\chapter{Fermions on the lattice}
\label{cha:fermions}
Formulating fermion discretizations is a challenging, but interesting aspect of lattice QCD. In this chapter, we focus on naive and staggered fermions~\cite{PhysRevD.11.395}, exploring how they relate to each other, where they differ, and what mathematical challenges they pose.\\
From this chapter on, I will drop the subscript $E$ from the Euclidean $\gamma$-matrices. Whenever it will be referred to the $\gamma$-matrices, the Euclidean ones are meant implicitly.
\section{Naive Fermions}
\label{sec:naive}
As discussed before the dynamics of the quarks, which are the fermionic fundamental particles of QCD, is described by the Euclidean Dirac equation
\begin{align}
    (\slashed \partial + m)\psi = 0,
\end{align}
which is the equation of motion of the Dirac action
\begin{align}
    S &= \int d^4 x \,\mathcal{L},\\
    \mathcal{L} &= \bar \psi (\slashed \partial + m)\psi. 
\end{align}
As mentioned before (Section~\ref{sec:lattice}) this action can be put on the lattice. If we ignore gauge field contribution, we can use a straightforward discretization scheme by replacing derivatives by symmetric finite differences and integrals by sums. The naive action for free quarks in QCD with all indices written explicitly is given by  
\begin{align}
\label{eq:naive_action}
    S^{\text{naive}} =  \sum_{n \in \Lambda}\sum_{a,b = 1}^3 \sum_{\alpha \beta = 1}^4\bar \psi_{\alpha}^a(n)\left(\sum_{\mu = 1}^4 \gamma_{\mu}^{\alpha\beta}\frac{\delta_{ab}\psi_{\beta}^b(n+\hat \mu)  -\delta_{ab}\psi_{\beta}^b(n-\hat \mu)}{2} + \delta^{\alpha \beta}\delta_{ab}m \psi^{b}_{\beta}(n) \right),
\end{align}
which is often written as a bilinear form
\begin{align}
      S^{\text{naive}} &= \sum_{n,m}\bar \psi(n) M^{\text{naive}}_{nm}\psi(m)\\
      M_{nm}^{\text{naive}} &= \frac{1}{2}\sum_{\mu = 1}^4\gamma_{\mu}\left(\delta_{n + \hat \mu,m} - \delta_{n - \hat \mu,m} \right) + m \delta_{nm}
\end{align}
with implicit Dirac and color indices~\cite{Zielinski:2016qxi}. In this discretization the fermion action provides a lot of interesting symmetries~\cite{Follana:2007uv}.
\subsection{Symmetries of the naive action}
\label{sec:naive_symmetries}
The homogeneous lattice structure of free Dirac fermions ensures that there are a lot of geometrical symmetry transformations that leave the action invariant.
\subsubsection*{Rotations and Inversions}
We start with the rotational symmetry whose symmetry group is often called \textit{special Würfel group} or $SW_4$, which consists of rotations of $\frac{\pi}{2}$ of all six coordinate planes. If one also allows spatial inversions, this group is extended by a factor of 2 and is called \textit{Würfel group} or $W_4$. In odd dimension a spatial inversion can be represented as a global sign in the defining representation, while in even dimensions a spatial inversion can be defined by an additional sign for only one axis. The concept of the Würfel group can be generalized to arbitrary dimensions similar to the generalization of continuous rotations to $O(n)$~\cite{doi:10.1063/1.525334,MANDULA198391}. In the Dirac representation the elementary rotations can be represented as 
\begin{align}
    R^{\mu\nu}:\, &\psi(n) \to \psi'(n') = \exp{i\frac{\pi}{4}\sigma_{\mu\nu}}\psi(n'),\\
     \left(R^{\mu\nu}\right)^{\dagger}:\,&\bar\psi(n) \to \bar\psi'(n') = \bar\psi(n')\exp{-i\frac{\pi}{4}\sigma_{\mu\nu}},
\end{align}
where $n'$ is the lattice point that is received by rotating $n$ in the $\mu\nu$-plane by an angle of $\frac{\pi}{2}$. The transformation $ \exp{i\frac{\pi}{4}\sigma_{\mu\nu}}$ rotates the $\gamma$-matrices in the same way~\cite{Mages:2015scv}. Note, that these matrices only build a projective representation (see Section~\ref{sec:proj_rep} of the symmetry group of a four dimensional hypercube, because of the identity
\begin{align}
    \left( \exp{i\frac{\pi}{4}\sigma_{\mu\nu}}\right)^4 = -1 \neq 1.
