Staggered Fermions
Maarten Golterman

Department of Physics and Astronomy, San Francisco State University,
San Francisco, CA 94132, USA

> These notes are based on a series of lectures on staggered fermions given at
> the Centre de Physique Théorique, Luminy, in Marseille, France, January
> 17-25, 2024.

##  I  Introduction

These lecture notes present the lattice formulation of vectorlike gauge
theories such as QED and QCD employing staggered fermions, which are
extensively used in numerical simulations of QCD. While many aspects of the
use of staggered fermions are discussed, these notes are not intended as a
comprehensive review. Several topics and alternative formulations are only
briefly mentioned, or not covered at all (see Sec. IX for pointers to
additional reading). Likewise, while important references on staggered
fermions and related topics have been included in the bibliography, the list
of references is not complete. Staggered fermions were first introduced in a
hamiltonian formalism [2, 3], but here we will only be concerned with the
euclidean theory. These notes complement the review of Ref. [4].

##  II  Staggered fermion action

In this section we begin with the naive fermion lattice action, from which we
derive the staggered fermion action through spin diagonalization. We then
derive the Feynman rules, show that the classical continuum limit reproduces
the expected gauge theory, and we list and discuss the symmetries of staggered
fermions. We end with the formulation of the staggered theory on the so-called
“taste basis.”

###  II.1  Naive fermions

The naive fermion action is given by

$S={1\over
2}\sum_{x\mu}\left(\overline{\psi}(x)\gamma_{\mu}U_{\mu}(x)\psi(x+\mu)-\overline{\psi}(x+\mu)\gamma_{\mu}U^{\dagger}_{\mu}(x)\psi(x)\right)+m\sum_{x}\overline{\psi}(x)\psi(x)\
.$ (1)

Points on the hypercubic lattice are labeled by $x$, and $x+\mu$ is the point
one lattice spacing away from $x$ in the $+\mu$ direction. Most of the time we
will set the lattice spacing $a=1$; if not, this point will be denoted as
$x+a\mu$. $\psi(x)$ is a Dirac spinor transforming in the fundamental
irreducible representation (irrep) of $SU(3)$, gauged by the link variables
$U_{\mu}(x)$. We are in four-dimensional euclidean space, and the Dirac
matrices $\gamma_{\mu}$ are hermitian. The parameter $m$ is the bare mass;
later on, we will generalize this to a mass matrix $M$ when we consider
$N_{f}>1$ flavors of staggered fermions. For now, we take $N_{f}=1$.

Let us obtain the free fermion propagator in momentum space. We define the
Fourier transform of $\psi$ and $\overline{\psi}$ by

$\psi(x)=\int_{p}e^{ipx}\tilde{\psi}(p)\
,\qquad\overline{\psi}(x)=\int_{p}e^{-ipx}\tilde{\overline{\psi}}(p)\ ,$ (2)

where

$\int_{p}\equiv\int_{-\pi}^{\pi}\frac{d^{4}p}{(2\pi)^{4}}\ .$ (3)

In momentum space, the free action, obtained by setting $U_{\mu}(x)=1$, is

$S_{\rm
free}=\int_{p}\sum_{\mu}\tilde{\overline{\psi}}(p)\,i\gamma_{\mu}\sin(p_{\mu})\tilde{\psi}(p)+m\int_{p}\tilde{\overline{\psi}}(p)\tilde{\psi}(p)\
.$ (4)

We see that the inverse propagator (at $m=0$) not only has a zero at $p=0$,
but at all momenta

$p=\pi_{A}\ ,$ (5)

where $\pi_{A}$ has components $0$ or $\pi$. There are 16 such vectors, and
the index $A$ thus runs from 1 to 16. For $p_{\mu}=\tilde{p}_{\mu}+\pi_{A}$,

$\sin(p_{\mu})=\sin(\tilde{p}_{\mu}+\pi_{A\mu})=S^{A}_{\mu}\sin(\tilde{p}_{\mu})\
,$ (6)

which defines the signs $S^{A}_{\mu}=e^{i\pi^{A}_{\mu}}$. Therefore, the
inverse propagator near $p=\pi_{A}$ for small $\tilde{p}$ equals

$S^{-1}(p)=\sum_{\mu}i\gamma_{\mu}S^{A}_{\mu}\tilde{p}_{\mu}+m\ ,$ (7)

showing that the continuum limit of the naive fermion action contains 16
relativistic fermions. These are the so-called fermion “species doublers.” In
particular, the matrices

$\gamma^{A}_{\mu}=\gamma_{\mu}S^{A}_{\mu}$ (8)

satisfy the Dirac algebra
$\\{\gamma_{\mu}^{A},\gamma_{\nu}^{A}\\}=2\delta_{\mu\nu}$, and are unitarily
equivalent to the original set. For the relativistic fermion near $p=\pi_{A}$
we can define

$\gamma_{5}^{A}=\gamma^{A}_{1}\gamma^{A}_{2}\gamma^{A}_{3}\gamma^{A}_{4}=S_{1}^{A}S_{2}^{A}S_{3}^{A}S_{4}^{A}\gamma_{1}\gamma_{2}\gamma_{3}\gamma_{4}=\pm\gamma_{5}\
,$ (9)

with the plus (minus) sign if the number of components of $\pi_{A}$ equal to
$\pi$ is even (odd). This implies that half of the doublers have axial charge
$+1$, and half of them have axial charge $-1$. The multiplet formed by the 16
species is thus anomaly free. This explains how the naive fermion action can
regulate a Dirac fermion while maintaining exact chiral symmetry (for $m=0$)
[5].

We note that the action (1) is invariant under hypercubic rotations in the
$\mu\nu$ plane with the matrix

$R_{1\over 2}=e^{{1\over 2}\omega\gamma_{\mu}\gamma_{\nu}}\ ,\qquad\mu<\nu$
(10)

with $\omega=\pi/2$ acting on the Dirac spinor $\psi$.111Note that the
generators ${1\over 2}\gamma_{\mu}\gamma_{\nu}$, $\mu<\nu$, are anti-
hermitian.

###  II.2  Spin diagonalization

We now carry out the unitary basis transformation

$\psi(x)=\gamma_{1}^{x_{1}}\gamma_{2}^{x_{2}}\gamma_{3}^{x_{3}}\gamma_{4}^{x_{4}}\chi(x)\
,\qquad\overline{\psi}(x)=\bar{\chi}(x)\gamma_{4}^{x_{4}}\gamma_{3}^{x_{3}}\gamma_{2}^{x_{2}}\gamma_{1}^{x_{1}}\
.$ (11)

This brings the action (1) into the form

$S={1\over
2}\sum_{x\mu}\eta_{\mu}(x)\left(\bar{\chi}(x)U_{\mu}(x)\chi(x+\mu)-\bar{\chi}(x+\mu)U^{\dagger}_{\mu}(x)\chi(x)\right)+m\sum_{x}\bar{\chi}(x)\chi(x)\
,$ (12)

with the phase factors $\eta_{\mu}(x)$ defined by

$\eta_{\mu}(x)=(-1)^{x_{1}+\dots+x_{\mu-1}}\ .$ (13)

We can make different choices in Eq. (11) by reordering the $\gamma$ matrices,
and this will lead to different results for the phase factors $\eta_{\mu}(x)$.
However, all essential properties of these phase factors that we will use
below are independent of this choice.

In Eq. (12) the spin matrices $\gamma_{\mu}$ have been diagonalized. We now
change the theory by dropping the spin index on the fields $\chi$ and
$\bar{\chi}$. This defines the staggered fermion action, with the “one-
component” fields222Of course, these fields still carry a color index and
possibly a flavor index, so “one-component” here only refers to spin. $\chi$
and $\bar{\chi}$ (for a discussion of the euclidean action and earlier
references, see Ref. [6]).

The reduction from the naive to the staggered theory reduces the number of
degrees of freedom by a factor 4, so one might expect the staggered theory to
only have 4 species doublers in the continuum limit. However, as this
reduction was obtained by spin diagonalization, it is not obvious that this is
the correct interpretation of the staggered theory. In order to demonstrate
this, one first should show that the action (12) has the correct classical
continuum limit, and then that quantum corrections do not change this
conclusion. If the theory is asymptotically free, so that the continuum limit
is taken at the gaussian fixed point, the latter question can be studied in
weak-coupling perturbation theory (WCPT), and we will do so in the next
chapter.

In order to see that the staggered action (12) may have a sensible continuum
limit, we will define several operators on the field $\chi$. First, we define

$T_{\pm\mu}:\ \chi(x)\to\eta_{\mu}(x)\chi(x\pm\mu)\ ,$ (14)

from which it follows that

$T_{\mu}T_{\nu}:\
\chi(x)\to\eta_{\mu}(x)\eta_{\nu}(x+\mu)\chi(x+\mu+\nu)=\varepsilon_{\mu\nu}(-1)^{x_{1}+\dots+x_{\mu-1}+x_{1}+\dots+x_{\nu-1}}\chi(x+\mu+\nu)\
,$ (15)

with

$\varepsilon_{\mu\nu}=\Biggl{\\{}\begin{array}[]{cc}+1\ ,&\mu\geq\nu\\\ -1\
,&\mu<\nu\end{array}\ .$ (16)

From this, it follows that

$\\{T_{\mu},T_{\nu}\\}:\ \chi(x)\to 2\delta_{\mu\nu}\chi(x+2\mu)\ .$ (17)

We see that the square of $T_{\mu}$ is a normal translation over two lattice
spacings in the $\mu$ direction. Moreover the $T_{\mu}$ themselves satisfy the
Dirac algebra modulo these translations by $2\mu$. The irreps of the group
generated by the $T_{\mu}$ live in momentum space, as $T_{\mu}$ involves a
translation. If we define $\hat{T}_{\mu}$ as the translation (14) modding out
the translations over $2\mu$, the $\hat{T}_{\mu}$ generate a 32-element finite
group isomorphic with that generated by the Dirac matrices $\gamma_{\mu}$. We
will denote this group as $\Gamma_{4}$.333Not to be confused with the matrix
$\Gamma_{4}$ that will be introduced later on. The context should help avoid
confusion.

This group has 16 one-dimensional irreps (obtained by mapping the matrices
$\gamma_{\mu}$ onto $\pm 1$ in all possible ways), and one four-dimensional
irrep generated by the $\gamma_{\mu}$. Since the 16 fields $\chi(x)$ within a
hypercube form a 16-dimensional representation of $\Gamma_{4}$, this
representation is reducible (but it will be irreducible under the full lattice
symmetry group, see Sec. II.4). We thus expect that a basis transformation
exists such that the $\hat{T}_{\mu}$ can be written as

$\hat{T}_{\mu}=\gamma_{\mu}\times 1_{4}\ ,$ (18)

with $1_{4}$ the $4\times 4$ unit matrix.

We expect similar transformations to exist of the form $1_{4}\times\xi_{\nu}$,
with $\xi_{\nu}$ also satisfying the Dirac algebra. Indeed, we can define
shifts [7, 8]

$S_{\pm\mu}:\ \chi\to\zeta_{\mu}(x)\chi(x\pm\mu)\ ,$ (19)

and require the new phase factors $\zeta_{\mu}(x)$ to be such that

$[S_{\mu},T_{\nu}]=0\ .$ (20)

This implies that

$\zeta_{\mu}(x)\eta_{\nu}(x+\mu)=\eta_{\nu}(x)\zeta_{\mu}(x+\nu)\ ,$ (21)

which can be rewritten as

$\zeta_{\mu}(x+\nu)=\eta_{\nu}(x)\eta_{\nu}(x+\mu)\zeta_{\mu}(x)=\varepsilon_{\mu\nu}\zeta_{\mu}(x)\
,$ (22)

and thus

$\zeta_{\mu}(x)=(-1)^{x_{\mu+1}+\dots+x_{4}}\ .$ (23)

With $S_{\mu}$ acting on the gauge fields as

$S_{\mu}:\ U_{\nu}(x)\to U_{\nu}(x+\mu)\ ,$ (24)

$S_{\mu}$ is a symmetry of the action (12), and

$\\{S_{\mu},S_{\nu}\\}:\ \chi(x)\to 2\delta_{\mu\nu}\chi(x+2\mu)\ .$ (25)

We note the similarity to Eq. (17). We will refer to this symmetry as shift
symmetry. It is a crucial symmetry for staggered fermions, as we will see
below. In contrast, the $T_{\mu}$ are not a symmetry of the theory, but they
are helpful for the interpretation of the theory: as we will see in more
detail below, they “generate” the spin matrices $\gamma_{\mu}$ needed to
interpret the continuum limit as a theory of four Dirac fermions.

It is instructive to recast the operators $T_{\mu}$ and $S_{\mu}$ in momentum
space. We first define a subset of the vectors $\pi_{A}$:

$\displaystyle\pi_{\eta_{1}}$ $\displaystyle=$ $\displaystyle(0,0,0,0)\ ,$
(26) $\displaystyle\pi_{\eta_{2}}$ $\displaystyle=$ $\displaystyle(\pi,0,0,0)\
,$ $\displaystyle\pi_{\eta_{3}}$ $\displaystyle=$ $\displaystyle(\pi,\pi,0,0)\
,$ $\displaystyle\pi_{\eta_{4}}$ $\displaystyle=$
$\displaystyle(\pi,\pi,\pi,0)\ ,$

in terms of which

$\eta_{\mu}(x)=e^{i\pi_{\eta_{\mu}}x}\ .$ (27)

We then have

$\eta_{\mu}(x)\chi(x+\mu)=\int_{p}e^{ipx+i\pi_{\eta_{\mu}}x+ip_{\mu}}\tilde{\chi}(p)=\int_{p}e^{ipx}e^{ip_{\mu}}\tilde{\chi}(p+\pi_{\eta_{\mu}})\
,$ (28)

where we used that $(\pi_{\eta_{\mu}})_{\mu}=0$ and that the exponential has a
periodicity $2\pi$ in each direction because $x_{\mu}$ is integer on the
lattice. We now define the full Brillouin zone (BZ) as

$-\frac{\pi}{2}<p_{\mu}\leq\frac{3\pi}{2}\ ,$ (29)

and the reduced BZ as

$-\frac{\pi}{2}<\tilde{p}_{\mu}\leq\frac{\pi}{2}\ .$ (30)

We write $p=\tilde{p}+\pi_{A}$ and define

$\phi_{A}(\tilde{p})=\chi(\tilde{p}+\pi_{A})\ .$ (31)

It follows that

$T_{\mu}:\ \phi_{A}(\tilde{p})=\chi(\tilde{p}+\pi_{A})\to
S^{A}_{\mu}e^{i\tilde{p}_{\mu}}\chi(\tilde{p}+\pi_{A}+\pi_{\eta_{\mu}})=S_{\mu}^{A}(\hat{\pi}_{\eta_{\mu}})_{AB}e^{i\tilde{p}_{\mu}}\phi_{B}(\tilde{p})\
,$ (32)

with

$(\hat{\pi}_{\eta_{\mu}})_{AB}=\overline{\delta}(\pi_{\eta_{\mu}}+\pi_{A}+\pi_{B})\
,$ (33)

and $\overline{\delta}(\pi_{A})=1$ for $\pi_{A}=0$ (mod $2\pi$) and zero
otherwise. Defining also

$(S_{\mu})_{AB}=S_{\mu}^{A}\delta_{AB}$ (34)

and

$\Gamma_{\mu}=S_{\mu}\hat{\pi}_{\eta_{\mu}}\ ,$ (35)

Eq. (32) can be rewritten as

$T_{\mu}:\ \phi(\tilde{p})\to e^{i\tilde{p}_{\mu}}\Gamma_{\mu}\phi(\tilde{p})\
.$ (36)

From Eq. (17) it then follows that

$\\{\Gamma_{\mu},\Gamma_{\nu}\\}=2\delta_{\mu\nu}\ ,$ (37)

i.e., the $16\times 16$ matrices $\Gamma_{\mu}$ obey the Dirac algebra.

Analogously, we can define

$\displaystyle\pi_{\zeta_{1}}$ $\displaystyle=$ $\displaystyle(0,\pi,\pi,\pi)\
,$ (38) $\displaystyle\pi_{\zeta_{2}}$ $\displaystyle=$
$\displaystyle(0,0,\pi,\pi)\ ,$ $\displaystyle\pi_{\zeta_{3}}$
$\displaystyle=$ $\displaystyle(0,0,0,\pi)\ ,$ $\displaystyle\pi_{\zeta_{4}}$
$\displaystyle=$ $\displaystyle(0,0,0,0)\ ,$

and

$\displaystyle(\hat{\pi}_{\zeta_{\mu}})_{AB}$ $\displaystyle=$
$\displaystyle\overline{\delta}(\pi_{\zeta_{\mu}}+\pi_{A}+\pi_{B})\ ,$ (39)
$\displaystyle\Xi_{\mu}$ $\displaystyle=$ $\displaystyle
S_{\mu}\hat{\pi}_{\zeta_{\mu}}\ ,$

to rewrite Eq. (19) in momentum space as

$S_{\mu}:\ \phi(\tilde{p})\to e^{i\tilde{p}_{\mu}}\Xi_{\mu}\phi(\tilde{p})\ ,$
(40)

with $\Xi_{\mu}$ another set of $16\times 16$ matrices. From Eqs. (20) and
(25), it follows that

$\\{\Xi_{\mu},\Xi_{\nu}\\}=2\delta_{\mu\nu}\
,\qquad[\Gamma_{\mu},\Xi_{\nu}]=0\ .$ (41)

As in Eq. (18), we can find a basis transformation such that

$\Gamma_{\mu}\Xi_{\nu}\to\gamma_{\mu}\times\xi_{\nu}\ ,$ (42)

with $\gamma_{\mu}$ and $\xi_{\nu}$ both sets of $4\times 4$ matrices
satisfying the Dirac algebra. We note that

$\displaystyle[S_{\mu},\hat{\pi}_{\eta_{\mu}}]=[S_{\mu},\hat{\pi}_{\zeta_{\mu}}]=0\
,$ $\displaystyle S_{\mu}=S_{\mu}^{T}\
,\quad\hat{\pi}_{\eta_{\mu}}=\hat{\pi}_{\eta_{\mu}}^{T}\
,\quad\hat{\pi}_{\zeta_{\mu}}=\hat{\pi}_{\zeta_{\mu}}^{T}\quad\Rightarrow\quad\Gamma_{\mu}=\Gamma_{\mu}^{T}\
,\quad\Xi_{\mu}=\Xi_{\mu}^{T}\ .$

For an explicit representation, see Ref. [8].

