Quaternionic Computing

U (g)

O(g )Og

S 2, j

1

S 3, k

1

S 3, k

1

S 2, j

1

Og j Ug j+1 Og I k+1 Og

k

Ug

Figure 4: Obtaining an expression for the (N+ 1)-ary circuit O (g) .O (1) O ( 2) O ( 3) O ( 4)

O1 U2 U3 U4 U1 U3 O1 O3 O2 O3 O4

Figure 5: Simulation of a quantum circuit by a real circuit. What is remarkable about this scheme, is that despite its simplicity, it gives precisely what we wanted, this is, that the nal operator OC be in some sense as similar as possible to the operator UC of the original circuit. In fact, we have the following third nice property of our simulation. Lemma 3. The inverse image of OC is precisely UC, i.e. OC= h(UC ). Proof. Because of the serialisation of Step 1, we have that UC= U (s) . . . U (2) U (1) . We use this and the group isomorphism properties of h from Lemma 1 to obtain the following expression for its image h(UC )= h(U (s) . . . U (1) )= h(U (s) ) . . . h(U (1) )s 1

=i= 0, g=s i

h(U (g) )

Quaternionic Computing

WecannowusetheexpressionofEquation17tosubstituteforU(g),

=h(Sg(Ug In 2)Sg) =h(Sg)h(Ug In 2)h(Sg)

SinceSgiscomposedonlyof0’sand1’s,wehavethatRe(Sg)=SgandIm(Sg)=0.Fur-

′thermore,wehavethatSg=I1 Sgfromtheirde nitioninEquations(17)and(22),and

thus,

=(I1 Sg)h(Ug In 2)(I1 Sg) ′′=Sgh(Ug In 2)Sg

However,thetensorproductisjustaformaloperation,anditsassociativitypropertyholdsevenwithatensorofoperatorslikeT.Hence,wehave

′′=Sg[T (Ug In 2)]Sg ′′=Sg[(T Ug) In 2]Sg ′′=Sg[h(Ug) In 2]Sg ′′=Sg(Og In 2)Sg

whichwiththepaddingexpressionofO(g)inEquation22 nallygives

=s 1 O(g)=OC.(23)

i=0,g=s i

3.3.4CircuitInitialisationandMeasurement

HavingdescribedhowtoconstructtherealcircuitC′fromtheoriginalcircuitC,westillhavetoaddresstheissueofhowtoinitialiseC′inStep4,andfurthermoreofhowtointerpretanduseitsmeasurementstosimulatetheinitialquantumalgorithminStep5.

Let|Ψ representtheinitialstategiventoC,andlet|Φ beitsimageunderUC,i.e.the nalstateofthecircuitbeforemeasurement.Ifwethinkbackofthetwohomomorphismsh0and

NNh1fromHCtoHC,inducedbyh,wehavetwologicalchoicesforinitialisingthecorrespondingrealcircuitOC,thestates|Ψ0 and|Ψ1 .Whichshouldwechoose,andineithercasewhatwilltheoutputlooklike?Theanswertothelatterquestionisgivenbythefollowinglemma.Lemma4.Theimagesof|Ψ0 and|Ψ1 intherealcircuitC′are

OC|Ψ0 =T0 |Φ =|Φ0 OC|Ψ1 =T1 |Φ =|Φ1

(24)(25)