\end{align}
This is related to the fact that the lattice fermions have a spin of $\frac{1}{2}$ like their continuum counterparts~\cite{Risch2024lecture}. The inversion on the other hand is represented with
\begin{align}
    I_s:\, &\psi(n) \to \psi'(n') = \gamma_4\psi(n'),\\
    I_s^\dagger :\,&\bar\psi(n) \to \bar\psi'(n') = \bar\psi(n')\gamma_4.
\end{align}
\subsubsection*{Shift symmetry}
The homogeneous structure of the free naive action provides a wide spectrum of shift symmetries. Let $S_{\mu}$ be the shift operator that is in the defining representation given by
\begin{align}
       S_\mu:\, &\psi(n) \to \psi'(n') =\psi(n+\hat \mu),\\
    S_\mu^\dagger :\,&\bar\psi(n) \to \bar\psi'(n') = \bar\psi(n-\hat \mu),
\end{align} from which can be concluded that the spinor structure is not affected by the shift symmetry transformations in contrast to the $W_4$-symmetry transformations. Now it is trivial to prove that the naive action is an invariant quantity with respect to this transformation.\\
Furthermore, there are power series of this shift that do not affect the action, to construct them we consider the bilinear $A(m) = \sum_{n}  \bar\psi(n)\psi(n+m)$ whose invariance for $m= \pm \hat{\mu}$ and $m = 0$ implies invariance of the action. Now consider power series of the shift operator (that are not canonically symmetry transformations):
\begin{align}
P(\alpha,S_\mu) \psi= \sum_{i = -\infty}^\infty \alpha_i S_\mu^i \psi(n) = \sum_{i = -\infty}^\infty \alpha_i \psi(n + i \hat \mu) 
\end{align}
This transforms the term $A(y)$ according to
\begin{align}
    A(m) = \sum_{n}  \bar\psi(n)\psi(n+m) \to \sum_n \sum_{ij}\alpha_i^*\alpha_j\bar\psi(n+i\hat \mu)\psi(n+m+j\hat \mu),
\end{align}
which makes it an invariant condition if
\begin{align}
    \sum_{a \in \mathbb{Z}}\alpha_{a+i}^*\alpha_a = \delta_{i0}.
\end{align}
Using the Fourier convolution theorem one can conclude that the elements of $\alpha$ are Fourier coefficients of a $U(1)$-valued functions on a compact interval:
\begin{align}
    \alpha_m = \frac{1}{2\pi}\int_0^{2\pi} dp\, e^{imp}\Tilde{\alpha}(p)\\
    \Tilde{\alpha}(p) \in \mathcal{L}([0,2\pi],U(1)).
\end{align}
There is a variety of function that fulfill this condition. For example, choosing $\Tilde{\alpha}$ as a constant, this results in the global $U(1)$ symmetry of the action. It is somewhat more interesting to choose 
\begin{align}
    \Tilde{\alpha}(p) = \exp{i\gamma p}.
\end{align}
Then $\alpha_m$ takes the form $\textrm{sinc}(\gamma - m)$, which is equal to $\delta_{\gamma,m}$ if $\gamma$ is an integer. So for $\gamma\in \mathbb{Z}$ this is the symmetry due to the shift $S_\mu^\gamma$. If $m$ is chosen not to be an integer we can use it as a generalization of discrete shifts to arbitrary (continuous) shifts gamma. In general, these types of symmetry transformations can be interpreted as diagonal unitary operators on the momentum modes~\cite{KILCUP1987493}:
\begin{align}
    \ket p \to \Tilde{\alpha}(p)\ket{p}.
