Relations for 2-qubit Clifford+T operator group

Transcription

1 Relations for -qubit Clifford+T operator group Peter Selinger and iaoning Bian Dalhousie University

2 Contents Some background The main theorem Proof of the main theorem Greylyn s theorem Presentation of a subgroup Choice of C, f, and h Reduction of equations

3 Clifford+T operators The class of Clifford+T operators is the smallest class of unitary operators that includes the operators ω = e iπ/4, = 1 ( ) ( ) ( ) , S = 0 i, T = 0 ω Z c = = Z = Z =, and is closed under composition and tensor product.

4 Clifford+T operator on qubits Notations for -qubit Clifford+T operator: T 0 = T I = T, T 1 = I T = T Similarly for 0, 1, S 0, S 1. We provide a presentation in terms of generators ω, Z c, T 0, T 1, 0, 1, S 0, S 1 and relations (main theorem) for -qubit Clifford+T operator group.

5 Why do we need relations?

6 Why do we need relations? For 1-qubit Clifford+T operators ([]) Exact synthesis algorithm Matsumoto-Amano normal form (T-optimal) TTSTTTSTSTTZ

7 Why do we need relations? For 1-qubit Clifford+T operators ([]) Exact synthesis algorithm Matsumoto-Amano normal form (T-optimal) TTSTTTSTSTTZ For n-qubit Clifford+T operators Exact synthesis Giles-Selinger algorithm ([1], but not T-optimal) No normal form so far ow to minimize the T-count? S T T T T T

8 The main theorem Theorem. The following set of relations is complete for -qubit Clifford+T circuits: ω 8 = 1 = 1 S 4 = 1 SSS = ω = S = S S = S S S S S = S S S S = = = S S S S S S S S S ω 1 S S S ω 1

9 The main theorem, continued: TT = S (TSS) = ω T = T T = T T S T T S T = ɛ T T T T T T = ɛ S T T T T T S T T T T T S T T S T T S T T T T T S T T T T S = ɛ S T

10 Clifford+T and U 4 (Z[ 1, i]) Theorem (Giles and Selinger, ariv: [1]). The -qubit Clifford+T operator group is the index subgroup of U 4 (Z[ 1, i]), consisting of operators with determinant ±1, ±i. ere, U 4 (Z[ 1, i]) is the group of unitary 4 4 matrices with entries in Z[ 1, i].

11 Greylyn s result Theorem (Greylyn, [4]). The group U 4 (Z[ 1, i]) can be presented by 16 generators and 13 equations. [i,j], [i,j], ω [k] (1 i < j 4, 1 k 4) ere, ω [k], and [i,j], [i,j] are one- and two-level operators, e.g.: ω [4] = , [,3] = ω

12 Greylyn s 13 relations Figure from Greylyn s master thesis ariv:

13 Proof idea of Greylyn s theorem 1. Build the Cayley graph of the group. Vertices = group elements, edges = generators. I

14 Proof idea of Greylyn s theorem 1. Build the Cayley graph of the group. Vertices = group elements, edges = generators. Cycles = relations. I

15 Proof idea of Greylyn s theorem 1. Build the Cayley graph of the group. Vertices = group elements, edges = generators. Cycles = relations.. The Giles-Selinger algorithm gives a canonical path from each group element to the identity. This forms a spanning tree. I

16 Proof idea of Greylyn s theorem, continued 3. Find finitely many relations of the form a b such that any arbitrary path can be transformed to the equivalent canonical path. By induction on the height of a and b.

17 Presentation of a subgroup We have Clifford+T U 4 (Z[ 1, i]). Greylyn s result gives us generators and relations for the bigger group. We face the following problem: Problem. Let be a subgroup of G, and suppose we have a presentation of G by generators and relations. Can we find a presentation of by generators and relations?

18 Presentation of a subgroup We have Clifford+T U 4 (Z[ 1, i]). Greylyn s result gives us generators and relations for the bigger group. We face the following problem: Problem. Let be a subgroup of G, and suppose we have a presentation of G by generators and relations. Can we find a presentation of by generators and relations? Example. G = A, B, C A, B, C, (BC) 3, (AC), (AB) 4 Let = AC, Y = BA. =, Y

19 Presentation of a subgroup We have Clifford+T U 4 (Z[ 1, i]). Greylyn s result gives us generators and relations for the bigger group. We face the following problem: Problem. Let be a subgroup of G, and suppose we have a presentation of G by generators and relations. Can we find a presentation of by generators and relations? Example. G = A, B, C A, B, C, (BC) 3, (AC), (AB) 4 Let = AC, Y = BA. =, Y, Y 4, (Y ) 3 Fortunately, there is a method for computing this.

20 Presentation of a subgroup Lemma. If (G 0, S) is a presentation of group G, and = 0 is a subgroup of G, and if C, f, and h are chosen as below, then ( 0, R) is a presentation of, where R consists of the following relations: (A) For each generator x 0, a relation x = g(f (x)) R; and (B) For each coset representative c C and each relation s = t S, a relation u = v R, where (u, d) = h(c, s), and (v, e) = h(c, t).

