aa r X i v : . [ qu a n t - ph ] M a y Simple derivation of the Weyl and Dirac quantum cellular automata
Philippe Raynal
1, 2 Centre for Quantum Technologies, National University of Singapore, Singapore University Scholars Programme, National University of Singapore, Singapore (Dated: May 29, 2017)We consider quantum cellular automata on a body-centred cubic lattice and provide a simplederivation of the only two homogenous, local, isotropic, and unitary two-dimensional automata [G.M. D’Ariano and P. Perinotti, Physical Review A 90, 062106 (2014)]. Our derivation relies on thenotion of Gram matrix and emphasises the link between the transition matrices that characterisethe automata and the body-centred cubic lattice: The transition matrices essentially are the matrixrepresentation of the vertices of the lattice’s primitive cell. As expected, the dynamics of these twoautomata reduce to the Weyl equation in the limit of small wave vectors and continuous time. Wealso briefly examine the four-dimensional case where we find two one-parameter families of automatathat reduce to the Dirac equation in a suitable limit.
I. INTRODUCTION
In the forties, Von Neumann proposed a discrete modelof computation where finite-state machines arranged ina grid, locally process information in discrete time-steps[1]. Von Neumann’s motivation to introduce cellular au-tomata was to obtain a complex behaviour from an arrayof simple processing units. Later, Feynman suggestedapplying the laws of quantum mechanics to computa-tion [2, 3]. This idea naturally led to the formalisationof quantum computation [4, 5] in general and QuantumCellular Automata (QCA) in particular [6–8]. In paral-lel to QCA, other models of quantum computation havebeen developed like circuit-based quantum computation[9], measurement-based quantum computation [10], oradiabatic quantum computation [11].Around the same time, conceptual links betweenphysics and computer science were examined. A famousexample is the relation between logical reversibility andphysical reversibility investigated by Landauer [12] andculminating with the development of reversible compu-tation [13, 14]. A more recent instance, motivated bythe believe that spacetime should ultimately be discrete,concerns the emergence of (continuous) dynamical laws ofphysics from (discrete) computation. This question wasfirst discussed by Suze [15] and later by Weehler [16]. Thefirst significant step towards emergent physical laws camefrom Bialynicki-Birula who considered a two-dimensionalQCA on a body-centred cubic lattice [17]. Bialynicki-Birula provided an example of a homogenous, linear, andunitary QCA and showed that the dynamic of any QCAthat fulfils these constraints necessarily reduces to theWeyl equation in the limit of small wave vectors andcontinuous time. He also showed that the simple cubiclattice admits no homogenous, linear, and unitary QCA.In passing, Bialynicki-Birula proposed a four-dimensionalQCA that reduces to the Dirac equation in the limit ofsmall wave vectors and continuous time. A couple ofyears latter, Meyer showed that there is no nontrivialhomogeneous, local and unitary one-dimensional cellu-lar automaton on a unidimensional lattice [18], a result later generalised to higher dimensions in [19] and [20].Recently, D’Ariano et al. extended Bialynicki-Birula’sand Meyer’s no-go results to the rhombohedral lattice.In addition, they showed that there are not one but twohomogeneous, local, linear, isotropic, and unitary QCAfor the body-centred cubic lattice, up to unitary conju-gation [21]. However, the derivation is rather lengthyand technical. In the present paper, we provide a shortand simple derivation of these two unitarily-inequivalentautomata. Our derivation uses the notion of Gram ma-trix and emphasises the relation between the transitionmatrices that describe the unitary evolution of the au-tomata and the body-centred cubic lattice: The transi-tion matrices essentially are the matrix representation ofthe vertices of the lattice’s primitive cell, here a centralvertex surrounded by a tetrahedron and its dual. Al-though we only consider the body-centred cubic latticein this paper, our technique can directly be used for allBravais lattices. Let us add here that the Klein-Gordonequation [22], Maxwell’s equations [23], Lorentz invari-ance [25, 26, 36] and curved spacetime [27, 28] have alsobeen investigated in the framework of QCA.This paper is organised as follows. In Section II,we recall the general framework to discuss QCA onan abstract lattice as established in [21]. In SectionIII, we focus our attention on two-dimensional homoge-nous, local, isotropic, and unitary automata on a body-centred cubic lattice and derive the only two unitarity-inequivalent solutions. This short and simple derivationis the main result of our paper. We briefly examine thefour-dimensional case in Section IV and find two one-parameter families of automata without completely solv-ing the case. These automata reduce to the Dirac equa-tion in a suitable limit. We conclude in Section V.
