Distinguished three-qubit ‘magicity’ via automorphisms of the split Cayley hexagon

Abstract

Disregarding the identity, the remaining 63 elements of the generalized three-qubit Pauli group are found to contain 12096 distinct copies of Mermin’s magic pentagram. Remarkably, 12096 is also the number of automorphisms of the smallest split Cayley hexagon. We give a few solid arguments showing that this may not be a mere coincidence. These arguments are mainly tied to the structure of certain types of geometric hyperplanes of the hexagon. It is further demonstrated that also an $(18_{2}, 12_{3})$-type of magic configurations, recently proposed by Waegell and Aravind (J Phys A Math Theor 45:405301, 2012), seems to be intricately linked with automorphisms of the hexagon. Finally, the entanglement properties exhibited by edges of both pentagrams and these particular Waegell–Aravind configurations are addressed.

Appendix

We shall show here why no edge of either a pentagram or a WA-configuration can feature a $W$-type of entanglement. In fact, we have the more general following statement:

Assume that $u_1$ and $u_2$ are two commuting operators of the generalized three-qubit Pauli group and let $v$ be a common eigenvector, then $v$ is either non-entangled, partially entangled or exhibits entanglement of a GHZ-type.

The proof can be carried out in two steps:

Operators involving the identity at least once

Let $u=U_1U_2U_3$ be an element of the generalized Pauli group where $U_i\in \{I, X, Y,Z\}$ and not all the three entries are simultaneously equal to the identity. There are $36$ elements with at least one of the $U_i$ being equal to $I$. Assume $U_1=I$ and denote by $E_u ^\lambda$ the eigenspace of the operator $u$ for the eigenvalue $\lambda$. Then the space of three qubits decomposes as $E_u^1\oplus E_u^{-1}$ and bases of the eigenspaces are given by :

$E_u^1$	$E_u^{-1}$
$|000\rangle +|0U_2(|0\rangle )U_3(|0\rangle )\rangle$	$|000\rangle -|0U_2(|0\rangle )U_3(|0\rangle )\rangle$
$|100\rangle +|1U_2(|0\rangle )U_3(|0\rangle )\rangle$	$|100\rangle -|1U_2(0\rangle )U_3(|0\rangle )\rangle$
$|01U_3(|1\rangle )\rangle +|0U_2(|1\rangle )1\rangle$	$|01U_3(|1\rangle )\rangle -|0U_2(|1\rangle )1\rangle$
$|11U_3(|1\rangle )\rangle +|1U_2(|1\rangle )1\rangle$	$|11U_3(|1\rangle )\rangle -|1U_2(|1\rangle )1\rangle$

An eigenvector $v$ can therefore be written as

$$\begin{aligned} v&= (\alpha |0\rangle +\beta |1\rangle )\otimes (|00\rangle \pm |U_2(|0\rangle )U_3(|0\rangle )\rangle )+(\gamma |0\rangle \\&+\delta |1\rangle )\otimes (|1U_3(|1\rangle )\rangle \pm |U_2(|1\rangle )1\rangle ) \end{aligned}$$

with $\alpha ,\beta , \gamma , \delta \in \mathbb C$. Such a vector is either non entangled, partially entangled or of GHZ-type. The same reasoning works if we consider $u=U_1IU_3$ or $u=U_1U_2I$.

Operators which do not involve the identity

There are $27$ operators $u_i=U^i _1U^i _2U^i _3$ such that $U^i_j\in \{X,Y,Z\}$. Let $u_1=U_1 ^1 U_2^1U_3^1$ and $u_2=U_1^2U_2^2 U_3^2$ two such operators which commute. Because of the commuting assumption we necessarily have a $j\in \{1,2,3\}$ such that $U_j^1=U_j^2$. Without loss of generality let us assume that $j=1$ and let $v$ be a common eigenvector of $u_1$ and $u_2$. The vector $v$ is also an eigenvector for $u=u_1u_2$. But by composition $u=u_1u_2=(U_1^1\times U_1^2)\otimes (U_2^1\times U_2^2)\otimes (U_3^1\times U_3^2)$, and $U_1^1\times U_1 ^2=I$ by hypothesis (the notation $\times$ stands for the usual product of matrices). Therefore $u=u_1u_2=U_2^3U_3^3$ where $U_2^3= U_2^1\times U_2^2$ and $U_3^3=U_3^1\times U_3^2$. In other words $u=u_1u_2$ is a Pauli operator of the three qubits system involving once the identity. By the previous step of our argument its eigenvector $v$ is either non-entangled, partially entangled or of GHZ-type.