A remarkable representation of the Clifford group Steve Brierley - PowerPoint PPT Presentation

A remarkable representation of the Clifford group Steve Brierley University of Bristol March 2012 Work with Marcus Appleby, Ingemar Bengtsson, Markus Grassl, David Gross and Jan-Ake Larsson

Outline ◮ Useful groups in physics ◮ The Zak basis for finite systems ◮ An application to SIC-POVMs

Heisenberg Groups Heisenberg groups can be defined in terms of upper triangular matrices   1 x φ 0 1 p   0 0 1 where x , p , φ are elements of a ring, R . ◮ R = R - a three dimensional Lie group whose Lie algebra includes the position and momentum commutator ◮ R = Z N - a finite dimensional Heisenberg group H ( N ) ◮ R = F p k - an alternative finite dimensional group

Finite and CV systems Finite systems CV systems Heisenberg group H ( N ) Heisenberg group H ( R ) elements are translations in elements are translations in discrete phase space ( N = p ) phase space H ( R ) | 0 � → Coherent states normalizer H ( N ) = C ( N ) normalizer H ( R ) = HSp The Clifford group The Affine Symplectic group C ( N ) ∼ HSp ∼ = H ( N ) ⋊ SL (2 , N ) = H ( R ) ⋊ SL (2 , R ) HSp | 0 � → Gaussian states Our new basis Zak basis

The Clifford group Write elements of H(N) as D ij = τ ij X i Z j i π N , X | j � = | j + 1 � and Z | j � = ω j | j � . where τ = − e Then the composition law is D ij D kl = τ kj − il D i + k , j + l The Clifford group is the normalizer of H ( N ) i.e. all unitary operators U such that UD ij U † = τ k ′ D i ′ , j ′

Definition of the basis The Heisenberg group H ( N ) is defined by ω = e 2 π i / N and generators X , Z with relations X N = Z N = 1 ZX = ω XZ , ◮ There is a unique unitary representation [Weyl] ◮ The standard representation is to choose Z to be diagonal. But suppose N = n 2 , then Z n X n = X n Z n ◮ There is a (maximal) abelian subgroup � Z n , X n � of order n 2 = N ◮ So choose a basis in which this special subgroup is diagonal

The entire Clifford group is monomial in the new basis Armchair argument: ◮ The Clifford group permutes the maximal abelian subgroups of H ( N ) ◮ It preserves the order of any element ◮ In dimension N = n 2 there is a unique maximal abelian subgroup where all of the group elements have order α n ◮ Hence the Clifford group maps � Z n , X n � to itself ◮ The basis elements are permuted and multiplied by phases Theorem: There exists a monomial representation of the Clifford group with H ( N ) as a subgroup if and only if the dimension is a square, N = n 2 .

The new basis In the Hilbert space H N = H n ⊗ H n , the new basis is � | r , s + 1 � if s + 1 � = 0 mod n X | r , s � = σ r | r , 0 � otherwise Z | r , s � = ω s | r − 1 , s � 2 π i 2 π i N and σ = e n . Indeed, where the phases are ω = e X n | r , s � = σ r | r , s � Z n | r , s � = σ s | r , s � We have gone “half-way” to the Fourier basis n − 1 1 � ω − ntr | nt + s � . | r , s � = √ n t =0 i.e. apply F n ⊗ I , where F n is the n × n Fourier matrix.

An Application to SIC POVMs A SIC is a set of N 2 vectors {| ψ i � ∈ C N } such that 1 |� ψ i | ψ j �| 2 = for i � = j N + 1 ◮ A 2-design with the minimal number of elements. ◮ A special kind of (doable) measurement. ◮ Potentially a “standard quantum measurement” (cf. quantum Bayesianism)

Zauner’s Conjecture SICs exist in every dimension ◮ Exact solutions in dimensions 2 − 16, 19, 24, 35 and 48 ◮ Numerical solutions in dimensions 2 − 67 They can be chosen so that they form an orbit of H ( N ) i.e. the SIC has the form D ij | ψ � and there is a special order 3 element of the Clifford group such that U z | ψ � = | ψ � ◮ All available evidence supports this conjecture

Images of SICs in the probability simplex | ψ � = ( ψ 00 , ψ 01 , ψ 10 , . . . ) T = ( √ p 00 , � � p 10 e i µ 10 , . . . ) T p 01 e i µ 01 , ↓ image w.r.t the basis prob vector = ( p 00 , p 01 , p 10 , . . . ) T Consider the equations � p rs q ru + sv � ψ | X nu Z nv | ψ � = r , s |� ψ | X nu Z nv | ψ �| 2 = � � p rs p r ′ s ′ q ( r − r ′ ) u +( s − s ′ ) v r , s r ′ , s ′ 1 � � p rs p r ′ s ′ q ( r − r ′ ) u +( s − s ′ ) v N + 1 = r , s r ′ , s ′ Take the Fourier transform 2 � p 2 rs = N + 1 r , s 1 � p rs p r + x , s + y = N + 1 for ( x , y ) � = (0 , 0) r , s

SICs are nicely aligned in the new basis Geometric interpretation: When the SIC is projected to the basis simplex, we see a regular simplex centered at the origin with N vertices. The new basis is nicely orientated...