II. QUANTUM CELLULAR AUTOMATA
We follow the general framework developed in [21] todescribe a homogenous, local, isotropic, and unitary au-tomaton on an abstract lattice. A QCA is defined as anumerable set G of identical cells that evolve in identi-cal discrete time-steps. Each cell is a quantum systemand the discrete evolution is unitary. We further focusour attention on the case of fermionic QCA, where thequantum system of each cell is described by a finite num-ber of fermionic field operators satisfying the usual anti-commutation relations [29] (See [8] for a formal defini-tion of a QCA in the qudit case.). We will later restrictthe abstract set G to the abelian group of translationsin a three-dimensional body-centred cubic lattice. Theassumption that the cells and the time-steps are all iden-tical represents our homogeneity assumption.Next, we define the finite set V g of neighbouring cellsof a given cell g and we assume the deterministic evolu-tion of the automaton to be local, that is, each cell g onlyinteracts with a finite number | V g | of neighbouring cellswithin a single step of computation. We now have a QCAon a graph Γ( V, E ) with vertex set V = { g ∈ G} and edgeset E = { ( g, g ′ ) | g ∈ G , g ′ ∈ V g } where a vector ψ g is as-sociated to each vertex and a transition function A g,g ′ isassociated to each directed edge ( g, g ′ ). The componentsof the vector ψ g are the field operators associated to thequantum system at site g . Furthermore, the homogeneityassumption means that all vertices are identical so that(1) the transition matrices A g,g ′ are independent of thevertex g and (2) each cell has the same number of neigh-bouring cells, that is, | V g | is independent of g . We are ledto introduce a multiplication law for the elements of G and define a subset S such that g ′ = gh for every neigh-bour g ′ of g . It follows that E = { ( g, gh ) | g ∈ G , h ∈ S } ,explicitly treating all neighbouring sets on equal footing.We also assume G to be closed under multiplication sothat G forms a group finitely generated by S . The subset S is often called the generating set of the group G in theframework of Cayley graphs [30]. By construction theinverse element h − of h must necessarily be in S as theconsideration of a primitive cell centred on g ′ immedi-ately reveals. In the following it will be useful to write S = S + ∪ S − , where S − denotes the set containing theinverse elements of S + . Note that this decomposition isnot unique. In this paper we only consider the case ofa body-centred cubic system, where the set S + containsthe vertices of a regular tetrahedron and S − containsthe vertices of the corresponding dual tetrahedron (seeFig. 1).Third, we assume the cellular automaton to be lin-ear. Physically, it means that no interaction betweenfermionic fields is allowed and we can focus on the dy-namics of a single particle. In fact, the term QuantumCellular Automata is often reserved for the dynamics ofinteracting fields while Quantum (Random) Walks (QW)is used for the dynamics of a single particle [31–33]. Asour general framework started with an abstract QCA, wewill keep the term QCA throughout this article, althoughwe ultimately examine a single particle. As often withQW [34], we can define the single-particle Hilbert space H = H G ⊗ H F where the space H G is used to describethe position of the particle while the space H F is used todescribe its internal degree of freedom. Since we are in- (a) (b) (c) FIG. 1: The primitive cell of the body-centred cubic latticecan be seen as a central vertex surrounded by a tetrahedronand its dual. (a) Graphical representation of the primitive cellof the body-centred cubic lattice with a central vertex and itseight neighbouring vertices (thick and dashed vectors). (b)Graphical representation of the tetrahedron generated by thefour vertices in S + (thick vectors). (c) Graphical representa-tion of the dual tetrahedron generated by the four vertices in S − (dashed vectors). terested in two-dimensional automata, we simply choose H F = C . Linearity means that the vector ψ g is the lin-ear combination of the linearly transformed vectors of itsneighbouring vertices and itself, at the preceding time-step. Therefore, the discrete-time evolution of the vector ψ g ( t ) at vertex g from step t to step t + 1 with transitionmatrices A g,g ′ can be written as ψ g ( t + 1) = X g ′ ∈ V g A g,g ′ ψ g ′ ( t ) + A g,g ψ g ( t ) , (1)where A g,g represents the transition matrix at the centreof the cell. Using the definition of the generating set S ,this evolution becomes ψ g ( t + 1) = X h ∈ S A g,gh ψ gh ( t ) + A g,g ψ g ( t ) . (2)Finally, the above local evolution can be rewritten us-ing the Hilbert space structure H = H G ⊗ C . Now, wesimply denote the transition matrices A g,gh as A h , theneutral element of G as e , and we introduce the shift op-erator T h in the right regular representation that shiftsa vertex | x i to the vertex T y | x i = | xy − i . It follows thatthe neighbouring vertex g ′ = gh is shifted by T h to thecentral vertex g . We finally obtain the evolution operator A such that ψ ( t + 1) = Aψ ( t ), ψ ( t ) in