\end{align}
\subsubsection{Taste symmetry}
Apart from this geometrical symmetries, there are symmetry transformations that cannot be derived from a defining representation of a transformation in $\mathbb{R}^4$. For $\xi\in \mathbb{Z}_2^4$, consider the transformation
\begin{align}
    B_{\xi}\psi(x) = (-)^{x\cdot \xi}\prod_{\rho = 1}^4(\gamma_5\gamma_{\rho})^{\xi_{\rho}}\psi(x),
\end{align}
that is a symmetry of the naive action~\cite{Follana:2006rc}. All of the $B_{\xi}$ transformations are unitary, but not all of them are Hermitian. In order to do this, one defines
\begin{align}
  \Tilde{B}_{\xi} =\begin{cases}iB_{\xi} \quad &\textrm{if $B_{\xi}$ anti-Hermitian,} \\
  B_{\xi} \quad &\textrm{if $B_{\xi}$ Hermitian,} \\
  \end{cases}
\end{align}
which are all Hermitian. It is very straightforward to prove that the set of $\Tilde{B}_{\xi}$ transformation forms a projective representation (see Section~\ref{sec:proj_rep}) of $\mathbb{Z}_2^4$. Using a central extension of this group (see Section~\ref{sec:central_extension}) one can consider
\begin{align}
    \{\pm \Tilde{B}_{\xi} \vert \xi \in \mathbb{Z}_2^4\},
\end{align}
which is indeed a group that is isomorphic to $\Gamma_{0,4}$.\\
Since all of the group elements are Hermitian, the group elements can be used as generators of a continuous 16 dimensional symmetry group:
\begin{align}
    \lbrace\exp \left(i\alpha_{\xi} \Tilde{B}_{\xi} \right)\vert \alpha_{\xi} \in \mathbb{R}\rbrace.
\end{align}
Furthermore, there are chiral symmetry transformations that do not affect the Dirac-part of the operator but generate a sign on the mass term:
\begin{align}
    A_{\xi} =  i^{\Vert \xi \Vert_1}B_{\xi}\gamma_{5}  = i^{\sum_{\mu} \vert\xi_{\mu}\vert}B_{\xi} \gamma_{5}.
\end{align}
It can be seen very easily that we have
\begin{align}
    A_{\bar \xi} = \pm B_{\xi} \epsilon(n).
\end{align}
Thus, they have the same structure constants.
We now can define the generators of left- and right-handed transformations by
\begin{align}
    B^R_{\xi} = \frac{1 + \epsilon(n)}{2}\Tilde B_{\xi},\\
    B^L_{\xi} = \frac{1 - \epsilon(n)}{2}\Tilde B_{\xi}.
\end{align}
We have commutation relations
\begin{align}
    [B^R_{\xi},B^R_{\zeta}] = \frac{1 + \epsilon(n)}{2}C_{\xi \zeta}^{\rho} \Tilde B_{\rho} = C_{\xi \zeta}^{\rho}B^R_{\rho},\\
     [B^L_{\xi},B^L_{\zeta}] = \frac{1 - \epsilon(n)}{2}C_{\xi \zeta}^{\rho} \Tilde B_{\rho} = C_{\xi \zeta}^{\rho}B^L_{\rho},\\
     [B^L_{\xi},B^R_{\zeta}] =  \frac{1 + \epsilon(n)}{2} \frac{1 - \epsilon(n)}{2}C_{\xi \zeta}^{\rho} \Tilde B_{\rho} = 0,
\end{align}
with $C_{\xi \zeta}^{\rho} \in \mathbb{R}$ being the structure constants of the $\tilde B_{\xi}$. The continuous transformations generated by the right-  and left-handed generators are
\begin{align}
\begin{split}
    U^R &= \exp \left(\sum_{\xi}\alpha_{\xi} \frac{1 + \epsilon(n)}{2}\Tilde B_{\xi}\right)  = \exp \left(\frac{1 + \epsilon(n)}{2}\sum_{\xi}\alpha_{\xi} \Tilde B_{\xi}\right)  
    = \sum_{k \geq 0} \frac{\left(\frac{1 + \epsilon(n)}{2}\sum_{\xi}\alpha_{\xi} \Tilde B_{\xi}\right)^k}{k!} \\
    &= 1 + \frac{1 + \epsilon(n)}{2}\sum_{k \geq 1} \frac{\left(\sum_{\xi}\alpha_{\xi} \Tilde B_{\xi}\right)^k}{k!} = 1 +\frac{1 + \epsilon(n)}{2}\left(\exp \left(\sum_{\xi}\alpha_{\xi} \Tilde B_{\xi}\right) -1\right)\\
    &= \frac{1 - \epsilon(n)}{2} + \frac{1 + \epsilon(n)}{2}\exp \left(\sum_{\xi}\alpha_{\xi} \Tilde B_{\xi}\right) 
\end{split}
\end{align}
and
\begin{align}
    U^L &=  \frac{1 + \epsilon(n)}{2} + \frac{1 - \epsilon(n)}{2}\exp \left(\sum_{\xi}\alpha_{\xi} \Tilde B_{\xi}\right).
