Ph219c/CS 219c
Due: Thursday 2 June 2022
3.1 Fibonacci anyons I
For the Fibonacci anyon model there is a trivial label (denoted 0) and a nontrivial label (denoted 1); the fusion rule is
1 × 1 = 0 + 1. (1) As discussed in Sec. 9.16 of the lecture notes, for this model there are
only two F -matrices that arise, which we will denote as
F0111 ≡F0 , F1111 ≡F1 . (2)
F0 is really the 1 × 1 matrix
(F0)ba = δa1δ1b , (3)
while F1 is a 2 × 2 matrix. The pentagon equation becomes
(Fc)da (Fa)cb = (Fd)ce (Fe)db (Fb)ea . (4)
Show that the general solution for F ≡ F1 is τ eiφ√τ
τ2+τ=1. (6)
3.2 Fibonacci anyons II
The 2 × 2 R-matrix that describes a counterclockwise exchange of two Fibonacci anyons has two eigenvalues — R0 for the case where the
F=e−iφ√τ −τ ,
phase convention), and τ = √5 − 1 /2 ≈ .618, which satisfies
(5) where eiφ is an arbitrary phase (which we can set to 1 with a suitable
Programming Help
total charge of the pair of anyons is trivial, and R1 for the case where the total charge is nontrivial. The hexagon equation becomes
Rc(F)caRa=(F)c0(F)0a+(F)c1R1(F)1a . (7) Using the expression for F found by solving the pentagon equation
(with eiφ set equal to 1), solve the hexagon equation for R, finding e4πi/5 0 τ √τ
R= 0 e−3πi/5 , F= √τ −τ . (8)
The only other solution for R is the complex conjugate of this one; this second solution really describes the same model, but with clock- wise and counterclockwise braiding interchanged. Therefore, an anyon model with the Fibonacci fusion rule really does exist, and it is essen- tially unique.
3.3 Ising anyons
The fusion rule for two Ising anyons is
σ × σ = 1 + ψ, (9)
where here 1 denotes the trivial charge and ψ denotes the fermion. By solving the pentagon and hexagon equations we can find the 2 × 2 R and F matrices
−iθ10 σ 11 1 R≡Rσσ=e 0 i , F≡Fσσσ=√2 1 −1 (10)
where eiθ = eiπ/8 is the topological spin of the σ particle. (This is actually one of eight possible solutions.)
Four σ anyons can fuse with trivial total charge in two distinct ways, and therefore can encode a qubit. Suppose the anyons are lined up in order 1234, numbered from left to right; in the standard basis state |0⟩, anyons 1 and 2 fuse to yield total charge 1, while in the standard basis state |1⟩, anyons 1 and 2 fuse to yield total charge ψ. Acting on this standard basis, the braid group generator σ1 (counterclockwise exchange of particles 1 and 2) is represented by R and the generator σ2 (counterclockwise exchange of particles 2 and 3) is represented by B = F RF . Verify that R and B satisfy the Yang-Baxter relation.
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