H , as A = X h ∈ S T h ⊗ A h + ⊗ A e . (3)Such an operator is often called the walk operator whilethe operator A e is often taken to be zero. In this paper,we do not assume A e to vanish.Next, we demand the automaton to be isotropic, thatis, all the directions on the lattice are equivalent. Thecorresponding mathematical definition is not trivial. Werequire the existence of a finite subgroup L of the auto-morphism group Aut( G ) of G , transitive over S + , and aunitary representation U of L such that, for all l in L , A = X h ∈ ˜ S T h ⊗ A h = X h ∈ ˜ S T l ( h ) ⊗ U l A h U † l , (4)where we include both S and the neutral element e inthe definition of the set ˜ S and where l ( h ) denotes thenew vertex after action of l on h . Note that the abovedefinition does not require the isotropy group L to in-clude all the symmetries of the lattice. To ensure thatall directions really are equivalent, an alternative defini-tion of isotropy would be to require the isotropy group tobe the full symmetry group of the lattice. Importantly,the above covariance of the automaton A implies the in-variance under conjugation of the two set of transitionmatrices A h and A h − , that is, A l ( h ± ) = U l A h ± U † l , (5)for all h in S + and h − in S − . Indeed, the choice of aregular representation implies that the T h ’s are linearlyindependent in the sense that if the action of a finite lin-ear combination of the T h ’s on any site | g i of H G vanishes,then each coefficient in the linear combination must van-ish. Consequently, the equality in Eq. (4) must hold termby term in the finite sum over ˜ S . Due to the transitiv-ity of L over S + , the two subsets S ± remain unchangedunder the action of the isotropic group L and we end upwith Eq.(5). It also follows that the transition matrix A at the centre of the cell must be invariant under conjuga-tion. If the unitary representation U of L is irreducible,then Schur’s Lemma ensures that A is proportional tothe identity matrix or zero. Let us emphasise here thatisotropy as defined in Eq. (4) is a very strong constraintindeed. Although the isotropy group L will not be apriori specified, covariance will be instrumental in char-acterising the QCA. Once the solutions are known, it willbe possible to find their symmetries and therefore to iden-tify the isotropy group L and its unitary representation U .So far, the construction holds for any countable group G finitely generated by a subset S . We now fix G to bethe abelian group of translations in the three-dimensionalbody-centred cubic lattice. This choice of group repre-sents our discrete three-dimensional lattice. Mathemat-ically, the choice of an abelian group is convenient asit allows to work in the Fourier space, also called thereciprocal lattice in crystallography. The irreducible rep-resentations of an abelian group are one-dimensional andunitary. Thus, we can now label the group elements of G as integer vectors in Z , denoted in bold letters, and usethe sum notation for the commutative group law so that T y | x i = | x − y i . (6)The periodicity of the lattice allows us to define the (con-tinuous) Fourier-transformed basis {| k i} of the (discrete)basis {| x i} as | k i = 1 p |B| X x ∈ Z e i k . x | x i (7)where |B| denotes the volume of the first Brillouin zone B of the body-centred cubic lattice. This first Brillouin zone takes the shape of a rhombic dodecahedron thatcomprises all wave vectors k such that in cartesian co-ordinates − π ≤ ± k x ± k y ≤ π , − π ≤ ± k x ± k z ≤ π ,and − π ≤ ± k y ± k z ≤ π [35]. Importantly, the shiftoperators T h are diagonal in the Fourier basis, that is, T h | k i = e i h . k | k i so that the unitary automaton takesthe form A = X h ∈ ˜ S Z B d k e i k . h | k ih k | ⊗ A h . (8)The final form of a homogenous, local, isotropic, andunitary automaton on a body-centred cubic lattice thenis A = Z B d k | k ih k | ⊗ A ( k ) (9)where the operator A ( k ) = X h ∈ ˜ S e i k . h A h (10)is unitary for all wave vectors in the first Brillouin zoneand the set of transition matrices A h , h in S + , the setof transition matrices A h − , and A are invariant underconjugation for some group of unitaries.Interestingly, the unitary property of the automata A further implies that the transition matrices must satisfythe conditions X h ∈ ˜ S A † h A h = X h ′′ = h − h ′ A † h ′ A h = 0 . (11)Indeed, since the operators T h are unitary, Eq. (3) impliesthat A † A = ⊗ X h A † h A h + X h ′′ = h − h ′ T h ′′ ⊗ A † h ′ A h , (12)where each term in the second sum has to vanish due tothe linear independence of the operators T h . Similarly,the condition AA † = X h ∈ ˜ S A h A † h = X h ′′ = h − h ′ A h ′ A † h = 0 . (13)Homogenous, linear, local, isotropic, and unitary au-tomata have another interesting property. Indeed, forany homogenous, linear, and local automaton, we have( ⊗ A † ( k = )) A = X h ∈ ˜ S T h ⊗ A ′ h , (14)where P h ∈ ˜ S A ′ h =
