Majorana Representation of Complex Vectors and Some of...

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

2/40

Ettore Majorana

Ettore Majorana

Born 5 August 1906, Catania

Died Unknown, missing since 1938; likely still alive in 1959.2 / 40

3/40

1. Introduction

An element of CP1 is represented by a point on S2.

This point is called the Bloch vector and the S2 is called the Bloch

sphere in physics.

We can visualize a 2-d “complex vector” by a unit vector in R3.

How do we visualize higher dimensional complex vectors?

“Majorana representation” makes it possible to visualize a vector in

Cd by d − 1 unit vectors in R3 (S2).

In this talk, we introduce how to obtain the Majorana representation

of |ψ⟩ ∈ Cd and introduce some of its applications to quantum

information and cold atom physics.

3 / 40

5/40

2. Bloch Vector

Pauli matrices (a set of generators of su(2))

σx =

(0 1

1 0

), σy =

(0 −i

i 0

), σz =

(1 0

0 −1

).

They represent the angular momentum vector of a spin.

Why θ/2?

⟨ψ|(n · σ)|ψ⟩ = n.

|ψ⟩ corresponds to a state in which a spin points the direction n on

average. It is natural to have the correspondence |ψ⟩ ⇔ n.

We write |ψ⟩ ∈ C2 whose Bloch vector is n ∈ S2 as |n⟩.

5 / 40

6/40

3. Majorana Representation of a vector in Cd

Majorana Representation

Tensor product of two 2-d irrep of SU(2); ⊗ = ⊕ .

Take the symmetric combination .

The representation space of is C3, which is identified with

Span(|00⟩, 1√2(|01⟩+ |10⟩), |11⟩). (Here |00⟩ = |0⟩ ⊗ |0⟩.)

Example (d = 3)

Take |ψ⟩ = |00⟩+ |11⟩ = (1, 0, 1)t ∈ C3, for example. Then

|ψ⟩ ∝ (|0⟩+ z1|1⟩)(|0⟩+ z2|1⟩) + (|0⟩+ z2|1⟩)(|0⟩+ z1|1⟩)

∝ |00⟩+ z1 + z2√2

1√2(|01⟩+ |10⟩) + z1z2|11⟩.

z1 + z2 = 0, z1z2 = 1 → z1 = i , z2 = −i .6 / 40

7/40

3. Majorana Representation of a vector in Cd

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 i3π/2 sin(π/4)|1⟩ →(θ, ϕ) = (π/2, 3π/2) → n = (0,−1, 0).

|0⟩+ i |1⟩ ∝ cos(π/4)|0⟩+ e iπ/2 sin(π/4)|1⟩ →(θ, ϕ) = (π/2, π/2) → n = (0, 1, 0).

We write |ψ⟩ ∈ C3 whose Majorana vectors are n1 and n2 as |n1, n2⟩.Note that |n1, n2⟩ = |n2, n1⟩.

7 / 40

8/40

4. Majorana Polynomials

Majorana Polynomials (d = 4)

Use (|000⟩, 1√3(|100⟩+ |010⟩+ |001⟩), 1√

3(|011⟩+ |101⟩+ |110⟩), |111⟩) as

a basis to represent |ψ⟩ ∈ C4.

|ψ⟩ = (1, c1, c2, c3)t

= |000⟩+ z1 + z2 + z3√3

1√3(|100⟩+ |010⟩+ |001⟩)

+z1z2 + z2z3 + z3z1√

3

1√3(|011⟩+ |101⟩+ |110⟩) + z1z2z3|111⟩.

Then z1, z2, z3 are solutions of M(z) = z3 −√3c1z

2 +√3c2z − c3 = 0

(Majorana polynomial).

For a general Cd , M(z) =d−1∑k=0

(−1)kck

√√√√(d − 1

k

)zd−1−k .