###  II.3  Feynman rules and classical continuum limit

Using Eq. (27), the free staggered action can be written in momentum space as

$S_{\rm free}={1\over
2}\int_{p}\int_{q}\tilde{\overline{\chi}}(q)\left(\sum_{\mu}\delta(p-q+\pi_{\eta_{\mu}})(e^{ip_{\mu}}-e^{-iq_{\mu}})+m\delta(p-q)\right)\tilde{\chi}(p)\
,$ (44)

from which we obtain the inverse propagator (using
$(\pi_{\eta_{\mu}})_{\mu}=0$ again)

$G^{-1}(p,-q)=\sum_{\mu}\delta(p-q+\pi_{\eta_{\mu}})i\sin(p_{\mu})+m\delta(p-q)\
,$ (45)

where $\delta$ is the periodic delta function with period $2\pi$ in each
direction. We now set

$p=\tilde{p}+\pi_{A}\ ,\qquad q=\tilde{q}+\pi_{B}\ ,$ (46)

to obtain

$\displaystyle G^{-1}(\tilde{p}+\pi_{A},-\tilde{q}-\pi_{B})$ $\displaystyle=$
$\displaystyle\sum_{\mu}\delta(\tilde{p}-\tilde{q}+\pi_{A}+\pi_{B}+\pi_{\eta_{\mu}})\,i\sin(p_{\mu})$
$\displaystyle\hskip 28.45274pt+m\delta(\tilde{p}-\tilde{q}+\pi_{A}+\pi_{B})$
$\displaystyle=$
$\displaystyle\delta(\tilde{p}-\tilde{q})\left(iS^{A}_{\mu}(\hat{\pi}_{\eta_{\mu}})_{AB}\sin(\tilde{p}_{\mu})+m\delta_{AB}\right)$
$\displaystyle\equiv$
$\displaystyle\delta(\tilde{p}-\tilde{q})S^{-1}_{AB}(\tilde{p})\ .$

The second line follows, because, from Eq. (30),
$-\frac{\pi}{2}<\tilde{p}_{\mu}-\tilde{q}_{\mu}<\frac{\pi}{2}$, with strict
inequalities. We thus find for the free propagator

$S(\tilde{p})=\left(\sum_{\mu}i\Gamma_{\mu}\sin(\tilde{p}_{\mu})+m\right)^{-1}\
,$ (48)

and the free action becomes

$S_{\rm
free}=\int_{\tilde{p}}\sum_{AB}\overline{\phi}_{A}(\tilde{p})\left(\sum_{\mu}i(\Gamma_{\mu})_{AB}\sin(\tilde{p}_{\mu})+m\delta_{AB}\right)\phi_{B}(\tilde{p})\
.$ (49)

This action is invariant under a group $U(4)$ generated by the $\Xi_{\mu}$
(all products of these matrices precisely give the generators of the group
$U(4)$), because this group commutes with $\Gamma_{\mu}$. Free staggered
fermions contain 4 degenerate species of Dirac spinors in the continuum limit.
We will refer to these 4 species as “tastes,” and we find that the free theory
has an exact $U(4)$ taste symmetry on the lattice. In the interacting theory,
the product of this $U(4)$ and lattice translations is broken to the discrete
symmetries $S_{\mu}$, cf. Eqs. (19), (24) and (40).

To obtain the one-gluon vertex, we expand

$U_{\mu}(x)=e^{igA_{\mu}(x)}=1+igA_{\mu}(x)-{1\over
2}\,g^{2}A_{\mu}^{2}(x)+\dots\ ,$ (50)

and Fourier transform

$A_{\mu}(x)=\int_{k}e^{ikx}\tilde{A}_{\mu}(k)\ .$ (51)

The one-gluon part of the action is thus

$S_{\rm
1-gluon}=ig\int_{p}\int_{q}\int_{k}\sum_{\mu}\tilde{\overline{\chi}}(q)\delta(p+k-q+\pi_{\eta_{\mu}}){1\over
2}(e^{ip_{\mu}}+e^{-iq_{\mu}})\tilde{A}_{\mu}(k)\tilde{\chi}(p)\ ,$ (52)

from which we read off the one-gluon vertex

$V^{m}_{\mu}(p,-q,k)=-ig\,\delta(p+k-q+\pi_{\eta_{\mu}})\,{1\over
2}(e^{ip_{\mu}}+e^{-iq_{\mu}})\,t_{m}\ ,$ (53)

where $\tilde{A}_{\mu}(k)=\sum_{m}\tilde{A}_{\mu}^{m}(k)t_{m}$, with $t_{m}$
the gauge group generators.

We can take the classical continuum limit by setting $p=\tilde{p}+\pi_{A}$,
$q=\tilde{q}+\pi_{B}$ with $\tilde{p}$, $\tilde{q}$ and $k$ small. Using

${1\over 2}(e^{ip_{\mu}}+e^{-iq_{\mu}})={1\over
2}(S_{\mu}^{A}e^{i\tilde{p}_{\mu}}+S_{\mu}^{B}e^{-i\tilde{q}_{\mu}})\approx{1\over
2}(S_{\mu}^{A}+S_{\mu}^{B})$ (54)

and

$\delta(\tilde{p}-\tilde{q}+k+\pi_{A}+\pi_{B}+\pi_{\eta_{\mu}})=\delta(\tilde{p}-\tilde{q}+k)(\hat{\pi}_{\eta_{\mu}})_{AB}$
(55)

for $\tilde{p}$, $\tilde{q}$ and $k$ small, the one-gluon vertex becomes

$V^{m}_{\mu}(\tilde{p}+\pi_{A},-\tilde{q}-\pi_{B},k)\approx-
ig\,\delta(\tilde{p}-\tilde{q}+k)(\Gamma_{\mu})_{AB}t_{m}\ ,$ (56)

which is the correct classical continuum limit. In particular, the classical
continuum limit is invariant under $U(4)_{\rm taste}$.

The two-gluon vertex can be read off using the $O(g^{2})$ part of the
expansion (50), with $\ell$ the momentum of the second gluon

$V^{m_{1}m_{2}}_{\mu_{1}\mu_{2}}(p,-q,k,\ell)={1\over
2}\,g^{2}\sum_{\mu}\delta(p+k+\ell-q+\pi_{\eta_{\mu}})\,{1\over
2}(e^{ip_{\mu}}-e^{-iq_{\mu}})\,\delta_{\mu\mu_{1}}\delta_{\mu\mu_{2}}\,{1\over
2}\\{t_{m_{1}},t_{m_{2}}\\}\ .$ (57)

Because of the minus sign between the momentum-dependent phase factors, it is
straightforward to show that this vertex vanishes in the classical continuum
limit.

We end this subsection with the observation that if we take $k_{\mu}=\pi$ in
Eq. (53), and use

${1\over 2}(e^{ip_{\mu}}+e^{-iq_{\mu}})={1\over
2}(e^{ip_{\mu}}-e^{-ip_{\mu}})=i\sin(p_{\mu})=iS_{\mu}^{A}\sin(\tilde{p}_{\mu})$
(58)

for this choice of $k$, we find that the one-gluon vertex for a gluon with
momentum $k_{\mu}=\pi$ is suppressed for small $\tilde{p}_{\mu}$. We will
return to this point in Sec. VIII. For the complete Feynman rules, we refer to
Ref. [8].

###  II.4  More symmetries

We have already identified one symmetry of the staggered action, the shifts
defined by Eqs. (19) and (24). There are more symmetries of course, which we
will discuss in this subsection. For more detail, see Ref. [8].

####  II.4.1  Hypercubic rotations

The action is invariant under hypercubic rotations, which are rotations over
an angle $\pi/2$ in each plane. Clearly, the action (49) is invariant under
rotations $R_{\rho\sigma}$ in the $\rho\sigma$ plane over an angle $\pi/2$
which takes $\tilde{p}_{\rho}\to\tilde{p}_{\sigma}$ and
$\tilde{p}_{\sigma}\to-\tilde{p}_{\rho}$, and

$\phi(\tilde{p})\to
e^{\frac{\pi}{4}\Gamma_{\rho}\Gamma_{\sigma}}\phi(R_{\rho\sigma}^{-1}\tilde{p})\qquad(\mbox{tentative})\
.$ (59)

However, this is in the free theory, which has more symmetry than the
interacting theory. Noting that the operators $T_{\mu}$ and $S_{\mu}$ are
expected to transform similarly under hypercubic rotations, we replace Eq.
(59) by

$\phi(\tilde{p})\to
e^{\frac{\pi}{4}\Gamma_{\rho}\Gamma_{\sigma}}e^{\frac{\pi}{4}\Xi_{\rho}\Xi_{\sigma}}\phi(R_{\rho\sigma}^{-1}\tilde{p})\
.$ (60)

Of course, the gauge fields have to be rotated accordingly, and Eq. (60) can
be translated back to position space. The form of rotations on the original
staggered fields $\chi(x)$ and $\bar{\chi}(x)$ as well as the links
$U_{\mu}(x)$ is given in Ref. [8]. Equation (60), along with the
transformation on the gauge fields, constitutes the hypercubic rotational
symmetry of the staggered action. These rotations form the discrete hypercubic
subgroup of a group $SO(4)$, which is the diagonal subgroup of $SO(4)_{\rm
spin}\times SO(4)_{\rm taste}$, with $SO(4)_{\rm taste}$ the unique $SO(4)$
subgroup of $SU(4)_{\rm taste}$.

The field $\phi$ transforms in a bosonic irrep of Eq. (60). However, we can
define a different group of lattice symmetries under which it transforms as a
fermion field. Since we have shift and rotational symmetry, we can define a
composite symmetry

$\tilde{R}_{\rho\sigma}=R_{\rho\sigma}S_{\rho}\ ,$ (61)

which can be visualized as rotations around the center of the hypercube based
at the origin. Using that

$R_{\rho\sigma}^{-1}S_{\rho}R_{\rho\sigma}=S_{\sigma}\ ,\qquad
R_{\rho\sigma}^{-1}S_{\sigma}R_{\rho\sigma}=-S_{\rho}\ ,$ (62)

one can prove that, modulo lattice translations over an even number of lattice
spacings in any direction,

$(\tilde{R}_{\rho\sigma})^{4}=-1\ ,$ (63)

i.e., that a rotation over $2\pi$ exists which is equal to $-1$, as one would
expect for a fermionic theory. In contrast, the fourth power of
$R_{\rho\sigma}$ equals $+1$. In the construction of the symmetry (61) shift
symmetry plays a key role. In momentum space, we find that

$\tilde{R}_{\rho\sigma}=e^{i\tilde{p}_{\rho}}e^{\frac{\pi}{4}\Gamma_{\rho}\Gamma_{\sigma}}\frac{1}{\sqrt{2}}(\Xi_{\rho}-\Xi_{\sigma})=-ie^{i\tilde{p}_{\rho}}e^{\frac{\pi}{4}\Gamma_{\rho}\Gamma_{\sigma}}e^{i\frac{\pi}{2}\frac{1}{\sqrt{2}}(\Xi_{\rho}-\Xi_{\sigma})}\
,$ (64)

using

$\left(\frac{1}{\sqrt{2}}(\Xi_{\rho}-\Xi_{\sigma})\right)^{2}=1\ .$ (65)

####  II.4.2  Axis reversal

Axis reversal $I_{\rho}$ in the $\rho$ direction takes $x_{\rho}\to-x_{\rho}$,
or $\tilde{p}_{\rho}\to-\tilde{p}_{\rho}$ in momentum space. Again, we expect
symmetry under the interchange $\Gamma_{\mu}\leftrightarrow\Xi_{\mu}$, and in
momentum space,

$\phi(\tilde{p})\to\Gamma_{\rho}\Gamma_{5}\Xi_{\rho}\Xi_{5}\phi(I_{\rho}\tilde{p})\
,$ (66)

where

$\Gamma_{5}=\Gamma_{1}\Gamma_{2}\Gamma_{3}\Gamma_{4}\
,\qquad\Xi_{5}=\Xi_{1}\Xi_{2}\Xi_{3}\Xi_{4}\ .$ (67)

This form follows because we need the matrix acting on $\phi$ to commute with
$\Gamma_{\mu}$ for $\mu\neq\rho$ and anticommute with $\Gamma_{\rho}$, because
$\sin(I_{\rho}\tilde{p}_{\rho})=-\sin(\tilde{p}_{\rho})$ (cf. Eq. (49)). In
position space, we have that [8]

$I_{\rho}:\quad\chi(x)\to(-1)^{x_{\rho}}\chi(I_{\rho}x)\ .$ (68)

We can define a parity transformation,

$I_{s}=I_{1}I_{2}I_{3}\ ,$ (69)

but we note that this involves a taste transformation:

$I_{s}:\ \phi(\tilde{p})\to\Gamma_{4}\Xi_{4}\phi(I_{s}\tilde{p})\ .$ (70)

Thus, “tasteless” parity $P$ can be defined by combining this with $S_{4}$,

$P:\ \phi(\tilde{p})\to e^{i\tilde{p}_{4}}\Gamma_{4}\phi(I_{s}\tilde{p})\ ,$
(71)

or, in position space,

$P:\
\chi(x)\to(-1)^{x_{1}+x_{2}+x_{3}}\chi(I_{s}x+4)=\eta_{4}(x)\chi(I_{s}x+4)\ .$
(72)

Of course, the gauge field also transforms non-trivially under axis reversal
[8].

####  II.4.3  Fermion number

Because the action is bilinear in $\chi$ and $\bar{\chi}$, it is invariant
under

$\chi\to e^{i\alpha}\chi\ ,\qquad\bar{\chi}\to\bar{\chi}e^{-i\alpha}\ .$ (73)

####  II.4.4  $U(1)_{\epsilon}$ symmetry

For $m=0$, the action is invariant under [5]

$\chi(x)\to e^{i\beta\epsilon(x)}\chi(x)\
,\qquad\bar{\chi}(x)\to\bar{\chi}(x)e^{i\beta\epsilon(x)}\ ,$ (74)

in which

$\epsilon(x)=(-1)^{x_{1}+x_{2}+x_{3}+x_{4}}=e^{i\pi_{\epsilon}x}\
,\qquad\pi_{\epsilon}=(\pi,\pi,\pi,\pi)\ .$ (75)

Since this is only a symmetry of the action for $m=0$, we can interpret this
symmetry as a chiral symmetry. (As always, this interpretation depends on the
form of the mass term. For instance, for a one-link mass term,
$U(1)_{\epsilon}$ is a vectorlike symmetry [8]!)

Using that

$T_{4}T_{3}T_{2}T_{1}S_{-1}S_{-2}S_{-3}S_{-4}:\ \chi(x)\to\epsilon(x)\chi(x)\
,$ (76)

and using the fact that $T_{\pm\mu}$ ($S_{\pm\mu}$) generates a $\Gamma_{\mu}$
($\Xi_{\mu}$), we find that

$\hat{\pi}_{\epsilon}=\Gamma_{5}\Xi_{5}\ ,$ (77)

and thus that in momentum space $U(1)_{\epsilon}$ symmetry takes the form

$\phi(\tilde{p})\to e^{i\beta\Gamma_{5}\Xi_{5}}\phi(\tilde{p})\
,\qquad\overline{\phi}(\tilde{p})\to\overline{\phi}(\tilde{p})e^{i\beta\Gamma_{5}\Xi_{5}}\
.$ (78)

Indeed, this looks like an non-singlet axial transformation.

####  II.4.5  Charge conjugation

The staggered action is invariant under charge conjugation $C_{0}$:

$\displaystyle\chi(x)\to i\epsilon(x)\bar{\chi}^{T}(x)\
,\qquad\bar{\chi}(x)\to i\epsilon(x)\chi^{T}(x)\ ,$ (79) $\displaystyle
U_{\mu}(x)\to U^{*}_{\mu}(x)\ ,$

where the transpose acts on the color (and possible flavor) indices. The
factors $i\epsilon(x)$ are needed to make both the mass term and the kinetic
term invariant under this symmetry. This implies that charge conjugation
involves a taste transformation; in momentum space it looks like

$\phi(\tilde{p})\to i\Gamma_{5}\Xi_{5}\overline{\phi}^{T}(-\tilde{p})\
,\qquad\overline{\phi}(\tilde{p})\to i\Gamma_{5}\Xi_{5}\phi^{T}(-\tilde{p})\
.$ (80)

The hope is, of course, that the lattice symmetries are sufficient to restore
the full $SO(4)_{\rm rot}\,\times\,SU(4)_{\rm taste}$ in the continuum limit
without any fine tuning of counter terms. We have noted already that the
hypercubic rotations (60) are a discrete subgroup of the diagonal $SO(4)_{\rm
diag}$ of $SO(4)_{\rm rot}\,\times\,SO(4)_{\rm taste}$, with $SO(4)_{\rm
taste}\,\subset\,SU(4)_{\rm taste}$. Likewise, the group $\Gamma_{4}$
generated by shifts (modulo even translations) is a discrete subgroup of
$SU(4)_{\rm taste}$. Finally, $U(1)_{\epsilon}$ provides one exact (non-
singlet, in taste space) chiral symmetry. At the classical level, we have seen
that indeed the continuum limit coincides with QCD. We will revisit this
question at the one-loop level in the next chapter, and in terms of the
Symanzik effective theory in Sec. IV.

###  II.5  Taste basis

Another basis that can sometimes be useful has been introduced a long time
ago, the so-called taste basis [9, 10]. We will review the construction here
for the free theory.

We define a unitary transformation

$\psi_{\alpha a}(y)=\frac{1}{2^{3/2}}\sum_{A}(\gamma_{A})_{\alpha
a}\chi(2y+A)\
,\qquad\overline{\psi}_{a\alpha}(y)=\frac{1}{2^{3/2}}\sum_{A}\bar{\chi}(2y+A)(\gamma_{A})^{\dagger}_{a\alpha}\
.$ (81)

Here $A$ runs over the 16 vectors with components $0$ or $1$. The $\psi$
fields live on a “coarse” lattice, labeled by lattice sites $y$, and

$\gamma_{A}=\gamma_{1}^{A_{1}}\gamma_{2}^{A_{2}}\gamma_{3}^{A_{3}}\gamma_{4}^{A_{4}}\
.$ (82)

The index $\alpha$ will be interpreted as a spin index, and the index $a$ as a
taste index below. In order to write the (free) one-component staggered
fermion action in terms of the fields $\psi$ and $\overline{\psi}$, we need to
invert Eq. (82). Using

${\rm tr}\,(\gamma_{B}^{\dagger}\gamma_{A})=4\delta_{AB}\ ,$ (83)

we find that

$\chi(2y+A)=\frac{1}{2^{5/2}}{\rm tr}\,(\gamma_{A}^{\dagger}\psi(y))\
,\qquad\bar{\chi}(2y+A)=\frac{1}{2^{5/2}}{\rm
tr}\,(\gamma_{A}\overline{\psi}(y))\ .$ (84)

We will also need the completeness relation

$\sum_{A}(\gamma_{A})_{\alpha
a}(\gamma_{A})^{\dagger}_{b\beta}=4\delta_{\alpha\beta}\delta_{ab}\ .$ (85)

Using this, it is straightforward to show that

$m\sum_{x}\bar{\chi}(x)\chi(x)=\frac{1}{8}m\sum_{y}\overline{\psi}(y)\psi(y)\
.$ (86)

The kinetic term is a little more involved:

$\displaystyle{1\over
2}\sum_{x\mu}\eta_{\mu}(x)(\bar{\chi}(x)\chi(x+\mu)-\bar{\chi}(x+\mu)\chi(x))$
$\displaystyle=$ $\displaystyle{1\over
2}\sum_{yA\mu}\eta_{\mu}(A)(\bar{\chi}(2y+A)\chi(2y+A+\mu)-\bar{\chi}(2y+A+\mu)\chi(2y+A))$
$\displaystyle=$ $\displaystyle{1\over
2}\sum_{yA\mu,A_{\mu}=0}\eta_{\mu}(A)(\bar{\chi}(2y+A)\chi(2y+A+\mu)-\bar{\chi}(2y+A+\mu)\chi(2y+A))$
$\displaystyle+{1\over
2}\sum_{yA\mu,A_{\mu}=1}\eta_{\mu}(A)(\bar{\chi}(2y+A)\chi(2y+A+\mu)-\bar{\chi}(2y+A+\mu)\chi(2y+A))$
$\displaystyle=$
$\displaystyle\frac{1}{2^{6}}\sum_{yA\mu,A_{\mu}=0}\eta_{\mu}(A)({\rm
tr}\,(\gamma_{A}\overline{\psi}(y)){\rm
tr}\,(\gamma_{A+\mu}^{\dagger}\psi(y))-{\rm
tr}\,(\gamma_{A+\mu}\overline{\psi}(y)){\rm
tr}\,(\gamma_{A}^{\dagger}\psi(y)))$
$\displaystyle+\frac{1}{2^{6}}\sum_{yA\mu,A_{\mu}=0}\eta_{\mu}(A)({\rm
tr}\,(\gamma_{A+\mu}\overline{\psi}(y)){\rm
tr}\,(\gamma_{A}^{\dagger}\psi(y+\mu))-{\rm
tr}\,(\gamma_{A}\overline{\psi}(y+\mu)){\rm
tr}\,(\gamma_{A+\mu}^{\dagger}\psi(y)))\ .$