1. In dimension d , the operator A ( k = ) is a dxd unitary transformation that is speci-fied by d real parameters. Therefore, the above equationdefines a d -parameter family of automata constructedfrom an automaton A such that P h ∈ ˜ S A h = ⊗ V . If, fur-thermore, the automaton is isotropic, then the operator ⊗ A † ( k = 0) is invariant under the unitary conjugationof the isotropy group L . It follows that if the repre-sentation of L is irreducible, then, by Schur’s lemma, A † ( k = 0) must be zero or a multiple of the iden-tity. If the automaton is also unitary, it cannot vanishand the multiplicative constant must be a phase suchthat A ( k = 0) reduces to the identity operator up toa physically-irrelevant phase. Therefore, if the isotropygroup L has an irreducible representation in two dimen-sions, the four-parameter family of QCA reduces to asingle QCA. Here we see again how strong the isotropyconstraint is. In the following, we do assume an automa-ton fulfilling the property A ( k = 0) =
1. In other words,we impose the constraintC : A + X h A h + A − h = . (15)Physically, this assumption is rather intuitive: If all di-rections are equivalent, it seems natural to require thatthe vertex remains unchanged when the wave vector ofthe particle is zero. This represents our first constraint.In the following, we will introduce three additional con-straints coming from the unitary constraint. These fourconstraints will be enough to determine fully the searchedautomata. III. DERIVATION OF THE WEYL AUTOMATA
We now start with the novelty of this article: A shortand simple derivation of the Weyl automata. The maindifference between D’Ariano and Perinotti’s derivation in[21] and that of the present paper is the use of Gram ma-trix. While D’Ariano and Perinotti express the transitionmatrices in various forms to tackle different constraints,we only consider their Pauli decomposition to make useof the notion of Gram matrix. It is this notion that allowsa shorter and perhaps more transparent derivation. An-other distinction concerns the transition matrix at thecenter of the body-centred cubic cell that we never as-sume to vanish.We call a homogenous, local, isotropic, and unitaryautomaton, a Weyl automaton as we already know thatsuch an automaton necessarily reduces to the Weyl equa-tion in the limit of small wave vectors and continuoustime [17]. To find such automata, we will examine theunitarity of the operator A ( k ) defined in Eq. (10). Butfirst, let us introduce two relevant Gram matrices. A. Gram matrix
The operator A ( k ) can be written as A ( k ) = A + X j e ik j A j + e − ik j A − j , (16)with k j = k . h j and where the eight vectors h j and h − j , j = 1 , , ,
4, correspond to the vertices of the regular tetrahedron S + and its dual S − . Note that we now enu-merate the vectors h as h j and transition matrices A h as A j to simplify the notations. The non-normalised vectorsof these two tetrahedra are defined in some orthonormalbasis as h = , h = − − , h = − − , h = − − , and h − j = − h j . (17)Our task is to find the nine 2x2 transition matrices A ± j and A in a simple and insightful manner.We recall here that the Gram matrix of a collection of n real vectors v is defined by the nxn symmetric matrix G such that G jk = v j . v k , with the usual definition ofthe dot product. In our case, the relevant vectors arecomplex, so we will define two Gram matrices denotedby G R and G C as follows G Rjk = v j . v k and G Cjk = v ∗ j . v k . (18)Note that only G C is positive semi-definite as the simpledot product in the definition of G R does not correspondto an inner product. Furthermore, for a set of four vec-tors, we can define a 3x4 matrix B whose columns arethe vectors v i s so that the two Gram matrices definedabove take the simple form G R = B t B and G C = B † B ,where t denotes transposition while † denotes Hermitianconjugation.The regular tetrahedron defined in Eq. (17) can be fullycharacterised in a coordinate-free manner by the relation h j . h k = 4 δ jk − , (19)where δ jk is the Kronecker delta. Interestingly, theserelations can be grouped in the Gram matrix G = T t T = − − − − − − − − − − − − (20)where T = − − − − − − (21)is the matrix containing the four vectors of the non-normalised tetrahedron. Finally, the four vectors of theregular tetrahedron in three-dimensional space verifiesthe simple linear-dependence relation P j =1 h j = . Inother words, the sum of all the terms in any row of theGram matrix G vanishes. B. Unitary constraint
The unitary constraint A ( k ) A † ( k ) = e ± ik i e ∓ ik j A ± i A †± j , e ∓ ik j A A †± j , e ± ik j A ± j A † , and A A † . These terms canbe grouped depending on their relative phase as givenin Eq. (13) and each group must vanish. Among thesegroups, three will suffice to characterise the automa-ton. More specifically, we choose the three groups withrelative phases e ik j , e i ( k i − k j ) , and e − ik j , respectively.Therefore, we end up with the three constraintsC : A j A †− j = 0 , (22)C : A i A † j + A − j A †− i = 0 , i = j, (23)C : A A † j + A − j A † = 0 , (24)and similarly for A † ( k ) A ( k ) = : A † j A − j = 0 , (25)C : A † i A j + A †− j A − i = 0 , i = j, (26)C : A † A j + A †− j A = 0 . (27)Together with the constraint C of Eq. (15) repeatedhere for convenienceC : A + X j A j + A − j = , (28)we now have a total of four constraints that we can startexploiting to find the Weyl automata. C. From two tetrahedra to one tetrahedron