Solutions to the SIC problem ◮ Dimension N = 2 2 is now trivial ◮ Dimension N = 3 2 can be solved on a blackboard ◮ Dimension N = 4 2 can be solved with a computer

Solving the SIC problem in dimension N = 2 2 First write the SIC fiducial as   ψ 00 ψ 01   | ψ � =   ψ 10   ψ 11 In the monomial basis, Zauner’s unitary is   0 1 0 0 0 0 1 0   U z =   1 0 0 0   0 0 0 1 Hence, Zauner’s conjecture, U z | ψ � = | ψ � , gives us  a  a   | ψ � =   a   be i θ

Solving the SIC problem in dimension N = 2 2 The SIC fiducial is  a  a   | ψ � =   a   be i θ Then we have two conditions, 3 a 2 + b 2 = 1 Norm = 1 ⇒ (1) 2 3 a 4 + b 4 = 2 � p 2 rs = N + 1 ⇒ (2) 5 Solving these equations gives √ √ � � 5 − 5 5 + 3 5 a = b = 20 20

Solving the SIC problem in dimension N = 2 2 Plugging these values into the equation |� ψ | X | ψ �| 2 = 1 5 Gives √ √ 3 − 5 + 1 + 5 cos 2 θ = 1 20 10 5 Hence θ = (2 λ + 1) π λ = 0 , 1 , 2 , 3 4 Conclusion: The Zauner eigenspace contains the fiducials  1  √ � 1 5 − 5   | ψ λ � =   1 20   √ � 5 e π i / 4 i λ 2 +

Solving the SIC problem in dimension N = 3 2 In the new basis, we can solve the SIC problem on the blackboard...

Solving the SIC problem in dimension N = 3 2 Zauner’s conjecture implies that − z 1 ω 7 | 1 , 1 � − z 2 ω | 2 , 2 � + z 3 ( ω 6 | 0 , 2 � + | 1 , 0 � + ω 8 | 2 , 1 � ) | ψ � = + z 4 ( ω 6 | 0 , 1 � + | 2 , 0 � + ω 5 | 1 , 2 � ) . z 1 = √ p 1 e i µ 0 z 2 = √ p 2 e − i µ 0 z 3 = √ p 3 e i µ 3 z 4 = √ p 4 e i µ 4

Solving the SIC problem in dimension N = 3 2 The absolute values: p 1 = a 1 + b 1 , p 2 = a 1 − b 1 , p 3 = a 3 + b 3 , p 4 = a 3 − b 3 √ √ √ 1 � � a 1 = 5 − s 0 5 3 + s 0 3 5 + 15 40 � � √ √ s 2 � b 1 = 15 15 + s 0 3 60 √ √ √ 1 � � a 3 = 15 + s 0 5 3 − s 0 3 5 − 15 120 √ √ √ � s 1 � � b 3 = 5 − 18 − s 0 7 3 + s 0 6 5 + 5 15 60 where s 0 = s 1 = s 2 = ± 1

Solving the SIC problem in dimension N = 3 2 The phases: � � 1 1 e i µ 0 = 2 + c 0 − is 1 2 − c 0 � 1 � � � 3 1 1 e i µ 3 q m 3 = − 2 − c 1 + c 2 + is 1 s 2 2 + c 1 − c 2 � 1 � � � 3 1 1 e i µ 4 q m 4 = − 2 − c 1 − c 2 + is 1 s 2 2 + c 1 + c 2 √ √ 1 � c 0 = 2(6 + s 0 3 − 15) 8 √ √ √ s 0 � c 1 = 9 − s 0 4 3 + s 0 3 5 − 2 15 8 √ √ √ s 1 s 0 � c 2 = 15( − 19 + s 0 12 3 − s 0 9 5 + 6 15) 24

Solving the SIC problem in dimension N = 4 2 The new basis allows us to solve the SIC problem in dimension 16... The solutions are given in a number field √ √ √ √ K = Q ( 2 , 13 , 17 , r 2 , r 3 , t 1 , t 2 , t 3 , t 4 , − 1) , of degree 1024, where � √ √ � r 2 = 221 − 11 r 3 = 15 + 17 √ √ � t 1 = 15 + (4 − 17) r 3 − 3 17 √ √ √ t 2 2 = ((3 − 5 17) 13 + (39 17 − 65)) r 3 √ √ √ + ((16 17 − 72) 13 + 936)) t 1 − 208 13 + 2288 √ √ � t 3 = 2 − 2 t 4 = 2 + t 3

Conclusion ◮ A basis were every element of the Clifford group is a monomial matrix ◮ The SICs are nicely orientated in the new basis ◮ The solutions to the SIC problem in dimensions 4, 9 and 16 are given entirely in terms of radicals, as expected (but not understood!) ◮ The result can be extended to non-square dimensions kn 2 ◮ Are there other applications in quantum information? N = n 2 : QIC vol 12, 0404 (2012), arXiv:1102.1268 N = kn 2 : in preparation