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This question is a follow-up to the previous QCSE question: "Are qudit graph states well-defined for non-prime dimension?". From the question's answer, it appears that there is nothing wrong in defining graph states using $d$-dimensional qudits, however, it seems that other definitional aspects of graph-states do not similarly extend to non-prime dimension.

Specifically, for qubit graph states, one key aspect to their prevalence and use is the fact that: any two graph states are local Clifford equivalent if and only if there is some sequence of local complementations that takes one graph to the other (for simple, undirected graphs). Needless to say, this is an incredibly useful tool in analyses of quantum error correction, entanglement and network architectures.

When considering $n$-qudit graph states, the equivalent graph is now weighted with adjacency matrix $A \in \mathbb{Z}_d^{n \times n}$, where $A_{ij}$ is the weight of edge $(i,j)$ (with $A_{ij}=0$ indicating no edge exists). In the qudit case, it was shown that LC equivalence can similarly be extended by the generalisation of local complementation ($\ast_a v$) and inclusion of an edge multiplication operation ($\circ_b v$), where: \begin{align} \ast_a v &: A_{ij} \mapsto A_{ij} + aA_{vi}A_{vj} \quad \forall\;\; i,j \in N_G(v), \;i \neq j \\ \circ_b v &: A_{vi} \mapsto bA_{vi} \quad\forall\;\; i \in N_G(v), \end{align} where $a, b = 1, \ldots, d-1$ and all arithmetic is performed modulo $p$.

Graphically, this is represented by the following operations (reproduced from Ref. 2):

However, if the graph state is defined on qudits of non-prime dimension, then we can see these operations (seem to) fail to represent LC-equivalence.

For example, take the qudit state $|G\rangle$ depicted the graph $G$ in Fig. 1, defined for qudit dimension $d=4$, and let $x=y=z=2$, such that $A_{12}=A_{13}=A_{14}=2$. In this case performing $\circ_2 1$ then $A_{1i} \mapsto 2 \times 2 = 4 \equiv 0 \bmod 4 \;\forall\; i$, and hence qudit $1$ is disentangled from all other qudits using only local operations. Clearly this is wrong and occurs because of the problem of zero divisors as mentioned in the previous questions' answer.

My question is: is there any set of graph operations that properly represent local Clifford equivalence for qudit graph states of non-prime dimension?

Note: I am primarily interested in operations that directly apply to a state's representation as a single weighted graph, rather than possible decompositions into multiple prime-dimensional graph states, as suggested in Sec. 4.3 of "Absolutely Maximally Entangled Qudit Graph States".

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  • $\begingroup$ Since you created the new tag graph-states, could you please write the tag wiki for it? Go here. Thank you. $\endgroup$ – Sanchayan Dutta Jul 5 '18 at 8:09
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It is incorrect to use modulo arithmetic in this context. Instead finite field arithmetic should be applied. In $\textrm{GF}(4) = \{0, 1, x, x^2\}$ where $x^2 = x + 1$ and conjugation of $a$ is defined as $\bar{a} = a^2$.

Addition, multiplication and conjugation tables are then as follows:

enter image description here

In this picture we have $0 \equiv 0$, $1 \equiv 1$, $2 \equiv x$, and $3 \equiv x^2$ such that $2 \times 2 = 3$ and so the apparent inconsistency does not occur.

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  • $\begingroup$ What definition of "2" are you using here? Absent any other convention for $\mathbb F = GF(4)$, I would presume $2 := 1_{\mathbb F} + 1_{\mathbb F} = 0_{\mathbb F} =: 0$, in which case $2\times 2 = 0$. $\endgroup$ – Niel de Beaudrap Jul 19 '18 at 11:11
  • $\begingroup$ I have added a clarifying edit :) $\endgroup$ – SLesslyTall Jul 19 '18 at 11:22

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