We express the 2x2 complex matrices A and A ± j interms of the three Pauli matrices σ , σ , and σ definedas σ = (cid:20) (cid:21) , σ = (cid:20) − ii (cid:21) , σ (cid:20) (cid:21) , (29)together with the identity matrix
1. We will see that inthe case of an isotropic automaton the transition matrices A ± j can be written as A j = α a j .σ ) and A − j = β − a ∗ j .σ ) , (30)where a j . a j = 1 , (31)and α and β are two complex numbers different fromzero. In other words, the transition matrices A ± j havenon vanishing traces and are such that a − j = − a ∗ j . Theimmediate consequence is that we only need to focus onthe four complex vectors a j as well as the two complex factors α and β to solve our problem. Let us now provethese claims.The two 2x2 complex matrix A j and A †− j can alwaysbe written as A j = a j a j .σ and A †− j = a ∗− j a ∗− j ..σ (32)with a ± j ∈ C , a ± j ∈ C . The two constraints C andC of Eqs. (22) and (25) yield the three conditions a j a ∗− j + a j . a ∗− j = 0 , (33) a ∗− j a j + a j a ∗− j ± i a j x a ∗− j = 0 . (34)There are three possible cases for the above system:First, none of the two matrices A j and A − j has a traceequal to zero. Second, only one matrix has a trace equalto zero. Third, both matrices have a trace equal to zero.In the first case, upon rescaling the complex vectors a j and a − j and using a conventional factor 1 /
2, the two ma-trices A j and A − j take the simple form given in Eqs. (30)and (31). Of course, the coefficient α ( β ) must be thesame for all transition matrices A j ( A − j ) since they areunitarily conjugated under the action of the isotropygroup and, therefore, their traces must be equal. Giventhe four complex vectors a j , we define the two Grammatrices G R and G C as G Rjk = a j . a k and G Cjk = a ∗ j . a k ,respectively. The full characterisation of these two Grammatrices will specify the four vectors a j , up to rotations.Such a freedom on the vector a j corresponds to the defi-nition of an automaton up to unitary conjugation. Inter-estingly, Eq. (31) already fixes the diagonal of the Grammatrix G R : All the diagonal entries of the Gram ma-trix G R must be equal to 1. Similarly, since A and A are unitarily conjugated, A † A and A † A must be uni-tarily conjugated too so that their traces must be equal,i.e. a ∗ . a = a ∗ . a . Therefore, our isotropic automaton issuch that for all j , a ∗ j . a j = c where c is a positive number:All the diagonal entries of the Gram matrix G C must beequal to c .In the second case, it immediately follows that the ma-trix with vanishing trace must completely vanish. Letus say that this vanishing matrix is A − j . Since all thetransition matrices A − j are unitarily conjugated, all ofthem must vanish. Thus, the constrains C and C ofEqs. (23) and (26) reduce to A i A † j = A † i A j = 0 and wemust have a j = − a ∗ i together with a i . a i = 1. Note thatthere is no other case as, by assumption, the matrices A i have a non-zero trace. Since the constrains C andC must be fulfilled for any pair i = j , all four vectors a j must necessarily be equal, imaginary, and such that a i . a ∗ j = − i, j (even for i = j ). This contra-dicts the fact that the Gram matrix G Cjk = a ∗ j . a k mustbe positive semi-definite.In the third case, all transition matrices have zerotrace. It follows from Eqs. (33) and (34) that a j . a ∗− j = 0and a j x a ∗− j = 0. Thus, there exists non-zero com-plex numbers λ j such that a ∗− j = λ j a j and a j . a j = 0.The constrain C now reads (1 + λ i λ ∗ j ) a i . a ∗ j = 0 and(1 − λ i λ ∗ j ) a i x a ∗ j = 0 such that necessarily a i . a ∗ j = 0,for all i = j . Thus, the corresponding Gram matrix G C contains only vanishing off-diagonal elements. Impor-tantly, G C is the Gram matrix of four vectors in threedimensions, therefore its determinant must vanish. Sinceall its diagonal elements must be identical because of theisotropy group, it follows that G C must entirely vanish sothat all transition matrices A ± j are zero. The constraintC of Eq.(28) directly implies that A and the automaton A ( k ) itself must be the identity matrix. This completesthe very special case where the traces of all transitionmatrices vanish.In conclusion, all the transition matrices A ± j have ei-ther zero or non-zero traces while a more asymmetricalcase is forbidden. We now focus on the non-trivial casewhere the transition matrices have non-vanishing tracesand can be written in the form given in Eqs. (30) and(31) with non-zero α and β . D. Characterisation of the Gram matrix G C The two constraints C and C will fix all the off-diagonal entries of the Gram matrix G C and most ofthe off-diagonal terms of the Gram matrix G R . Let usnow proceed with solving the constraint C . The identity A i A † j + A − j A †− i = 0, i = j reads( | α | + | β | )(1 + a i . a ∗ j ) = 0 , (35)( | α | − | β | )( a i + a ∗ j + i a i x a ∗ j ) = 0 . (36)Similarly, the identity A † i A j + A †− j A − i = 0, i = j reads( | α | + | β | )(1 + a i . a ∗ j ) = 0 , (37)( | α | − | β | )( a i + a ∗ j − i a i x