A remarkable representation of the Clifford group

Transcription

1 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

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

3 Heisenberg Groups Heisenberg groups can be defined in terms of upper triangular matrices 1 x φ 0 1 p 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

4 Finite and CV systems Finite systems Heisenberg group H(N) elements are translations in discrete phase space (N = p) CV systems Heisenberg group H(R) elements are translations in 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) = H(N) SL(2, N) HSp = H(R) SL(2, R) HSp 0 Gaussian states Our new basis Zak basis

5 The Clifford group Write elements of H(N) as D ij = τ ij X i Z j where τ = e iπ N, X j = j + 1 and Z j = ω j j. 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

6 Definition of the basis The Heisenberg group H(N) is defined by ω = e 2πi/N and generators X, Z with relations ZX = ωxz, X N = Z N = 1 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

7 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 αn 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.

8 The new basis In the Hilbert space H N = H n H n, the new basis is { r, s + 1 if s mod n X r, s = σ r r, 0 otherwise Z r, s = ω s r 1, s where the phases are ω = e 2πi N and σ = e 2πi n. Indeed, 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 r, s = 1 n 1 ω ntr nt + s. n t=0 i.e. apply F n I, where F n is the n n Fourier matrix.

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

10 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

11 Images of SICs in the probability simplex ψ = (ψ 00, ψ 01, ψ 10,...) T = ( p 00, p 01 e iµ01, p 10 e iµ10,...) T image w.r.t the basis prob vector = (p 00, p 01, p 10,...) T Consider the equations ψ X nu Z nv ψ = r,s p rs q ru+sv ψ X nu Z nv ψ 2 = p rs p r s r,s r,s 1 N + 1 = p rs p r s r,s r,s Take the Fourier transform r,s r,s p 2 rs = 2 N + 1 p rs p r+x,s+y = 1 N + 1 q(r r )u+(s s )v q(r r )u+(s s )v for (x, y) (0, 0)

12 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...

13 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

14 Solving the SIC problem in dimension N = 2 2 First write the SIC fiducial as ψ = ψ 00 ψ 01 ψ 10 ψ 11 In the monomial basis, Zauner s unitary is U z = Hence, Zauner s conjecture, U z ψ = ψ, gives us ψ = a a a be iθ

15 Solving the SIC problem in dimension N = 2 2 The SIC fiducial is ψ = Then we have two conditions, a a a be iθ Norm = 1 3a 2 + b 2 = 1 (1) p 2 rs = 2 N + 1 3a4 + b 4 = 2 5 (2) Solving these equations gives 5 5 a = 20 b =

16 Solving the SIC problem in dimension N = 2 2 Plugging these values into the equation ψ X ψ 2 = 1 5 Gives cos 2 θ = Hence (2λ + 1)π θ = λ = 0, 1, 2, 3 4 Conclusion: The Zauner eigenspace contains the fiducials ψ λ = e πi/4 i λ

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

18 Solving the SIC problem in dimension N = 3 2 Zauner s conjecture implies that ψ = z 1 ω 7 1, 1 z 2 ω 2, 2 + z 3 (ω 6 0, 2 + 1, 0 + ω 8 2, 1 ) +z 4 (ω 6 0, 1 + 2, 0 + ω 5 1, 2 ). z 1 = p 1 e iµ 0 z 2 = p 2 e iµ 0 z 3 = p 3 e iµ 3 z 4 = p 4 e iµ 4

19 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 a 1 = 1 ( 5 s s ) b 1 = s 2 ( ) s a 3 = 1 ( 15 + s s ) b 3 = s ( s s ) where s 0 = s 1 = s 2 = ±1

20 Solving the SIC problem in dimension N = 3 2 The phases: e iµ = 2 + c 0 is 1 2 c 0 ( e iµ 3 = q m c 1 + c 2 + is 1 s c 1 c 2 ( e iµ 4 = q m c 1 c 2 + is 1 s c 1 + c 2 ) 1 3 ) 1 3 c 0 = 1 2(6 + s ) 8 c 1 = s 0 9 s s c 2 = s 1s ( 19 + s s )

21 Solving the SIC problem in dimension N = 4 2 The new basis allows us to solve the SIC problem in dimension 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 221 r 2 = 11 r3 = t 1 = 15 + (4 17)r t 2 2 = ((3 5 17) 13 + ( ))r 3 + (( ) ))t t 3 = 2 2 t 4 = 2 + t 3

22 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: N = kn 2 : in preparation