8 / 40

9/40

5. Inner Product of Complex Vectors in terms of Majorana

Vectors (d = 2)

Inner Product (d = 2)

Let |ψk⟩ = |nk⟩ ∈ C2 (k = 1, 2).

Then

|⟨n1|n2⟩|2 =1

2(1 + n1 · n2)

|⟨ψ1|ψ2⟩|2 = 1 → n1 = n2.

|⟨ψ1|ψ2⟩|2 = 0 → n1 = −n2.

|⟨ψ1|ψ2⟩|2 = 1/2 → n1 · n2 = 0 (MUB)

|⟨ψ1|ψ2⟩|2 = 1/3 → n1 · n2 = −1/3 (SIC)

If the set {nk} is equiangular in R3, {|nk⟩} is equiangular in C2.

9 / 40

10/40

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 Cd are MUBs if

|⟨ψ(1)j |ψ(2)

k ⟩|2 = 1/d for all 1 ≤ j , k ≤ d . A set of bases are mutually

unbiased if every pair among them is MUBs.

Definition: Symmetric Informationally Complete Positive

Operator-Valued Measures (SIC-POVM)

A set of d2 normalized vectors {|ψk⟩}1≤k≤d2 is a SIC-POVM if it satisfies

|⟨ψj |ψk⟩|2 =1

d + 1(j = k).

10 / 40

11/40

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 |ψ1⟩ = |n1, n2⟩ and |ψ2⟩ = |m1, m2⟩. Then

|⟨m1, m2|n1, n2⟩|2 =2F − (1− n1 · n2)(1− m1 · m2)

(3 + n1 · n2)(3 + m1 · m2),

where F = (1 + n1 · m1)(1 + n2 · m2) + (1 + n1 · m2)(1 + n2 · m1).

Important Cases

|⟨ψ1|ψ2⟩|2 = 1 → F − (1 + n1 · n2)(1 + m1 · m2)− 4 = 0.

|⟨ψ1|ψ2⟩|2 = 0 → 2F − (1− n1 · n2)(1− m1 · m2) = 0.

|⟨ψ1|ψ2⟩|2 = 1/3 → 3F − 2(n2 · n2)(m2 · m2)− 6 = 0.

|⟨ψ1|ψ2⟩|2 = 1/4 → 8F − 5(n1 · n2)(m1 · m2) + n1 · n2 + m1 · m2 − 13 = 0.

11 / 40

12/40

5. Inner Product of Complex Vectors (d = 4)

Question: How about d = 4? (3 Majorana vectors for each |ψ1,2⟩ ∈ C4).

|n1, n2, n3⟩ =∑σ∈S3

|nσ(1)⟩ ⊗ |nσ(2)⟩ ⊗ |nσ(3)⟩ →

⟨n1, n2, n3|n1, n2, n3⟩ = 6(n1 · n2 + n1 · n3 + n2 · n3 + 3)

We want to obtain

|⟨n1, n2, n3|m1, m2, m3⟩|2

and its higher-dimensional generalizations.

12 / 40

13/40

6. Application to SIC-POVM

SIC-POVM = Symmetric Informationally Complete Positive

Operator-Valued Measures.

Definition

A set of d2 normalized vectors {|ψk⟩}1≤k≤d2 is a SIC-POVM if it satisfies

|⟨ψj |ψk⟩|2 =1

d + 1(j = k).

It is easy to showd2∑k=1

|ψk⟩⟨ψk | = dId .

Zauner’s conjecture; SIC-POVM exsit for all Cd .

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

14/40


Example (d = 2)

Recall that |⟨n1|n2⟩|2 = 12(1 + n1 · n2).

Take a tetrahedron in R3 with verticies (M-vectors):

v1 = (0, 0, 1)t , v2 = (sin θ0, 0, cos θ0)t ,

v3 = (sin θ0 cos(2π/3), sin θ0 sin(2π/3), cos θ0)t ,

v4 = (sin θ0 cos(4π/3), sin θ0 sin(4π/3), cos θ0)t , where cos θ0 = −1/3.