a ∗ j ) = 0 . (38)Since α and β are non zero, it immediately follows that a i . a ∗ j = − , (39)such that the Gram matrix G C is of the form G C = c − − − − c − − − − c − − − − c . (40)Therefore, G C is real and symmetric and it can be diag-onalised using the rotation matrix R = 12 − − − − − − (41)which is nothing but the matrix T of Eq. (21) with anadditional first row of ones and a normalising factor 1 / G C then are { c, c, c, − c } . Since G C is the Gram matrix of four vectors in threedimensions, its determinant must vanish. In other words, c can only be -1 or 3. The number c = a i . a ∗ i beingpositive, we end up with G C = − − − − − − − − − − − − , (42)which is nothing but the Gram matrix G of the tetra-hedron given in Eq. (20). In other words, we have G C = T t T . This identity suggests a link between thetransition matrices A j and the regular tetrahedron. Be-fore turning our attention to the off-diagonal elements of G R , let us note that the two equations Eqs. (36) and (38)imply | α | = | β | . (43)If we assume otherwise, the four vectors a j would all beequal, so that a i . a ∗ j = a i . a ∗ i ≥
0. This would contradictthe additional requirement that a i . a ∗ j = − A must vanish. Using the decomposition of Eq. (30) for thetransition matrices A ± j together with A = a a .σ ,we can rewrite the constraint C in Eq. (28) as( a + 2( α + β )) a + α X j a j − β X j a ∗ j .σ = . (44)It immediately follows that a + 2( α + β ) = 1 , (45) a k . a + α X j a k . a j = 0 (46)since P j a k . a ∗ j = 0, that is, the sum of all the terms inany row of the Gram matrix G C vanishes. We now needthe constraint C that corresponds to the three identities α ∗ a + βa ∗ + ( α ∗ a − β a ∗ ) . a ∗ j = 0 , (47) α ∗ a + β a ∗ + ( α ∗ a − βa ∗ ) a ∗ j = 0 , (48)( α ∗ a + β a ∗ ) x a j = 0 . (49)In particular, Eq. (48) implies α ∗ a . a ∗ j + β a ∗ . a ∗ j + ( α ∗ a − βa ∗ ) = 0 (50)since a j . a j = 1. It follows from this equation togetherwith Eq. (47) that a = a . a j = a ∗ . a j = 0 . (51)Now, we can insert these results in Eqs. (45) and (46) toobtain α + β = 12 (52) X j a k . a j = 0 . (53)The later identity means that the sum of all the terms inany row of the Gram matrix G R vanishes. We will see inthe next subsection that the four vectors a j span a three-dimensional space and therefore a . a j = 0 in Eq. (51)implies a = 0 so that A must vanish. E. Characterisation of the Gram matrix G R We now are ready to characterise completely G R and A and to find the Weyl automata. To do so, we donot need to solve any further the constraints C or C .Instead, we simply exploit Eq. (53). These are four linearconstraints for six unknowns since the diagonal terms ofthe Gram matrix G R are already fixed to 1 as given inEq. (31). Therefore we finds that G R only depends ontwo unknowns, say x = a . a and y = a . a , and it takesthe form G R = x − − x − y yx y − − x − y − − x − y y xy − − x − y x . (54)Next, we observe that the Gram matrix G R can be diag-onalised by the very same rotation R as G C . Its eigen-values are { , x ) , − x + y ) , y ) } such that itcan be written as G R = T t (1 + x ) / − ( x + y ) / y ) / T. (55)This identity leaves little freedom to the unkonwns x and y . Indeed, by definition of the two Gram matri-ces G R and G C , there exists a 3x4 matrix B such that G R = B t B and G C = B † B . Since we already know that G C = T t T with T t = T † , there must exist a 3x3 unitarymatrix Z such that B = ZT and we can write the Grammatrix G R as G R = T t Z t ZT or simply G R = T t W T where W is a 3x3 symmetric unitary matrix. If we de-note the diagonal matrix in Eq.(55) as D , we now have T t DT = T t W T . We can multiply each side of this iden-tity by
T / T t / D = W . Since D is diagonal while W is unitary, thesetwo matrices can only be equal if they are diagonal withonly phases on the diagonal. In other words, we simplyhave | (1 + x ) / | = | ( x + y ) / | = | (1 + y ) / | = 1.It follows that the pair ( x, y ) must necessarily be equalto (1 , − , − , G R are G R = −
31 1 − − − , (56) G R = − − − − , (57) G R = − − −
31 1 − . (58)For the sake of completeness, we write the Gram matrices G C and G R in terms of the matrix T containing the fourvectors h j of the regular tetrahedron. They are G C = T T T,G R = T T − T,G R = T T − T, (59) G R = T T − T. Therefore, there are six possible matrices B whosecolumns are the searched vectors a j . They are B ± = ± i T, (60) B ± = ± i
00 0 1 T, (61) B ± = ± i T . (62)We notice that B j + = B ∗ j − , j = 1 , ,
3. Importantly, wenow know that the four vectors a j span a three dimen-sional space. Therefore, we can come back to Eq. (51)and conclude that a . a j = 0 is only possible when a = 0so that A must vanish.Importantly, the six automata corresponding to thesesix sets of vectors a j are not all unitarily-inequivalent.To see that, let us find their spectra. Since the modulusof the determinant of a unitary matrix must be equal to1 and we already know that | α | = | β | , a direct calculationof the determinant of A ( k ) shows that | α | = | β | = 18 . (63)As we also know from Eq. (52) that α + β = 1 /