\end{align}
\subsection{The doubling problem}
It is very straightforward to diagonalize the naive action matrix $M^{\text{naive}}_{mn}$. The Lagrange density of the action is homogeneous in space and therefore should conserve the conjugate momenta, which translates in the fact that $M^{\text{naive}}$ is diagonal in momentum space. The Fourier transformation of equation \eqref{eq:naive_action} is given by:
\begin{align}
    \int \frac{dp}{2\pi} \int \frac{dq}{2\pi} \sum_{a,b = 1}^3 \sum_{\alpha, \beta = 1}^4\bar \psi_{\alpha}^a(p)\Tilde M\psi_{\beta}^b(q)=\\
   \int \frac{dp}{2\pi} \int \frac{dq}{2\pi} \sum_{a,b = 1}^3 \sum_{\alpha, \beta = 1}^4\bar \psi_{\alpha}^a(p)\left(i\sum_{\mu = 1}^4 \gamma_{\mu}^{\alpha\beta} \sin p_{\mu} + m\delta_{\alpha \beta} \right)\delta_{ab}\delta(p-q)\psi_{\beta}^b(q),
\end{align}
The Matrix $\Tilde M$ can be inverted easily:
\begin{align}
    \tilde M^{-1} = \frac{m\delta^{\alpha \beta} - i \sum_{\mu}\gamma_{\mu}^{\alpha \beta}\sin p_{\mu}}{m^2 + \sum_{\mu}\sin^2 p_{\mu} }\delta_{ab}\delta(p-q).
\end{align}
The free propagator for naive quarks is then given by the backwards transformation of the inverse Matrix in momentum representation. It is common knowledge that poles in the massless propagator correspond to physical states. Since the lattice momenta $p_{\mu}$ are taken from the interval $[0,2\pi)$, there are 16 momenta corresponding to a physical state:
\begin{align}
\label{eq:doublermomenta}
    (0,0,0,0),\, (0,0,0,\pi),\, , ...\, (\pi,\pi,\pi,\pi).
\end{align}
Therefore, 15 modes correspond to unphysical states, called the \textit{doublers}~\cite{GattringerLang}. The doublers are directly related to the symmetry transformations given in Section~\ref{sec:naive_symmetries}. Let $p = (p_1,p_2,p_3,p_4)$ with $p_i \in \{0,1\}$ be one of the momenta from Equation~\eqref{eq:doublermomenta} in units of $\pi$ and $B_\xi$ with $\xi \in \mathbb{Z}_2^4$ one of the taste transformations. Then this transformation acts in the following way:
\begin{align}
    B_\xi p \equiv p + \xi \mod 2.
\end{align}
So the physical interpretation of this observation is the following: The naive action describes 16 quarks of the same mass instead of one. Therefore the action is symmetric when interchanging the quarks. The corresponding symmetry transformations are the $B_\xi$ that are discussed before. 
\subsubsection{The No-go theorem}
One might think that there is a way to get rid of the doublers without any disadvantages, but this is not the case:\\
If $D$ be a massless discrete Dirac operator, then at least one of the following statements is not true~\cite{Varnhorst2023lecture}:
\begin{enumerate}
    \item $D$ has only one zero-mode. This is equivalent to the statement that $D$ has no doublers.
    \item $D$ is \textit{local}, which means that there are constants $C,c>0$ such that 
    \begin{align}
      \vert  D_{mn}\vert < C \exp\left( - c\vert n - m\vert\right).