Corresponding complex vectors: |ψ1⟩ = |0⟩, |ψ2⟩ =√

13 |0⟩+

√23 |1⟩,

|ψ3⟩ =√

13 |0⟩+ e i2π/3

√23 |1⟩, |ψ4⟩ =

√13 |0⟩+ e i4π/3

√23 |1⟩.

They satisfy |⟨ψj |ψk⟩|2 = 1/3 (j = k).

14 / 40

15/40


Example (d = 2)

SIC-POVM is also found with the Weyl-Heisenberg group

Djk = −ωjk/2X jZ k (0 ≤ j , k ≤ d − 1), ω = e2πi/d ,

where X |ej⟩ = |ej+1⟩,Z |ej⟩ = ωj |ej⟩.Take n = (1, 1, 1)t/

√3 → |ψ1⟩ = cos(θ0/2)|0⟩+ e iπ/4 sin(θ0/2)|1⟩,

where θ0 = arccos(1/√3).

|ψ2⟩ := D10|ψ1⟩ ∝ sin(θ0/2)|0⟩+ e−iπ/4 cos(θ0/2)|1⟩,|ψ3⟩ := D01|ψ1⟩ ∝ cos(θ0/2)|0⟩ − e iπ/4 sin(θ0/2)|1⟩,|ψ4⟩ := D11|ψ1⟩ ∝ sin(θ0/2)|0⟩ − e−iπ/4 cos(θ0/2)|1⟩.The set {|ψk⟩}1≤k≤4 is a SIC-POVM.

This construction works for any d provided that the fiducial vector

|ψ1⟩ is found. (This is the most difficult part!)

15 / 40

16/40


Example (d = 3): Appleby’s SIC

D. M. Appleby, SIC-POVM and the extended Clifford group, J. Math.Phys. 46, 052107 (2005).

Group C3 Majorana 1 Majorana 2

v1 = (0, e−it ,−e it) a1 = (π, 0) a2 = (θ0,π2 − 2t)

1 v2 = (0, e−itω,−e itω2) a1 = (π, 0) a2 = (θ0,5π6 − 2t)

v3 = (0, e−itω2,−e itω) a1 = (π, 0) a2 = (θ0,π6 − 2t)

v4 = (−e it , 0, e−it) a1 = (π2 , t −π2 ) a2 = (π2 , t +

π2 )

2 v5 = (−e itω2, 0, e−itω) a1 = (π2 , t +5π6 ) a2 = (π2 , t −

π6 )

v6 = (−e itω, 0, e−itω2) a1 = (π2 , t +7π6 ) a2 = (π2 , t +

π6 )

v7 = (e−it ,−e it , 0) a1 = (0, 0) a2 = (π − θ0,π2 − 2t)

3 v8 = (e−itω,−e itω2, 0) a1 = (0, 0) a2 = (π − θ0,5π6 − 2t)

v9 = (e−itω2,−e itω, 0) a1 = (0, 0) a2 = (π − θ0,π6 − 2t)

where θ0 = cos−1(1/3), t ∈ [0, π/6].16 / 40

17/40


Example (d = 3): Aravind-1 SIC

Junjiang Le, Worcester Polytechnic Institute bachelor thesis (2017).

Group C3 Majorana 1 Majorana 2

v1 = (1, 0,−1) a1 = (π/2, 0) a2 = (π/2, π)

1 v2 = (1, 0,−ω) a1 = (π/2, π/3) a2 = (π/2, 4π/3)

v3 = (1, 0,−ω2) a1 = (π/2, 2π/3) a2 = (π/2, 5π/3)

v4 = (1, e iϕ1 , 0) a1 = (0, 0) a2 = (π − θ0, ϕ1)

