I am looking to take a classical non-negative real valued network and generalize it to the quantum case for processing. A network is given by an adjacency matrix, essentially edge weights $e_{ij}$ for $1 \le i < j \le n$ where $n$ is the number of vertices where $e_{ij} \in \mathbb{R}_{\ge 0}$. I'd like to promote $e_{ij}$ to an element of the Bloch sphere $\mathbb{S}^3$, so $e_{ij} \in \mathbb{C}^2$ such that $||e_{ij}||^2=|e_{ij}^0|^2+|e_{ij}^1|^2=1$, or in other words $e_{ij} = e_{ij}^0|0\rangle + e_{ij}^1|1\rangle$.

In addition, these networks are unlabeled, meaning that for a permutation $\sigma \in \Sigma_n$ which permutes nodes (inducing an edge permutation) and corresponding permutation matrix $P_{\sigma}$, two networks are equivalent under the group action $\sigma \cdot \{e_{ij}\} \sim \{e_{ij}\}$, or in adjacency matrix terms $P_{\sigma}AP_{\sigma}^{-1} \sim A$. One can represent an unlabeled network $e$ as an equivalence class, namely $\Sigma_n \cdot e$. That is, we take all $n!$ permutations of the network.

Generalizing to the quantum case, we obtain a lift $\left(\mathbb{S}^3\right)^N \rightarrow \mathbb{R}_{\ge 0}^{N} \rightarrow \{0,1\}^N$. The first map is $e_{ij} \mapsto |e_{ij}^k|^2$ where $k=0$ or $k=1$. The second map is the semi-ring morphism $\mathbb{R}_{\ge 0} \rightarrow \{0,1\}$ where $N = \binom{n}{2}$ is the number of edges of a complete graph.

I'm using Qiskit and have successfully sampled a three qubit circuit.

Is this how people usually represent quantum networks, by putting qubits (one imagines a 3-sphere stretched between two nodes) on the edges?

  • $\begingroup$ Such kind of representation is unfamiliar to me. If the qubits are edges, what do the vertices represent? $\endgroup$ Sep 18, 2023 at 14:53
  • $\begingroup$ Clasically, edges represent "connection exists" or "connection strength". I suppose these are "connection unobserved" where the observation is quantum mechanical. $\endgroup$ Sep 18, 2023 at 14:58
  • $\begingroup$ If $e_{i,j}$ represents the state of some qubit, then what is $i$ and what is $j$? Instead, if $e_{i,j}$ represents a connection (between qubits $i$ and $j$), then how can $e$ as well be a qubit? $\endgroup$ Sep 18, 2023 at 15:03
  • $\begingroup$ $i$, $j$ are labels/addresses, the same way that if you have $N$ qubits then you can label them $f_k$ for $1 \le k \le N$ and you have the product state $\mathcal{V}^{\otimes N}$ and your states are linear combinations of $2^N$ states $|f_1 \ldots f_N \rangle$. $k$ could be shorthand for the location of the physical bits. We're always modeling something. Classically, the "labels" $i$, $j$ could be labeling intersections of a road network. I suppose here the totally pure states $|0\rangle$ and $|1\rangle$ would touch $i$, $j$. $\endgroup$ Sep 18, 2023 at 15:14


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.