    \end{align}
    One has to note that the constant $c$ should not depend on the lattice spacing.
    \item $D$ has the correct continuum limit, i.e
    \begin{align}
        \lim_{a \to 0} D = \slashed \partial.
    \end{align}
    \item $D$ respects chiral symmetry, i.e.,
    \begin{align}
        \{D, \gamma_5\} = 0.
    \end{align}
\end{enumerate}
\section{Staggered fermions}
\label{sec:stagg}
Staggered fermions can be introduced to reduce the number of doublers by a factor of four. This is still not consistent with the theory we want to describe, but will allow us to save a lot of simulation time by not computing 12 out of 16 identical fermions. The remaining three unwanted fermions can be removed by a technique called rooting that will be mentioned later on.\\
\subsection{The staggered transformation}
In order to construct staggered fermions from naive fermions, the unitary basis transformation
\begin{align}
    \Gamma(x) = \prod_{i = 1}^4 \gamma_i^{x_i}\quad    \Gamma^{\dagger}(x) = \prod_{i = 4}^1 \gamma_i^{x_i}
\end{align}
is defined. The basis transformation is used to defined the staggered fields $\chi$ and $\bar \chi$ as:
\begin{align}
\psi^\alpha(x) = \Gamma^{\alpha \beta}(x)\chi_\beta(x)\\
    \bar\psi^\alpha(x) = \bar \chi_\beta(x)\left( \Gamma^{\dagger}\right)^{ \beta \alpha}(x)
\end{align}
With this redefinition the free naive action in equation \eqref{eq:naive_action} becomes:
\begin{align}
    S = \sum_{x \mu}\bar\chi_\alpha(x)\left(\eta_\mu(x)\delta^{\alpha \beta}\frac{\chi_\beta(x + \hat \mu) -\chi_\beta(x - \hat \mu)  }{2} + m \delta^{\alpha \beta} \chi_\beta(x)\right)\\
    \eta_\mu(x) = (-)^{\sum_{\nu <\mu}x_\nu}
\end{align}
By inspecting this expression one will notice that there is no mixing between the spinor components. It is therefore sufficient to work with only one of those which reduces the degrees of freedom and therefore the number of doublers by a factor of four. This feature is useful in the physical interpretation of the lattice fermions (because we got rid of 12 unphysical modes) as well as in terms of computational effort~\cite{Staggered_fermions,PhysRevD.15.1111,PhysRevD.16.3031}.
So the staggered action (which is one fourth of the naive action) is given by
\begin{align}
    S^{st} = \bar \chi M^{st}\chi,\\
    M^{st} = \eta_\mu(x)\frac{U_{x,\mu}\delta_{x+\hat\mu,y} - U_{x-\mu,\mu}^\dagger \delta_{x-\hat\mu,y}}{2} + m \delta_{x,y},
\end{align}
where the gauge links are already included, such that the action is gauge invariant.
\subsection{Symmetry properties of staggered fermions}
\label{sec:stagg_symm}
The (free) staggered action comes along with multiple symmetries. Averaged over all gauge field configurations, these symmetries also holds for the interacting case~\cite{KILCUP1987493}. Before listing the most important symmetries, a few definitions have to be done:
\begin{align}
    \zeta_\mu(x) = (-)^{\sum_{\nu >\mu}x_\nu}\\
    \epsilon(x) = (-)^{\sum_\mu x_\mu}\\
    S_R(x) = \frac{1+\zeta_\mu(x)\zeta_\nu(x) - \eta_\mu(x)\eta_\nu(x) + \zeta_\mu(x)\zeta_\nu(x)\eta_\mu(x)\eta_\nu(x)}{2},
\end{align}
with $R$ being a rotation in the $\mu\nu$-plane.
