The stabilizer generators of the general five-qudit codeJ5, 1, 3Kdtakes the same form in all dimensions
[100, 41]. It is usually presented in terms of the conventional generalised Pauli operators of Eq. (2.2), as shown in Tab. (2.2). It is easy to see that these generators commute and ¯X and ¯Z form logical operators. Based on these stabilizers we will study the distillation map of Eq. (2.24) for the four Bloch compo- nents of a general input qutrit state. Notice that the decoding of a stabilizer code is not unique, but one of an equivalence class of unitaries—a coset of the Clifford group—which are all equally valid choices.
Distillation [[5, 1, 3]]3
Two types of Magic states
H− states |H±
H2− states|Φ0,π
Parity-Checker
Converts magic into
‘plus’ states |ψ+
Equatorialization
Converts plus states into ‘phase’ states
Non-Clifford Gate Teleportation |Φ0,π
Figure 2.3: An outline of the different qutrit protocols in Secs. 2.4 and 2.5 and how they are related. In Sec. 2.4 we discover two types of magic states distillable by theJ5, 1, 3K3, the so-calledH−states ∣H±⟩ and H2−states ∣φ⟩. We then consider two sub-protocols in Sec. 2.5, the Parity-Checker and Equatorialization, that produce a suitable magic states (the phase states ∣Φ0,π⟩) which are then used to implement a qutrit non-Clifford gate.
g1= X Z Z−1 X−1 I g2= I X Z Z−1 X−1 g3= X−1 I X Z Z−1 g4= Z−1 X−1 I X Z ¯ X = Z Z Z Z Z ¯ Z = X X X X X
Table 2.2: The stabilizer generators of the five-qudit codeJ5, 1, 3Kd.
The choice of decoding will affect the iterative distillation behaviour. The decoding specified by the logical operators in Tab. (2.2) is the canonical one, though we found the behaviour was simplified by following each iterate with the following additional correction Clifford unitaryUc:
Uc= ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 1 ω ω ω2 ω ω2 ω2 ω2 ω ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ , (2.25)
where this maps the qutrit density operatorρ(α) ≡ ρ(α1,0, α0,1, α1,1, α1,2)such that:
g1= σ1,0 σ0,1 σ0,−1 σ−1,0 σ0,0 g2= σ0,0 σ1,0 σ0,1 σ0,−1 σ−1,0 g3= σ−1,0 σ0,0 σ1,0 σ0,1 σ0,−1 g4= σ0,−1 σ−1,0 σ0,0 σ1,0 σ0,1 σL1,0= σ0,1 σ0,1 σ0,1 σ0,1 σ0,1 σL 0,1= σ1,0 σ1,0 σ1,0 σ1,0 σ1,0
Table 2.3: The stabilizer generators of the five-qudit code and the qutrit logical operators expressed in theσj,k notation.
Without this corrective Clifford one observes a cycling behaviour throughout the distillation process, we will discuss this behaviour at the end of Sec. (2.4.3).
We will now compute the exact distillation map on the Bloch components after a single round of distillation. We start by expressing the stabilizer generators and the logical operators of theJ5, 1, 3K3in
terms of theσj,k operators as shown in Tab. (2.3). The input to the distillation protocol is five identical
copies of a qutrit stateρ(α). Using Eq. (2.21), the input state is expressed as: ρ(α)⊗5= 1
35 ∑
(j,k)∈Z5 3
αj1...j5,k1...k5σj1...j5,k1...k5, (2.27)
Using Eq. (2.22), a successful measurement of the four stabilizer generators with outcome ‘+1’ corre- sponds to the following projector:
Π = 1 34 ∑ q∈Z4 3 gq1 1 g q2 2 g q3 3 g q4 4 . (2.28)
Substituting the stabilizer generators in Tab. (2.3) into the above expression and using the composition law in Eq. (2.4), the projector becomes
Π = 1 81q∑∈Z4 3 (σ(q 1−q3),(−q4)⊗σ(q2−q4),(q1)⊗σ(q3),(−q1+q2) ⊗σ(−q1+q4),(−q2+q3)⊗σ(−q2),(−q3+q4)). (2.29)
We can simplify the notation of the above expression by using Eq. (2.5), which removes the tensor product sign: Π = 1 81q∑∈Z4 3 σ(q1−q3)1(q2−q4)2(q3)3(−q1+q4)4(−q2)5,(−q4)1(q1)2(−q1+q2)3(−q2+q3)4(−q3+q4)5. (2.30)
The distillation map in Eq. (2.24) can be put into a simpler form: αoutj,k = tr(Πρ⊗nΠ†σ¯† j,k) tr(ρ⊗nΠ) = tr(ρ⊗nΠ¯σ−j,−k) tr(ρ⊗nΠ) . (2.31)
where in the last step Eq. (2.13),Π = Π†, [¯σ
j,k, Π] = 0 and the cyclic property of the trace were used.