We now use that

$\displaystyle\sum_{A,A_{\mu}=0}\eta_{\mu}(A)(\gamma_{A+\mu})_{\alpha
a}(\gamma_{A}^{\dagger})_{b\beta}$ $\displaystyle=$
$\displaystyle\sum_{A}\eta_{\mu}(A)(\gamma_{A+\mu})_{\alpha
a}(\gamma_{A}^{\dagger})_{b\beta}{1\over 2}((1+(-1)^{A_{\mu}})$
$\displaystyle={1\over
2}\sum_{A}(\gamma_{\mu})_{\alpha\gamma}(\gamma_{A})_{\gamma
a}(\gamma^{\dagger}_{A})_{b\beta}+{1\over
2}\sum_{A}(-1)^{A_{\mu}}\eta_{\mu}(A)\zeta_{\mu}(A)(\gamma_{A})_{\alpha
c}(\gamma_{\mu})_{ca}(\gamma^{\dagger}_{A})_{b\beta}$ $\displaystyle={1\over
2}\sum_{A}(\gamma_{\mu})_{\alpha\gamma}(\gamma_{A})_{\gamma
a}(\gamma^{\dagger}_{A})_{b\beta}+{1\over
2}\sum_{A}\epsilon(A)(\gamma_{A})_{\alpha
c}(\gamma_{\mu})_{ca}(\gamma^{\dagger}_{A})_{b\beta}$ $\displaystyle={1\over
2}\sum_{A}(\gamma_{\mu})_{\alpha\gamma}(\gamma_{A})_{\gamma
a}(\gamma^{\dagger}_{A})_{b\beta}+{1\over
2}\sum_{A}(\gamma_{5})_{\alpha\gamma}(\gamma_{A})_{\gamma
d}(\gamma_{5})_{dc}(\gamma_{\mu})_{ca}(\gamma^{\dagger}_{A})_{b\beta}$
$\displaystyle=2(\gamma_{\mu})_{\alpha\beta}\delta_{ab}+2(\gamma_{5})_{\alpha\beta}(\gamma_{5}\gamma_{\mu})_{ba}\
,$

and, likewise,

$\sum_{A,A_{\mu}=0}\eta_{\mu}(A)(\gamma_{A})_{\alpha
a}(\gamma_{A+\mu}^{\dagger})_{b\beta}=2(\gamma_{\mu})_{\alpha\beta}\delta_{ab}-2(\gamma_{5})_{\alpha\beta}(\gamma_{5}\gamma_{\mu})_{ba}\
.$ (89)

This allows us to write Eq. (II.5) as

$\displaystyle\frac{1}{2^{4}}{1\over 2}\sum_{y\mu}\Biggl{(}{\rm
tr}\,(\overline{\psi}(y)\gamma_{\mu}\psi(y+\mu)-\overline{\psi}(y+\mu)\gamma_{\mu}\psi(y))$
(90) $\displaystyle\hskip 36.98866pt+{\rm
tr}\,(\overline{\psi}(y)\gamma_{5}\psi(y+\mu)\gamma_{5}\gamma_{\mu}+\overline{\psi}(y+\mu)\gamma_{5}\psi(x)\gamma_{5}\gamma_{\mu}-2\overline{\psi}(y)\gamma_{5}\psi(y)\gamma_{5}\gamma_{\mu})\Biggr{)}\
.$

We see that in this form, the free action is that of a naive fermion, with the
addition of a spin ($\gamma_{5}$) and taste ($\gamma_{5}\gamma_{\mu}$) non-
diagonal Wilson-like mass term, which removes the doublers. The fermion field
$\psi$ is thus undoubled, and shows the explicit four-fold degeneracy of
staggered fermions in the continuum limit.

One can also translate the shift symmetry to the new basis. One obtains, in
the free theory, that

$S_{\mu}:\ \psi(y)\to{1\over 2}(\psi(x)+\psi(x+\mu))\gamma_{\mu}+{1\over
2}\gamma_{5}\gamma_{\mu}(\psi(x)-\psi(x+\mu))\gamma_{5}\approx\psi(y)\gamma_{\mu}\
,$ (91)

where the approximate equality on the right applies in the continuum limit. We
can identify a taste matrix $\xi_{\mu}=\gamma_{\mu}^{T}$ acting on the taste
index of the field $\psi$.

It is tempting to gauge Eq. (90) on the coarse lattice, instead of starting
from the one-component staggered action. However, it is clear that doing this
breaks shift symmetry,444Gauging also Eq. (91) does not make it into a
symmetry of the gauged version of Eq. (90). and we would thus end up with a
different interacting theory, with fewer lattice symmetries. We will see in
the next chapter that this would undo a key property of staggered fermions,
which is that the mass renormalizes multiplicatively.

##  III  One loop calculations

In this section, we calculate the one-loop vacuum polarization, and
demonstrate that it satisfies the Ward–Takahashi identity (WTI), which follows
from the gauge invariance of the lattice theory. We also consider the one-loop
fermion self energy, and show how staggered symmetries limit possible counter
terms. These examples illustrate the conjecture that lattice QCD (or QED) with
staggered fermions has a well-defined relativistic continuum limit, with no
need for additional counter terms in the lattice action.

###  III.1  Vacuum polarization

In Fig. 1 we show the two diagrams contributing at one loop to the vacuum
polarization. For simplicity, we consider the case of QED with gauge coupling
$g$. Using the Feynman rules of Sec. II.3, the tadpole diagram is equal to

$\Pi^{\rm
tadpole}_{\mu\nu}(k,\ell)=-g^{2}\,\delta_{\mu\nu}\int_{pq}\delta(q-p+k-\ell+\pi_{\eta_{\mu}})\,{1\over
2}\left(e^{i(q+k-\ell)_{\mu}}-e^{-iq_{\mu}}\right)G(p,-q)\ .$ (92)

Setting $p=\tilde{p}+\pi_{A}$ and $q=\tilde{q}+\pi_{B}$ this becomes

$-g^{2}\delta_{\mu\nu}e^{\frac{i}{2}(k-\ell)_{\mu}}\\!\\!\int_{\tilde{p}\tilde{q}}\sum_{AB}\delta(\tilde{q}-\tilde{p}+k-\ell+\pi_{A}+\pi_{B}+\pi_{\eta_{\mu}})i\sin(\tilde{q}_{\mu}+\mbox{$\small{{1\over
2}}$}(k_{\mu}-\ell_{\mu}))S_{\mu}^{B}G(\tilde{p}+\pi_{A},-\tilde{q}-\pi_{B})\
.$ (93)

We now use that (cf. Eq. (II.3))

$G(\tilde{p}+\pi_{A},-\tilde{q}-\pi_{B})=\delta(\tilde{p}-\tilde{q})S_{AB}(\tilde{p})\
,$ (94)

which sets the combination $\tilde{q}-\tilde{p}$ in
$\delta(\tilde{q}-\tilde{p}+k-\ell+\pi_{A}+\pi_{B}+\pi_{\eta_{\mu}})$ equal to
zero. Then, we use that $k$ and $\ell$, being external momenta, are taken
small, so that we can factorize

$\delta(k-\ell+\pi_{A}+\pi_{B}+\pi_{\eta_{\mu}})=\delta(k-\ell)(\hat{\pi}_{\eta_{\mu}})_{AB}\
,$ (95)

and use Eq. (35) to arrive at

$\displaystyle\Pi^{\rm tadpole}_{\mu\nu}(k)$ $\displaystyle=$
$\displaystyle-g^{2}\delta_{\mu\nu}\int_{\tilde{q}}{\rm
tr}\,\left(\Gamma_{\mu}\frac{-i\sum_{\kappa}\Gamma_{\kappa}\sin(\tilde{q}_{\kappa})+m}{\sum_{\kappa}\sin^{2}(\tilde{q}_{\kappa})+m^{2}}\right)i\sin(\tilde{q}_{\mu})$
$\displaystyle=$
$\displaystyle-16g^{2}\delta_{\mu\nu}\int_{\tilde{q}}\frac{\sin^{2}(\tilde{q}_{\mu})}{\sum_{\kappa}\sin^{2}(\tilde{q}_{\kappa})+m^{2}}\
,$

where we omitted $\delta(k-\ell)$. We see that $\Pi^{\rm tadpole}_{\mu\nu}(k)$
does not actually depend on $k$, as one would expect for a tadpole diagram.

$\mu$$\nu$$\to k$$\to\ell$$p$$q$$\mu$$\nu$$q^{\prime}$$p^{\prime}$$p$$q$$\to
k$$\to\ell$ Figure 1: The two diagrams contributing to the one-loop vacuum
polarization. The left panel shows the tadpole diagram, the right panel the
sunset diagram.

The other diagram in Fig. 1 is

$\displaystyle\Pi^{\rm sunset}_{\mu\nu}(k,\ell)$ $\displaystyle=$
$\displaystyle
g^{2}\int_{pqp^{\prime}q^{\prime}}\delta(q^{\prime}-p+k+\pi_{\eta_{\mu}})\,{1\over
2}\left(e^{i(q^{\prime}+k)_{\mu}}+e^{-iq^{\prime}_{\mu}}\right)G(p,-q)$
$\displaystyle\hskip
34.14322pt\times\delta(q-p^{\prime}-\ell+\pi_{\eta_{\nu}})\,{1\over
2}\left(e^{i(q-\ell)_{\nu}}+e^{-iq_{\nu}}\right)G(p^{\prime},-q^{\prime})\ .$

Integrating over $p^{\prime}$ and $q^{\prime}$ and writing
$p=\tilde{p}+\pi_{A}$, $q=\tilde{q}+\pi_{B}$ makes this equal to

$\displaystyle\Pi^{\rm sunset}_{\mu\nu}(k,\ell)$ $\displaystyle=$
$\displaystyle g^{2}\int_{\tilde{p}\tilde{q}}\sum_{AB}{1\over
2}S_{\mu}^{A}\left(e^{i\tilde{p}_{\mu}}+e^{-i(\tilde{p}-k)_{\mu}}\right)\delta(\tilde{p}-\tilde{q})S_{AB}(\tilde{p})$
$\displaystyle\times{1\over
2}S_{\nu}^{B}\left(e^{i(\tilde{q}-\ell)_{\nu}}+e^{-i\tilde{q}_{\nu}}\right)G(\tilde{q}-\ell+\pi_{\eta_{\nu}}+\pi_{B},-\tilde{p}+k+\pi_{\eta_{\mu}}+\pi_{A})\
,$

where we used the inverse of Eq. (II.3). Using that we can set
$\tilde{q}=\tilde{p}$ because of the $\delta(\tilde{p}-\tilde{q})$ inside the
integrand, we use Eqs. (45) to write

$\displaystyle
G^{-1}(\tilde{p}-\ell+\pi_{\eta_{\nu}}+\pi_{B},-\tilde{p}+k+\pi_{\eta_{\mu}}+\pi_{A})=$
(99) $\displaystyle\hskip
28.45274pt\sum_{\kappa}\delta(-\ell+k+\pi_{\eta_{\kappa}}+\pi_{\eta_{\nu}}+\pi_{B}+\pi_{\eta_{\mu}}+\pi_{A})iS_{\kappa}^{\eta_{\nu}}S_{\kappa}^{B}\sin(\tilde{p}_{\kappa}-\ell_{\kappa})$
$\displaystyle\hskip
28.45274pt+m\delta(-\ell+k+\pi_{\eta_{\nu}}+\pi_{B}+\pi_{\eta_{\mu}}+\pi_{A})\
.$

Now, using that $k$ and $\ell$ are small,

$\displaystyle\delta(-\ell+k+\pi_{\eta_{\nu}}+\pi_{B}+\pi_{\eta_{\mu}}+\pi_{A})$
$\displaystyle=$
$\displaystyle\delta(k-\ell)\overline{\delta}(\pi_{\eta_{\nu}}+\pi_{B}+\pi_{\eta_{\mu}}+\pi_{A})$
$\displaystyle=$
$\displaystyle\delta(k-\ell)\sum_{C}\overline{\delta}(\pi_{\eta_{\nu}}+\pi_{B}+\pi_{C})\overline{\delta}(\pi_{\eta_{\mu}}+\pi_{A}+\pi_{C})$
$\displaystyle=$
$\displaystyle\delta(k-\ell)(\hat{\pi}_{\eta_{\nu}}\hat{\pi}_{\eta_{\mu}})_{BA}\
,$

and, likewise,

$\displaystyle\delta(-\ell+k+\pi_{\eta_{\kappa}}+\pi_{\eta_{\nu}}+\pi_{B}+\pi_{\eta_{\mu}}+\pi_{A})S_{\kappa}^{\eta_{\nu}}S_{\kappa}^{B}=$
(101) $\displaystyle\hskip
28.45274pt=\delta(k-\ell)\overline{\delta}(\pi_{\eta_{\nu}}+\pi_{B}+\pi_{\eta_{\kappa}}+\pi_{\eta_{\mu}}+\pi_{A})S_{\kappa}^{\eta_{\nu}}S_{\kappa}^{B}$
$\displaystyle\hskip
28.45274pt=\delta(k-\ell)\sum_{CD}\overline{\delta}(\pi_{\eta_{\nu}}+\pi_{B}+\pi_{D})\overline{\delta}(\pi_{\eta_{\kappa}}+\pi_{D}+\pi_{C})\overline{\delta}(\pi_{\eta_{\mu}}+\pi_{C}+\pi_{A})S_{\kappa}^{D}$
$\displaystyle\hskip
28.45274pt=\delta(k-\ell)(\hat{\pi}_{\eta_{\nu}}\Gamma_{\kappa}\hat{\pi}_{\eta_{\mu}})_{BA}\
.$

We thus obtain

$G(\tilde{p}-\ell+\pi_{\eta_{\nu}}+\pi_{B},-\tilde{p}+k+\pi_{\eta_{\mu}}+\pi_{A})=\delta(k-\ell)(\hat{\pi}_{\eta_{\mu}}S(\tilde{p}-\ell)\hat{\pi}_{\eta_{\nu}})_{BA}\
.$ (102)

Combining with the factors $S_{\nu}^{B}$ and $S_{\mu}^{A}$ in Eq. (III.1), we
arrive at

$\Pi^{\rm
sunset}_{\mu\nu}(k,\ell)=g^{2}e^{\frac{i}{2}(k_{\mu}-\ell_{\nu})}\delta(k-\ell)\int_{\tilde{p}}{\rm
tr}\,\left(S(\tilde{p})\Gamma_{\mu}S(\tilde{p}-k)\Gamma_{\nu}\right)\cos(\tilde{p}_{\mu}-\mbox{$\small{{1\over
2}}$}k_{\mu})\cos(\tilde{p}_{\nu}-\mbox{$\small{{1\over 2}}$}k_{\nu})\ .$
(103)

Using

${\rm
tr}\,(\Gamma_{\kappa}\Gamma_{\nu}\Gamma_{\lambda}\Gamma_{\mu})=16(\delta_{\kappa\nu}\delta_{\lambda\mu}-\delta_{\kappa\lambda}\delta_{\mu\nu}+\delta_{\kappa\mu}\delta_{\lambda\nu})\
,$ (104)

this equals

$\displaystyle\Pi^{\rm
sunset}_{\mu\nu}(k)=16g^{2}e^{\frac{i}{2}(k_{\mu}-k_{\nu})}$ (105)
$\displaystyle\int_{\tilde{p}}\frac{\delta_{\mu\nu}(\sum_{\kappa}\sin(\tilde{p}_{\kappa})\sin(\tilde{p}_{\kappa}-k_{\kappa})+m^{2})-\sin(\tilde{p}_{\mu})\sin(\tilde{p}_{\nu}-k_{\nu})-\sin(\tilde{p}_{\nu})\sin(\tilde{p}_{\mu}-k_{\mu})}{(\sum_{\kappa}\sin^{2}(\tilde{p}_{\kappa})+m^{2})(\sum_{\lambda}\sin^{2}(\tilde{p}_{\lambda}-k_{\lambda})+m^{2})}$
$\displaystyle\phantom{\int_{\tilde{p}}}\times\cos(\tilde{p}_{\mu}-\mbox{$\small{{1\over
2}}$}k_{\mu})\cos(\tilde{p}_{\nu}-\mbox{$\small{{1\over 2}}$}k_{\nu})\ ,$

where again we omitted $\delta(k-\ell)$. Formally, in the continuum we would
have obtained one quarter of this result (with
$\sin(\tilde{p}_{\mu})\to\tilde{p}_{\mu}$, $\cos(\tilde{p}_{\mu}-{1\over
2}k_{\mu})\to 1$, etc.), confirming that there are four species doublers
(tastes) in the staggered theory.

Defining

$\Pi_{\mu\nu}(k)=\Pi^{\rm sunset}_{\mu\nu}(k)+\Pi^{\rm tadpole}_{\mu\nu}(k)\
,$ (106)

we first show that this vanishes at $k=0$, thus demonstrating that the tadpole
diagram removes the quadratic divergence present in the sunset diagram. At
$k=0$,

$\Pi^{\rm
sunset}_{\mu\nu}(0)=16g^{2}\int_{\tilde{p}}\left(-2\frac{\sin(\tilde{p}_{\mu})\sin(\tilde{p}_{\nu})\cos(\tilde{p}_{\mu})\cos(\tilde{p}_{\nu})}{(\sum_{\kappa}\sin^{2}(\tilde{p}_{\kappa})+m^{2})^{2}}+\delta_{\mu\nu}\frac{\cos^{2}(\tilde{p}_{\mu})}{\sum_{\kappa}\sin^{2}(\tilde{p}_{\kappa})+m^{2}}\right)\
.$ (107)

Using

$-2\,\frac{\sin(\tilde{p}_{\mu})\cos(\tilde{p}_{\mu})}{(\sum_{\kappa}\sin^{2}(\tilde{p}_{\kappa})+m^{2})^{2}}=\frac{\partial}{\partial\tilde{p}_{\mu}}\frac{1}{\sum_{\kappa}\sin^{2}(\tilde{p}_{\kappa})+m^{2}}$
(108)

and partially integrating the first term, one can show that $\Pi^{\rm
sunset}_{\mu\nu}(0)$ is equal to minus $\Pi^{\rm tadpole}_{\mu\nu}(0)$. There
are no boundary terms, because the integrand is periodic in every direction
with period $\pi$.

