Pauli Matrices
Definition
The following three $2 \times 2$ complex matrices are called the Pauli matrices.
$$ \sigma_{1} = \sigma_{x} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \quad \sigma_{2} = \sigma_{y} = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}, \quad \sigma_{3} = \sigma_{z} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \tag{1} $$
Explanation
By property (c) below, they algebraically form a basis of $\mathfrak{su}(2)$, and physically they are used to describe spin operators.
Properties
(a) They are unitary matrices as well as Hermitian matrices. $I$ is the identity matrix.
$$ \sigma_{i}^{\ast}\sigma_{i} = I, \quad \sigma_{i}^{\ast} = \sigma_{i} $$
Since they are unitary and Hermitian, the following holds.
$$ \sigma_{i}^{2} = I $$
(b) The determinant is $-1$ and the trace is $0$.
$$ \det (\sigma_{i}) = -1, \quad \tr (\sigma_{i}) = 0 $$
Moreover, since the eigenvalues of a Hermitian matrix are real and the determinant is the product of the eigenvalues, the eigenvalues of the Pauli matrices are $\pm 1$.
(c) The Pauli matrices can be expressed as follows.
$$ \sigma_{i} = \begin{bmatrix} \delta_{i3} & \delta_{i1} - i\delta_{i2} \\ \delta_{i1} + i\delta_{i2} & -\delta_{i3} \end{bmatrix} $$
Here $\delta_{ij}$ is the Kronecker delta.
(d) The following holds. The Einstein notation is used from here on.
$$ \sigma_{i}\sigma_{j} = \delta_{ij}I + i\epsilon_{ijk}\sigma_{k} $$
$\epsilon_{ijk}$ is the Levi-Civita symbol.
(e) The following commutation relations hold.
$$ \left[ \sigma_{i}, \sigma_{j} \right] = 2i\epsilon_{ijk} \sigma_{k},\qquad \left\{ \sigma_{i}, \sigma_{j} \right\} = 2\delta_{ij}I $$
Proof
(a)
That they are Hermitian is obvious from $(1)$. They are unitary since the following is satisfied.
$$ \begin{align*} \sigma_{1}^{\ast}\sigma_{1} &= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] \sigma_{2}^{\ast}\sigma_{2} &= \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] \sigma_{3}^{\ast}\sigma_{3} &= \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \end{align*} $$
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(c)
The following holds.
$$ \begin{align*} \begin{bmatrix} \delta_{{\color{red}1}3} & \delta_{{\color{red}1}1} - i\delta_{{\color{red}1}2} \\ \delta_{{\color{red}1}1} + i\delta_{{\color{red}1}2} & -\delta_{{\color{red}1}3} \end{bmatrix}&= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \sigma_{1} \\[1em] \begin{bmatrix} \delta_{{\color{red}2}3} & \delta_{{\color{red}2}1} - i\delta_{{\color{red}2}2} \\ \delta_{{\color{red}2}1} + i\delta_{{\color{red}2}2} & -\delta_{{\color{red}2}3} \end{bmatrix}&= \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} = \sigma_{2} \\[1em] \begin{bmatrix} \delta_{{\color{red}3}3} & \delta_{{\color{red}3}1} - i\delta_{{\color{red}3}2} \\ \delta_{{\color{red}3}1} + i\delta_{{\color{red}3}2} & -\delta_{{\color{red}3}3} \end{bmatrix}&= \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \sigma_{3} \end{align*} $$
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(d)
We use the formula from (c). For convenience, set the following.
$$ u_{i} = \delta_{i1} + i\delta_{i2} \implies \sigma_{i} = \begin{bmatrix} \delta_{i3} & \overline{u_{i}} \\ u_{i} & -\delta_{i3} \end{bmatrix} $$
Then $\sigma_{i}\sigma_{j}$ is as follows.
$$ \sigma_{i}\sigma_{j} = \begin{bmatrix} \delta_{i3} & \overline{u_{i}} \\ u_{i} & -\delta_{i3} \end{bmatrix} \begin{bmatrix} \delta_{j3} & \overline{u_{j}} \\ u_{j} & -\delta_{j3} \end{bmatrix} = \begin{bmatrix} \delta_{i3}\delta_{j3} + \overline{u_{i}}u_{j} & \delta_{i3}\overline{u_{j}} - \overline{u_{i}}\delta_{j3} \\ u_{i}\delta_{j3} - \delta_{i3}u_{j} & u_{i}\overline{u_{j}} + \delta_{i3}\delta_{j3} \end{bmatrix} $$
Here $\overline{u_{i}}u_{j}$ is as follows.