2 v5 = (1, ωe iϕ1 , 0) a1 = (0, 0) a2 = (π − θ0, 2π/3 + ϕ1)

v6 = (1, ω2e iϕ1 , 0) a1 = (0, 0) a2 = (π − θ0, 4π/3 + ϕ1)

v7 = (0, 1, e iϕ2) a1 = (π, 0) a2 = (θ0, ϕ2)

3 v8 = (0, 1, ωe iϕ2) a1 = (π, 0) a2 = (θ0, 2π/3 + ϕ2)

v9 = (0, 1, ω2e iϕ2) a1 = (π, 0) a2 = (θ0, 4π/3 + ϕ2)

where θ0 = cos−1(1/3), ϕ1, ϕ2 ∈ [0, π/6].

17 / 40

18/40


18 / 40

19/40


Aravind-1 reduces to Appleby’s SIC when ϕ1 = ϕ2 = t.

Question: Is Aravind-1 more general than Appleby’s SIC?Gram matrix is

1 14

(1 − i

√3)

14

(1 + i

√3)

12

12

12

− 12e−iϕ2 1

4

(1 + i

√3)e−iϕ2 1

4

(1 − i

√3)e−iϕ2

14

(1 + i

√3)

1 14

(1 − i

√3)

12

12

12

14

(1 − i

√3)e−iϕ2 − 1

2e−iϕ2 1

4

(1 + i

√3)e−iϕ2

14

(1 − i

√3)

14

(1 + i

√3)

1 12

12

12

14

(1 + i

√3)e−iϕ2 1

4

(1 − i

√3)e−iϕ2 − 1

2e−iϕ2

12

12

12

1 14

(1 − i

√3)

14

(1 + i

√3)

eiϕ12

eiϕ12

eiϕ12

12

12

12

14

(1 + i

√3)

1 14

(1 − i

√3)

14i(i +

√3)e iϕ1 1

4i(i +

√3)e iϕ1 1

4i(i +

√3)e iϕ1

12

12

12

14

(1 − i

√3)

14

(1 + i

√3)

1 − 18

(2 + 2i

√3)e iϕ1 − 1

8

(2 + 2i

√3)e iϕ1 − 1

8

(2 + 2i

√3)e iϕ1

− eiϕ22

14

(1 + i

√3)e iϕ2 1

4

(1 − i

√3)e iϕ2 e−iϕ1

2− 1

4i(−i +

√3)e−iϕ1 1

4i(i +

√3)e−iϕ1 1 1

4

(1 − i

√3)

14

(1 + i

√3)

14

(1 − i

√3)e iϕ2 − eiϕ2

214

(1 + i

√3)e iϕ2 e−iϕ1

2− 1

4i(−i +

√3)e−iϕ1 1

4i(i +

√3)e−iϕ1 1

4

(1 + i

√3)

1 14

(1 − i

√3)

14

(1 + i

√3)e iϕ2 1

4

(1 − i

√3)e iϕ2 − eiϕ2

2e−iϕ1

2− 1

4i(−i +

√3)e−iϕ1 1

4i(i +

√3)e−iϕ1 1

4

(1 − i

√3)

14

(1 + i

√3)

1

It seems ϕ1 and ϕ2 are independent parameters. Is it true?

19 / 40

20/40


ϕ1, ϕ2 → ϕ1 + ϕ2

UD =

1 0 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0 0

0 0 1 0 0 0 0 0 0

0 0 0 1 0 0 0 0 0

0 0 0 0 1 0 0 0 0

0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 e−iϕ2 0 0

0 0 0 0 0 0 0 e−iϕ2 0

0 0 0 0 0 0 0 0 e−iϕ2

combines ϕ1 and ϕ2 as

20 / 40

21/40


ϕ1, ϕ2 → ϕ1 + ϕ2

UDG (ϕ1, ϕ2)U†D = G (ϕ1 + ϕ2)

1 14

(1 − i

√3)

14

(1 + i

√3)

12

12

12

− 12

14

(1 + i

√3)