Next, we demonstrate that the WTI is satisfied, which on the lattice takes the
form

$\sum_{\mu}(1-e^{-ik_{\mu}})\Pi_{\mu\nu}(k)=0\ .$ (109)

Using that

$(1-e^{-ik_{\mu}})e^{\frac{i}{2}k_{\mu}}\Gamma_{\mu}\cos(\tilde{p}_{\mu}-{1\over
2}k_{\mu})=S^{-1}(\tilde{p})-S^{-1}(\tilde{p}-k)\ ,$ (110)

one can simplify

$\displaystyle\sum_{\mu}(1-e^{-ik_{\mu}})\Pi^{\rm sunset}_{\mu\nu}(k)$
$\displaystyle\\!\\!=\\!\\!$ $\displaystyle
16g^{2}e^{-\frac{i}{2}k_{\nu}}\int_{\tilde{p}}{\rm
tr}\,(\Gamma_{\nu}(S(\tilde{p}-k)-S(\tilde{p}))\cos(\tilde{p}_{\nu}-\mbox{$\small{{1\over
2}}$}k_{\nu})$
$\displaystyle=-16ig^{2}e^{-\frac{i}{2}}\int_{\tilde{p}}\left(\frac{\sin(\tilde{p}_{\nu}-k_{\nu})}{\sum_{\kappa}\sin^{2}(\tilde{p}_{\kappa}-k_{\kappa})+m^{2}}-\frac{\sin(\tilde{p}_{\nu})}{\sum_{\kappa}\sin^{2}(\tilde{p}_{\kappa})+m^{2}}\right)\cos(\tilde{p}_{\nu}-\mbox{$\small{{1\over
2}}$}k_{\nu})$
$\displaystyle=-16ig^{2}e^{-\frac{i}{2}k_{\nu}}\int_{\tilde{p}}\frac{\sin(\tilde{p}_{\nu})}{\sum_{\kappa}\sin^{2}(\tilde{p}_{\kappa})+m^{2}}(\cos(\tilde{p}_{\nu}+\mbox{$\small{{1\over
2}}$}k_{\nu})-\cos(\tilde{p}_{\nu}-\mbox{$\small{{1\over 2}}$}k_{\nu})$
$\displaystyle=16g^{2}(1-e^{-ik_{\nu}})\int_{\tilde{p}}\frac{\sin^{2}(\tilde{p}_{\nu})}{\sum_{\kappa}\sin^{2}(\tilde{p}_{\kappa})+m^{2}}\
,$

where we again used that the integrand is periodic in $\tilde{p}_{\mu}$ with
period $\pi$. This equals $-\sum_{\mu}(1-e^{-ik_{\mu}})\Pi^{\rm
tadpole}_{\mu\nu}(k)$, thus proving Eq. (109).

Finally, in this subsection, we will indicate how one may actually go about
completing the calculation of the vacuuum polarization at one-loop. We start
from the expression

$\displaystyle\Pi_{\mu\nu}(k)=16g^{2}e^{\frac{i}{2}(k_{\mu}-k_{\nu})}$ (112)
$\displaystyle\int_{\tilde{p}}\Biggl{(}\frac{\delta_{\mu\nu}(\sum_{\kappa}\sin(\tilde{p}_{\kappa})\sin(\tilde{p}_{\kappa}-k_{\kappa})+m^{2})-\sin(\tilde{p}_{\mu})\sin(\tilde{p}_{\nu}-k_{\nu})-\sin(\tilde{p}_{\nu})\sin(\tilde{p}_{\mu}-k_{\mu})}{(\sum_{\kappa}\sin^{2}(\tilde{p}_{\kappa})+m^{2})(\sum_{\lambda}\sin^{2}(\tilde{p}_{\lambda}-k_{\lambda})+m^{2})}$
$\displaystyle\phantom{\int_{\tilde{p}}}\times\cos(\tilde{p}_{\mu}-\mbox{$\small{{1\over
2}}$}k_{\mu})\cos(\tilde{p}_{\nu}-\mbox{$\small{{1\over 2}}$}k_{\nu})$
$\displaystyle\phantom{\int_{\tilde{p}}}-\frac{\delta_{\mu\nu}(\sum_{\kappa}\sin^{2}(\tilde{p}_{\kappa})+m^{2})-2\sin(\tilde{p}_{\mu})\sin(\tilde{p}_{\nu})}{(\sum_{\kappa}\sin^{2}(\tilde{p}_{\kappa})+m^{2})^{2}}\cos(\tilde{p}_{\mu})\cos(\tilde{p}_{\nu})\Biggr{)}\
,$

where the last line, obtained by setting $k=0$ in the middle two lines, is
equal to the tadpole contribution because $\Pi_{\mu\nu}(0)=0$. This form is
useful because it shows explicitly that $\Pi_{\mu\nu}(k)$ is logarithmically
divergent; the last line explicitly subtracts the quadratic divergence present
in $\Pi^{\rm sunset}_{\mu\nu}(k)$. Moreover, by shifting $\tilde{p}-{1\over
2}k\to\tilde{p}$, which is allowed in an integral that is only logarithmically
divergent, one can show that the integral is quadratic in $k$ in the continuum
limit (modulo, of course, a logarithmic dependence on $k$).

Returning to Eq. (112), we split the integral into two regions:

$\int_{\tilde{p}}=\int_{|\tilde{p}|<\delta}+\int_{|\tilde{p}|>\delta}\ ,$
(113)

where we take

$|ak|\sim|am|\ll\delta\ll 1\ ,$ (114)

and we take the limit $a\to 0$ first, followed by $\delta\to 0$. (In Eq. (114)
we wrote $k$ and $m$ in physical units, temporarily restoring the lattice
spacing $a$.) We will refer to these two regions as the “inner” and “outer”
regions.

In the inner region, we may replace the integrand by its covariant form:

$\displaystyle\Pi^{\rm inner}_{\mu\nu}(k)$ $\displaystyle=$ $\displaystyle
16g^{2}\int_{|\tilde{p}|<\delta}\Biggl{(}\frac{\delta_{\mu\nu}(\tilde{p}(\tilde{p}-k)+m^{2})-\tilde{p}_{\mu}(\tilde{p}_{\nu}-k_{\nu})-\tilde{p}_{\nu}(\tilde{p}_{\mu}-k_{\mu})}{(\tilde{p}^{2}+m^{2})((\tilde{p}-k)^{2}+m^{2})}$
$\displaystyle\phantom{16g^{2}\int_{|\tilde{p}|<\delta}\Biggl{(}}-\frac{\delta_{\mu\nu}(\tilde{p}^{2}+m^{2})-2\tilde{p}_{\mu}\tilde{p}_{\nu}}{(\tilde{p}^{2}+m^{2})^{2}}\Biggr{)}$
$\displaystyle=$
$\displaystyle\frac{1}{3\pi^{2}}g^{2}(k_{\mu}k_{\nu}-\delta_{\mu\nu}k^{2})\log(\delta^{2})+\mbox{finite\
terms}\ .$

Since $\Pi_{\mu\nu}(k)$ does not depend on $\delta$, the same logarithmic
divergence (up to a sign) also has to be present in the outer region integral.
We handle this by adding and subtracting

$\frac{1}{3}g^{2}(k_{\mu}k_{\nu}-\delta_{\mu\nu}k^{2})\int_{|p|>\delta}\frac{1}{(\sum_{\mu}\sin^{2}({1\over
2}p_{\mu}))^{2}}\ ,$ (116)

because, as we will show,

$I\equiv\lim_{\delta\to
0}\left(\int_{|p|>\delta}\frac{1}{(\sum_{\mu}\sin^{2}({1\over
2}p_{\mu}))^{2}}+\frac{1}{\pi^{2}}\log(\delta^{2})\right)$ (117)

is finite and can be explicitly calculated. This means that the sum of Eq.
(116) and the inner region is finite, and thus that the difference of the
outer region integral and Eq. (116) is finite as well. Since, in the outer
region, $|ak|\sim|am|\ll\delta<|\tilde{p}|$, we can expand the subtracted
outer-region integral in powers of $ak_{\mu}$ and $am$, keep those powers that
survive in the continuum limit, and then take $\delta$ to zero, leading to
purely numerical integrals that can be computed numerically. All non-analytic
dependence on $k$ and $m$ resides in the inner-region integral.

What remains to be done is to calculate Eq. (117). We begin rewriting this as

$\lim_{\delta\to 0}\left(\lim_{m\to
0}\left(\int_{p}\frac{1}{(\sum_{\mu}\sin^{2}({1\over
2}p_{\mu})+\frac{1}{4}m^{2})^{2}}-\int_{|p|<\delta}\frac{16}{(p^{2}+m^{2})^{2}}\right)+\frac{1}{\pi^{2}}\log(\delta^{2})\right)\
.$ (118)

The integral over the components $p_{\mu}$ of $p$ in the first integral runs
from $-\pi\to\pi$. We use

$\displaystyle\int_{p}\frac{1}{\sum_{\mu}\sin^{2}({1\over
2}p_{\mu})+\frac{1}{4}m^{2}}$ $\displaystyle=$ $\displaystyle
2\int_{0}^{\infty}dx\int_{p}e^{-{1\over 2}xm^{2}-2x\sum_{\mu}\sin^{2}({1\over
2}p_{\mu})}$ $\displaystyle=$ $\displaystyle 2\int_{0}^{\infty}dx\,e^{-{1\over
2}xm^{2}}e^{-4x}I_{0}^{4}(x)\ ,$

where $I_{0}(x)$ is the modified Bessel function, and hence

$\displaystyle\int_{p}\frac{1}{(\sum_{\mu}\sin^{2}({1\over
2}p_{\mu})+\frac{1}{4}m^{2})^{2}}$ $\displaystyle=$
$\displaystyle-4\,\frac{\partial}{\partial
m^{2}}\int_{p}\frac{1}{\sum_{\mu}\sin^{2}({1\over
2}p_{\mu})+\frac{1}{4}m^{2}}$ $\displaystyle=$ $\displaystyle
4\int_{0}^{\infty}dx\,x\,e^{-{1\over 2}xm^{2}}e^{-4x}I_{0}^{4}(x)\ .$

Furthermore,

$-16\int_{|p|<\delta}\frac{1}{(p^{2}+m^{2})^{2}}=-16\int_{|p|<1}\frac{1}{(p^{2}+\frac{m^{2}}{\delta^{2}})^{2}}=-\frac{1}{\pi^{2}}\log\frac{\delta^{2}}{m^{2}}+O\left(\frac{m^{2}}{\delta^{2}}\right)\
,$ (121)

showing that the double limit in Eq. (118) is finite. We then calculate the
integral on the left of Eq. (121) in a way similar to the calculation in Eq.
(III.1), obtaining

$-16\int_{|p|<\delta}\frac{1}{(p^{2}+m^{2})^{2}}=-\frac{1}{\pi^{2}}\int_{0}^{\infty}\frac{dx}{x}\,e^{-{1\over
2}x\frac{m^{2}}{\delta^{2}}}\left(1-(1+{1\over 2}x)e^{-{1\over 2}x}\right)\ .$
(122)

We can now combine all terms in Eq. (118), take first $m$ and then $\delta$ to
zero, obtaining

$I=4\int_{0}^{\infty}dx\,x\left(e^{-4x}I_{0}^{4}(x)-\frac{1}{4\pi^{2}x^{2}}\left(1-(1+{1\over
2}x)e^{-{1\over 2}x}\right)\right)=0.4855321\ .$ (123)

###  III.2  Fermion self energy

The one-loop fermion self energy can be calculated similarly. The steps are
similar to those followed in Sec. III.1 (for details, see Ref. [8]), and here
we just give the result (in Feynman gauge):

$\displaystyle\Sigma(\tilde{p})$ $\displaystyle=$ $\displaystyle
g^{2}\sum_{\mu}\int_{k}{1\over
2}(1+\cos(k_{\mu}+2\tilde{p}_{\mu}))\Gamma_{\mu}S(k+\tilde{p})\Gamma_{\mu}D(k)$
$\displaystyle-g^{2}\sum_{\mu}\Gamma_{\mu}\,i\sin(\tilde{p}_{\mu})\int_{k}D(k)\
,$ $\displaystyle D(k)$ $\displaystyle=$
$\displaystyle\frac{1}{4\sum_{\mu}\sin^{2}({1\over 2}k_{\mu})}\ .$

The term on the second line comes from a gauge-field tadpole diagram. This
result can be further evaluated by splitting into inner and outer regions,
etc., as we did for the vacuum polarization. $\Sigma(\tilde{p})$ is a
$16\times 16$ matrix, and thus a linear combination of products of the
matrices $\Gamma_{\mu}$ and $\Xi_{\nu}$. In the continuum limit, non-analytic
dependence on $\tilde{p}$ and $m$ is the same as we would obtain in any other
regularization, and comes from the inner-region integral.

By dimensional analysis, the most general contact term one can find in the
continuum limit has the form

$C\equiv\tilde{p}_{\mu}X_{\mu}+\frac{1}{a}\,Y_{0}+mY\ ,$ (125)

where again we restored the lattice spacing, and where $X_{\mu}$, $Y_{0}$ and
$Y$ are $16\times 16$ matrices. We will now see how lattice symmetries
restrict the form of $C$. First, shift symmetry implies that $C$ should
commute with all $\Xi_{\nu}$, and therefore $X_{\mu}$, $Y_{0}$ and $Y$ cannot
contain any $\Xi_{A}$.

Then (hypercubic) rotational symmetry implies that $X_{\mu}$ has to be a
vector, which implies that $X_{\mu}\propto\Gamma_{\mu}$ or
$X_{\mu}\propto\Gamma_{\mu}\Gamma_{5}$, while both $Y_{0}$ and $Y$ have to be
proportional to the $16\times 16$ unit matrix or $\Gamma_{5}$. Reflection
symmetry $I_{s}$ of Eq. (70) excludes $\Gamma_{5}$ from appearing in any of
these, and thus we find that

$C=\alpha\,\tilde{p}_{\mu}\Gamma_{\mu}+\beta\,\frac{1}{a}+\gamma\,m\ ,$ (126)

with $\alpha$, $\beta$ and $\gamma$ constants. Finally, we use that for $m=0$
we have $U(1)_{\epsilon}$ symmetry, which implies that, for $m=0$, $C$ should
anti-commute with $\Gamma_{5}\Xi_{5}$. This sets $\beta=0$, and we thus see
that $m$ renormalizes multiplicatively, while the contact terms take on the
form expected in the continuum theory [8] (with coefficients specific to the
lattice regularization, of course).

We note that, in this argument, shift symmetry played a crucial role. Suppose
we were to gauge the staggered theory on the taste basis, with action (90)
plus (86) on the coarse lattice, losing shift symmetry in the process because
the gauge fields $U_{\mu}(y)$ would now live on the coarse lattice. We would
expect a counter term of the form

$\sum_{\mu}{\rm
tr}\,(\overline{\psi}(y)\gamma_{5}\psi(y)\gamma_{5}\gamma_{\mu})$ (127)

because of the form of the “Wilson” term in Eq. (90). On our momentum basis,
this translates into a counter term
$Y_{0}\propto\sum_{\mu}\Gamma_{5}\Xi_{5}\Xi_{\mu}$. Indeed, this anti-commutes
with $\Gamma_{5}\Xi_{5}$, so it is not excluded by $U(1)_{\epsilon}$ symmetry!
And, indeed, such a counter term appears at one loop in the theory on the
taste basis if it is gauged on the coarse lattice [11]. One can verify that
the action on the taste basis is invariant under hypercubic rotations of the
form (64). In particular,

$\tilde{R}^{-1}_{\rho\sigma}\left(\sum_{\mu}\Gamma_{5}\Xi_{5}\Xi_{\mu}\right)\tilde{R}_{\rho\sigma}=\sum_{\mu}\Gamma_{5}\Xi_{5}\Xi_{\mu}\
,$ (128)

showing that a counter term of the form (127) is consistent with rotational
symmetry.

##  IV  Symanzik effective theory for staggered fermions

The Symanzik effective theory (SET) is a continuum effective theory that
reproduces correlation functions of the underlying lattice theory expanded in
powers of the lattice spacing, for momenta $\Lambda_{\rm QCD}\ll|p|\ll 1/a$,
where $p$ is the typical momentum of these correlation functions.555In these
notes we ignore anomalous dimensions. However, these could become important
with high precision computations near the continuum limit [12]. In this
chapter, we discuss the SET for QCD with $N_{f}$ flavors of staggered
fermions, for $N_{f}>1$. The SET is formulated in terms of continuum quark
fields $q_{i}$ ($i$ is the flavor index; we leave spin and color indices
implicit) and gluons $A_{\mu}$, with field strength $G_{\mu\nu}$ [13, 14],
because at these momenta quarks and gluons are still the relevant degrees of
freedom.

At leading order, the SET consists of all 4-dimensional operators consistent
with the lattice symmetries. As we have seen in the previous chapter, the
fermion mass renormalizes multiplicatively, and there are no dimension-3
operators.666In the previous chapter, we demonstrated this only to one loop,
but the symmetry arguments we used generalize to higher loops. As we have
seen, a crucial role in arriving at this conclusion is played by shift and
$U(1)_{\epsilon}$ symmetries. Dimensional analysis, gauge invariance, and
staggered symmetries imply that the dimension-4 operators in the SET are just
those defining continuum QCD.

Before we go on to discuss operators with dimension $d$ larger than 4 (which
are thus multiplied by $a^{d-4}$, with $a$ the lattice spacing), we make a
general observation on how shift symmetry works in the SET, using the fact
that this is a continuum theory, which, by assumption, is invariant under
continuous translation symmetry [15]. In momentum space, a shift symmetry
takes the form

$\phi(p)\to S_{\mu}\phi(p)=e^{iap_{\mu}}\Xi_{\mu}\phi(p)\ ,$ (129)

where $-\pi/(2a)<p\leq\pi/(2a)$ is the physical momentum of the field $\phi$
on which $S_{\mu}$ acts, and we made the lattice spacing dependence explicit.
The SET is invariant under shift symmetry because the underlying lattice
theory is. At the same time, it is also invariant under continuous
translations of the form

$\phi(p)\to e^{ipr}\phi(p)\ ,$ (130)

where $r$ is the vector over which we translate the field. Combining these two
symmetries, choosing $r_{\mu}=-a$, $r_{\nu\neq\mu}=0$, we find that the SET is
invariant under

$\phi(p)\to\Xi_{\mu}\phi(p)\ .$ (131)

This is convenient, as this symmetry does not mix operators of different
dimensionality in the SET. Below we will make extensive use of Eq. (131), and
we will still refer to this symmetry as “shift” symmetry. For additional
arguments supporting this claim, diagramatically or using the taste basis, see
Ref. [15].

In the discussion below, we will assume that a basis transformation has been
carried out such that

$\Gamma_{\mu}\to\gamma_{\mu}1_{4}\ ,\qquad\Xi_{\mu}\to 1_{4}\xi_{\mu}\ ,$
(132)

where we omit the direct-product sign (i.e.,
$\gamma_{\mu}1_{4}=\gamma_{\mu}\times 1_{4}$, etc.), and we will work with the
spin matrices $\gamma_{\mu}$ and the taste matrices $\xi_{\mu}$, both in the
4-dimensional irrep of the Dirac algebra in four dimensions. We will also drop
the index $4$ on the $4\times 4$ unit matrix; in fact, we will mostly not
write the unit matrix at all, when no confusion is possible.

###  IV.1  Dimension-5 operators

We begin with operators of dimension 5, which would be of order $a$ in the
Symanzik expansion [16]. These operators can at most contain one bilinear in
$q_{i}$ and $\overline{q}_{i}$. They cannot contain any non-trivial taste
matrix, because of the symmetry (131). Bilinears of the form

$\overline{q}_{i}\sigma_{\mu\nu}G_{\mu\nu}q_{i}\
,\qquad\overline{q}_{i}D_{\mu}D_{\mu}q_{i}$ (133)

(where $D_{\mu}$ is the color-covariant derivative) are ruled out immediately
by $U(1)_{\epsilon}$ symmetry. In order to consider other dimension-5
operators, we first define

$\displaystyle q_{Li}\\!\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!\\!{1\over 2}(1-\gamma_{5}\xi_{5})q_{i}\ ,\qquad
q_{Ri}={1\over 2}(1+\gamma_{5}\xi_{5})q_{i}\ ,$ (134)
$\displaystyle\overline{q}_{Li}\\!\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!\\!\overline{q}_{i}{1\over 2}(1+\gamma_{5}\xi_{5})\
,\qquad\overline{q}_{Ri}=\overline{q}_{i}{1\over 2}(1-\gamma_{5}\xi_{5})\ .$

The lattice theory has a $U(N_{f})_{L}\times U(N_{f})_{R}$ symmetry,777Note
that this symmetry is smaller than the $U(4N_{f})_{L}\times U(4N_{f})_{R}$
symmetry of the continuum limit. under which these fields transform as

$\displaystyle q_{L}\to V_{L}q_{L}\ ,\qquad$ $\displaystyle q_{R}\to
V_{R}q_{R}\ ,$
$\displaystyle\overline{q}_{L}\to\overline{q}_{L}V_{L}^{\dagger}\ ,\qquad$
$\displaystyle\overline{q}_{R}\to\overline{q}_{R}V_{R}^{\dagger}\ ,$

with

$V_{L}=e^{\frac{i}{2}\alpha^{a}T^{a}(1-\gamma_{5}\xi_{5})}\ ,\qquad
V_{R}=e^{\frac{i}{2}\beta^{a}T^{a}(1+\gamma_{5}\xi_{5})}\ ,$ (136)

in which $T^{a}$ are the $U(N_{f})$ generators. This symmetry is a
generalization of $U(1)\times U(1)_{\epsilon}$ symmetry for one flavor to the
case of $N_{f}$ flavors. Note that other symmetries, such as shift and
hypercubic symmetry, do not enlarge with the presence of more flavors, because
the gauge field transforms non-trivially under these symmetries.

