Higher-dimensional performance of port-based teleportation

Port-based teleportation (PBT) is a variation of regular quantum teleportation that operates without a final unitary correction. However, its behavior for higher-dimensional systems has been hard to calculate explicitly beyond dimension d = 2. Indeed, relying on conventional Hilbert-space representations entails an exponential overhead with increasing dimension. Some general upper and lower bounds for various success measures, such as (entanglement) fidelity, are known, but some become trivial in higher dimensions. Here we construct a graph-theoretic algebra (a subset of Temperley-Lieb algebra) which allows us to explicitly compute the higher-dimensional performance of PBT for so-called “pretty-good measurements” with negligible representational overhead. This graphical algebra allows us to explicitly compute the success probability to distinguish the different outcomes and fidelity for arbitrary dimension d and low number of ports N, obtaining in addition a simple upper bound. The results for low N and arbitrary d show that the entanglement fidelity asymptotically approaches N/d2 for large d, confirming the performance of one lower bound from the literature.


Supplementary Derivation S1
If the teleportation is successful, Bob will discard all of his ports except the ith port B i which is the unknown quantum state σ in C . We derive the mathematical representation of the entanglement fidelity for PBT according to Satoshi Ishizaka and Tohya Hiroshima's initial article 1 .
In the last step, we use the notation: , and the entanglement fidelity should be: Term Φ DC |Φ DB i can be translated into: With Eq. (3), Eq. (2) can be translated into: In the last step, we use

Supplementary Derivation S2
For convenience, we will denote σ i AC as σ i in the following paragraphs. For N=2, we have obtained the following equation rules in the paper (denote Swap i j as S i j ).
When N = 2, ρ = σ 1 + σ 2 . We express √ ρ as the combination of the four states like Eq. (6), and use relation ρ = ( √ ρ) 2 to solve α and γ. As σ 1 and σ 2 share the same position in the algebra, we give them the same coefficient and so do S 12 σ 1 and S 12 σ 2 .

Supplementary Derivation S3
When we consider the linear relation between diagonal matrices, it equals to consider the column vector (C ρ ) formed by the diagonal elements of the matrices. If we assume all the none-zero different diagonal elements of matrices D ρ (the corresponding diagonal matrices of ρ) is {r 1 , r 2 , r 3 , . . . , r n }, then we can prove {C ρ , (C ρ ) 2 , (C ρ ) 3 , . . . , (C ρ ) n } is independent of each other and (C ρ ) n+1 is a linear combination of this set. We denote the matrix {C ρ , (C ρ ) 2 , (C ρ ) 3 , . . . , (C ρ ) n } by δ , and expand its determinant in Eq. (10). The determinant in the second step of Eq. (10) is the Vandermonde determinant 2 of order n which equals to ∏ 1≤ j<i≤n (r i − r j ). As all the elements r i are none-zero and different, the final result of Eq. (10) must be none-zero. Thus, matrix δ is full rank and {C ρ , (C ρ ) 2 , (C ρ ) 3 , . . . , (C ρ ) n } must be independent of each other.
Next, we will prove (C ρ ) n+1 is a linear combination of the lower power of (C ρ ). As matrix δ is full rank, Eq.(11) must have one none-zero solution which means (C ρ ) n+1 is the linear combination of set {(C ρ ) i } (1 ≤ i ≤ n). As column vector C ρ equals to the diagonal form of matrix ρ, we can replace it by ρ and obtain expression ρ n+1 = a 1 ρ + a 2 ρ 2 + . . . + a n ρ n which equals to the minimal polynomial ρ n = a 1 + a 2 ρ + . . . + a n ρ n−1 .

Supplementary Derivation S4
We will give more details about the calculations method we use for larger N. Although we use N = 2 as an example to explain the method, the method is general. We firstly calculate ρ, ρ 2 , ρ 3 , . . . based on the permutation theory. We find all kinds of conjugacy classes already appear in ρ 2 , which means the algebra is close at ρ 2 . Thus, we can solve ρ 3 = a 1 ρ + a 2 ρ 2 based on their permutation expressions and the solution is a 1 = 1 − d 2 , a 2 = 2d. Therefore, the minimal polynomial of ρ is ρ 2 = 1 − d 2 + 2dρ whose solutions are ρ's different nonzero eigenvalues d − 1 and d + 1. Using these eigenvalues, we can built up the support matrix of ρ, ρ 2 and directly calculate the support matrix of ρ −1/2 . According to the support matrix, ρ −1/2 = b 1 ρ + b 2 ρ 2 while b 1 and b 2 are expressions about d. As long as we insert the permutation expression of ρ back into this relation we can get the permutation expression of ρ −1/2 . Finally, we can calculate ρ −1/2 σ 1 ρ −1/2 σ 1 and its trace based on the permutation expression. The result is (d + √ d 2 − 1)/2d which agrees with the result in the Supplementary Derivation S2. Because all the calculation is based on the permutation expression, so the essential part is to figure out how the permutation states interact with each other. As there is an entanglement element in the permutation, the two permutation cannot directly multiply with each other like traditional permutation. We will denote Permutation i as P i and explain its special multiply rules.
Combining with the discussion in the paper, we know (P i σ i ) * (P j σ j ) is the general form of the multiplication which equals to P k σ j . To figure out P k , we only need change P j into a new permutation and directly multiply P i with the new P j like traditional permutation. To figure out the new P j , we obtain the following three rules through analysing the graphical algebra. Notice, every parenthesis represents a circle of all the numbers in it; and A, B,C, D represent a series of number.
(1) If i and j are in different circles, then we only need to find which number will arrive i and connect it to the number after j, like Eq. (12).
(2) If i and j are in the same circle, then cut out the part from i to j and make it a separate circle and remind all the other unchanged, like Eq. (13).
(3) If the number before i is j, then add d, like Eq. (14). In fact, this is the condition that the graphical algebra will produce a circle.
We write all the codes in Mathematica based on the above method. During calculating the minimal polynomial, we use the conjugacy class to simplify the expression of ρ, ρ 2 , . . . , and during calculating the trace of ρ −1/2 σ 1 ρ −1/2 σ 1 we have to return to the full terms format. The number of terms making up the state for N = 2, 3, 4, 5 separately is 4, 18, 96, 600 while the corresponding conjugacy class number is 2, 4, 7, 12. So conjugacy class theory can exponentially simplify our calculation in some steps.