The remaining task is to substitute Eqs. (2.27) and (2.30) into Eq. (2.31) to calculate the distillation map on the Bloch components.
Let us start by evaluating tr(ρ⊗5Π). Recall that all the σj,k operators are traceless except for the
identity operatorσ0,0. Therefore, the only terms that will survive in tr(ρ⊗5Π) are the coefficients of the
identity operator. We get the identity operator inρ⊗5Π when the σj,k operators inρ⊗5and theσj′,k′ in
Π have the opposite subscripts (i.e. σj,kσj′,k′ = σ0,0 if and only ifj = −j′ andk = −k′). As a result,
tr(ρ⊗5Π) will be the sum of all the Bloch components that are the coefficient of the σj,koperators such
that the subscripts (j, k) are the negative of the subscripts in Eq. (2.29). In fact, since the summation is over all the elements of the ring Z43it is possible to multiply all the subscripts by (−1) without changing the actual value of the summation. Hence, tr(ρ⊗5Π) can be compactly expressed as follows
tr(ρ⊗5Π) = 1 81q∑∈Z4
3
α(q1−q3)1(q2−q4)2(q3)3(−q1+q4)4(−q2)5,(−q4)1(q1)2(−q1+q2)3(−q2+q3)4(−q3+q4)5. (2.32) In a similar way, we can express tr(ρΠ¯σ−j,−k)for all four logical operators. For example, in the case of evaluating the output Bloch componentαout1,0, Eq. (2.31) becomes
αout1,0 = tr(ρ⊗5Π¯σ−1,0) tr(ρ⊗5Π) , (2.33) with tr(ρ⊗5Π¯σ−1,0)given by tr(ρ⊗5Π¯σ−1,0) = 1 81q∑∈Z4 3 α(q1−q3)1(q2−q4)2(q3)3(−q1+q4)4(−q2)5,(−q4+1)1(q1+1)2(−q1+q2+1)3(−q2+q3+1)4(−q3+q4+1)5. (2.34)
We have evaluated the expressions for the four output Bloch components. However, writing them out in terms ofαj,knotation is cumbersome. Therefore, for clarity, we will relabel the four qutrit Bloch
components as follows (α1,0, α0,1, α1,1, α1,2) ≡ (A, B, C, D). For example, tr(ρ⊗5Π) is given in Eq. (2.35), where the subscriptr represent the number of the distillation rounds with r = 0 corresponding to the initial input state.
tr (ρ⊗5Π) =1
81(1 + 10 (∣Ar∣
2
+ ∣Dr∣2) (∣Br∣2+ ∣Cr∣2) +5(Br2A∗rCr∗2+Dr2A∗2r B∗r+
Dr(A2rDrCr∗+Br2Cr2) +Br∗2(ArCr2+Cr∗2D∗r) +D∗2r (A2rBr+CrA∗2r ))). (2.35) Furthermore, it can be shown that the resultant expressions for the four output Bloch components can compactly be expressed in terms of the following single function:
F (A, B, C, D) =1 81(B
5
r+10Br(DrA∗r+Br∗) (ArCr∗+CrDr∗) +5(ArCr2∣Ar∣2+Dr2(ArB∗2r +Cr) + A∗2r (Br∗2Dr∗+Cr∗) +Dr∗2(A2r+DrCr∗2) + ∣Cr∣4B∗r))/tr (ρ⊗5Π) . (2.36) Based on this function the distillation map can be expressed as
Ar+1=F (A, B, C, D), (2.37)
Br+1=F (B∗, A, D, C∗), (2.38)
Cr+1=F (A∗, C, B, D), (2.39)
Dr+1=F (B∗, D∗, A∗, C). (2.40)
These four expressions represents the complete distillation map, as the remaining four components are simply the complex conjugates of these expressions. However, the above expressions do not incor- porate the additional corrective CliffordUc. We need to ensure that the map in Eq. (2.26) is applied after
every iteration. This can easily be achieved in our formalism by the appropriate relabelling as follows:
Ar+1=F (D∗, C, B∗, A), (2.41)
Br+1=F (C∗, D∗, A, B), (2.42)
Cr+1=F (D, B∗, C, A), (2.43)
which is the corrected distillation map.
In order to calculate the fixed points of this map analytically, one would have to solve the above simultaneous complex multi-variable polynomials of order5. It is known from the famous Abel-Ruffini theorem that there is no algebraic solution for a general polynomial of order five or above. Therefore, the best way to discover the fixed points of the distillation is through numerical means. We started with initial statesρ(A, B, C, D) for certain Bloch components and computed the above expressions for a number of iterations, and observed whether there is a convergence toward a fixed point. If a fixed point corresponds to a non-stabilizer pure state, then it is a magic state. We identify two qualitatively different families of magic states. Firstly, those that are in the Hadamard plane, which satisfyHρH†=ρ.
Secondly, we investigate the distillation of an interesting set of states outside the Hadamard plane. Each of these studies has its own merits. All quantum states can be mapped onto the Hadamard plane and so this is the study of most generic value.