21 C, f, and h Coset representatives C x 0 f (x), some sequence in G 0 s.t. [f (x)] = x Define a map (where w and d satisfy cy = [w]d) h : C G 0 0 C (c, y) (w, d) Extend h to h : C G 0 0 C h(c 0, y 1 y...y n ) = (w 1 w...w n, c n ) Define g : G 0 [y1...y n] 0 given by g(u) = v iff h(1, u) = (v, 1)

22 Choice of C, f, and h C = {1, ω [4] } f is defined by x f (x) x f (x) 0 [1,3] [0,] 1 [,3] [0,1] S 0 ω[3] ω [] S 1 ω[3] ω [1] Z c ω[3] 4 ω ω [0] ω [1] ω [] ω [3] T 0 ω [] ω [3] T 1 ω [3] ω [1] Choice of h, using the following abbreviations Swap =, T = T 7, C 0 = 0 Z c 0 0 = 0 S 0 S 0 0, 1 = 1 S 1 S 1 1, S = S 3, C 1 = 1 Z c 1

23 Choice of C, f, and h y h(1, y) h(ω [4], y) [0,1] ( 0 C 1 0, 1) ( 0 C 1 0, ω [4] ) [0,] (Swap 0 C 1 0 Swap, 1) (Swap 0 C 1 0 Swap, ω [4] ) [0,3] (C 0 0 C 1 0 C 0, 1) (C 0 0 T 1 C 1 T 1 0C 0, ω [4] ) [1,] (C 0 1 C 1 1 C 0, 1) (C 0 1 C 1 1 C 0, ω [4] ) [1,3] (SwapC 1 Swap, 1) (SwapT 1 C 1 T 1 Swap, ω [4]) [,3] (C 1, 1) (T 1 C 1 T 1, ω [4]) [0,1] ( 0 S 1 1T 1 C 1T 1 1 S 1 0, 1) ( 0 S 1 1T 1 C 1T 1 1 S 1 0, ω [4] ) [0,] (Swap 0 S 1 1T 1 C 1T 1 1 S 1 0 Swap, 1) (Swap 0 S 1 1T 1 C 1T 1 1 S 1 0 Swap, ω [4] ) [0,3] (C 0 0 S 1 1T 1 C 1T 1 1 S 1 0 C 0, 1) (C 0 0 T 1 S 1 1T 1 C 1T 1 1 S 1 T 1 0C 0, ω [4] ) [1,] (C 0 1 S 1 1T 1 C 1T 1 1 S 1 1 C 0, 1) (C 0 1 S 1 1T 1 C 1T 1 1 S 1 1 C 0, ω [4] ) [1,3] (SwapS 1 1T 1 C 1T 1 1 S 1 Swap, 1) (SwapT 1 S 1 1T 1 C 1T 1 1 S 1 T 1 Swap, ω [4]) [,3] (S 1 1T 1 C 1T 1 1 S 1, 1) (T 1 S 1 1T 1 C 1T 1 1 S 1 T 1, ω [4]) ω [0] (C 0 0 T 1 C 1T 1 C 1 0 C 0, ω [4] ) (C 0 0 T 0 0 C 0, 1) ω [1] (SwapT 1 C 1T 1 C 1 Swap, ω [4] ) (SwapT 0 Swap, 1) ω [] (T 1 C 1T 1 C 1, ω [4] ) (T 0, 1) ω [3] (ɛ, ω [4] ) (T 1 T 0 C 1 T 1 C 1, 1)

24 Reduction of equations Apply this lemma, we get = 54 equations, all very long.

25 Reduction of equations Apply this lemma, we get = 54 equations, all very long. We already know some obvious equations: All Clifford equations Obvious Clifford+T equations TT = S (TSS) = ω T = T T = T

26 Reduction of equations Apply this lemma, we get = 54 equations, all very long. We already know some obvious equations: All Clifford equations Obvious Clifford+T equations TT = S (TSS) = ω T = T T = T After automatic reduction, we have 40 left After manual reduction, we have 3 left

27 T S T T S T = ɛ T T T T T T = ɛ S T T T T T S T T T T T S T T S T T S T T T T T S T T T T S = ɛ S T

28 Sketch of the automated reduction Following Gosset, Kliuchnikov, Mosca, and Russo ([3]), we define, for any Pauli operators P, Q: R(P Q) = 1 + ω I + 1 ω (P Q).

29 Sketch of the automated reduction Following Gosset, Kliuchnikov, Mosca, and Russo ([3]), we define, for any Pauli operators P, Q: R(P Q) = 1 + ω I + 1 ω (P Q). Then every Clifford+T operator can be written (not uniquely) as R(P 1 Q 1 ) R(P k Q k ) C, where P j, Q j are Pauli and C is Clifford.

30 Sketch of the automated reduction Following Gosset, Kliuchnikov, Mosca, and Russo ([3]), we define, for any Pauli operators P, Q: R(P Q) = 1 + ω I + 1 ω (P Q). Then every Clifford+T operator can be written (not uniquely) as R(P 1 Q 1 ) R(P k Q k ) C, where P j, Q j are Pauli and C is Clifford. We can use the obvious equations to convert any Clifford+T operator to this form. Also, R(P Q) and R(P Q ) commute iff P Q and P Q commute. Using these techniques, most of the 54 equations can be automatically proven.

31 This concludes the proof of the main theorem! Theorem. The following set of relations is complete for -qubit Clifford+T circuits: Clifford equations [5] TT = S (TSS) = ω T = T T = T T S T T S T = ɛ T T T T T T = ɛ S T T T T T S T T T T T S T T S T T S T T T T T S T T T T S = ɛ S T

32 References B. Giles and P. Selinger. Exact synthesis of multiqubit Clifford+T circuits. Physical Review A, 87(3):0333, 013. B. Giles and P. Selinger. Remarks on Matsumoto and Amano s normal form for single-qubit Clifford+T operators. ariv preprint ariv: , 013. D. Gosset, V. Kliuchnikov, M. Mosca, and V. Russo. An algorithm for the t-count. Quantum Information & Computation, 14(15-16): , 014. S. E. M. Greylyn. Generators and relations for the group U 4 (Z[ 1, i]). ariv preprint ariv: , 014. P. Selinger. Generators and relations for n-qubit Clifford operators.

33 Thank You!