Majorana Representation of Complex Vectors and Some of Applications - PowerPoint PPT Presentation

Majorana Representation of Complex Vectors and Some of Applications Mikio Nakahara and Yan Zhu Department of Mathematics Shanghai University, China April 2019 @Shanghai Jiao Tong University 1/40

Ettore Majorana Ettore Majorana Born 5 August 1906, Catania Died Unknown, missing since 1938; likely still alive in 1959. 2/40 2 / 40

1. Introduction An element of C P 1 is represented by a point on S 2 . This point is called the Bloch vector and the S 2 is called the Bloch sphere in physics. We can visualize a 2-d “complex vector” by a unit vector in R 3 . How do we visualize higher dimensional complex vectors? “Majorana representation” makes it possible to visualize a vector in C d by d − 1 unit vectors in R 3 ( S 2 ). In this talk, we introduce how to obtain the Majorana representation of | ψ ⟩ ∈ C d and introduce some of its applications to quantum information and cold atom physics. 3/40 3 / 40

2. Bloch Vector Bloch Vector An element of C P 1 ; | ψ ⟩ = cos θ 2 | 0 ⟩ + e i φ sin θ 2 | 1 ⟩ , where ( ) ( ) 1 0 | 0 ⟩ = and | 1 ⟩ = . 0 1 n = (sin θ cos ϕ, sin θ sin ϕ, cos θ ) ∈ S 2 ; Bloch vector. | ψ ⟩ ⇔ ˆ In quantum mechanics, a state is represented by a “complex vector” where | ψ ⟩ ∼ e i α | ψ ⟩ . This is not a vector but an element of C P n for some n . 4/40 4 / 40

2. Bloch Vector Pauli matrices (a set of generators of su (2)) ( ) ( ) ( ) 0 1 0 − i 1 0 σ x = , σ y = , σ z = . 1 0 0 0 − 1 i They represent the angular momentum vector of a spin. Why θ/ 2? ⟨ ψ | (ˆ n · ⃗ σ ) | ψ ⟩ = ˆ n . | ψ ⟩ corresponds to a state in which a spin points the direction ˆ n on average. It is natural to have the correspondence | ψ ⟩ ⇔ ˆ n . We write | ψ ⟩ ∈ C 2 whose Bloch vector is ˆ n ∈ S 2 as | ˆ n ⟩ . 5/40 5 / 40

3. Majorana Representation of a vector in C d Majorana Representation Tensor product of two 2-d irrep of SU(2); ⊗ = ⊕ . Take the symmetric combination . is C 3 , which is identified with The representation space of 1 Span( | 00 ⟩ , 2 ( | 01 ⟩ + | 10 ⟩ ) , | 11 ⟩ ). (Here | 00 ⟩ = | 0 ⟩ ⊗ | 0 ⟩ .) √ Example ( d = 3) Take | ψ ⟩ = | 00 ⟩ + | 11 ⟩ = (1 , 0 , 1) t ∈ C 3 , for example. Then | ψ ⟩ ∝ ( | 0 ⟩ + z 1 | 1 ⟩ )( | 0 ⟩ + z 2 | 1 ⟩ ) + ( | 0 ⟩ + z 2 | 1 ⟩ )( | 0 ⟩ + z 1 | 1 ⟩ ) | 00 ⟩ + z 1 + z 2 1 ∝ √ √ ( | 01 ⟩ + | 10 ⟩ ) + z 1 z 2 | 11 ⟩ . 2 2 z 1 + z 2 = 0 , z 1 z 2 = 1 → z 1 = i , z 2 = − i . 6/40 6 / 40