Mass terms are not invariant under this symmetry, but can be made invariant by
making the $N_{f}\times N_{f}$ mass matrix ${\cal M}$ into a spurion field
(i.e., a non-dynamical field) that transforms as

${\cal M}\to V_{R}{\cal M}V_{L}^{\dagger}\ ,$ (137)

so that the 4-dimensional staggered mass term

$\bar{\chi}_{R}(x){\cal M}\chi_{L}(x)+\bar{\chi}_{L}(x){\cal
M}^{\dagger}\chi_{R}(x)$ (138)

is invariant (with the symmetry $U(N_{f})_{L}\times U(N_{f})_{R}$ acting on
the staggered fields $\chi$ and $\bar{\chi}$ as in Eq. (136) with
$\gamma_{5}\xi_{5}$ replaced by $\epsilon(x)$). The idea is to use the spurion
form of ${\cal M}$ to see what terms can show up in the SET, and then, once we
have found all such terms, set ${\cal M}$ equal to the physical mass matrix
$M$. The dimension-4 mass term in the SET,

$\overline{q}_{R}{\cal M}q_{L}+\overline{q}_{L}{\cal M}^{\dagger}q_{R}$ (139)

is also invariant. We are now ready to return to dimension-5 quark bilinears
containing one or more masses. Using the transformations (IV.1) and (137), the
symmetry $U(N_{f})_{L}\times U(N_{f})_{R}$ rules out terms of the form

$\overline{q}M\hbox to0.0pt{\hbox{$\mskip 1.0mu/$}\hss}\\!Dq\
,\qquad\overline{q}M^{2}q\ ,\qquad{\rm tr}\,(M)G_{\mu\nu}G_{\mu\nu}\ .$ (140)

This exhausts all possible dimension-5 contributions to the SET, and we thus
conclude that dependence on the lattice spacing starts at order $a^{2}$, i.e.,
at dimension 6. We note that no use of equations of motion or field
redefinitions was made to rule out all dimension-5 operators, and our
conclusion thus holds for off-shell correlation functions as well.

We end this subsection with a comment on the taste basis of Sec. II.5. The
action (90) has a Wilson term, which is, in fact, a dimension-5 operator.
However, this action is invariant under shift symmetry in the form (91), but
not under the symmetry $\psi\to\psi\gamma_{\mu}$, which is the continuum form
equivalent to Eq. (131). As it is the latter symmetry which is relevant for
restricting the form of the SET, there is thus no contradiction.

###  IV.2  Dimension-6 operators

In this section, we will discuss the order-$a^{2}$ part of the SET, i.e., all
dimension-6 operators that contribute, with the focus on operators that break
$U(4)_{\rm taste}$ (which is, of course, only a symmetry of the continuum
limit). These operators were constructed in Refs. [16, 17] (see also Ref.
[18]).

There are no taste-breaking bilinears. This follows from the fact that any
bilinear has to be invariant under the discrete group $\Gamma_{4}$, generated
by the $\xi_{\mu}$. (There exist taste-invariant bilinear operators, such as
$\overline{q}\,\hbox to0.0pt{\hbox{$\mskip 1.0mu/$}\hss}\\!D^{3}q$, see the
end of this subsection.)

Many 4-quark operators can be constructed, and if we require them to be
invariant under $SU(N_{f})$, they take the form888They can be color-mixed or
-unmixed.

$(\overline{q}_{i}Xq_{i})(\overline{q}_{j}Yq_{j})\ ,$ (141)

in which sums over $i$ and $j$ are implied, and where $X=\gamma_{A}\xi_{B}$
and $Y=\gamma_{C}\xi_{D}$, with

$\displaystyle\gamma_{A}\ ,\gamma_{C}$ $\displaystyle\in$ $\displaystyle\\{1,\
\gamma_{\mu},\ i\gamma_{\mu}\gamma_{\nu},\ i\gamma_{5}\gamma_{\mu},\
\gamma_{5}\\}\ ,$ (142) $\displaystyle\xi_{B}\ ,\xi_{D}$ $\displaystyle\in$
$\displaystyle\\{1,\ \xi_{\mu},\ i\xi_{\mu}\xi_{\nu},\ i\xi_{5}\xi_{\mu},\
\xi_{5}\\}\ ,$

in which $\mu<\nu$. Since no mass matrix can appear in these operators,
$U(N_{f})_{L}\times U(N_{f})_{R}$ symmetry implies that both
$\gamma_{A}\xi_{B}$ and $\gamma_{C}\xi_{D}$ have to anti-commute with
$\gamma_{5}\xi_{5}$, which, in turn, implies that

$\displaystyle\gamma_{A}\in\\{1\ \mbox{(S)},\ i\gamma_{\mu}\gamma_{\nu}\
\mbox{(T)},\ \gamma_{5}\ \mbox{(P)}\\}\quad$
$\displaystyle\Rightarrow\quad\xi_{B}\in\\{\xi_{\mu}\ \mbox{(V)},\
i\xi_{5}\xi_{\mu}\ \mbox{(A)}\\}\ ,$
$\displaystyle\gamma_{A}\in\\{\gamma_{\mu}\ \mbox{(V)},\
i\gamma_{5}\gamma_{\mu}\ \mbox{(A)}\\}\quad$
$\displaystyle\Rightarrow\quad\xi_{B}\in\\{1\ \mbox{(S)},\
i\xi_{\mu}\xi_{\nu}\ \mbox{(T)},\ \xi_{5}\ \mbox{(P)}\\}\ .$

Here we denoted the various tensor structures with S (scalar), T (tensor), P
(pseudo-scalar), V (vector) and A (axial-vector), for later convenience. Both
bilinears thus have to be “odd,” which means that the spin-taste matrix has to
have an odd number of Lorentz indices. Translating this back to lattice
operators implies that the corresponding lattice bilinears have to be odd-link
operators, since each $\gamma_{\mu}$ or $\xi_{\mu}$ is generated by a shift
over one lattice spacing in the $\mu$ direction.

Next, $\Gamma_{4}$ symmetry implies that $\xi_{B}=\xi_{D}$, because, if
$\xi_{B}$ were not equal to $\xi_{D}$, we could find a $\xi_{X}$ such that
$\xi_{B}\xi_{X}=-\xi_{X}\xi_{B}$ but $\xi_{D}\xi_{X}=+\xi_{X}\xi_{D}$, and the
operator would not be invariant under $q\to\xi_{X}q$,
$\overline{q}\to\overline{q}\xi_{X}$. Our operators thus have the form

$(\overline{q}_{i}\gamma_{A}\xi_{B}q_{i})(\overline{q}_{j}\gamma_{C}\xi_{B}q_{j})\
,$ (144)

with the constraint (IV.2). A similar argument sets $\gamma_{A}=\gamma_{C}$.
Rotations over 180 degrees take the form (cf. Eq. (60))

$q\to\gamma_{\rho}\gamma_{\sigma}\xi_{\rho}\xi_{\sigma}q\
,\quad\rho\neq\sigma\ ,$ (145)

and if we combine these with shifts in the $\rho$ and $\sigma$ directions, it
follows that the SET has to be invariant under

$q(x)\to\gamma_{\rho}\gamma_{\sigma}q(R^{-1}_{\rho\sigma}(\pi)x),\quad\rho\neq\sigma\
.$ (146)

Likewise, we can combine $I_{s}$ of Eq. (70) with a shift in the 4th
direction, resulting in a SET symmetry

$q(x)\to\gamma_{4}q(I_{s}x)\ .$ (147)

Together, these imply that $\gamma_{A}$ has to be equal to $\gamma_{C}$ (with
Eq. (147) excluding $\gamma_{C}=\gamma_{A}\gamma_{5}$). Our 4-fermion
operators thus take the form

$(\overline{q}_{i}\gamma_{A}\xi_{B}q_{i})(\overline{q}_{j}\gamma_{A}\xi_{B}q_{j})\
,$ (148)

with the proviso that $\gamma_{A}\xi_{B}$ has to be odd. It remains to
consider rotations over 90 degrees. In order to make sure our 4-fermion
operators are invariant under these, we need to sum over Lorentz indices, to
make all operators into scalars under 90-degree rotations. There will be two
types of operators [16]: those operators which are invariant not only under
hypercubic rotations, but also under $SO(4)$, which we will refer to as “type
A,” and those which are not invariant under $SO(4)$, but only under hypercubic
rotations, “type B.” To show how this works, it is easiest to give some
examples. Examples of type-A operators are

$\displaystyle\sum_{\mu}(\bar{\gamma}_{i}\xi_{\mu}q_{i})(\overline{q}_{j}\xi_{\mu}q_{j})\
,$ (149)
$\displaystyle\sum_{\mu,\nu\neq\rho}(\overline{q}_{i}\gamma_{\mu}\,i\xi_{\nu}\xi_{\rho}q_{i})(\overline{q}_{j}\gamma_{\mu}\,i\xi_{\nu}\xi_{\rho}q_{j})\
.$

These operators are both invariant under $SO(4)$. The first of these two
operators we refer to as $S\times V$, since $\gamma_{A}$ is scalar and
$\xi_{B}$ is vector; the second we refer to as $V\times T$, because
$\gamma_{A}$ is vector and $\xi_{B}$ is tensor. An example of a type-B
operator is

$\sum_{\mu\neq\nu}(\overline{q}_{i}\gamma_{\mu}\,i\xi_{\mu}\xi_{\nu}q_{i})(\overline{q}_{j}\gamma_{\mu}\,i\xi_{\mu}\xi_{\nu}q_{j})\
.$ (150)

Because the index $\mu$ is repeated four times, this operator is invariant
under hypercubic rotations, but not under $SO(4)$. We may refer to this
operator as $V_{\mu}\times T_{\mu}$, where the repeated $\mu$ reminds us of
the four times repeated index $\mu$ in Eq. (150). This type can only occur if
either $\gamma_{A}$ or $\xi_{B}$ is tensor (and thus the other is vector or
axial-vector).

In fact, the type-B operator of our example can be taken to be

$\sum_{\mu\neq\nu}\left((\overline{q}_{i}\gamma_{\mu}\,i\xi_{\mu}\xi_{\nu}q_{i})(\overline{q}_{j}\gamma_{\mu}\,i\xi_{\mu}\xi_{\nu}q_{j})-(\overline{q}_{i}\gamma_{\mu}\,i\xi_{\mu}\xi_{\nu}\xi_{5}q_{i})(\overline{q}_{j}\gamma_{\mu}\,i\xi_{\mu}\xi_{\nu}\xi_{5}q_{j})\right)\
,$ (151)

because the orthogonal combination

$\displaystyle\sum_{\mu\neq\nu}\left((\overline{q}_{i}\gamma_{\mu}\,i\xi_{\mu}\xi_{\nu}q_{i})(\overline{q}_{j}\gamma_{\mu}\,i\xi_{\mu}\xi_{\nu}q_{j})+(\overline{q}_{i}\gamma_{\mu}\,i\xi_{\mu}\xi_{\nu}\xi_{5}q_{i})(\overline{q}_{j}\gamma_{\mu}\,i\xi_{\mu}\xi_{\nu}\xi_{5}q_{j})\right)$
$\displaystyle=$
$\displaystyle\sum_{\mu\neq\nu}\left((\overline{q}_{i}\gamma_{\mu}\,i\xi_{\mu}\xi_{\nu}q_{i})(\overline{q}_{j}\gamma_{\mu}\,i\xi_{\mu}\xi_{\nu}q_{j})+(\overline{q}_{i}\gamma_{\mu}\,i\xi_{\rho}\xi_{\sigma}q_{i})(\overline{q}_{j}\gamma_{\mu}\,i\xi_{\rho}\xi_{\sigma}q_{j})_{\rho\sigma\neq\mu\nu}\right)$
$\displaystyle=$
$\displaystyle\sum_{\mu,\rho\neq\sigma}(\overline{q}_{i}\gamma_{\mu}\,i\xi_{\rho}\xi_{\sigma}q_{i})(\overline{q}_{j}\gamma_{\mu}\,i\xi_{\rho}\xi_{\sigma}q_{j})\
,$

which is type A. We see that the requirement that the 4-quark operators are
invariant under the staggered symmetry group leads to a huge reduction in the
number of 4-quark operators. With $\gamma_{A}$, $\xi_{B}$, $\gamma_{C}$ and
$\xi_{D}$ unrestricted, there are $16^{4}=65536$ choices!

For $N_{f}=1$ there is a subtlety, based on the fact that the flavor symmetry
group is just $U(1)\times U(1)_{\epsilon}$. It is straightforward to verify
that, e.g.,

$(\overline{q}q)(\overline{q}q)-(\overline{q}\gamma_{5}\xi_{5}q)(\overline{q}\gamma_{5}\xi_{5}q)$
(153)

is invariant under this group. Using our shorthand notation, we can label this
operator as $S\times S-P\times P$. The fermion bilinears in this combination
are even, and this operator thus does not fit into the general pattern we
established above. However, it can be made to fit, by a Fierz transformation
[16], as we will now show.

A Fierz transformation is based on identities of the form

$(\gamma_{A})_{\alpha\beta}(\gamma_{A})_{\gamma\delta}=\sum_{B}c(A,B)(\gamma_{B})_{\alpha\delta}(\gamma_{B})_{\gamma\beta}\
,$ (154)

with numerical coefficients $c(A,B)$ that depend on $A$ and $B$. For instance,

$1_{\alpha\beta}1_{\gamma\delta}=\frac{1}{4}\sum_{B}(\gamma_{B})_{\alpha\delta}(\gamma_{B})_{\gamma\beta}\
.$ (155)

Similar identities hold, of course, for the $\xi_{A}$. Using the abbreviations
$S$, $V$, $T$, $A$ and $P$ introduced above, we can use Eq. (155) for both the
$\gamma$ and $\xi$ matrices to write

$\displaystyle S\times S$ $\displaystyle=$
$\displaystyle\frac{1}{16}(S+V+T+A+P)\times(S+V+T+A+P)\ ,$ (156)
$\displaystyle P\times P$ $\displaystyle=$
$\displaystyle\frac{1}{16}(S-V+T-A+P)\times(S-V+T-A+P)\ .$

Subtracting these two, we find that

$\displaystyle S\times S-P\times P=\frac{1}{8}\Bigl{(}S\times V+V\times
S+T\times V+V\times T+P\times V+V\times P$ $\displaystyle\hskip
56.9055pt+S\times A+A\times S+T\times A+A\times T+P\times A+A\times P\Bigr{)}\
,$ (157)

which allows us to write the operator (153) in the general form we derived
above for arbitrary $N_{f}$. Of course, Fierzing interchanges color-mixed and
color-unmixed operators, but both types occur in the general analysis as well.

There are also dimension-6 operators with just one quark bilinear. Staggered
and flavor symmetries restrict these to the list999The operator
$\sum_{\mu}\overline{q}_{i}(D_{\mu}^{2}\,\hbox to0.0pt{\hbox{$\mskip
1.0mu/$}\hss}\\!D-\,\hbox to0.0pt{\hbox{$\mskip
1.0mu/$}\hss}\\!DD_{\mu}^{2})q_{i}$ has negative charge parity [16].

$\displaystyle\overline{q}_{i}\,\hbox to0.0pt{\hbox{$\mskip
1.0mu/$}\hss}\\!D^{3}q_{i}\
,\quad\sum_{\mu}\overline{q}_{i}(D_{\mu}^{2}\,\hbox to0.0pt{\hbox{$\mskip
1.0mu/$}\hss}\\!D+\,\hbox to0.0pt{\hbox{$\mskip
1.0mu/$}\hss}\\!DD_{\mu}^{2})q_{i}\
,\quad\sum_{\mu}\overline{q}_{i}D_{\mu}\,\hbox to0.0pt{\hbox{$\mskip
1.0mu/$}\hss}\\!DD_{\mu}q_{i}\
,\quad\sum_{\mu}\overline{q}_{i}\gamma_{\mu}D_{\mu}^{3}q_{i}\ ,$
$\displaystyle\overline{q}_{i}M\,\hbox to0.0pt{\hbox{$\mskip
1.0mu/$}\hss}\\!D^{2}q_{i}\ ,\quad\overline{q}_{i}MD_{\mu}^{2}q_{i}\
,\quad\overline{q}_{i}M^{2}\,\hbox to0.0pt{\hbox{$\mskip
1.0mu/$}\hss}\\!Dq_{i}\ ,\quad\overline{q}_{i}M^{3}q_{i}\ .$ (158)

In particular, the fermion bilinear cannot contain any taste matrix, because
of shift symmetry. The operators containing the mass matrix $M$ can be
constructed using Eqs. (IV.1) and (137). For example, the last operator is
obtained from

$\overline{q}_{Ri}{\cal M}{\cal M}^{\dagger}{\cal
M}q_{Li}+\overline{q}_{Li}{\cal M}^{\dagger}{\cal M}{\cal M}^{\dagger}q_{Ri}\
,$ (159)

and setting ${\cal M}={\cal M}^{\dagger}=M$. Finally, there are purely gluonic
dimension-6 operators, shared with other discretizations of QCD, which we do
not list here.

###  IV.3  Operators with dimension larger than 6

We briefly discuss what happens beyond dimension 6, i.e., at order $a^{3}$ and
higher.

####  IV.3.1  Dimension 7

Let us start with dimension-7 operators. We again restrict ourselves to
operators containing quark fields (no purely gluonic operators with odd
dimension exist). Dimension-7 operators, if they exist, can contain one or two
quark bilinears.

First, consider operators with only one quark bilinear. Starting from the list
in Eq. (IV.2), we need to add one mass dimension, which can be done by adding
an extra covariant derivative $D_{\mu}$, or an extra mass matrix $M$. However,
if we insert a $D_{\mu}$, Lorentz indices still have to balance, and the only
way to do that is to remove or add a $\gamma_{\mu}$ inside the bilinear. But
this would change the properties of this bilinear under $U(N_{f})_{L}\times
U(N_{f})_{R}$, so this does not work. An analogous argument excludes adding a
mass matrix $M$ (using again a spurion-based argument).

Essentially the same argument applies to operators with two quark bilinears.
Inserting a $D_{\mu}$ leads to an unmatched Lorentz index, or a change in the
symmetry properties of one fermion bilinear, making it inconsistent with
$U(N_{f})_{L}\times U(N_{f})_{R}$. A similar argument applies with inserting a
mass matrix $M$. If we replace one of the two bilinears
$\overline{q}_{i}Xq_{i}$ with $\overline{q}_{i}MYq_{i}$, we need to choose $Y$
such that $\overline{q}_{i}MYq_{i}$ is invariant under $U(N_{f})_{L}\times
U(N_{f})_{R}$ (after promoting $M$ to the spurions ${\cal M}$ and ${\cal
M}^{\dagger}$). But this implies that $Y\neq X$, and thus shift and/or lattice
rotation/reflection symmetries would be broken. We conclude that no
dimension-7 operators exist that are part of the SET.

