Are there any 3 or more Hermitian solutions to the problem: $\alpha_i^2=1$, $\{\alpha_i, \alpha_j \}=2$

Question

I’m trying to generate some matrices which are similar to Pauli’s but with the following anti-commutation relation

$$\{\alpha_i, \alpha_j\}=\alpha_i \alpha_j + \alpha_j \alpha_i = 2 \tag{1}$$

And

$$\alpha_i^2=1 \tag{2}$$

2 trivial solutions to this can be immediately spotted: first the scalar number 1 and any unitary matrix and its inverse. The last solution is boring because it would give me only 2 possible solutions.

From $(1)$ we know the solutions should have the form $$\alpha_i=\left(\begin{array}{cc}a&b\\c&-a\end{array}\right)\tag{3}$$ With the condition that $a^2+bc=1$

Using a similar procedure found here I made an initial choice by setting $a=1 , b=c=0$ (Pauli’s $\sigma_z$ matrix) which them, gave me the following solutions

$$\alpha_1 = \left(\begin{array}{cc}1 & 0 \\0 & -1 \\\end{array}\right) \tag{4}$$ Plus $$\alpha_2 = \left(\begin{array}{cc}1 & 1 \\0 & -1 \\\end{array}\right) \tag{5}$$ $$\alpha_3 = \left(\begin{array}{cc}1 & -1 \\0 & -1 \\\end{array}\right) \tag{6}$$ Or $(4)$ plus $$\alpha_2 = \left(\begin{array}{cc}1 & 0 \\1 & -1 \\\end{array}\right) \tag{7}$$ $$\alpha_3 = \left(\begin{array}{cc}1 & 0 \\-1 & -1 \\\end{array}\right) \tag{8}$$

My obvious problem is that these matrices are not Hermitian. I believe that my initial choice led to this result, so I tried several other choices but none seems to work. At most I found 2 solutions. Therefore, my question is:

Are there any 3 or more Hermitian solutions to this problem?

EDIT As said in a previous comment $\alpha_i$, $\alpha_j$ are $n \times n$ matrices, not necessarily $2 \times 2$.

doetoe · Accepted Answer · 2021-06-18T00:26:36.803

Let's start completely abstractly, and consider the abstract algebra over $\mathbb C$ generated by symbols $\sigma_1,\sigma_2,\sigma_3$ subject to your relations: $\{\sigma_i,\sigma_j\} = 2$, in particular $\sigma_i^2 = 1$.

It is not hard to see that this algebra is generated as a vector space by the 8 elements $1, \sigma_1, \sigma_2, \sigma_3, \sigma_1\sigma_2, \sigma_1\sigma_3, \sigma_2\sigma_3, \sigma_1\sigma_2\sigma_3$, all other monomials in the $\sigma_i$ can be reduced to sums of these using the relations. Let's call these elements $e_1,\ldots,e_8$.

Left-multiplication by any element $\tau$ in this algebra is a linear transformation, and the regular representation of this algebra in this basis is the matrix algebra thus obtained. The matrices corresponding to the $\sigma_i$ are an $8\times 8$-example of what you're looking for.

Concretely, what $\sigma_1$ does on the basis is $e_1\mapsto e_2, e_2\mapsto e_1, e_3\mapsto e_5, e_4\mapsto e_6, e_5\mapsto e_3, e_6\mapsto e_4, e_7\mapsto e_8, e_8\mapsto e_7$. In matrix form

$$\sigma_1 = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ \end{pmatrix}$$

Likewise you will get matrices for $\sigma_2$ and $\sigma_3$, and they will satisfy the relations.

For example, for $\sigma_2$ we have $e_1 = 1 \mapsto \sigma_2 = e_3$, and $e_2 = \sigma_1\mapsto \sigma_2\sigma_1 = 2 - \sigma_1\sigma_2 = 2e_1 - e_5$, etc, giving us the matrix form

$$\sigma_2 = \begin{pmatrix}0 & 2 & 1 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & -1 & 0 & 0 & 0\\1 & 0 & 0 & 0 & 2 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 2 & 1 & 0\\0 & -1 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & 0 & -1\\0 & 0 & 0 & 1 & 0 & 0 & 0 & 2\\0 & 0 & 0 & 0 & 0 & -1 & 0 & 0\end{pmatrix}$$

and in the same way we find

$$\sigma_3 = \begin{pmatrix}0 & 2 & 2 & 1 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & -2 & -1 & 0 & 0\\0 & 0 & 0 & 0 & 2 & 0 & -1 & 0\\1 & 0 & 0 & 0 & 0 & 2 & 2 & 0\\0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\0 & -1 & 0 & 0 & 0 & 0 & 0 & -2\\0 & 0 & -1 & 0 & 0 & 0 & 0 & 2\\0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\end{pmatrix}.$$

Note that there may be smaller representations.

In case you'd like to do some computations, here are the matrices in SymPy format:

from sympy import Matrix, eye
I = eye(8)
s1 = Matrix([[0, 1, 0, 0, 0, 0, 0, 0],
             [1, 0, 0, 0, 0, 0, 0, 0],
             [0, 0, 0, 0, 1, 0, 0, 0],
             [0, 0, 0, 0, 0, 1, 0, 0],
             [0, 0, 1, 0, 0, 0, 0, 0],
             [0, 0, 0, 1, 0, 0, 0, 0],
             [0, 0, 0, 0, 0, 0, 0, 1],
             [0, 0, 0, 0, 0, 0, 1, 0]])
s2 = Matrix([[0,  2, 1, 0,  0,  0, 0,  0],
             [0,  0, 0, 0, -1,  0, 0,  0],
             [1,  0, 0, 0,  2,  0, 0,  0],
             [0,  0, 0, 0,  0,  2, 1,  0],
             [0, -1, 0, 0,  0,  0, 0,  0],
             [0,  0, 0, 0,  0,  0, 0, -1],
             [0,  0, 0, 1,  0,  0, 0,  2],
             [0,  0, 0, 0,  0, -1, 0,  0]])
s3 = Matrix([[0,  2,  2, 1,  0,  0,  0,  0],
             [0,  0,  0, 0, -2, -1,  0,  0],
             [0,  0,  0, 0,  2,  0, -1,  0],
             [1,  0,  0, 0,  0,  2,  2,  0],
             [0,  0,  0, 0,  0,  0,  0,  1],
             [0, -1,  0, 0,  0,  0,  0, -2],
             [0,  0, -1, 0,  0,  0,  0,  2],
             [0,  0,  0, 0,  1,  0,  0,  0]])

Note that to find Hermitian matrices more has to be done.

Are there any 3 or more Hermitian solutions to the problem: $\alpha_i^2=1$, $\{\alpha_i, \alpha_j \}=2$

1 Answers1