14

(1 − i

√3)

14

(1 + i

√3)

1 14

(1 − i

√3)

12

12

12

14

(1 − i

√3)

− 12

14

(1 + i

√3)

14

(1 − i

√3)

14

(1 + i

√3)

1 12

12

12

14

(1 + i

√3)

14

(1 − i

√3)

− 12

12

12

12

1 14

(1 − i

√3)

14

(1 + i

√3)

12e i(ϕ1+ϕ2) 1

2e i(ϕ1+ϕ2) 1

2e i(ϕ1+ϕ2)

12

12

12

14

(1 + i

√3)

1 14

(1 − i

√3)

14i(i +

√3)e i(ϕ1+ϕ2) 1

4i(i +

√3)e i(ϕ1+ϕ2) 1

4i(i +

√3)e i(ϕ1+ϕ2)

12

12

12

14

(1 − i

√3)

14

(1 + i

√3)

1 − 14i(−i +

√3)e i(ϕ1+ϕ2) − 1

4i(−i +

√3)e i(ϕ1+ϕ2) − 1

4i(−i +

√3)e i(ϕ1+ϕ2)

− 12

14

(1 + i

√3)

14

(1 − i

√3)

12e−i(ϕ1+ϕ2) 1

4

(−1 − i

√3)e−i(ϕ1+ϕ2) 1

4

(−1 + i

√3)e−i(ϕ1+ϕ2) 1 1

4

(1 − i

√3)

14

(1 + i

√3)

14

(1 − i

√3)

− 12

14

(1 + i

√3)

12e−i(ϕ1+ϕ2) 1

4

(−1 − i

√3)e−i(ϕ1+ϕ2) 1

4

(−1 + i

√3)e−i(ϕ1+ϕ2) 1

4

(1 + i

√3)

1 14

(1 − i

√3)

14

(1 + i

√3)

14

(1 − i

√3)

− 12

12e−i(ϕ1+ϕ2) 1

4

(−1 − i

√3)e−i(ϕ1+ϕ2) 1

4

(−1 + i

√3)e−i(ϕ1+ϕ2) 1

4

(1 − i

√3)

14

(1 + i

√3)

1

21 / 40

22/40


Zhu’s Invariants

This is also confirmed by evaluating the Zhu’s invariants.

H. Zhu, SIC POVM and Clifford groups in prime dimensions, J. Phys.

A: Math and Theor. 43, 305305.(2010).

Let Πk = |ψk⟩⟨ψk | be the projection operator to |ψk⟩ ∈ SIC. Then

Γjkl := tr(ΠjΠkΠl) is invariant under U(d) transformations of

|ψj⟩, |ψk⟩, |ψl⟩. The set {Γjkl} is invariant under phase changes and

permutations of the SIC vectors.

Note that tr(Πk) = 1, tr(ΠjΠk) = 1/(d + 1).

tr(ΠjΠkΠl) = (1/√d + 1)3e i(αjk+αkl+αlj ). The phase has the

information.

22 / 40

23/40


Zhu’s Invariants

For Aravind-1, it is shown that

Value of the phase Multiplicity

0 10

π/3 36

2π/3 10

π 2

−(2ϕ1 − ϕ2) 10

2π/3− (2ϕ1 − ϕ2) 8

−2π/3− (2ϕ1 − ϕ2) 8

The phases appear only as a combination 2ϕ1 − ϕ2, showing there is

only one phase degree of freedom.

23 / 40

24/40


Why SIC?

A SIC set {|ψk}1≤k≤d2 is symmetric since these vectors are

distributed uniformly in Cd (or CPd−1).

They are informatinally complete since the measurements of

Πk = |ψk⟩⟨ψk | for all 1 ≤ k ≤ d2 completely determine the quantum

state of the system.