3. Majorana Representation of a vector in C d Example ( d = 3) | ψ ⟩ = | 00 ⟩ + | 11 ⟩ = (1 , 0 , 1) t = S ( | 0 ⟩ − i | 1 ⟩ , | 0 ⟩ + i | 1 ⟩ ). | 0 ⟩ − i | 1 ⟩ ∝ cos( π/ 4) | 0 ⟩ + e i 3 π/ 2 sin( π/ 4) | 1 ⟩ → ( θ, ϕ ) = ( π/ 2 , 3 π/ 2) → ˆ n = (0 , − 1 , 0). | 0 ⟩ + i | 1 ⟩ ∝ cos( π/ 4) | 0 ⟩ + e i π/ 2 sin( π/ 4) | 1 ⟩ → ( θ, ϕ ) = ( π/ 2 , π/ 2) → ˆ n = (0 , 1 , 0). We write | ψ ⟩ ∈ C 3 whose Majorana vectors are ˆ n 1 and ˆ n 2 as | ˆ n 1 , ˆ n 2 ⟩ . Note that | ˆ n 1 , ˆ n 2 ⟩ = | ˆ n 2 , ˆ n 1 ⟩ . 7/40 7 / 40

4. Majorana Polynomials Majorana Polynomials ( d = 4) 1 1 Use ( | 000 ⟩ , 3 ( | 100 ⟩ + | 010 ⟩ + | 001 ⟩ ) , 3 ( | 011 ⟩ + | 101 ⟩ + | 110 ⟩ ) , | 111 ⟩ ) as √ √ a basis to represent | ψ ⟩ ∈ C 4 . (1 , c 1 , c 2 , c 3 ) t | ψ ⟩ = | 000 ⟩ + z 1 + z 2 + z 3 1 = √ √ ( | 100 ⟩ + | 010 ⟩ + | 001 ⟩ ) 3 3 + z 1 z 2 + z 2 z 3 + z 3 z 1 1 √ √ ( | 011 ⟩ + | 101 ⟩ + | 110 ⟩ ) + z 1 z 2 z 3 | 111 ⟩ . 3 3 √ √ Then z 1 , z 2 , z 3 are solutions of M ( z ) = z 3 − 3 c 1 z 2 + 3 c 2 z − c 3 = 0 (Majorana polynomial). � d − 1 ( ) � d − 1 ∑ ( − 1) k c k � z d − 1 − k . For a general C d , M ( z ) = � k k =0 8/40 8 / 40

5. Inner Product of Complex Vectors in terms of Majorana Vectors ( d = 2) Inner Product ( d = 2) n k ⟩ ∈ C 2 ( k = 1 , 2). Let | ψ k ⟩ = | ˆ Then n 2 ⟩| 2 = 1 |⟨ ˆ n 1 | ˆ 2(1 + ˆ n 1 · ˆ n 2 ) |⟨ ψ 1 | ψ 2 ⟩| 2 = 1 → ˆ n 1 = ˆ n 2 . |⟨ ψ 1 | ψ 2 ⟩| 2 = 0 → ˆ n 1 = − ˆ n 2 . |⟨ ψ 1 | ψ 2 ⟩| 2 = 1 / 2 → ˆ n 1 · ˆ n 2 = 0 (MUB) |⟨ ψ 1 | ψ 2 ⟩| 2 = 1 / 3 → ˆ n 1 · ˆ n 2 = − 1 / 3 (SIC) n k } is equiangular in R 3 , {| ˆ n k ⟩} is equiangular in C 2 . If the set { ˆ 9/40 9 / 40

MUB and SIC Definition: Mutually Unbiased Bases (MUBs) Two ON bases, {| ψ (1) k ⟩} 1 ≤ k ≤ d and {| ψ (2) k ⟩} 1 ≤ k ≤ d of C d are MUBs if |⟨ ψ (1) | ψ (2) k ⟩| 2 = 1 / d for all 1 ≤ j , k ≤ d . A set of bases are mutually j unbiased if every pair among them is MUBs. Definition: Symmetric Informationally Complete Positive Operator-Valued Measures (SIC-POVM) A set of d 2 normalized vectors {| ψ k ⟩} 1 ≤ k ≤ d 2 is a SIC-POVM if it satisfies 1 |⟨ ψ j | ψ k ⟩| 2 = ( j ̸ = k ) . d + 1 10/40 10 / 40

