# Lattice Field Theory: Fermions

My in-progress, constantly changing notes on lattice field theory. This post is about Fermions on the lattice.

## Canonical Fermion Quantisation

In the canonical quantisation formalism, one introduces creation and annihilation operators which obey anticommutation relations:

$\{a_i^{+}, a_j\} = \delta_{ij}$

with all other anticommutators being zero (operators fully anticommuting). Therefore the square of any operator is zero.

The multifermion states are elements of the Fock space, which is built on top of the vacuum state $$\ket{0}$$, which is the unique vector destroyed by every annihilation operator.

### Jordan-Wigner Representation

A representation for the annihilation and creation operators can be produced by taking direct products (direct vector space sum) of Pauli matrices.

The Pauli matrices are given by

$\sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}, \ \sigma_2 = \begin{pmatrix} 0 & i \\ -i & 0 \\ \end{pmatrix}, \ \sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}$

and

$\sigma_{\pm} = \frac12\left( \sigma_1 \pm i\sigma_2 \right).$

Labelling the particular subspaces of the direct sum by $$i_1, i_2$$, and the particular Pauli matrix by $$m_1, m_2$$, we have the commutation relation

$\left[\sigma_{i_1m_1},\sigma_{i_2m_2}\right] = 2i\delta_{i_1i_2}\epsilon_{m_1m_2m_3}\sigma_{i_1m_3}.$

If we choose the basis in the $$N$$-dimensional space where all of the $$\sigma_3$$ Pauli matrices are diagonal, we can denote the eigenvalues by $$\xi = \pm 1$$, and then we have the eigenvalue equation

$\sigma_{i_j3}\ket{\xi_{i_1},\xi_{i_2}, \ldots, \xi_{i_N}} = \xi_{i_j}\ket{\xi_{i_1},\xi_{i_2}, \ldots, \xi_{i_N}}$

where in particular the vacuum state is given by the state $$\ket{0} = \ket{-1, -1, \ldots, -1}$$.

Using this, there is a representation of the fermionic creation and annihilation operators as

$a^+_i = (-1)^{N-i} \sigma_{(N)3}\sigma_{(N-1)3}\cdots\sigma_{(i+1)3}\sigma_{(i)+}$

and

$a_i = (-1)^{N-i} \sigma_{(N)3}\sigma_{(N-1)3}\cdots\sigma_{(i+1)3}\sigma_{(i)-}$

with $$\sigma_{\pm}$$ defined as above.

### A Two-State Fermion System

Let us consider the simple two-state fermionic Hamiltonian

$H_2 = m(a_1^+a_1 + a_2^+a_2) + K(a_1^+a_2 + a_2^+a_1)$

By ‘solving the system’ we mean determining the partition function

$Z = \textup{Tr}\ e^{-\beta H_2}.$

One way to do this is to change to the basis where $$H_2$$ is diagonal, and then (taking care with the anticommuting operators) we obtain

$Z = (1 + e^{-\beta(m+K)})(1+e^{-\beta(m-K)}).$

Another way of doing this, which is analogous to the path integral of bosonic systems is to divide the imaginary time interval into $$T$$ equal parts:

$Z = \textup{Tr}\ e^{-\beta H_2} = \lim\limits_{T \to \infty}\textup{Tr}\ (1 - \Delta\beta H_2)^T$

where $$\Delta\beta = \beta/T$$.

Inserting a complete set of states between each factor and taking the trace explicitly gives us

$\lim\limits_{T \to \infty}\textup{Tr}\ (1 - \Delta\beta H_2)^T = \sum\limits_{\xi^{(1)}\xi^{(2)}\cdots\xi^{(T)}} \braket{\xi^{(1)}\vert 1-\Delta\beta H_2\vert\xi^{(2)}} \braket{\xi^{(2)}\vert 1-\Delta\beta H_2\vert\xi^{(3)}}\cdots \braket{\xi^{(T)}\vert 1-\Delta\beta H_2\vert\xi^{(1)}}$

where the sum over $$\xi^{(1)}$$ is the one taking the trace.