A quantum state is given by a matrix ρ, which is (i) Hermitian (ii)

nonnegative and (iii) trρ = 1. So it is expanded in terms of d2

generators of u(d); ρ = 1d Id +

∑d2−1k=1 ckTk , where {Tk} is the set of

traceless Hermitian generators of su(d).

ρ is competely determined by the measurement outcomes

xk = tr(Πkρ) for 1 ≤ k ≤ d2 − 1.

24 / 40

25/40


Example (d = 2)

Suppose there is a quantum state

ρ =

(a b + ic

b − ic d

)

where a, b, c , d ∈ R are not known.

By measuring Πk of the Weyl-Heisenberg example, we obtain

x1 =16

((3 +

√3)a+ 2

√3b − 2

√3c −

√3d + 3d

),

x2 =16

(−(√

3− 3)a+ 2

√3b + 2

√3c +

(3 +

√3)d),

x3 =16

((3 +

√3)a− 2

√3b + 2

√3c −

√3d + 3d

),

x4 =16

(−(√

3− 3)a− 2

√3b − 2

√3c +

(3 +

√3)d).

These equations can be inverted and a, b, c , d are completely fixed by

{xk}1≤k≤4.

25 / 40

26/40

7. Application to Cold Atoms

K. Turev, T Ollikainen, P. Kuopanportti, M. Nakahara, D. Hall and M Mottonen,

New J. Phys., 20 (2018) 055011.

Cold Atoms

Atoms at very low temperature behaves as a single entity described

by a single complex vector field |Ψ(r)⟩.Here we are interested in |Ψ(r)⟩ that belongs to the 5-d irrep of

SU(2). We write |Ψ(r)⟩ = e iφ(r)√

n(r)|ξ(r)⟩, where|ξ(r)⟩ = (ξ2, ξ1, ξ0, ξ−1, ξ−2)

t , ⟨ξ|ξ⟩ = 1.

The energy of this system is

E (|Ψ⟩) =∫

n2(r)

2[c1|S(r)|2 + c2|A20(r)|2]dr,

where S = ⟨ξ|F|ξ⟩ and A20 =1√5(2ξ2ξ−2 − 2ξ1ξ−1 + ξ20).

F = (Fx ,Fy .Fz) is the 5-d irrep of su(2) generator.26 / 40

27/40


27 / 40

28/40


Cold Atoms

When c1 > 0, c2 < 0, E is minimized by the biaxial nematic (BN)

state |ξ⟩BN = (1, 0, 0, 0, 1)t/√2. (|S| = 0, |A20| = 1/

√5).

When c1 > 0, c2 > 0, E is minimized by the cyclic (C) state

|ξ⟩C = (√1/3, 0, 0,

√2/3, 0)t . (|S| = 0, |A20| = 0)).

Majorana Representation

y

z

x y

z

x

C2

C2

C4

'

C2

C3

''

They correspond to the (meta)stable solutions of the Thomson problem.28 / 40

29/40


Cold Atoms

The “state” is specified by the orientation of a square (BN) and the

tetrahedron (C).

For BN, the state is specified by GBN = U(1)× SO(3)/D4, where D4

is the dihedral group of order 4.

For C, the state is specified by GC = U(1)× SO(3)/T , where T is

the tetrahedral group.

We look at what kind of topologically nontrivial structure exists in

this system.

29 / 40

30/40


Homotopy Group

Maps Sn → M is classified by the homotopy group πn(M).

Examples: π1(S1) ≃ π1(U(1)) ≃ Z, π3(S2) ≃ Z (Hopf fibration).

R3 is compactified to S3 by identifying infinite points (one-point

compactification).

Then maps S3 → G is classified by the homotopy group π3(G ), where

G = GC or G = GBN.

It turns out that π3(GC) ≃ π3(GBN) ≃ Z. This nontrivial structure is

called the Skyrmion.

Becuase of the factors D4 and T , the map sweeps G many times as

S3 = R3 ∩ {∞} is scanned.

