higher dimensional Pauli matrices?

Question

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 $\det(M) = u^4 + v^4 + x^4 + y^4 + mess$

Answer 1

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 $\{q_i,q_j,q_k\}=6\delta_{ijk}$.

Answer 2

Pauli matrices are essentially 4-dimensional split-complex numbers. Split-complex numbers can be only $2^n$-dimensional.

Particularly, the 8-dimensional split-complex space can be represented with real matrices as:

$\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right);\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{array} \right);\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{array} \right);\left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{array} \right);$ $\left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array} \right);\left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ \end{array} \right);\left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \\ \end{array} \right);\left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{array} \right).$

Alternatively, you can take a look at Triplex numbers, but they are not the same thing.

Answer 3

Since the original post I now think the cross terms are desirable, and not something to be eliminated. Consequently, I have settled on the below set of matrices (along with the identity matrix): $$ T_1 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} T_2 = \begin{bmatrix} 0 & 0 & w^2 \\ w & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} T_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & w & 0 \\ 0 & 0 & w^2 \end{bmatrix} T_4 = \begin{bmatrix} 0 & 0 & w \\ w^2 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} $$ And the second set are: $$ \tau_1 = T_2 T_3 = \begin{bmatrix} 0 & 0 & w \\ w & 0 & 0 \\ 0 & w & 0 \end{bmatrix} \tau_2 = T_3 T_1 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & w \\ w^2 & 0 & 0 \end{bmatrix} \tau_3 = T_1 T_2 = \begin{bmatrix} w & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & w^2 \end{bmatrix} \tau_4 = \begin{bmatrix} 0 & w^2 & 0 \\ 0 & 0 & w \\ 1 & 0 & 0 \end{bmatrix} $$ with the properties: $$r = p_1 T_1 + p_2 T_2 + p_3 T_3 + p_4 T_4$$ and $$r^3 = J_3(p_1, p_2, p_3) + p_4^3$$ And the second set have similar properties: $$s = q_1 \tau_1 + q_2 \tau_2 + q_3 \tau_3 + q_4 \tau_4$$ and $$s^3 = J_3(q_1, q_2, q_3) + q_4^3$$ where $$J_3(a,b,c) = a^3 + b^3 + c^3 - 3 a b c$$ is the determinant of a 3x3 circulant matrix. And $w$ is the principle third root of unity with the standard property $w^3 = +1$. The 3x3 circulant matrix is given by $$ C = \begin{bmatrix} a & b & c \\ c & a & b \\ b & c & a \end{bmatrix} $$