\begin{enumerate}
    \item The first symmetry is the symmetry of the staggered action under rotations of $\pi/2$ in the $\mu\nu$-plane. It is similar to the corresponding symmetry transformation for naive fermions, but the pre-factor that mixes the spinor-components is replaced by a location-dependent pre-factor:
    \begin{align}
    R^{\mu\nu}:\, &\chi(n) \to \chi'(n') = S_R(n')\chi(n')
\end{align}
\item Of course staggered fermions also have a shift symmetry, but since the staggered action comes along with a position-dependent pre-factor, this has to be respected in the transformation rules:
\begin{align}
        S^{\mu}:\, &\chi(n) \to \chi'(n') = \zeta_\mu(n)\psi(n')
\end{align}
\item The staggered phase $\eta_\mu(x)$ takes somehow the place of the $\gamma$-matrices, when transforming the naive action to the staggered one. This also happens when considering the spatial inversion symmetry:
\begin{align}
          I_s:\, &\chi(n) \to \chi'(n') = \eta_4(n)\psi(n')
\end{align}
\item Next there is the charge conjugation symmetry that has a local pre-factor in the staggered formulation and interchanged particles and antiparticles:
\begin{align}
    C_0:\, &\chi(n) \to \epsilon(n)\bar\chi(n)\\
    &\bar\chi(n) \to -\epsilon(n)\chi(n)
\end{align}
\item The Baryon number is the only global symmetry in this list. It multiplies the quarks with a $U(1)_V$-phase factor:
\begin{align}
    B:\, \chi(n) \to g\chi(n)\quad g \in U(1)
\end{align}
\item The axial hyper-charge is explicitly broken for $m \neq 0$. The quarks are multiplied by an $U(1)_A$-phase factor.
\begin{align}
    H:\, \chi(n) \to g^\epsilon(n)\quad g \in U(1)
\end{align}
    \end{enumerate}

It has to be noted that all of the symmetries with local pre-factors are involutions expected for the shifts. If the shift operator is squared, the pre-factors cancel and what remains is the shift by two lattice units. Therefore, in momentum space, this operator $S_\mu^2 \eqqcolon T_\mu$ can be represented as
\begin{align}
    T_\mu \ket{p} = \exp(2\mathrm{i}p_\mu)\ket{p}.
\end{align}
Next we want to take the square root of this operator which leaves two choices for each eigenvalue. We decide to take the one with the with positive real part which ensures that the momenta are in the first Brillouin zone:
\begin{align}
    T^{1/2}_\mu\ket{p} = \exp(\mathrm{i}p_\mu)\ket{p}\quad p_\mu \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right]
\end{align}
Using this definition, we can disentangle shift and taste from the symmetry operators by defining
\begin{align}
    \Xi_\mu = T_\mu^{-1/2}S_\mu.
\end{align}
In the defining representation the $\Xi_\mu$ operators fulfill the anticommutation relations of the euclidean $\gamma$-matrices, thus generating the 32-element group $\Gamma_{4,0}$~\cite{KILCUP1987493}. Also, the taste transformations anticommute with shifts in perpendicular directions and they are related to rotations via
\begin{align}
    R_{\mu\nu}\Xi_\nu R_{\mu\nu}^{-1}\Xi_\nu = 1.
\end{align}
This means that a rotation of the taste operators by an angular of $\pi$ will not bring in a sign as it would for vectors like the shift operator.\\
The total symmetry group of staggered fermions\footnote{We ignore the baryon number symmetry because its trivial and the axial symmetry because it is not exact.} is then given as
\begin{align}
    \mathcal{G} = \{R_{\mu\nu},I_s,C_0,\Xi_\mu,T_\mu^{1/2}\}.
\end{align}
If we consider only particles at rest, $T_\mu^{1/2}$ can be ignored (or represented trivially) and in combination with our knowledge on how the Würfel group acts on taste and charge transformations, the symmetry group at rest can be written as the semi-direct product of two groups:
\begin{align}
    \mathcal{G}_0 = \{C_0,\Xi_\mu\}\rtimes \{R_{\mu\nu},I_s\} = \Gamma_{4,1}\rtimes W_4.
\end{align}
$\Gamma_{4,1}$ is the extension of $\Gamma_{4,0}$ by the generating element $C_0$ that anticommutes with the other generators $\Xi_\mu$ and has the property $C_0^2 = -1$. If we want to consider particles that are not at rest, we have to add back the translation group. For simplicity, we use a finite lattice $N$ lattice site in each direction and periodic boundary conditions. Then the total symmetry group is given by~\cite{KILCUP1987493}
\begin{align}
    \mathcal{G} = \{\mathbb{Z}_N^4\}\rtimes\mathcal{G}_0.
