaard. Proof. Ay the resuts eading u to Lea.6 not just for the hadaard H(, ), but for a H(, ) with (, ), which give a tota of such Hadaards. That taes care of the ony if condition. Now for the if. H are the (, ) are the utuay unbiased grou-ie Hadaard atrices. Define i = i. Define i H i. Thus: j = H(, )j, = j H So H(, ) = H. If we noraize the first row and the first coun of this hadaard we ay define a new hadaard: H(, ). Define: U = H(, ) n, n n n U is a secia case where can can choose any H(, ) to buid the diagonas (since these atrices are aways diagona with resect to basis, and hence onoia). For a fixed we have a cass of d couting atrices. Each cass contains I because H(, ) n, = (due to the noraization). Reoving I fro d atrices for each vaue of. Assuing that there are hadaards the above give sets of d atrices. Each set is ceary orthogona due to being constructed with diagonas fro hadaards (which due to unitarity have orthonora couns and rows). Due to sharing an eigenbasis they coute. Mebers fro different casses are orthogona as a consequence of having utuay unbiased eigenbases and being traceess. The ast roerty we need to show is that these atrices are a onoias. See that: Un = Z(, ) Z (, ), Z(, ) n where Z(, ) n is the n th coun of H(, ) as a diagona unitary. Due to the way we derived H fro H, we can see that Z(, ) n = Z (, ) Z(, ) n c (n, ), thus: And hence: U n Z(, ) n = Z(, ) n Z(, ) c(n, ) = Z(, ) Z (, ), Z(, ) n = c (, )c(, ) Z(, ) Z (, ), Z(, ) n = c (, )c(, ) Z(, ), Z(, ) n = c (, )c(, )δ,n+ Which is ceary onoia, thus we have coete the roof. 4

5 Definition.8. We say that sets of d couting traceess onoia orthogona unitaries are coetey onoia if they are onoia in a the eigenbases. Theore.9. We have (d ) atrices eeting definition.8 if and ony if there exist utuay unbiased Hadaards which are a equivaent to the sae grou-ie Hadaard. Proof. This roof wi foow Lea.4. In that roof we saw that the atrices ζ(α) were equivaent to Z(, ) which were the couns. In this roof, we woud instead see that resut for Z(i, j), since the atrices are onoia in a bases. But the diagonaization ζ(α) does not change in different bases, so a the Hadaards ust be equivaent to ζ(α). More over, the unitaries theseves ust be isoorhic to the sae grou. Conjecture.0. A onoia sets are coetey onoia. Conjecture.. A sets of axiay couting onoias are nice. The revious conjecture, if true, eans that onoias cannot beat the reduce to rie owers aroach. It shoud be ossibe to rove or disrove both of the above rather easiy, but I have not tried yet. 5