30 / 40

31/40


Shankar Skyrmion

R. Shankar, J. Physique 38 1405 (1977)

GShankar = SO(3) is swept twice as S3 is scanned.31 / 40

32/40


Skyrmions (BN)

GBN = U(1)× SO(3)/D4

GBN = U(1)× SO(3)/D4 is swept 16 times as S3 is scanned once. 32 / 40

33/40


Skyrmions (C)

GC = U(1)× SO(3)/T

GC = U(1)× SO(3)/T is swept 24 times as S3 is scanned once. 33 / 40

34/40

8. Summary

|ψ⟩ ∈ Cd can be visualized by d − 1 Majorana vectors in S2.

It has many applications in quantum information theory, such as

MUBs and SIC-POVM.

Topologically nontrivial structures in cold atoms system are visualized

by making use of Majorana representation.

Other related subjects; anticoherent state, spherical t-designs, the

Thomason problmes and so on.

Your input to physics is welcome!

34 / 40

35/40

謝謝

Thank you very much for your attention!

35 / 40

36/40

ABCDEFG

36 / 40

37/40

Note on SIC-POVM

Let H be a d-dimensional Hilbert space.

Definition (POVM)

A set of Hermitian operators {Ek}nk=1 on H is a positive operator-valued

measure (POVM) if Ek ≥ 0 (1 ≤ k ≤ n) and∑

k Ek = I (completeness

relation).

The probability of observing the outcome k is p(k) = tr(ρEk). p(k) ≥ 0,∑k p(k) = 1.

Definition (Informationary Complete)

A POVM is IC if any unknown quantum state ρ (mixed in general) is

completely fixed by {p(k)}1≤k≤n.

The space of d-dim. Hermitian operators is a d2-dim. real vector space

with the inner product ⟨Hj ,Hk⟩ = tr(HjHk). IC POVM must contain at

least d2 elements (n ≥ d2). 37 / 40

38/40

Note on SIC-POVM

Definition (SIC-POVM)

A SIC-POVM is a POVM with d2 elements {aΠk}d2

k=1, where a ∈ R is fixed

later. Πk is a rank-1 projection operator satisfying tr(ΠjΠk) = c ,∀j = k,

where c ∈ R is a constant (symmetric) to be fixed later.

SIC-POVM is informationally complete. The set {Πk} is linearly

independent: Suppose∑

k akΠk = 0 (∗). Multiply both sides by Πj and

take trace → aj + c∑

k =j ak = 0. Taking trace of (∗) →∑

k ak = 0.

Since c = 1, it follows aj = 0 for all j . There are d2 linearly independent

elements in SIC-POVM, which shows it is IC.

38 / 40

39/40

Note on SIC-POVM

From∑

k aΠk = I, it follows thatd2∑

j ,k=1

ΠjΠk =

d2∑k=1

Πk

2

= I/a2.

Taking trace of both sides, it follows d2 + (d4 − d2)c = d/a2 (1).

Since {Πk} is a linearly independent set, it can expand I asI =

∑d2

k=1 dkΠk . By taking trace, it follows d =∑

k dk . By taking trace

after multiplying Πj , it follows 1 = dj + c∑

k =j dk , from which it follows

dj = (1− cd)/(1− c)(2). By solving (1) and (2), we obtain dj = 1/d and

c = 1/(d + 1). Moreover, it shows the constant a = dj = 1/d .

39 / 40

40/40

Note on SIC-POVM

Definition (SIC-POVM 2)

A set of normalized vectors {|ψk⟩}d2

k=1 is called SIC, SIC vectors or

SIC-POVM if it satisfies

|⟨ψj |ψk⟩|2 =1

d + 1(j = k).

(Note that one can write Πk = |ψk⟩⟨ψk | → tr(ΠjΠk) = |⟨ψj |ψk⟩|2).

40 / 40

Majorana Representation of Complex Vectors and Some of...

Documents

Transcript of Majorana Representation of Complex Vectors and Some of...