####  IV.3.2  Dimension 8

In contrast, many operators of dimension 8 exist, and they are easily
constructed from dimension-6 operators, e.g., by inserting $D_{\mu}^{2}$ or
$\hbox to0.0pt{\hbox{$\mskip 1.0mu/$}\hss}\\!D^{2}$, or by multiplying with
${\rm tr}\,(M^{2})$. However, all continuum symmetries are already broken down
to the staggered symmetry group at the level of dimension-6 operators (because
of type-B operators!), so no new symmetry breaking occurs at order $a^{4}$
[19].101010In particular, no new taste spurions occur, because the number of
quark bilinears is still restriced to two. See next section.

####  IV.3.3  Odd dimension $\geq 9$

At dimension 9, we should consider operators with three quark bilinears.
Because of dimensional reasons, these cannot contain any derivatives or
masses. Each of the bilinears needs to be invariant under $U(N_{f})_{L}\times
U(N_{f})_{R}$, which implies that each bilinear has to be odd—it has an odd
number of Lorentz indices. But this is impossible, because all Lorentz indices
need to be contracted because of hypercubic invariance. In general, any
operator, at any dimension, has to have an even number of Lorentz indices.

What about dimension-9 operators with only one quark bilinear? Either the
bilinear is odd, which means it has 1 or 3 Lorentz indices, or, if it contains
a mass matrix $M$, it carries an even number of Lorentz indices, because of
$U(N_{f})_{L}\times U(N_{f})_{R}$ symmetry. (If there are two mass matrix
insertions, it has to be odd again; if there are three mass matrices, it has
to be even again, etc.)

The rest of the operator has to be made out of $k$ appearances of $D_{\mu}$
and/or $\ell$ appearances of $G_{\kappa\lambda}$. This part of the operator
thus has dimension $d=k+2\ell$, and it has $k+2\ell$ mod 2 Lorentz indices. If
the quark bilinear has dimension 3 (no $M$), the complete operator thus has
dimension $d=3+k+2\ell$, and it has $1+k+2\ell$ mod 2 Lorentz indices. For odd
$d$, we thus need $k+2\ell$ even, but that implies that the number of Lorentz
indices is odd. Likewise, if the bilinear contains one factor of $M$, we need
$d=4+k+2\ell$ to be odd, and thus that $k+2\ell$ is odd. But now the quark
bilinear needs to be even, so again the number of Lorentz indices is odd. This
argument generalizes to operators with more insertions of $M$.

A very similar argument takes care of dimension-9 operators with two quark
bilinears. The general pattern is that the dimension of the gluonic part to be
added is always such that the number of Lorentz indices is odd. The conclusion
appears to be that no operators with odd dimension, and thus an odd power of
the lattice spacing, can appear in the SET.

###  IV.4  What causes taste breaking?

Finally, in this chapter, let us have a closer look at what causes taste-
breaking operators to appear in the SET. As they first appear at dimension 6
through 4-quark operators, we expect them to come from quark-quark scattering,
as depicted in Fig. 2. Indeed, when the gluon momentum exchanged between the
two quarks has momentum $ak=\pi_{E}\neq 0$, with $\pi_{E}$ one of the momentum
vectors with components equal to 0 or $\pi$, this short-distance gluon
exchange generates taste-breaking 4-quark operators, as we will show below.
Note that in this figure, we restored the lattice spacing $a$.

$ak=\pi_{E}$$a\tilde{p}^{\prime}+\pi_{C}$$a\tilde{q}^{\prime}+\pi_{D}$$a\tilde{p}+\pi_{A}$$a\tilde{q}+\pi_{B}$
Figure 2: Exchange of a gluon with momentum equal to $\pi_{E}$ between two
staggered quarks, in momentum space.

In Feynman gauge, the gluon propagator $D(k)$ in Eq. (III.2), for such
momenta, is equal to

$D_{\mu\nu}(k=\pi_{E}/a)=\frac{\delta_{\mu\nu}}{\sum_{\mu}\frac{4}{a^{2}}\sin^{2}({1\over
2}ak_{\mu})}\Bigg{|}_{k=\pi_{E}/a}=\frac{a^{2}}{4n}\,\delta_{\mu\nu}\ ,$ (160)

where $n$ is the number of components of $\pi_{E}$ equal to $\pi$. Using Eq.
(53), and working in Feynman gauge, the diagram is thus equal to (retaining
the explicit lattice spacing $a$ only in the prefactor and omitting color
generators)

$\displaystyle-g^{2}\frac{a^{2}}{4n}\sum_{\mu}\delta(\tilde{p}+\pi_{A}-\tilde{p}^{\prime}+\pi_{C}+\pi_{E}+\pi_{\eta_{\mu}}){1\over
2}(S_{\mu}^{A}e^{i\tilde{p}_{\mu}}+S_{\mu}^{C}e^{-i\tilde{p}^{\prime}_{\mu}})$
$\displaystyle\phantom{-g^{2}\frac{a^{2}}{4n}\sum_{\mu}}\times\delta(\tilde{q}+\pi_{B}-\tilde{q}^{\prime}+\pi_{D}+\pi_{E}+\pi_{\eta_{\mu}}){1\over
2}(S_{\mu}^{B}e^{i\tilde{q}_{\mu}}+S_{\mu}^{D}e^{-i\tilde{q}^{\prime}_{\mu}})$
$\displaystyle=$
$\displaystyle-g^{2}\frac{a^{2}}{4n}\sum_{\mu}\delta(\tilde{p}-\tilde{p}^{\prime})\delta(\tilde{q}-\tilde{q}^{\prime}){1\over
2}(S_{\mu}^{A}e^{i\tilde{p}_{\mu}}+S_{\mu}^{C}e^{-i\tilde{p}_{\mu}}){1\over
2}(S_{\mu}^{B}e^{i\tilde{q}_{\mu}}+S_{\mu}^{D}e^{-i\tilde{q}_{\mu}})$
$\displaystyle\phantom{-g^{2}\frac{a^{2}}{4n}\sum_{\mu}}\times\overline{\delta}(\pi_{A}+\pi_{C}+\pi_{E}+\pi_{\eta_{\mu}})\overline{\delta}(\pi_{B}+\pi_{D}+\pi_{E}+\pi_{\eta_{\mu}})$
$\displaystyle\approx$
$\displaystyle-g^{2}\frac{a^{2}}{4n}\frac{1}{4}\sum_{\mu}\delta(\tilde{p}-\tilde{p}^{\prime})\delta(\tilde{q}-\tilde{q}^{\prime})\Biggl{(}(\hat{\pi}_{E}\Gamma_{\mu})_{CA}(\hat{\pi}_{E}\Gamma_{\mu})_{DB}+(\Gamma_{\mu}\hat{\pi}_{E})_{CA}(\Gamma_{\mu}\hat{\pi}_{E})_{DB}$
$\displaystyle\phantom{-g^{2}\frac{a^{2}}{4n}\frac{1}{4}\sum_{\mu}}+(\hat{\pi}_{E}\Gamma_{\mu})_{CA}(\Gamma_{\mu}\hat{\pi}_{E})_{DB}+(\Gamma_{\mu}\hat{\pi}_{E})_{CA}(\hat{\pi}_{E}\Gamma_{\mu})_{DB}\Biggr{)}\
,$

where in the last expression we approximated $e^{i\tilde{p}_{\mu}}\approx 1$,
etc., and we used

$\overline{\delta}(\pi_{A}+\pi_{C}+\pi_{E}+\pi_{\eta_{\mu}})=\sum_{F}\overline{\delta}(\pi_{E}+\pi_{C}+\pi_{F})\overline{\delta}(\pi_{\eta_{\mu}}+\pi_{F}+\pi_{A})=(\hat{\pi}_{E}\hat{\pi}_{\eta_{\mu}})_{CA}=(\hat{\pi}_{\eta_{\mu}}\hat{\pi}_{E})_{CA}$
(162)

and likewise for
$\overline{\delta}(\pi_{B}+\pi_{D}+\pi_{E}+\pi_{\eta_{\mu}})$. (We recall that
$(\pi_{\eta_{\mu}})_{\mu}=0$.)

We now use that

$\hat{\pi}_{E}\in\\{\Gamma_{\mu}\Xi_{\mu},\
\Gamma_{\mu}\Gamma_{\nu}\Xi_{\mu}\Xi_{\nu},\
\Gamma_{\mu}\Gamma_{5}\Xi_{\mu}\Xi_{5},\ \Gamma_{5}\Xi_{5}\\}\ .$ (163)

This is easy to see, because in these combinations, the factors $S_{\mu}$ in
Eqs. (35) and (39) cancel, so
$\Gamma_{\mu}\Xi_{\mu}=\hat{\pi}_{\eta_{\mu}}\hat{\pi}_{\zeta_{\mu}}$, etc.,
and hence the matrices in Eq. (163) generate “pure” $\hat{\pi}$ matrices, a
total of 15 of them. Equation (IV.4) thus generates the 4-quark operators with
spin-taste structure

$\displaystyle\hat{\pi}_{E}=$ $\displaystyle\Gamma_{\mu}\Xi_{\mu}:\qquad$
$\displaystyle S\times V\ ,$
$\displaystyle\Gamma_{\nu}\Xi_{\nu}|_{\nu\neq\mu}:\qquad$ $\displaystyle
T_{\nu}\times V_{\nu}\ ,$
$\displaystyle\Gamma_{\mu}\Gamma_{\nu}\Xi_{\mu}\Xi_{\nu}:\qquad$
$\displaystyle V_{\nu}\times T_{\nu}\ ,$
$\displaystyle\Gamma_{\rho}\Gamma_{\sigma}\Xi_{\rho}\Xi_{\sigma}|_{\rho,\sigma\neq\mu}:\qquad$
$\displaystyle A_{\nu}\times T_{\rho\sigma}\ (\nu\neq\rho,\sigma)\ ,$
$\displaystyle\Gamma_{\mu}\Gamma_{5}\Xi_{\mu}\Xi_{5}:\qquad$ $\displaystyle
P\times A\ ,$
$\displaystyle\Gamma_{\nu}\Gamma_{5}\Xi_{\nu}\Xi_{5}|_{\nu\neq\mu}:\qquad$
$\displaystyle T_{\rho\sigma}\times A_{\nu}\ (\nu\neq\rho,\sigma)\ ,$
$\displaystyle\Gamma_{5}\Xi_{5}:\qquad$ $\displaystyle A\times P\ .$

As expected, these are all odd operators, because the $U(N_{f})_{L}\times
U(N_{f})_{R}$ symmetry is respected by the kinetic term in the staggered
action from which the Feynman rule used above derives. We note that we do not
obtain all 4-quark operators from this diagram, but other processes, for
instance with more than one gluon exchanged, can occur. The scale of the
coupling at the quark-gluon vertex in Fig. 2 is that of the exchanged gluon,
and we thus expect taste breaking to be suppressed by a factor of order
$\alpha_{s}(\pi/a)$ (with $\alpha_{s}$ the strong coupling). This already
follows from the fact that the free theory has exact $U(4)$ taste symmetry, so
internal gluons are needed to produce taste-breaking operators.

##  V  Staggered chiral perturbation theory

The continuum limit of QCD with $N_{f}$ staggered quarks has $4N_{f}$ quarks
in the continuum limit, with flavor symmetry group $U(1)\times
SU(4N_{f})_{L}\times SU(4N_{f})_{R}$. It follows that this theory is expected
to have $(4N_{f})^{2}-1$ Nambu–Goldstone bosons (NGBs). Of course, the number
of NGBs is too large, and this is why, in practical numerical computations,
the 4th root of the staggered determinant is taken, in order to reduce the
number of quark degrees of freedom by a factor of 4.111111The staggered tastes
can also be interpreted as physical flavors, and an action can be constructed
that completely breaks the taste degeneracy in the continuum limit [8].
However, this construction leads to a practical problems, and has not been
used in practice. While we will not review the arguments supporting the “4th-
root trick,” we can accommodate this trick in both WCPT and chiral
perturbation theory (ChPT) by employing the replica trick. (For a review of
the rooting trick and further references, see Ref. [20]. See also Ref. [15].)

We introduce $n_{r}$ replicas of each staggered fermion, so that now we have
$N=4n_{r}N_{f}$ quarks in the continuum limit, and we thus expect
$(4n_{r}N_{f})^{2}-1$ NGBs. Setting, at the end of any calculation (in WCPT or
ChPT) $n_{r}=\frac{1}{4}$ thus yields effectively $N_{f}^{2}-1$ NGBs, i.e.,
the desired number. We will thus develop staggered ChPT (SChPT) for $N$
continuum quarks, and see in an example how the replica trick works. Of
course, if we wish to consider unrooted staggered quarks, we would set
$n_{r}=1$.

###  V.1  Construction of the SChPT lagrangian

The chiral lagrangian is, like the SET, a continuum effective theory, with
lattice spacing dependence taken into account in terms of an expansion in
powers of $a^{2}$. It is formulated in terms of a non-linear field

$\Sigma=e^{\frac{i}{f}\pi}\ ,$ (165)

where $\pi$ is a traceless, hermitian $4n_{r}N_{f}\times 4n_{r}N_{f}$ matrix
describing the $(4n_{r}N_{f})^{2}-1$ NGBs. The field $\Sigma$ transforms as

$\Sigma_{ia,jb}\sim q_{Lia}\overline{q}_{Rjb}\ ,$ (166)

with $i,\ j=1,\dots,n_{r}N_{f}$ replica-flavor indices, and $a,\ b=1,\dots,4$
taste indices. In this chapter, the left-handed and right-handed projections
are defined as in continuum QCD:

$\displaystyle q_{L}\\!\\!\\!$ $\displaystyle=$ $\displaystyle\\!\\!\\!{1\over
2}(1-\gamma_{5})q\ ,\qquad q_{R}={1\over 2}(1+\gamma_{5})q\ ,$ (167)
$\displaystyle\overline{q}_{L}\\!\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!\\!\overline{q}{1\over 2}(1+\gamma_{5})\
,\qquad\overline{q}_{R}=\overline{q}{1\over 2}(1-\gamma_{5})\ .$

We embed the taste matrices $\xi_{A}$ into $4n_{r}N_{f}\times 4n_{r}N_{f}$
matrices by simply repeating the $4\times 4$ matrices $\xi_{A}$ $n_{r}N_{f}$
times along the main diagonal, thus constructing a block-diagonal matrix
$\xi_{A}^{(n_{r}N_{F})}$. However, we will drop the superscript $(n_{r}N_{F})$
below, when no confusion is possible. For instance, since $\Sigma$ is a
$4n_{r}N_{f}\times 4n_{r}N_{f}$ matrix, we have that

$\xi_{A}\Sigma\equiv\xi_{A}^{(n_{r}N_{F})}\Sigma\ .$ (168)

For reviews of continuum ChPT and applications to various discretizations of
QCD, see Refs. [21, 22].

At leading order (LO) the chiral lagrangian is given by

${\cal L}_{LO}=\frac{1}{8}f^{2}\,{\rm
tr}\,(\partial_{\mu}\Sigma^{\dagger}\partial_{\mu}\Sigma)-\frac{1}{4}Bf^{2}\,{\rm
tr}\,({\cal M}^{\dagger}\Sigma+{\cal M}\Sigma^{\dagger})\ .$ (169)

The trace is over replica/flavor and taste indices (cf. Eq. (166)). This
lagrangian is invariant under

$\Sigma\to V_{L}\Sigma V^{\dagger}_{R}\ ,\qquad{\cal M}\to V_{L}{\cal
M}V_{R}^{\dagger}\ ,$ (170)

with now $V_{L}\in SU(N)_{L}$ and $V_{R}\in SU(N)_{R}$, with $N=4n_{r}N_{f}$.
As before, ${\cal M}$ is a mass spurion, to be set equal to the quark mass
matrix $M$ after the construction of the chiral lagrangian. $f$ and $B$ are
low-energy constants (LECs), to be determined by matching with the underlying
theory, QCD.121212$f$, of course, is the pion decay constant, normalized such
that at LO, $f_{\pi}=f=130.4$ MeV. We will choose the quark mass matrix $M$ to
be diagonal and positive. The first four diagonal elements will be equal to
the up quark mass $m_{u}$, the next four will be equal to the down quark mass
$m_{d}$, etc. (As before, the four tastes produced by each staggered fermion
are degenerate in the continuum limit.)

To incorporate terms of order $a^{2}$ into ChPT, we use a spurion trick just
like for the mass term in Eq. (169). It is easiest to explain this through
some examples. Our first example starts from the following operator appearing
in the SET lagrangian:

$a^{2}(\overline{q}\xi_{5\nu}q)(\overline{q}\xi_{5\nu}q)=a^{2}(\overline{q}_{R}\xi_{5\nu}q_{L}+\overline{q}_{L}\xi_{5\nu}q_{R})^{2}\to(\overline{q}_{R}X_{R}q_{L}+\overline{q}_{L}X_{L}q_{R})^{2}\
,$ (171)

where $\xi_{5\nu}=i\xi_{5}\xi_{\nu}$ and $X_{L,R}$ are new spurions
transforming as

$X_{L}\to V_{L}X_{L}V^{\dagger}_{R}\ ,\qquad X_{R}\to
V_{R}X_{R}V^{\dagger}_{L}\ .$ (172)

We can take $X_{R}=X_{L}^{\dagger}$, but this is not necessary. To recover the
4-quark operator we started from, we have to set

$X_{L}=X_{R}=a\xi_{5\nu}\ .$ (173)

The next step is simply to write down all operators involving the non-linear
field $\Sigma$ and two spurions $X_{L}$ and/or $X_{R}$, such that these
operators are invariant under Eqs. (170) and (172). The possibilities are

$\displaystyle{\rm tr}\,(X_{L}\Sigma^{\dagger})\,{\rm tr}\,(X_{R}\Sigma)\ ,$
(174) $\displaystyle\left({\rm
tr}\,(X_{L}\Sigma^{\dagger})\right)^{2}+\left({\rm
tr}\,(X_{R}\Sigma)\right)^{2}\ ,$ $\displaystyle{\rm
tr}\,(X_{L}\Sigma^{\dagger}X_{L}\Sigma^{\dagger})+{\rm tr}\,(X_{R}\Sigma
X_{R}\Sigma)\ .$

The plus signs follow from parity symmetry (Eq. (173) requires parity to
exchange $X_{L}$ and $X_{R}$).131313We note that parity
$q(x)\to\gamma_{4}\xi_{4}q(I_{s}x)$ is a symmetry of the SET. Setting the
spurions equal to their physical values, Eq. (173), yields the operators

$\displaystyle a^{2}{\rm tr}\,(\xi_{5\nu}\Sigma^{\dagger})\,{\rm
tr}\,(\xi_{5\nu}\Sigma)\ ,$ (175) $\displaystyle a^{2}\left({\rm
tr}\,(\xi_{5\nu}\Sigma^{\dagger})\right)^{2}+a^{2}\left({\rm
tr}\,(\xi_{5\nu}\Sigma)\right)^{2}\ ,$ $\displaystyle a^{2}{\rm
tr}\,(\xi_{5\nu}\Sigma^{\dagger}\xi_{5\nu}\Sigma^{\dagger})+a^{2}{\rm
tr}\,(\xi_{5\nu}\Sigma\xi_{5\nu}\Sigma)\ .$

Each of these operators should be added to Eq. (169) with an arbitary new
coefficient, thus introducing new LECs into the chiral lagrangian.

