I'm trying to find $3 \times 3$ matrices with some similarity to Pauli matrices.
I have some candidates, but they are not perfect. I'm not sure if perfect versions exist, and that is my question.
Recall for Pauli matrices $p_k$:
$\{p_i,p_j\} = p_ip_j + p_jp_i = 2 \delta_{ij}$
I'm looking for matrices that satisfy:
$\{a_i,a_j,a_k\} = 3! \delta_{ijk}$
where $\{A,B,C\}$ is the sum over all permutations of three symbols:
$\{A,B,C\} = ABC + BCA + CAB + ACB + CBA + BAC$
My candidate matrices are:
$a_1 = \begin{bmatrix}
0 & 0 & 1\\
1 & 0 & 0\\
0 & 1 & 0\\
\end{bmatrix}$
$a_2 = \begin{bmatrix}
0 & 0 & a\\
b & 0 & 0\\
0 & c & 0\\
\end{bmatrix}$
$a_3 = \begin{bmatrix}
0 & a & 0\\
0 & 0 & b\\
c & 0 & 0\\
\end{bmatrix}$
$a_4 = \begin{bmatrix}
a & 0 & 0\\
0 & b & 0\\
0 & 0 & c\\
\end{bmatrix}$
where $a,b,c$ are the three third roots of unity (and have the property $a + b + c = 0$).
And of course, unlike Pauli matrices, the eigenvalues of these matrices are not real (they are $a,b,c$), which is expected since they are permutation matrices with $a^3 = I$.
Now, they have the desired property of:
$\{a_1,a_1,a_1\} = 3\\
\{a_1,a_1,a_2\} = 0\\
\{a_1,a_2,a_3\} = 0\\
\{a_1,a_3,a_4\} = 0\\ \cdots$
However:
$\{a_2,a_3,a_4\} = -3I$
So my question is, are there a set of 4 matrices that satisfy
$\{a_i,a_j,a_k\} = 3! \delta_{ijk}$
We have 3 such matrices $a_1,a_2,a_3$, but do there exist 4?
I'm 90% sure the answer is no, but I'd like to be 100%.
And we can extend the problem to higher dimensions:
For example, the $d = 4$ case. Define:
$\{A,B,C,D\}$ as the sum over all permutations of four symbols.
How many matrices can we find that satisfy:
$\{a_i,a_j,a_k,a_l\} = 4! \delta_{ijkl}$
It suspect it might be 3 for all higher d.
Now, where did this problem come from? Consider:
$M_1 = u*a_1 + v*a_2 + w*a_3$ such that $M_1^3 = (u^3 + v^3 + w^3)I$
And $M_2 = u*a_1 + v*a_2 + x*a_3 + y*a_4$ such that $M_2^3 = (u^3 + v^3 + x^3 + y^3)I$
And their determinants:
$\det(M_1) = u^3 + v^3 + w^3\\
\det(M_2) = u^3 + v^3 + x^3 + y^3 -3vxy$
So an alternative definition of the problem, do there exist matrices, not equal to the identity matrix, $a_1,a_2,a_3,a_4$ such that $\det(M_2)$ doesn't have the cross term? Similarly for higher $d$, where the determinant gets even messier! eg: $d = 4, M = u*b_1 + v*b_2 + x*b_3 + y*b_4$ and $\det(M) = u^4 + v^4 + x^4 + y^4 + mess$
Best Answer
Not a full answer but a suggestion that is a bit too long to fit into a comment.
The most important property about Pauli matrices is not the one you give but the following $$ p_ip_j=\sum_k\varepsilon_{ijk}\;p_k$$ where $\varepsilon_{ijk}=\pm1$ if $(i,j,k)$ is a permutation of $(1,2,3)$. The sign $\pm1$ is the signature of the permutation. In all other cases, $\varepsilon_{ijk}=0$. It is easy to check that the property $\{p_i,p_j\}=2\delta_{ij}$ is a direct consequence of this more fundamental property.
So you should try to look for permutation matrices satisfying $$ q_iq_jq_k=\varepsilon_{ijk\ell}q_\ell$$ and they will immediately verify $\{q_i,q_j,q_k\}=6\delta_{ijk}$.