5. Inner Product of Complex Vectors ( d = 3) P. K. Aravind, MUBs and SIC-POVMs of a spin-1 system from the Majorana approach , arXiv:1707.02601 (2017). Proposition Let | ψ 1 ⟩ = | ˆ n 1 , ˆ n 2 ⟩ and | ψ 2 ⟩ = | ˆ m 1 , ˆ m 2 ⟩ . Then n 2 ⟩| 2 = 2 F − (1 − ˆ n 1 · ˆ n 2 )(1 − ˆ m 1 · ˆ m 2 ) |⟨ ˆ m 1 , ˆ m 2 | ˆ n 1 , ˆ , (3 + ˆ n 1 · ˆ n 2 )(3 + ˆ m 1 · ˆ m 2 ) where F = (1 + ˆ n 1 · ˆ m 1 )(1 + ˆ n 2 · ˆ m 2 ) + (1 + ˆ n 1 · ˆ m 2 )(1 + ˆ n 2 · ˆ m 1 ) . Important Cases |⟨ ψ 1 | ψ 2 ⟩| 2 = 1 → F − (1 + ˆ n 1 · ˆ n 2 )(1 + ˆ m 1 · ˆ m 2 ) − 4 = 0 . |⟨ ψ 1 | ψ 2 ⟩| 2 = 0 → 2 F − (1 − ˆ n 1 · ˆ n 2 )(1 − ˆ m 1 · ˆ m 2 ) = 0. |⟨ ψ 1 | ψ 2 ⟩| 2 = 1 / 3 → 3 F − 2(ˆ n 2 · ˆ n 2 )( ˆ m 2 · ˆ m 2 ) − 6 = 0. |⟨ ψ 1 | ψ 2 ⟩| 2 = 1 / 4 → 8 F − 5(ˆ n 1 · ˆ n 2 )( ˆ m 1 · ˆ m 2 ) + ˆ n 1 · ˆ n 2 + ˆ m 1 · ˆ m 2 − 13 = 0. 11/40 11 / 40

5. Inner Product of Complex Vectors ( d = 4) Question: How about d = 4? (3 Majorana vectors for each | ψ 1 , 2 ⟩ ∈ C 4 ). ∑ | ˆ n 1 , ˆ n 2 , ˆ | ˆ n σ (1) ⟩ ⊗ | ˆ n σ (2) ⟩ ⊗ | ˆ n 3 ⟩ = n σ (3) ⟩ → σ ∈ S 3 ⟨ ˆ n 1 , ˆ n 2 , ˆ n 3 | ˆ n 1 , ˆ n 2 , ˆ n 3 ⟩ = 6(ˆ n 1 · ˆ n 2 + ˆ n 1 · ˆ n 3 + ˆ n 2 · ˆ n 3 + 3) We want to obtain m 3 ⟩| 2 |⟨ ˆ n 1 , ˆ n 2 , ˆ n 3 | ˆ m 1 , ˆ m 2 , ˆ and its higher-dimensional generalizations. 12/40 12 / 40

6. Application to SIC-POVM SIC-POVM = Symmetric Informationally Complete Positive Operator-Valued Measures. Definition A set of d 2 normalized vectors {| ψ k ⟩} 1 ≤ k ≤ d 2 is a SIC-POVM if it satisfies 1 |⟨ ψ j | ψ k ⟩| 2 = ( j ̸ = k ) . d + 1 d 2 ∑ It is easy to show | ψ k ⟩⟨ ψ k | = dI d . k =1 Zauner’s conjecture; SIC-POVM exsit for all C d . Existence of SIC-POVM is proved algebraically for some d and is shown numerically for some d but a formal proof of this conjecture is still lackinig. 13/40 13 / 40