2, wenecessarily have α = 1 ± i β ∗ , (64)leading to twelve possible automata. The spectra ofthese automata immediately follows from their traces.It turns out that there are only two different spectra { e iw ± , e − iw ± } , wherecos w ± = (cos k x cos k y cos k z ± sin k x sin k y sin k z ) , (65) k = { k x , k y , k z } . The ± sign in the above formula onlydepends on the choice α = (1 ± i ) /
4. Two unitarily-inequivalent automata can conveniently be chosen to be A ( k ) and A † ( − k ), where the first automaton A ( k ) isdefined with α = (1 + i ) / a j of any B i ± in Eqs. (60)-(62).This completes the derivation of the homogenous, lo-cal, isotropic, and unitary two-dimensional automaton.We have found two unitarily-inequivalent automata A ( k ) = X j e ik j A j + e − ik j A − j (66)and A † ( − k ), constructed from the four vectors h j of theregular tetrahedron with a multiplicative phase i in frontof one of their three components. Since the transitionmatrices are of the form A j ∝ ( a j .σ ) where the vec-tors a j essentially are the vectors h j , we say that thetransition matrices A ± j essentially are the matrix rep-resentation of the vertices of the lattice’s primitive cell.Finally, let us remember that we assumed the existence ofan irreducible representation of the isotropy group in twodimensions. This isotropy group L turns out to be thegroup of rotations of angle π around the three coordinateaxes, a rather simple group whose unitary irreducible rep-resentation U in two dimensions is { , σ , σ , σ } . F. Continuum limit
As proven in [17], the found automata should reduceto the Weyl equation in the limit of small wave vectors k and continuous time. We can quickly verify it here.Indeed, we have for | k | << A ( k ) = α X j (1 + ik j ) 12 ( a j .σ ) (67)+ α ∗ X j (1 − ik j ) 12 ( − a ∗ j .σ ) . We can choose the automata A ( k ) to be given by α = (1 + i ) / a j of any B j − inEqs. (60)-(62). Then, a direct calculation shows that thetwo unitary-inequivalent automata A ( k ) and A † ( − k ) re-duce to A ( k ) = i k .σ. (68) If, instead, one chooses α = (1 + i ) / a j of any B j + , we would obtain the aboveequation for the wave vector − k and rotated by an ele-ment of the isotropy group. Finally, to obtain the contin-uum limit of our automata, we define the Hermitian oper-ator H ( k ) of the unitary transformation A ( k ) = e − iH ( k ) for the discrete time evolution | ψ ( t + 1) i = A ( k ) | ψ ( t ) i sothat for a small-amplitude Hamiltonian A = − iH ( k ), H ( k ) = − k .σ . We then identify H ( k ) with the Hamil-tonian H of the unitary transformation U ( t ) = e − i H t forthe continuous time evolution i∂ t ψ ( t ) = H ψ ( t ). It fol-lows that H = − k .σ . Therefore, in the limit of smallwave vector and continuous time, the two homogenous,local, isotropic, and unitary two-dimensional automata A ( k ) and A † ( − k ) reduce to the Weyl equation in mo-mentum space i∂ t ψ = − k .σψ. (69)Let us add for the sake of completeness that there aretwo Weyl equations discussed in the literature and of-ten called right- and left-handed Weyl equations. Theywould correspond to the two Weyl automata A ( k ) and A † ( k ), not A † ( − k ). In fact, in the four-dimensional casebelow we will find a QCA that corresponds to the cou-pling of two Weyl automata, one right-handed and oneleft-handed, as one would expect from the Dirac equa-tion. Note also that rigorous continuum limits for quan-tum walks were discussed in [36]. IV. DERIVATION OF THE DIRAC AUTOMATA
With the two-dimensional case completed, we brieflyturn our attention to the four-dimensional case. Four-dimensional homogenous, local, isotropic, and unitaryautomata on a three-dimensional lattice have been in-vestigated in [17] and [21]. Starting from a special formof four-dimensional automata, two families were found,both of them reducing to the Dirac equation in a suitablelimit. In this paper, we use a different starting point butend up with the same two families of, so called, Diracautomata. More specifically, D’Ariano and Perinotti in[21] started from two Weyl automata that they coupled.The coupling was then determined by the unitary con-straint. In the present paper, we will try to solve theunitary constraint by restricting the possible solutions toa simple form inspired by the two-dimensional solutionsbut without assuming two Weyl automata. After somealgebra, we will find two coupled Weyl automata.Given the large dimension of the space where the so-lutions are to be found (i.e. sixteen real dimensions), weonly consider a special class of automata. In the two-dimensional case, we discovered that the transition ma-trices A ± j can essentially be thought as the matrix rep-resentation of the vertices of the lattice’s primitive cell.Therefore, we keep the same structure but now use a four-dimensional representation of the Pauli algebra to findsolution automata in the four-dimensional case. Such arepresentation can be achieved with the help of the fourfour-dimensional Gamma matrices: We simply replacethe two-dimensional Pauli matrices σ i by the product γ i γ , where the four Gamma matrices in the Weyl repre-sentation are defined as γ = (cid:20) (cid:21) , γ i = (cid:20) σ i − σ i (cid:21) . (70)Importantly, we also have γ γ such that our two-dimensional automata A ( k ) given in Eq. (16) can be di-rectly written in four dimensions. In the following, wedenote all relevant four-dimensional matrices with a B instead of an A . In other words, we now consider thematrices B ( k ), B ± j , and B instead of A ( k ), A ± j , and A . We write B ( k ) = B + α X j e ik j