Before we go on to our next example, we note that we could proceed slightly
differently. Instead of the spurions $X_{L,R}$ we could first write

$a^{2}(\overline{q}\xi_{5\nu}q)(\overline{q}\xi_{5\nu}q)=a^{2}\left((\overline{q}_{R}\xi_{5\nu}q_{L})^{2}+(\overline{q}_{R}\xi_{5\nu}q_{L})(\overline{q}_{L}\xi_{5\nu}q_{R})+(\overline{q}_{L}\xi_{5\nu}q_{R})^{2}\right)\
,$ (176)

and introduce ${\cal O}(a^{2})$ spurions

$\displaystyle X_{LL}$ $\displaystyle\to$ $\displaystyle
a^{2}\xi_{5\nu}\times\xi_{5\nu}\ ,$ (177) $\displaystyle X_{LR}$
$\displaystyle\to$ $\displaystyle a^{2}\xi_{5\nu}\times\xi_{5\nu}\ ,$
$\displaystyle X_{RR}$ $\displaystyle\to$ $\displaystyle
a^{2}\xi_{5\nu}\times\xi_{5\nu}\ ,$

for each of the three terms on the right-hand side of Eq. (176), thus working
only with spurions of order $a^{2}$. The resulting terms in the chiral
lagrangian will be the same, so instead we will use the simpler ${\cal O}(a)$
spurions above, with the extra rule that all terms need to contain two of the
spurions $X_{L}$ and $X_{R}$.

Our second example starts from

$a^{2}(\overline{q}\gamma_{\mu}\xi_{5}q)(\overline{q}\gamma_{\mu}\xi_{5}q)=a^{2}(\overline{q}_{L}\gamma_{\mu}\xi_{5}q_{L}+\overline{q}_{R}\gamma_{\mu}\xi_{5}q_{R})^{2}\to((\overline{q}_{L}\gamma_{\mu}Y_{L}q_{L}+\overline{q}_{R}\gamma_{\mu}Y_{R}q_{R})^{2}\
,$ (178)

where we introduced two new spurions $Y_{L}$ and $Y_{R}$ transforming as

$Y_{L}\to V_{L}Y_{L}V_{L}^{\dagger}\ ,\qquad Y_{R}\to
V_{R}Y_{R}V_{R}^{\dagger}\ .$ (179)

Their physical values are

$Y_{L}=Y_{R}=a\xi_{5}\ .$ (180)

Again, we need the rule that any term in the chiral lagrangian should contain
two $Y$ spurions, conforming with Eq. (178). The only operator that can be
constructed from $\Sigma$ and two $Y$-spurions is

${\rm tr}\,(Y_{L}\Sigma Y_{R}\Sigma^{\dagger})\to a^{2}{\rm
tr}\,(\xi_{5}\Sigma\xi_{5}\Sigma^{\dagger})\ ,$ (181)

which thus introduces one new LEC into the chiral lagrangian.

Performing this procedure for all dimension-6 operators we constructed in Sec.
IV.2 leads to an order-$a^{2}$ “potential”

$a^{2}{\cal V}=a^{2}({\cal U}+{\cal U}^{\prime})$ (182)

to be added to Eq. (169), with [16, 18]

$\displaystyle-{\cal U}$ $\displaystyle=$ $\displaystyle C_{1}\,{\rm
tr}\,(\xi_{5}\Sigma\xi_{5}\Sigma^{\dagger})+{1\over
2}C_{3}\sum_{\nu}\left({\rm
tr}\,(\xi_{\nu}\Sigma\xi_{\nu}\Sigma)+\mbox{h.c.}\right)$
$\displaystyle+{1\over 2}C_{4}\sum_{\nu}\left({\rm
tr}\,(\xi_{5\nu}\Sigma\xi_{5\nu}\Sigma)+\mbox{h.c.}\right)+C_{6}\sum_{\mu<\nu}{\rm
tr}\,(\xi_{\mu\nu}\Sigma\xi_{\mu\nu}\Sigma^{\dagger})\ ,$ $\displaystyle-{\cal
U}^{\prime}$ $\displaystyle=$
$\displaystyle\frac{1}{4}C_{2V}\sum_{\nu}\left(({\rm
tr}\,(\xi_{\nu}\Sigma))^{2}+\mbox{h.c.}\right)+\frac{1}{4}C_{2A}\sum_{\nu}\left(({\rm
tr}\,(\xi_{5\nu}\Sigma))^{2}++\mbox{h.c.}\right)$ $\displaystyle+{1\over
2}C_{5V}\sum_{\nu}{\rm tr}\,(\xi_{\nu}\Sigma)\,{\rm
tr}\,(\xi_{\nu}\Sigma^{\dagger})+{1\over 2}C_{5A}\sum_{\nu}{\rm
tr}\,(\xi_{5\nu}\Sigma)\,{\rm tr}\,(\xi_{\nu}\Sigma^{\dagger})\ ,$

in which $\xi_{\mu\nu}=i\xi_{\mu}\xi_{\nu}$. We find that the operators
$S\times V$, $S\times A$, $P\times V$, $P\times A$, $T\times V$, and $T\times
A$ are like example 1 above, and contribute to the LECs $C_{3}$, $C_{4}$,
$C_{2V,A}$ and $C_{5V,A}$. The operators $V\times P$, $A\times P$, $V\times
T$, and $A\times T$ are like example 2 above, and contribute to $C_{1}$ and
$C_{6}$. The operators $V\times S$ and $A\times S$ do not break taste
symmetry, and thus do not contribute any new operators to the chiral
lagrangian. However, they do lead to order-$a^{2}$ corrections to LECs already
present in the continuum lagrangian. The lagrangian obtained by adding ${\cal
V}$ to Eq. (169) defines staggered ChPT (SChPT) at lowest-order.

Operators like $V_{\mu}\times T_{\mu}$, i.e., operators of type B, do
contribute at order $a^{2}$ to the SET, but not at this order to SChPT. The
reason is that operators that break $SO(4)$ (while preserving hypercubic
rotations) can occur only at higher orders in SChPT. For example one can make
a hypercubic invariant $\sim\sum_{\mu}\partial_{\mu}^{4}$ which is not an
$SO(4)$ invariant. Such an operator is $O(a^{2}p^{4})$ in ChPT power counting
(the $a^{2}$ has to be present because the continuum theory respects $SO(4)$
rotational invariance), and thus always of higher order than the LO
lagrangian. This has an important consequence: To order $a^{2}$, the physics
of NGBs is invariant under the $SO(4)_{\rm taste}$ subgroup of $SU(4)_{\rm
taste}$, and not just under the smaller staggered symmetry group [16]! Other
examples are operators of the schematic form
$\sim\sum_{\mu}\partial_{\mu}^{2}\xi_{\mu}\xi_{\mu}$ or
$\sum_{\mu}\xi_{\mu}\xi_{\mu}\xi_{\mu}\xi_{\mu}$, which also respect only the
staggered hypercubic invariance. We will discuss such examples toward the end
of this subsection; such operators are also higher order than the LO
lagrangian.

This point brings us to the issue of power counting. SChPT has three small
parameters, the typical momentum of physical low-energy pions, the quark
masses in $M_{ia,jb}=m_{i}\delta_{ij}\delta_{ab}$, and the parameter $a^{2}$.
Since for physical momenta, $p^{2}=-m_{\pi}^{2}$, and, as we will see shortly,
$m_{\pi}^{2}\sim m_{i}$, the first two expansion parameters are
related.141414However, for an alternative scenario, which appears not to be
realized in nature, see Ref. [23]. But $a^{2}$ and the $m_{i}$ are in
principle independent, and it depends on the actual numerical computations
carried out in lattice QCD which power counting is most appropriate. Most
staggered computations have NGB (squared-mass) taste splittings comparable in
size to the NGB (squared) masses. We will thus assume a power counting

$a^{2}\Lambda_{\rm QCD}^{2}\sim\frac{m_{i}}{\Lambda_{\rm
QCD}}\sim\frac{p^{2}}{\Lambda_{\rm QCD}^{2}}\ .$ (184)

Of course, if the lattice spacing is small enough, it may be possible to
ignore lattice spacing artifacts, in which case one can use continuum ChPT. On
the other hand, if the lattice spacing is too large, SChPT may break down,
even if the masses $m_{i}$ are small enough.

We can now look at the NGB masses predicted by LO SChPT. We write the pion
matrix in Eq. (165) as

$\pi_{ij}=\sum_{A}\xi_{A}\pi^{A}_{ij}\ ,\qquad\xi_{A}\in\\{\xi_{5},\
\xi_{5\mu},\ \xi_{\mu\nu},\ \xi_{\mu},\ 1\\}$ (185)

and expand the LO lagrangian to second order in the pion fields. This allows
us to read off the masses at lowest order. For $i\neq j$ we find

$m_{Aij}^{2}=B(m_{i}+m_{j})+a^{2}\Delta(\xi_{A})\ ,$ (186)

with

$\displaystyle\Delta(\xi_{5})$ $\displaystyle=$ $\displaystyle 0\ ,$ (187)
$\displaystyle\Delta(\xi_{5\mu})$ $\displaystyle=$
$\displaystyle\frac{16}{f^{2}}(C_{1}+3C_{3}+C_{4}+3C_{6})\ ,$
$\displaystyle\Delta(\xi_{\mu\nu})$ $\displaystyle=$
$\displaystyle\frac{16}{f^{2}}(2C_{3}+2C_{4}+4C_{6})\ ,$
$\displaystyle\Delta(\xi_{\mu})$ $\displaystyle=$
$\displaystyle\frac{16}{f^{2}}(C_{1}+C_{3}+3C_{4}+3C_{6})\ ,$
$\displaystyle\Delta(1)$ $\displaystyle=$
$\displaystyle\frac{16}{f^{2}}(4C_{3}+4C_{4})\ .$

We see that there is a degeneracy predicted by the accidental $SO(4)$ symmetry
that is present in the chiral lagrangian at LO, i.e., to order $a^{2}$. In the
next section, we will see that the flavor non-diagonal NGBs form eight
different multiplets (instead of the five seen in Eq. (187)) if taste symmetry
is fully broken. We also note that the $\xi_{5}$ NGBs are massless when all
quark masses vanish. This is the NGB associated with the exactly conserved
axial currents for the $U(N_{f})_{L}\times U(N_{f})_{R}$ symmetry group of
Eqs. (IV.1) and (136).

The LECs in the double-trace part of the potential, ${\cal U}^{\prime}$,
contribute only to flavor-diagonal NGB masses. To quadratic order,

${\cal U}^{\prime}_{\rm
quad}=\frac{32}{f^{2}}(C_{2V}-C_{5V})\sum_{ij}\pi^{V}_{ii}\pi^{V}_{jj}+\frac{32}{f^{2}}(C_{2A}-C_{5A})\sum_{ij}\pi^{A}_{ii}\pi^{A}_{jj}\
,$ (188)

leading to corresponding contributions to (the squares of) their masses. It is
important to note that, even if one restricts oneself to external pions that
are flavor off-diagonal, flavor diagonal pions still contribute to loops. For
examples, see Ref. [18].

If any of the $\Delta(\xi_{A})$ would be negative, this could induce
flavor/taste symmetry breaking when the quark masses $m_{i}$ are small enough.
However, QCD inequalities (see e.g. Ref. [24]) prevent this from happening.
There is, however, no argument that the combinations $C_{2V,A}-C_{5V,A}$
cannot be negative. For further discussion, see Ref. [25]. In practice no
spontaneous breaking of flavor/taste symmetries has been observed.

At the next order in SChPT, $SO(4)_{\rm taste}$ breaks down to $\Gamma_{4}$,
and rotational symmetry breaks down to the hypercubic group. We give a few
examples [19]. A single insertion of a 4-quark operator can lead to a ChPT
operator of the form

$a^{2}\sum_{\mu}{\rm
tr}\,(\Sigma\partial_{\mu}\Sigma^{\dagger}\xi_{\mu}\Sigma^{\dagger}\partial_{\mu}\Sigma\xi_{\mu})=\frac{4a^{2}}{f^{2}}\sum_{\mu}{\rm
tr}\,(\partial_{\mu}\pi\xi_{\mu}\partial_{\mu}\pi\xi_{\mu})+\dots\ ,$ (189)

in which the first $\xi_{\mu}$ comes from a spurion like $X_{L}$ and the
second from a spurion like $X_{R}$. This operator leads to direction
dependence of the dispersion relation. This contribution appears at order
$a^{2}p^{2}$, i.e., at next-to-leading order (NLO).

Another example comes from a double insertion of two 4-quark operators, and
takes the form

$a^{4}\sum_{\mu}\sum_{\nu\neq\mu}{\rm
tr}\,(\xi_{\mu\nu}\Sigma\xi_{\mu}\Sigma\xi_{\mu\nu}\Sigma^{\dagger}\xi_{\mu}\Sigma^{\dagger})\
,$ (190)

with the taste matrices coming from spurions $Y_{L}$, $X_{R}$, etc. Because of
the four spurions, this contribution appears at order $a^{4}$, which in our
power counting is NLO. At tree level, it only contributes to processes with
four or more pions. This is easy to see: if we set the last $\Sigma^{\dagger}$
equal to one, we can use $\xi_{\mu}\xi_{\mu\nu}=i\xi_{\nu}$ to write the
operator as

$a^{4}\sum_{\mu}\sum_{\nu\neq\mu}i\,{\rm
tr}\,(\xi_{\nu}\Sigma\xi_{\mu}\Sigma\xi_{\mu\nu}\Sigma^{\dagger})\ ,$ (191)

which has $SO(4)$ taste symmetry. Similar simplifications happen when one of
the other non-linear fields is set equal to one. At next-to-next-to-leading
order (NNLO) this operator does contribute to two-point functions through
loops. For a complete construction of the NLO SChPT lagrangian, see Ref. [19].

Finally, in this section, we discuss whether the accidental $SO(4)$ symmetry
at order $a^{2}$ also occurs for other hadrons than pions. As an example, we
consider taste breaking for the $\rho$ meson [16], introducing a “heavy”
$\rho$ field $\rho_{\mu}$ [26]. In the infinite-mass limit, its 4-velocity
$v_{\mu}$ is fixed. We first choose $v_{\mu}$ to be a spurion, transforming
like a vector under hypercubic rotations, and once the effective lagrangian
for the $\rho$ mass has been constructed, we set $v_{4}=1$ and $\vec{v}=0$,
choosing the $\rho$’s restframe. Using this and staggered symmetries, we can
write down the most general quadratic term in the field $\rho_{\mu
ij}=\sum_{A}\xi_{A}\rho^{A}_{\mu ij}$:

$\displaystyle{\cal L}_{\rho\rm-mass}$ $\displaystyle=$ $\displaystyle
R_{1}\sum_{\mu}{\rm tr}\,(\rho_{\mu}^{\dagger}\rho_{\mu})+R_{2}\sum_{\mu}{\rm
tr}\,(\rho_{\mu}^{\dagger}\xi_{5}\rho_{\mu}\xi_{5})+R_{3}\sum_{\mu\nu}{\rm
tr}\,(\rho_{\mu}^{\dagger}\xi_{\nu}\rho_{\mu}\xi_{\nu})$ $\displaystyle+$
$\displaystyle R_{4}\sum_{\mu\nu}{\rm
tr}\,(\rho_{\mu}^{\dagger}\xi_{5\nu}\rho_{\mu}\xi_{5\nu})+R_{5}\sum_{\mu\
\nu<\kappa}{\rm
tr}\,(\rho_{\mu}^{\dagger}\xi_{\nu\kappa}\rho_{\mu}\xi_{\nu\kappa})+R_{6}\sum_{\mu}{\rm
tr}\,(\rho_{\mu}^{\dagger}\xi_{\mu}\rho_{\mu}\xi_{\mu})$ $\displaystyle+$
$\displaystyle R_{7}\sum_{\mu}{\rm
tr}\,(\rho_{\mu}^{\dagger}\xi_{5\mu}\rho_{\mu}\xi_{5\mu})+R_{8}\sum_{\mu\neq\nu}{\rm
tr}\,(\rho_{\mu}^{\dagger}\xi_{\mu\nu}\rho_{\mu}\xi_{\mu\nu})+R_{9}\sum_{\mu}{\rm
tr}\,(\rho_{\mu}^{\dagger}\xi_{4}\rho_{\mu}\xi_{4})$ $\displaystyle+$
$\displaystyle R_{10}\sum_{\mu}{\rm
tr}\,(\rho_{\mu}^{\dagger}\xi_{45}\rho_{\mu}\xi_{45})+R_{11}\sum_{\mu\nu}{\rm
tr}\,(\rho_{\mu}^{\dagger}\xi_{4\nu}\rho_{\mu}\xi_{4\nu})+R_{12}\sum_{\mu}{\rm
tr}\,(\rho_{\mu}^{\dagger}\xi_{4\mu}\rho_{\mu}\xi_{4\mu})\ .$

For example, the term with $R_{9}$ is obtained from

$\sum_{\mu\nu}{\rm
tr}\,(\rho_{\mu}^{\dagger}\xi_{\nu}\rho_{\mu}\xi_{\nu})v_{\nu}v_{\nu}\big{|}_{v=(0,0,0,1)}=\sum_{\mu}{\rm
tr}\,(\rho_{\mu}^{\dagger}\xi_{4}\rho_{\mu}\xi_{4})\ .$ (193)

There are 11 different quark bilinear operators that create a $\rho$ with the
form

$\displaystyle\overline{q}\gamma_{k}\xi_{A}q\ ,$ (194)
$\displaystyle\xi_{A}=(\xi_{5},\xi_{m},\xi_{4k},\xi_{m5},\xi_{45},\xi_{k5},\xi_{m4},\xi_{k4},\xi_{km},\xi_{\ell
m})\ ,$

where $k\neq\ell\neq m\neq k$ [27]. Each of these transforms in an irrep of
the relevant staggered symmetry group (see next section). Thus, the 11
operators with LECs $R_{2}$ to $R_{12}$ in Eq. (V.1) completely break the
degeneracy (the operator with LEC $R_{1}$ gives the same mass to all
tastes).151515It appears that the term with LEC $R_{12}$ was missed in Ref.
[16].

###  V.2  Example: Electromagnetic current correlator at NLO

As an example of SChPT at work, we will calculate the electromagnetic (EM)
current two-point function to one loop, which is the leading order for this
correlation function in ChPT. We usually refer to this as NLO, since it
involves a loop, but there is no tree-level contribution, as we will see.

We will use SChPT with two flavors, and $n_{r}$ replicas. This theory has
$4n_{r}$ up quarks, and $4n_{r}$ down quarks in the continuum limit; we will
consider the isospin-symmetric case with $m_{u}=m_{d}$. At the end of the
calculation, we will set $n_{r}=1/4$, thus implementing the rooting trick. In
the continuum limit, this will yield the expected result, which, at this
order, is that one would obtain in scalar QED (where the complex scalar is the
charged pion). This validates, in this example, the rooting trick.

In two-flavor QCD, the continuum EM current for $n_{r}$ replicas of two light
flavors is

$j_{\mu}=\sum_{i}\left(\frac{2}{3}\,\overline{u}_{i}\gamma_{\mu}u_{i}-\frac{1}{3}\,\overline{d}_{i}\gamma_{\mu}d_{i}\right)\
,$ (195)

where the index $i$ is now the replica index.

