The Jordan-Wigner transformation turns fermionic creation and annihilation operators into a combination of spin operators. I had a hard time understanding how it works until yesterday I came across the notes in this place. The notes list the definitions and the algebra of the “string+spin” operators. The notes themselves, however, still didn’t make me fully understand why things work like this.
But fortunately, I worked out some Kronecker products (or tensor products) of spin operators when I was developing the fishbonett package. The combination of the matrix forms of the Kronecker products and the notes I mentioned hit the spot!
Let’s start from a very simple case, two spin-less fermions in two states which we label as state $a$ and state $b$. There’re two sets of fermions creation and annihilation operators, $c^\dagger_a, c_a$ and $c^\dagger_b, c_b$. We will use the occupation-number basis $\ket{11},\ket{10}, \ket{01},\ket{00}$ in which the first number is the occupation number in the $a$ site, the second number is the $b$ site. The effects of operators $c_a,c_a^\dagger$ on those basis vectors are
\[\begin{array}{cccc} c_a \ket{00}=0 & c_a \ket{01}=0 & c_a \ket{10}=\ket{00} & c_a \ket{11}=\ket{01}&\\ c_a^\dagger \ket{00}=0 & c_a^\dagger \ket{01}=\ket{11} & c_a^\dagger \ket{10}=0 & c_a^\dagger \ket{11}=0&\\ \end{array} \]
and the results for $c_b,c_b^\dagger$ are
\[\begin{array}{cccc} c_b \ket{00}=0 & c_b \ket{01}=0 & c_b \ket{10}=\ket{00} & c_b \ket{11}= - \ket{10}&\\ c_b^\dagger \ket{00}=\ket{01} & c_b^\dagger \ket{01}=0 & c_b^\dagger \ket{10}=-\ket{11} & c_b^\dagger \ket{11}=0&\\ \end{array} \]
Now we can calculate the matrix elements of the second-quantized operators and give the matrix representations of them:
\[ \begin{matrix} c_a & \ket{11} & \ket{10} & \ket{01} & \ket{00} \\ \ket{11} & 0 & 0 & 0 & 0 \\ \ket{10} & 0 & 0 & 0 & 0 \\ \ket{01} & 1 & 0 & 0 & 0 \\ \ket{00} & 0 & 1 & 0 & 0 \end{matrix} \]
A bit of explanation about the notation. $\ket{0}$ represents the unoccupied state in $a$ but also is the up-spin state $\ket{\uparrow}$ for the Pauli matrices. $\ket{1}$ is the down-spin state $\ket{\downarrow}$ and the occupied state in $b$.
I am familiar with the matrix representations of the tensor products of Pauli matrices, so I immediately know that the matrix above is the tensor product $\sigma_-\otimes \sigma_0 $. On the lower-left is the identity matrix multiplied by the factor of 1 in $\sigma_-$. And not surprisingly, the matrix representations of other second-quantized operators $c_a^\dagger, c_b, c_b^\dagger$ can also be written as tensor products of some of the Pauli matrices.
To complete the story, I list the matrices for $c_a^\dagger, c_b, c_b^\dagger$ according to the matric elements we got at the start and the corresponding tensor products. To check the correctness of the matrices, we can verify that one gets $c^\dagger_a,c_b^\dagger$ if $c_a ,c_b$ are hermitian conjugated.
\[ \begin{matrix} c_a^\dagger & \ket{11} & \ket{10} & \ket{01} & \ket{00} \\ \ket{11} & 0 & 0 & 1 & 0 \\ \ket{10} & 0 & 0 & 0 & 1 \\ \ket{01} & 0 & 0 & 0 & 0 \\ \ket{00} & 0 & 0 & 0 & 0 \end{matrix}\quad \sigma_+\otimes \sigma_0 \]
\[ \begin{matrix} c_b & \ket{11} & \ket{10} & \ket{01} & \ket{00} \\ \ket{11} & 0 & 0 & 0 & 0 \\ \ket{10} & 1 & 0 & 0 & 0 \\ \ket{01} & 0 & 0 & 0 & 0 \\ \ket{00} & 0 & 0 & -1 & 0 \end{matrix}\quad \sigma_z\otimes \sigma_- \]
\[ \begin{matrix} c_b^\dagger & \ket{11} & \ket{10} & \ket{01} & \ket{00} \\ \ket{11} & 0 & 1 & 0 & 0 \\ \ket{10} & 0 & 0 & 0 & 0 \\ \ket{01} & 0 & 0 & 0 &-1 \\ \ket{00} & 0 & 0 & 0 & 0 \end{matrix}\quad \sigma_z\otimes \sigma_+ \]
This is exactly what the Jordan-Wigner transformation does. Sometimes the matrix representation is really helpful for understanding quantum mechanics.