Jordan–Wigner transformation

The Jordan–Wigner transformation is a transformation that maps spin operators onto fermionic creation and annihilation operators. It was proposed by Pascual Jordan and Eugene Wigner for one-dimensional lattice models, but now two-dimensional analogues of the transformation have also been created. The Jordan–Wigner transformation is often used to exactly solve 1D spin-chains such as the Ising and XY models by transforming the spin operators to fermionic operators and then diagonalizing in the fermionic basis.

This transformation actually shows that at least in some cases with one spatial dimension, the distinction between spin-1/2 particles and fermions is nonexistent.

Analogy between spins and fermions

In what follows we will show how to map a 1D spin chain of spin-1/2 particles to fermions.

Take spin-1/2 Pauli operators acting on a site $j$ of a 1D chain, $\sigma_{j}^{+}, \sigma_{j}^{-}, \sigma_{j}^{z}$. Taking the anticommutator of $\sigma_{j}^{+}$ and $\sigma_{j}^{-}$, we find $\{\sigma_{j}^{+},\sigma_{j}^{-}\} = 1$, as would be expected from fermionic creation and annihilation operators. We might then be tempted to set

$\sigma_{j}^{+} =( \sigma_{j}^{x}+i\sigma_{j}^{y})/2 = f_{j}^{\dagger}$
$\sigma_{j}^{-} = (\sigma_{j}^{x}-i\sigma_{j}^{y} )/2= f_{j}$
$\sigma_{j}^{z} = f_j^{\dagger}f_j - \frac{1}{2}.$

Now, we have the correct same-site fermionic relations $\{f_{j}^{\dagger}, f_{j}\}=1$, however, on different sites, we have the relation $[f_{j}^{\dagger},f_{k}] = 0$, where $j \neq k$, and so spins on different sites commute unlike fermions which anti-commute. We must remedy this before we can take the analogy very seriously.

A transformation which recovers the true fermion commutation relations from spin-operators was performed in 1928 by Jordan and Wigner. This is a special example of a Klein transformation. We take a chain of fermions, and define a new set of operators

$a_{j}^{\dagger} = e^{+i\pi \sum_{k=1}^{j-1}f^{\dagger}_k f_k} f^{\dagger}_j$
$a_{j} = e^{-i\pi \sum_{k=1}^{j-1}f^{\dagger}_k f_k} f_j$
$a_j^{\dagger} a_j - \frac{1}{2} = f^{\dagger}_j f_j - \frac{1}{2}.$

They differ from the above only by a phase $e^{\pm i\pi \sum_{k=1}^{j-1}f^{\dagger}_k f_k}$. The phase is determined by the number of occupied fermionic modes in modes $k=1,\ldots,j-1$ of the field. The phase is equal to $+1$ if the number of occupied modes is even, and $-1$ if the number of occupied modes is odd. This phase is often expressed as

$e^{\pm i\pi \sum_{k=1}^{j-1}f^{\dagger}_k f_k}=\prod_{k=1}^{j-1}e^{\pm i\pi f^{\dagger}_k f_k}=\prod_{k=1}^{j-1}(1-2f^{\dagger}_k f_k).$

Where the last equality makes use of the fact that $f^{\dagger}_k f_k\in\{0, 1\}.$

The transformed spin operators now have the appropriate fermionic commutation relations

$\{a_i^\dagger, a_j\}=\delta_{i,j},\{a_i^\dagger, a_j^\dagger\}=0, \{a_i, a_j\}=0.$

The inverse transformation is given by

$a^{\dagger}_j = e^{+i\pi \sum_{k=1}^{j-1}a^{\dagger}_k a_k} \sigma_j^+$
$a_j = e^{-i\pi \sum_{k=1}^{j-1}a^{\dagger}_k a_k} \sigma_j^-$
$a^\dagger_j a_j = \sigma_j^z+\frac{1}{2}.$

Note that the definition of the fermionic operators is nonlocal with respect to the bosonic operators because we have to deal with an entire chain of operators to the left of the site the fermionic operators are defined with respect to. This is also true the other way around. This is an example of a 't Hooft operator, which is a disorder operator instead of an order operator. This is also an example of an S-duality.