\end{align}
In the application in Chapter~\ref{cha:Exploratory} we will consider the symmetry group at a fixed time-slice,which changed the group to
\begin{align}
        \mathcal{H} = \{\mathbb{Z}_N^3\}\rtimes\Gamma_{4,1}\rtimes W_3 =  \{\mathbb{Z}_N^3\}\rtimes [SW_4 \times \Gamma_{2,2}]/\mathbb{Z}_2.
\end{align}
\subsection{Momentum, spin and taste operators for staggered fermions}
In the studies in the second part of this work, not only resting scalar fields are considered, but also a large variety of operators. Thus, I want to recap shortly how one can implement the different types of operators. Firstly, the are non-zero momenta. Let us consider a resting staggered field $\chi(x)$ that is put into a functional:
\begin{align}
    F[\chi, p = 0] = \sum_x \mathcal{A}_x \chi_x.
\end{align}
If we want to align a momentum $q$ to the staggered field, we include a complex phase depending on the position $x$ of the quark field:
\begin{align}
    F[\chi, p = q] = \sum_x \mathcal{A}_x \chi_x e^{\mathrm{i}q\cdot x}.
\end{align}
It is important that the components of $p$, i.e., $p_\mu$, are a multiple of $\frac{2\pi}{L_\mu}$ if the lattice has finite extension $L_\mu$ in $\mu$-direction. From the construction it can be seen directly that $p_\mu \equiv p_\mu + 2\pi$ and thus, the momentum is chosen from the interval $(-\pi,\pi]$. In the case of naive fermions (Section~\ref{sec:naive}) it was shown that momenta that are multiples of $\pi$ can be interpreted as different tastes and not as higher momenta. During the staggered construction we could get rid of some of those unwanted tastes. However, three unphysical tastes remain~\cite{GattringerLang}.\\
The construction of taste and spin structure for free quarks is explained in some detail in the appendix of~\cite{Follana:2006rc}, but what is more interesting here is the taste and spin structure of mesons. A meson is a bound state consisting of a quark $\chi$ and an antiquark $\bar\chi$. Each of them has four spin and four taste components that are decomposed in the following way:
\begin{align}
\mathbf{\bar 4}_s \otimes \mathbf{4}_s &= \mathbf{1}_s\oplus \mathbf{4}_s\oplus\mathbf{6}_s\oplus\mathbf{4}_s\oplus\mathbf{1}_s  \\
\mathbf{\bar 4}_t \otimes \mathbf{4}_t &= \mathbf{1}_t\oplus \mathbf{4}_t\oplus\mathbf{6}_t\oplus\mathbf{4}_t\oplus\mathbf{1}_t  
\end{align} 
This decomposition is exactly the number of choices there are to pick out $i$ out of 4 elements, what is exactly how meson in the continuum are defined. There, one puts $i$ distinct $\gamma$-matrices between the quark fields, i.e., $\bar \psi\left( \prod_\mu\gamma_\mu^{n_\mu}\right) \psi$ with in $n \in \mathbb{Z}_2^4$. However, due to the staggered construction we have no $\gamma$-matrices acting on spinor components but there is implicit spin and taste. A staggered meson with a spinor structure of $n \in \mathbb{Z}_2^4$ and with a taste structure of $s \in \mathbb{Z}_2^4$ is given by
\begin{align}
    j_n^s(x) = \beta_n^s(x)\bar \chi(x)\chi(x + \delta x_{sn}),
\end{align}
with $\delta x_{sn}\equiv (s+n) \mod 2 $ and 
\begin{align}
    \beta_n^s(x) = (-)^{x\cdot(s^< + n^>) + n\cdot(s+n)^<},
\end{align}
where the superscript $<$ denotes that the components of the vector are the cumulative sums of the elements with a smaller index (modulo 2) and the superscript $>$ denotes that the components of the vector are the cumulative sums of the elements with a larger index (modulo 2)~\cite{Golterman:1985dz,Follana:2006rc}~.