PhyslibAlpha.Relativity.PauliMatrices.Anticommuting

Let $\iota$ be an index set. A family of $2 \times 2$ complex matrices $\tau: \iota \to \text{Mat}(2, \mathbb{C})$ is said to be anticommuting if for all $a, b \in \iota$, the relation $$\tau_a \tau_b + \tau_b \tau_a = 2 \delta_{ab} I$$ holds, where $I$ is the $2 \times 2$ identity matrix and $\delta_{ab}$ is the Kronecker delta (equal to $1$ if $a = b$ and $0$ otherwise). This implies that each matrix in the family squares to the identity ($\tau_a^2 = I$) and any two distinct matrices anticommute ($\tau_a \tau_b = -\tau_b \tau_a$).

Let $\iota$ be an index set and $\tau: \iota \to \text{Mat}(2, \mathbb{C})$ be an anticommuting family of $2 \times 2$ complex matrices. For any $a \in \iota$, the square of the matrix $\tau_a$ is the identity matrix $I$, i.e., $\tau_a^2 = I$.

Let $\tau: \iota \to \text{Mat}(2, \mathbb{C})$ be an anticommuting family of $2 \times 2$ matrices, which by definition satisfies $\tau_a \tau_b + \tau_b \tau_a = 2 \delta_{ab} I$ for all $a, b \in \iota$. For any two distinct indices $a, b \in \iota$ (where $a \neq b$), the matrices satisfy the anticommutation relation $\tau_b \tau_a = -(\tau_a \tau_b)$.

For an anticommuting family of $2 \times 2$ complex matrices $\tau$ indexed by $\iota$, the trace of the product of any two matrices $\tau_a$ and $\tau_b$ is given by $$\text{tr}(\tau_a \tau_b) = 2\delta_{ab}$$ where $\delta_{ab}$ is the Kronecker delta (equal to $1$ if $a = b$ and $0$ otherwise). This means the trace is $2$ when $a = b$ and $0$ when $a \neq b$.

Let $\{\tau_k\}_{k \in \iota}$ be an anticommuting family of $2 \times 2$ complex matrices satisfying the relation $\tau_a \tau_b + \tau_b \tau_a = 2 \delta_{ab} I$ for all $a, b \in \iota$. If there exist two distinct indices $a, b \in \iota$ such that $a \neq b$, then the trace of the matrix $\tau_a$ is zero ($\text{tr}(\tau_a) = 0$).

Let $\iota$ be a finite index set with at least two elements. Let $\{\tau_a\}_{a \in \iota}$ be an anticommuting family of $2 \times 2$ complex matrices such that $\tau_a \tau_b + \tau_b \tau_a = 2 \delta_{ab} I$ for all $a, b \in \iota$, where $I$ is the $2 \times 2$ identity matrix and $\delta_{ab}$ is the Kronecker delta. Then the set of matrices $\{I\} \cup \{\tau_a\}_{a \in \iota}$ is linearly independent over $\mathbb{C}$.

Let $\iota$ be a finite index set and $\{\tau_i\}_{i \in \iota}$ be a family of $2 \times 2$ complex matrices such that they satisfy the anticommutation relation $\tau_a \tau_b + \tau_b \tau_a = 2 \delta_{ab} I$ for all $a, b \in \iota$, where $I$ is the identity matrix and $\delta_{ab}$ is the Kronecker delta. Then the number of elements in the family is at most 3, i.e., $|\iota| \leq 3$.

The family of Pauli matrices $\sigma_1, \sigma_2, \sigma_3$ is anticommuting. That is, for any indices $a, b \in \{1, 2, 3\}$, the matrices satisfy the relation $$\sigma_a \sigma_b + \sigma_b \sigma_a = 2 \delta_{ab} I$$ where $I$ is the $2 \times 2$ identity matrix and $\delta_{ab}$ is the Kronecker delta. This implies that each Pauli matrix squares to the identity ($\sigma_a^2 = I$) and any two distinct Pauli matrices anticommute ($\sigma_a \sigma_b = -\sigma_b \sigma_a$).

The maximum size of an anticommuting family of $2 \times 2$ complex matrices is 3. More precisely, let $S$ be the set of natural numbers $n$ such that there exists a family of $n$ matrices $\{\tau_a\}_{a=1}^n \in \text{Mat}(2, \mathbb{C})$ satisfying the anticommutation relation $$\tau_a \tau_b + \tau_b \tau_a = 2 \delta_{ab} I$$ for all $a, b \in \{1, \dots, n\}$, where $I$ is the $2 \times 2$ identity matrix and $\delta_{ab}$ is the Kronecker delta. Then 3 is the greatest element of $S$. This implies that while the three Pauli matrices $\sigma_1, \sigma_2, \sigma_3$ form such a family, no family of four or more such matrices exists.