# higher dimensional Pauli matrices?

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

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 $\{q_i,q_j,q_k\}=6\delta_{ijk}$.
• that makes sense. Though you are missing a $\delta_{ijk}I$ term. I find the {a,b,c} anticommutator version easier to verify in mathematica though. – Garry Apr 5 '16 at 11:57
• OK. Your equation implies: $q_1q_2q_3 = q_4$ So use my $a_1, a_2, a_3$ from above, and try this candidate as the fourth matrix: $a_5 = a_1a_2a_3$ Then set $X = u*a_1 + v*a2 + x*a3 + y*a5$ and take the determinant: $$Det(X) = u^3 + v^3 + x^3 + y^3 -3uvy$$ So we still have the cross term. – Garry Apr 7 '16 at 9:42