12 ( a j .γγ ) (71)+ β X j e − ik j
12 ( − a ∗ j .γγ ) . We choose to keep the four vectors a ∗− j = − a j and | α | = | β | from the two-dimensional results. However, α , β , and B are no longer known. These choices define aspecial class of automata, so that there may exist moregeneral solutions than those we are about to find.It is convenient to multiply the transition matrices B and B j on the left by the full-rank matrix γ . This willsimplify our calculations. We now have B ′ ( k ) = B ′ + X j e ik j B ′ j + X j e − ik j B ′− j . with B ′ j = α/ γ − a j .γ ) and B ′− j = β/ γ + a ∗ j .γ ).Interestingly, we see that α ∗ B ′− j = βB ′ † j since a − j = − a ∗ j while γ † i = − γ i and γ † = γ . Second, the transitionmatrix B ′ must satisfy the constraint C , so that wehave β ∗ B ′ B ′ j + αB ′ j B ′ † = 0 , (72) αB ′ † B ′ j + β ∗ B ′ j B ′ = 0 . (73)An obvious solution to this system is β ∗ = α togetherwith B ′ = ir r is a real number. To find α and β ,we use the unitary constraint B ′ ( k ) B ′ † ( k ) = k . When the wave vector k vanishes, we find r + 16Re( α ) = 1. When we choose k = ( π/ , π/ , π/ t , we obtain r + 16Im( α ) = 1. Itimmediately follows that α = s (1 ± i ) /
4, where s is a realnumber such that r + s = 1. Therefore, we obtain theautomata B ′ ( k , s, ± ) = ± i p − s s X j e ik j B ′ j + X j e − ik j B ′− j , (74) where B ′ j = α/ γ − a j .γ ), B ′− j = α ∗ / γ + a ∗ j .γ ), α = (1 ± i ) / s is a real parameter. Note that wenow use α to denote (1 ± i ) /
4, instead of s (1 ± i ) / s appears explicitly in Eq. (74). Upon re-multiplying B ′ ( k , s, ± ) by γ on the left, these automata finally takethe form B ( k , s, ± ) = (cid:20) sA ( k ) ± i √ − s ± i √ − s sA † ( k ) (cid:21) . (75)In fact, one can go from an automata with parameter+ √ − s to another with parameter −√ − s by con-jugation with the unitary transformation γ = iγ γ γ γ that anti-commutes with all four Gamma matrices. Fur-thermore, the automaton B ( k , s, ± ) is defined up to aphysically-irrelevant global phase, so that we can alwayschoose s to be positive. Therefore, we can restrict our-selves to automata of the form B ( k , s ) = (cid:20) s A ( k ) i √ − s i √ − s s A † ( k ) (cid:21) , (76)where s is positive. We notice that the above automa-ton B ( k , s ) corresponds to the coupling of two Weylautomata, one right-handed and one left-handed, asdiscussed at the end of Section III. As in the two-dimensional case, these automata can have two differentspectra depending on the choice α = (1 ± i ) / A ( k ), leading us to two unitarily-inequivalentone-parameter families of automata. Indeed, the spec-tra of the automata in Eq. (76) are { e iω ± , e − iω ± } , eacheigenvalues with multiplicity two, wherecos ω ± = s (cos k x cos k y cos k z ± sin k x sin k y sin k z ) . (77)Here again, the ± sign corresponds to the choice α = (1 ± i ) /
4. Finally, a convenient choice for thetwo one-parameter families of automata is B ( k , s ) and B † ( − k , s ) where the automaton A ( k ) is defined with α = (1 + i ) / a j of any B j − . In conclusion, we have found two unitary-inequivalent one-parameter families of four-dimensionalhomogenous, local, isotropic, and unitary automata. Forsmall wave vectors k and large parameter s ≈
1, or equiv-alently a small positive parameter r = √ − s << B ( k , r ) = i k .γγ + irγ . (78)In the limit of continuous time, we obtain the Dirac equa-tion in momentum space iγ ∂ t Ψ = k .γ Ψ + r Ψ , (79)where it is tempting to interpret retrospectively the smallpositive parameter r as the mass of a free particle withstate Ψ in C . V. CONCLUSION
In this paper, we have provided a simple derivation ofthe two-dimensional homogenous, local, isotropic, and0unitary automata on a body-centred cubic lattice, usingthe notion of Gram matrix. Our derivation emphasisesthe link between the automata and the underlyinglattice. Indeed, the transition matrices that characterisethe QCA essentially are the matrix representation ofthe vertices of the lattice’s primitive cell. We have alsoproven that the transition matrix at the centre of thebody-centred cubic cell must be zero. We have foundtwo unitarily-inequivalent automata that reduce to theWeyl equation in the limit of small wave vectors andcontinuous time. We have also briefly examined thefour-dimensional case to find two unitarily-inequivalentone-parameter families of automata that both reduce tothe Dirac equation in the limit of continuous time, small wave vectors, and small positive parameter r . Moreefforts will be required to solve completely the three andfour-dimensional cases for a single particle. The case ofinteracting fields remains untouched. Acknowledgments
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