$$ \begin{align*} \overline{u_{i}}u_{j} &= (\delta_{i1} - i\delta_{i2})(\delta_{j1} + i\delta_{j2}) \\ &= \delta_{i1}\delta_{j1} + \delta_{i2}\delta_{j2} + i\left( \delta_{i1}\delta_{j2} - \delta_{i2}\delta_{j1} \right) \\ &= \delta_{i1}\delta_{j1} + \delta_{i2}\delta_{j2} + i\epsilon_{ij3} \end{align*} $$ $$ \begin{align*} u_{i}\overline{u_{j}} &= (\delta_{i1} + i\delta_{i2})(\delta_{j1} - i\delta_{j2}) \\ &= \delta_{i1}\delta_{j1} + \delta_{i2}\delta_{j2} + i\left( \delta_{i2}\delta_{j1} - \delta_{i1}\delta_{j2} \right) \\ &= \delta_{i1}\delta_{j1} + \delta_{i2}\delta_{j2} - i\epsilon_{ij3} \end{align*} $$
The last equality holds by the property of the Levi-Civita symbol $\epsilon_{ij3} = \delta_{i1}\delta_{j2} - \delta_{i2}\delta_{j1}$. Then the $(1,1)$ entry is simplified as follows.
$$ \begin{align*} \delta_{i3}\delta_{j3} + \overline{u_{i}}u_{j} &= \delta_{i1}\delta_{j1} + \delta_{i2}\delta_{j2} + \delta_{i3}\delta_{j3} + i\epsilon_{ij3} = \delta_{ij} + i\epsilon_{ij3} \\ u_{i}\overline{u_{j}} + \delta_{i3}\delta_{j3} &= \delta_{i1}\delta_{j1} + \delta_{i2}\delta_{j2} + \delta_{i3}\delta_{j3} - i\epsilon_{ij3} = \delta_{ij} - i\epsilon_{ij3} \end{align*} $$
Here the property of the Kronecker delta $\delta_{ik}\delta_{jk} = \delta_{ij}$ was used. The off-diagonal entries can be obtained in the same way. Consider the $(1,2)$ entry.
$$ \begin{align*} \delta_{i3}\overline{u_{j}} - \overline{u_{i}}\delta_{j3} &= \delta_{i3}(\delta_{j1} - i\delta_{j2}) - (\delta_{i1} - i\delta_{i2})\delta_{j3} \\ &= \left( \delta_{i3}\delta_{j1} - \delta_{i1}\delta_{j3} \right) - i\left( \delta_{i3}\delta_{j2} - \delta_{i2}\delta_{j3} \right) \\ &= \epsilon_{ij2} + i\epsilon_{ij1} \end{align*} $$
Since the $(2,1)$ entry is the conjugate of the $(1,2)$ entry, $\sigma_{i}\sigma_{j}$ is as follows.
$$ \begin{align*} \sigma_{i}\sigma_{j} &= \begin{bmatrix} \delta_{ij} + i\epsilon_{ij3} & \epsilon_{ij2} + i\epsilon_{ij1} \\ -\epsilon_{ij2} + i\epsilon_{ij1} & \delta_{ij} - i\epsilon_{ij3} \end{bmatrix} \\ &= \begin{bmatrix} \delta_{ij} + i\epsilon_{ijk}\delta_{k3} & \epsilon_{ijk}\delta_{k2} + i\epsilon_{ijk}\delta_{k1} \\ -\epsilon_{ijk}\delta_{k2} + i\epsilon_{ijk}\delta_{k1} & \delta_{ij} - i\epsilon_{ijk}\delta_{k3} \end{bmatrix} \\ &= \delta_{ij}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + i\epsilon_{ijk}\begin{bmatrix} \delta_{k3} & \delta_{k1} - i\delta_{k2} \\ \delta_{k1} + i\delta_{k2} & -\delta_{k3} \end{bmatrix} \\ &= \delta_{ij}I + i\epsilon_{ijk}\sigma_{k} \end{align*} $$
Here, in the second equality, the formula $\epsilon_{ijk}\delta_{k\ell} = \epsilon_{ij\ell}$ was used.
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(e)
Let us swap $i$ and $j$ in (d). Since the Levi-Civita symbol changes sign when two indices are swapped, $\epsilon_{jik} = -\epsilon_{ijk}$, and for the Kronecker delta, $\delta_{ji} = \delta_{ij}$.
$$ \sigma_{j}\sigma_{i} = \delta_{ji}I + i\epsilon_{jik}\sigma_{k} = \delta_{ij}I - i\epsilon_{ijk}\sigma_{k} $$
Subtracting this equation from (d) gives the commutator, and adding them gives the anticommutator; in each case only one term survives.
$$ \begin{align*} \left[ \sigma_{i}, \sigma_{j} \right] &= \sigma_{i}\sigma_{j} - \sigma_{j}\sigma_{i} = \left( \delta_{ij}I + i\epsilon_{ijk}\sigma_{k} \right) - \left( \delta_{ij}I - i\epsilon_{ijk}\sigma_{k} \right) = 2i\epsilon_{ijk}\sigma_{k} \\[1em] \left\{ \sigma_{i}, \sigma_{j} \right\} &= \sigma_{i}\sigma_{j} + \sigma_{j}\sigma_{i} = \left( \delta_{ij}I + i\epsilon_{ijk}\sigma_{k} \right) + \left( \delta_{ij}I - i\epsilon_{ijk}\sigma_{k} \right) = 2\delta_{ij}I \end{align*} $$
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