6. Application to SIC-POVM Example ( d = 2) n 2 ⟩| 2 = 1 Recall that |⟨ ˆ n 1 | ˆ 2 (1 + ˆ n 1 · ˆ n 2 ). Take a tetrahedron in R 3 with verticies (M-vectors): v 1 = (0 , 0 , 1) t , v 2 = (sin θ 0 , 0 , cos θ 0 ) t , v 3 = (sin θ 0 cos(2 π/ 3) , sin θ 0 sin(2 π/ 3) , cos θ 0 ) t , v 4 = (sin θ 0 cos(4 π/ 3) , sin θ 0 sin(4 π/ 3) , cos θ 0 ) t , where cos θ 0 = − 1 / 3. √ √ 1 2 Corresponding complex vectors: | ψ 1 ⟩ = | 0 ⟩ , | ψ 2 ⟩ = 3 | 0 ⟩ + 3 | 1 ⟩ , √ √ √ √ 1 3 | 0 ⟩ + e i 2 π/ 3 2 1 3 | 0 ⟩ + e i 4 π/ 3 2 | ψ 3 ⟩ = 3 | 1 ⟩ , | ψ 4 ⟩ = 3 | 1 ⟩ . They satisfy |⟨ ψ j | ψ k ⟩| 2 = 1 / 3 ( j ̸ = k ). 14/40 14 / 40

6. Application to SIC-POVM Example ( d = 2) SIC-POVM is also found with the Weyl-Heisenberg group D jk = − ω jk / 2 X j Z k (0 ≤ j , k ≤ d − 1), ω = e 2 π i / d , where X | e j ⟩ = | e j +1 ⟩ , Z | e j ⟩ = ω j | e j ⟩ . √ 3 → | ψ 1 ⟩ = cos( θ 0 / 2) | 0 ⟩ + e i π/ 4 sin( θ 0 / 2) | 1 ⟩ , n = (1 , 1 , 1) t / Take ˆ √ where θ 0 = arccos(1 / 3). | ψ 2 ⟩ := D 10 | ψ 1 ⟩ ∝ sin( θ 0 / 2) | 0 ⟩ + e − i π/ 4 cos( θ 0 / 2) | 1 ⟩ , | ψ 3 ⟩ := D 01 | ψ 1 ⟩ ∝ cos( θ 0 / 2) | 0 ⟩ − e i π/ 4 sin( θ 0 / 2) | 1 ⟩ , | ψ 4 ⟩ := D 11 | ψ 1 ⟩ ∝ sin( θ 0 / 2) | 0 ⟩ − e − i π/ 4 cos( θ 0 / 2) | 1 ⟩ . The set {| ψ k ⟩} 1 ≤ k ≤ 4 is a SIC-POVM. This construction works for any d provided that the fiducial vector | ψ 1 ⟩ is found. (This is the most difficult part!) 15/40 15 / 40

6. Application to SIC-POVM Example ( d = 3): Appleby’s SIC D. M. Appleby, SIC-POVM and the extended Clifford group , J. Math. Phys. 46 , 052107 (2005). C 3 Group Majorana 1 Majorana 2 v 1 = (0 , e − it , − e it ) a 2 = ( θ 0 , π a 1 = ( π, 0) 2 − 2 t ) a 2 = ( θ 0 , 5 π v 2 = (0 , e − it ω, − e it ω 2 ) 1 a 1 = ( π, 0) 6 − 2 t ) v 3 = (0 , e − it ω 2 , − e it ω ) a 2 = ( θ 0 , π a 1 = ( π, 0) 6 − 2 t ) v 4 = ( − e it , 0 , e − it ) a 1 = ( π 2 , t − π a 2 = ( π 2 , t + π 2 ) 2 ) v 5 = ( − e it ω 2 , 0 , e − it ω ) 2 , t + 5 π a 1 = ( π a 2 = ( π 2 , t − π 2 6 ) 6 ) v 6 = ( − e it ω, 0 , e − it ω 2 ) a 1 = ( π 2 , t + 7 π a 2 = ( π 2 , t + π 6 ) 6 ) a 2 = ( π − θ 0 , π v 7 = ( e − it , − e it , 0) a 1 = (0 , 0) 2 − 2 t ) v 8 = ( e − it ω, − e it ω 2 , 0) a 2 = ( π − θ 0 , 5 π 3 a 1 = (0 , 0) 6 − 2 t ) v 9 = ( e − it ω 2 , − e it ω, 0) a 2 = ( π − θ 0 , π a 1 = (0 , 0) 6 − 2 t ) where θ 0 = cos − 1 (1 / 3) , t ∈ [0 , π/ 6]. 16/40 16 / 40