Considering the term $$\braket{\xi^{(i)}\vert 1-\Delta\beta H_2\vert\xi^{(i+1)}}$$ by itself, there are $$6$$ nonzero contributions. These can be represented graphically by considering the initial and final states with the same number of occupied states (because the Hamiltonian cannot change the number of occupied states).

Considering many such terms multiplied together, we get something which is analogous to a bosonic path integral. This analogy can be extended by using the technology of integrals over Grassmann variables.

## Grassmann Numbers

For each fermion state, we introduce a pair of Grassmann variables $$(\eta_i,\overline{\eta}_j)$$. These are totally anticommuting, so instead of the algebra of the fermionic operators where there is a nonzero anticommutator, here all anticommutators vanish:

$\{\eta_i,\eta_j\} = \{ \overline{\eta}_i,\overline{\eta}_j\} = \{\overline{\eta}_i, \eta_j\} = 0.$

In particular, the square of any Grassmann number is zero.

Due to this, the most general function of a single Grassmann variable is linear:

$F(\eta) = F_0 + F_1\eta.$

Since we consider pairs of Grassmann variables, the most general function of a pair of variables is

$F(\overline{\eta},\eta) = F^{00} + F^{01} \eta + F^{10} \overline{\eta} + F^{11}\overline{\eta}\eta.$

We define the integral over a Grassmann variable to be

$\int d\eta\,d\overline{\eta}\,F(\overline{\eta},\eta) = F^{11}.$

It can be shown that $$\{d\overline{\eta},d\eta\} = 0$$, so we also have

$\int d\overline{\eta}\,d\eta\,F(\overline{\eta},\eta) =- F^{11}.$

A simple example is the integral

$\int d\overline{\eta}\,d\eta\, e^{-\lambda\overline{\eta}\eta} = \int d\overline{\eta}\,d\eta\, (1 + \eta\overline{\eta}\lambda) = \lambda.$

## Grassmannian Coherent States

A coherent state is defined as an eigenstate of the annihilation operator. For this to hold for fermions, we need that the eigenvalues are themselves Grassmann numbers.

A general fermionic coherent state is defined as

$\ket{\eta} = e^{(a^\dagger, \eta)}\ket{0}$

where we have introduced the notation $$(a^\dagger, \eta) = \sum_i a_i^\dagger \eta_i$$ where $$i$$ runs over all possible fermion states.

Due to the fact that $$[a_i^\dagger \eta_i, a_j^\dagger \eta_j] = 0$$, we can rewrite this as the product

$\fcolorbox{black}{lavender}{  \ket{\eta} = \prod_{i}(1 + a_i^\dagger \eta_i)\ket{0}.  }$

For two different coherent states, we have that

$\braket{\eta|\eta'} = e^{(\overline{\eta},\eta')}.$

In addition it is straightforward to prove that for a general bilinear operator $$(a^\dagger, Aa)$$ the matrix element is

$\braket{\eta|(a^\dagger, Aa)|\eta'} = (\overline{\eta}, A\eta') \braket{\eta|\eta'}.$

A result that will be very useful is the matrix element of the exponential of a bilinear operator

$\fcolorbox{black}{lavender}{ \braket{\eta|\exp{(a^\dagger, Aa)}|\eta'} = \exp{(\overline{\eta}, e^A\eta')}. }$

We will show this by expanding the exponential:

\begin{aligned} \braket{\eta|\exp{(a^\dagger, Aa)}|\eta'} &= \braket{\eta|1 + (a^\dagger, Aa) + \frac12 (a^\dagger, Aa)^2 + \ldots |\eta'}\\ &= \braket{\eta|\eta'} + \braket{\eta|(a^\dagger, Aa)|\eta'} + \frac12 \braket{\eta|(a^\dagger, Aa)^2|\eta'} + \ldots \end{aligned}

For the first and second terms we have simply

$\braket{\eta|\eta'} = e^{(\overline{\eta},\eta')}$

and

$\braket{\eta|(a^\dagger, Aa)|\eta'} = (\overline{\eta}, A\eta') \braket{\eta|\eta'}$

as before.

The third term gives (after normal-ordering via anticommutation relations)