To couple SChPT to electromagnetism, we introduce the covariant derivative

$D_{\mu}\Sigma=\partial_{\mu}\Sigma-i\ell_{\mu}\Sigma+i\Sigma r_{\mu}\ ,$
(196)

where $\ell_{\mu}$ and $r_{\mu}$ are external vector fields gauging the groups
$U(N)_{L}$ and $U(N)_{R}$, respectively, with $N=8n_{r}$ (since have now
specialized to $N_{f}=2$). To obtain the coupling of the pions to the EM
field, we take $\ell_{\mu}=r_{\mu}=Qv_{\mu}$, with

$Q=\left(\begin{array}[]{cc}\frac{2}{3}I&0\\\
0&-\frac{1}{3}I\end{array}\right)\ ,$ (197)

with $I$ the $4n_{r}\times 4n_{r}$ unit matrix. We thus have

$\displaystyle D_{\mu}\Sigma$ $\displaystyle=$
$\displaystyle\partial_{\mu}\Sigma-iv_{\mu}[Q,\Sigma]\ ,$ (198)
$\displaystyle(D_{\mu}\Sigma)^{\dagger}$ $\displaystyle=$
$\displaystyle\partial_{\mu}\Sigma^{\dagger}-iv_{\mu}[Q,\Sigma^{\dagger}]\ .$

The kinetic term in the chiral lagrangian becomes

${\cal L}_{\rm kin}=\frac{f^{2}}{8}{\rm
tr}\,\left((D_{\mu}\Sigma)^{\dagger}D_{\mu}\Sigma\right),$ (199)

and the ChPT form of the EM current (to lowest order) is obtained from the
terms linear in $v_{\mu}$:

$\displaystyle j_{\mu}$ $\displaystyle=$ $\displaystyle\frac{f^{2}}{8}\,{\rm
tr}\,\left(-i[Q,\Sigma^{\dagger}]\partial_{\mu}\Sigma-i\partial_{\mu}\Sigma^{\dagger}[Q,\Sigma]\right)$
$\displaystyle=$ $\displaystyle-\frac{i}{4}\,{\rm
tr}\,\left(Q(\pi\partial_{\mu}\pi-\partial_{\mu}\pi\pi)\right)+\dots\ ,$

where in the second line we expanded $\Sigma=1+\frac{i}{f}\pi+\dots$. Writing
the pion fields as

$\pi=\left(\begin{array}[]{cc}\frac{1}{\sqrt{2}}\pi^{0}&\pi^{+}\\\
\pi^{-}&-\frac{1}{\sqrt{2}}\pi^{0}\end{array}\right)\ ,$ (201)

and using $\pi^{\pm}_{ij}=\sum_{A}\xi_{A}\pi^{\pm A}_{ij}$, this can be
written as

$j_{\mu}=i\left(\pi^{-A}_{ij}\partial_{\mu}\pi^{+A}_{ji}-\pi^{+A}_{ij}\partial_{\mu}\pi^{-A}_{ji}\right)\
.$ (202)

It is straightforward to check that this current is conserved by using the
equation of motion $\raisebox{-1.72218pt}{\large$\Box$}\pi^{\pm
A}_{ij}=m_{A}^{2}\pi^{\pm A}_{ij}$. This current corresponds to the conserved
staggered current

$\displaystyle J_{\mu}^{\rm cons}$ $\displaystyle=$ $\displaystyle{1\over
2}\frac{2}{3}\left(\bar{\chi}_{u}(x)\eta_{\mu}(x)U_{\mu}(x)\chi_{u}(x+\mu)+\bar{\chi}_{u}(x+\mu)\eta_{\mu}(x)U^{\dagger}_{\mu}(x)\chi_{u}(x)\right)$
$\displaystyle-{1\over
2}\frac{1}{3}\left(\bar{\chi}_{d}(x)\eta_{\mu}(x)U_{\mu}(x)\chi_{d}(x+\mu)+\bar{\chi}_{d}(x+\mu)\eta_{\mu}(x)U^{\dagger}_{\mu}(x)\chi_{d}(x)\right)\
.$

The conserved staggered current is not the only discretization of the EM
current we may consider. Another example, which is not conserved on the
lattice, is the local current

$J_{\mu}^{\rm
local}=\frac{2}{3}\,\bar{\chi}_{u}(x)\eta_{\mu}(x)\zeta_{\mu}(x)\chi_{u}(x)-\frac{1}{3}\,\bar{\chi}_{d}(x)\eta_{\mu}(x)\zeta_{\mu}(x)\chi_{d}(x)\
.$ (204)

In Eqs. (V.2) and (204) we left the replica index implicit. The continuum
limit of $J_{\mu}^{\rm local}$ is161616We use $J$ for lattice currents, and
$j$ for continuum currents.

$j_{\mu}^{\rm
local}=\frac{2}{3}\,\overline{u}_{i}\gamma_{\mu}\xi_{\mu}u_{i}-\frac{1}{3}\,\overline{d}_{i}\gamma_{\mu}\xi_{\mu}d_{i}=\overline{q}\gamma_{\mu}Q\xi_{\mu}^{(2n_{r})}q\
,$ (205)

where in the second step we grouped the $u_{i}$ and $d_{i}$ quarks together
into the field $q$. To translate this current into ChPT, we set
$\ell_{\mu}=r_{\mu}=Q\xi_{\mu}^{(2n_{r})}v_{\mu}$. In general, we can define
“tasteful” currents (as opposed to the conserved current, which is tasteless)
by generalizing this to

$j_{\mu}^{A}=\frac{2}{3}\,\overline{u}_{i}\gamma_{\mu}\xi_{A}u_{i}-\frac{1}{3}\,\overline{d}_{i}\gamma_{\mu}\xi_{A}d_{i}=\overline{q}\gamma_{\mu}Q\xi_{A}^{(2n_{r})}q\
,$ (206)

with the local current corresponding to $\xi_{A}=\xi_{\mu}$, and the conserved
current corresponding to $\xi_{A}=1$. In SChPT, the corresponding currents are

$\displaystyle j_{\mu}^{A}$ $\displaystyle=$
$\displaystyle-i\,\frac{f^{2}}{4}\,{\rm
tr}\,\left(Q\xi_{A}^{(2n_{r})}[\Sigma^{\dagger},\partial_{\mu}\Sigma]\right)$
$\displaystyle=$ $\displaystyle-\frac{i}{4}\,{\rm
tr}\,\left(Q\xi_{A}^{(2n_{r})}[\pi,\partial_{\mu}\pi]\right)+\dots\ ,$

where in the second line we again expanded in the pion fields. Instead of Eq.
(201), we will now write

$\pi=\left(\begin{array}[]{cc}U&\pi^{+}\\\ \pi^{-}&D\end{array}\right)\ ,$
(208)

which, as we will see, is convenient. In terms of the fields introduced in Eq.
(208),

$j_{\mu}^{A}=-\frac{i}{6}\,{\rm
tr}\,\left(\xi_{A}\left(U{\overleftrightarrow{\partial}_{\\!\\!\\!\mu}}U+\pi^{+}{\overleftrightarrow{\partial}_{\\!\\!\\!\mu}}\pi^{-}\right)\right)+\frac{i}{12}\,{\rm
tr}\,\left(\xi_{A}\left(\pi^{-}{\overleftrightarrow{\partial}_{\\!\\!\\!\mu}}\pi^{+}+D{\overleftrightarrow{\partial}_{\\!\\!\\!\mu}}D\right)\right)\
.$ (209)

We used the notation

$f{\overleftrightarrow{\partial}_{\\!\\!\\!\mu}}g=f\partial_{\mu}g-(\partial_{\mu}f)g\
.$ (210)

Note that in these expressions, we have replaced $\xi_{A}^{(n_{r})}$ by
$\xi_{A}$, for simplicity. If $\xi_{A}=1$, we have that

${\rm tr}\,(U{\overleftrightarrow{\partial}_{\\!\\!\\!\mu}}U)=0\ ,\qquad{\rm
tr}\,(D{\overleftrightarrow{\partial}_{\\!\\!\\!\mu}}D)=0\ ,\qquad{\rm
tr}\,(\pi^{-}{\overleftrightarrow{\partial}_{\\!\\!\\!\mu}}\pi^{+})=-{\rm
tr}\,(\pi^{+}{\overleftrightarrow{\partial}_{\\!\\!\\!\mu}}\pi^{-})\ ,$ (211)

and we recover Eq. (202). If $\xi_{A}\neq 1$,

${\rm
tr}\,(\xi_{A}U{\overleftrightarrow{\partial}_{\\!\\!\\!\mu}}U)=\sum_{BC}{\rm
tr}\,(\xi_{A}[U^{B}\xi_{B},\partial_{\mu}U^{C}\xi_{C}]=\sum_{BC}{\rm
tr}\,(\xi_{A}[\xi_{B},\xi_{C}])U^{B}_{ij}\partial_{\mu}U^{C}_{ji}\neq 0$ (212)

in general, so the terms with $U$ and $D$ in Eq. (209) do not vanish. However,
if $\xi_{B}=1$, the commutator $[\xi_{B},\xi_{C}]=[1,\xi_{C}]=0$, and thus the
contribution with $U^{B}=U^{S}$ (with $U^{S}$ the coefficient of $1$ in
$U=\sum_{B}\xi_{B}U^{B}$) does in fact vanish, and hence the neutral singlet
fields $U^{S}$ and $D^{S}$ do not contribute to the current (209). Since there
is no mixing between any of the $U^{A}$ and any of the $D^{B}$ for $A,\ B\neq
S$, we can work on the basis of Eq. (208). We have that

$\displaystyle\partial_{\mu}j_{\mu}^{A}$ $\displaystyle=$
$\displaystyle-\frac{i}{6}\sum_{BC}{\rm
tr}\,(\xi_{A}[\xi_{B},\xi_{C}])U^{B}_{ij}m_{\pi C}^{2}U^{C}_{ji}+\dots$
$\displaystyle=$ $\displaystyle-\frac{i}{12}\sum_{BC}{\rm
tr}\,(\xi_{A}[\xi_{B},\xi_{C}])U^{B}_{ij}(m_{\pi C}^{2}-m_{\pi
B}^{2})U^{C}_{ji}+\dots\neq 0\ ,$

we see that indeed the current $j_{\mu}^{A}$ is also not conserved in SChPT,
unless $\xi_{A}=1$. Of course, in the continuum limit, where all pions become
degenerate become equal, all currents are conserved.

We note that

$Q={1\over 2}\tau_{3}+\frac{1}{6}Y={1\over 2}\left(\begin{array}[]{cc}1&0\\\
0&-1\end{array}\right)+\frac{1}{6}\left(\begin{array}[]{cc}1&0\\\
0&1\end{array}\right)\ .$ (214)

Since only the commutator $[Q,\Sigma]$ appears in the conserved current, the
hypercharge part $Y$ drops out, and thus in SChPT, the conserved EM current is
the isospin-1 current. For the other currents, with a non-trivial $\xi_{A}$,
the situation is different, as we will discuss shortly.

First, let us return to the $\xi_{A}=1$ current $j_{\mu}^{S}=j_{\mu}$, in the
continuum, at the quark level. With $\langle\dots\rangle_{U_{\mu}}$ denoting
the average over the gauge fields

$\langle(\overline{q}(x)\gamma_{\mu}{1\over
2}\tau_{3}q(x))(\overline{q}(y)\gamma_{\nu}{1\over
2}\tau_{3}q(y))\rangle=-{1\over 2}\langle{\rm
tr}\,(\gamma_{\mu}S(x,y;U_{\mu})\gamma_{\nu}S(y,x;U_{\mu})\rangle_{U_{\mu}}\
,$ (215)

when $m_{u}=m_{d}$, because any quark-disconnected parts cancel. Here
$S(x,y;U_{\mu})$ is the quark propagator (up or down) in a gauge-field
background. The trace is over spin, taste and replica indices. We also have

$\langle(\overline{q}(x)\gamma_{\mu}{1\over
2}\tau_{3}q(x))(\overline{q}(y)\gamma_{\nu}\frac{1}{6}Yq(y))\rangle=0\ ,$
(216)

and, finally,

$\displaystyle\langle(\overline{q}(x)\gamma_{\mu}\frac{1}{6}Yq(x))(\overline{q}(y)\gamma_{\nu}\frac{1}{6}Yq(y))\rangle$
$\displaystyle=$ $\displaystyle-\frac{1}{18}\langle{\rm
tr}\,(\gamma_{\mu}S(x,y;U_{\mu})\gamma_{\nu}S(y,x;U_{\mu})\rangle_{U_{\mu}}$
$\displaystyle+\frac{1}{9}\langle{\rm tr}\,(\gamma_{\mu}S(x,x;U_{\mu}))\,{\rm
tr}\,(\gamma_{\nu}S(y,y;U_{\mu}))\rangle_{U_{\mu}}\ .$

The hypercharge current two-point function is the sum of a quark-connected
(first line) and quark-disconnected (second line) part.

It follows that the light-quark connected part of the EM current two-point
function is equal to

$-\frac{5}{9}\langle{\rm
tr}\,(\gamma_{\mu}S(x,y;U)\gamma_{\nu}S(y,x;U)\rangle_{U}=\frac{10}{9}\times\langle(\overline{q}(x)\gamma_{\mu}{1\over
2}\tau_{3}q(x))(\overline{q}(y)\gamma_{\nu}{1\over 2}\tau_{3}q(y))\rangle\ ,$
(218)

i.e., the light-quark connected part of the EM current two-point function is
equal to $10/9$ times the $I=1$ current two-point function.

Now consider a similar exercise for the current $j_{\mu}^{A}$ with
$\xi_{A}\neq 1$. In this case, the relations (215) and (V.2) hold, with a
$\xi_{A}$ inserted next to every $\gamma_{\mu}$ and $\gamma_{\nu}$. For this
current, there is no quark-disconnected part, because ${\rm tr}\,(\xi_{A})=0$!
Therefore the $j_{\mu}^{A}$ two-point function is equal to its light-quark
connected part, and, in the continuum limit, the light-quark connected parts
of all currents, including the one with $\xi_{A}$, are equal. Therefore, in
the continuum limit,

$\langle j_{\mu}^{A}(x)j_{\nu}^{A}(y)\rangle\big{|}_{A\neq
S}=\frac{10}{9}\langle j_{\mu}^{S}(x)j_{\nu}^{S}(y)\rangle\ .$ (219)

We thus expect to reproduce this relation in the continuum limit of SChPT as
well.

Next, we carry out the actual calculation, beginning with the $U$ part of the
current (209). As we will be interested in

$C^{A}(t)=\frac{1}{3}\sum_{i=1}^{3}\sum_{\vec{x}}\langle
j_{i}^{A}(\vec{x},t)j_{i}^{A}(0)\rangle\ ,$ (220)

we calculate, using $U_{k\ell}(x)=\sum_{X}\xi_{X}U^{X}_{k\ell}(x)$ (where $k$
and $\ell$ are replica indices),

$\displaystyle\langle j_{\mu}^{A}(x)j_{\nu}^{A}(y)\rangle_{U\mbox{\tiny-
part}}$ $\displaystyle=$
$\displaystyle-\frac{1}{36}\sum_{XX^{\prime}YY^{\prime}}\langle(U^{X}_{mn}(x){\overleftrightarrow{\partial}_{\\!\\!\\!\mu}}U^{X^{\prime}}_{nm}(x))(U^{Y}_{k\ell}(y){\overleftrightarrow{\partial}_{\\!\\!\\!\mu}}U^{Y^{\prime}}_{\ell
k}(y))\rangle$
$\displaystyle\phantom{-\frac{1}{36}\sum_{XX^{\prime}YY^{\prime}}}\times\,{\rm
tr}\,(\xi_{A}\xi_{X}\xi_{X^{\prime}})\,{\rm
tr}\,(\xi_{A}\xi_{Y}\xi_{Y^{\prime}})$ $\displaystyle=$
$\displaystyle-\frac{n_{r}^{2}}{36}\int\frac{dp_{4}}{2\pi}\int\frac{dq_{4}}{2\pi}\frac{1}{V}^{2}\sum_{\vec{p},\vec{q}}\sum_{XX^{\prime}}\frac{e^{ip(x-y)}e^{iq(y-x)}}{(p^{2}+m_{X}^{2})(q^{2}+m_{X^{\prime}}^{2})}(p_{\mu}+q_{\mu})^{2}$
$\displaystyle\phantom{-}\times\Bigl{(}{\rm
tr}\,(\xi_{A}\xi_{X}\xi_{X^{\prime}})\,{\rm
tr}\,(\xi_{A}\xi_{X}\xi_{X^{\prime}})-{\rm
tr}\,(\xi_{A}\xi_{X}\xi_{X^{\prime}})\,{\rm
tr}\,(\xi_{A}\xi_{X^{\prime}}\xi_{X})\Bigr{)}\ ,$

where $V$ is the spatial volume (we take the time direction infinite), and

$\langle
U^{X}_{mn}(x)U_{k\ell}^{Y}(y)\rangle=\delta_{XY}\delta_{m\ell}\delta_{nk}\int\frac{dp_{4}}{2\pi}\frac{1}{V}\sum_{\vec{p}}\frac{e^{ip(x-y)}}{p^{2}+m_{X}^{2}}\
.$ (222)

Setting $\mu=i$, summing over $\vec{x}$, $\vec{y}$ and $i$, and using

$\displaystyle\int\frac{dp_{4}}{2\pi}\frac{e^{ip_{4}t}}{p^{2}+m_{X}^{2}}$
$\displaystyle=$
$\displaystyle\frac{1}{2E_{X}(\vec{p})}\,e^{-E_{X}(\vec{p})|t|}\ ,$ (223)
$\displaystyle E_{X}(\vec{p})$ $\displaystyle=$
$\displaystyle\sqrt{{\vec{p}}^{2}+m_{X}^{2}}\ ,$

we obtain

$\displaystyle 3C^{A}(t)_{U\mbox{\tiny-part}}$ $\displaystyle=$
$\displaystyle-\frac{n_{r}^{2}}{36V}\sum_{\vec{p}}\sum_{XX^{\prime}}\frac{e^{-(E_{X}(\vec{p})+E_{X^{\prime}}(\vec{p}))|t|}}{E_{X}(\vec{p})E_{X^{\prime}}(\vec{p})}\,{\vec{p}}^{2}$
$\displaystyle\phantom{-}\times\Bigl{(}{\rm
tr}\,(\xi_{A}\xi_{X}\xi_{X^{\prime}})\,{\rm
tr}\,(\xi_{A}\xi_{X}\xi_{X^{\prime}})-{\rm
tr}\,(\xi_{A}\xi_{X}\xi_{X^{\prime}})\,{\rm
tr}\,(\xi_{A}\xi_{X^{\prime}}\xi_{X})\Bigr{)}\ .$

We will refer to the first double trace over taste matrices as the “unmixed”
part, and to the second double trace as the “mixed” part. We note that for
$\xi_{A}=1$ this vanishes, consistent with the fact that there is no $U$
contribution to the conserved current. For the $D$ contribution, we get $1/4$
of this, and thus the sum of the $U$ and $D$ contributions is equal to
$36\times\left(1/36+1/144\right)=5/4$ times Eq. (V.2).

It remains to add the contribution from the charged-pion part of the current.
One can verify that choosing the term
$\pi^{+}{\overleftrightarrow{\partial}_{\\!\\!\\!\mu}}\pi^{-}$ in both
currents yields 1 times the mixed term in Eq. (V.2), choosing the term
$\pi^{-}{\overleftrightarrow{\partial}_{\\!\\!\\!\mu}}\pi^{+}$ in both
currents yields $1/4$ times the mixed term in Eq. (V.2), while the cross terms
yields $-1$ times the unmixed term in Eq. (V.2). Adding all contributions
leads to

$\displaystyle 3C^{A}(t)$ $\displaystyle=$
$\displaystyle\frac{n_{r}^{2}}{144V}\sum_{\vec{p}}\sum_{XX^{\prime}}\frac{e^{-(E_{X}(\vec{p})+E_{X^{\prime}}(\vec{p}))|t|}}{E_{X}(\vec{p})E_{X^{\prime}}(\vec{p})}\,{\vec{p}}^{2}$
$\displaystyle\times\Bigl{(}10{\rm tr}\,(\xi_{A}\xi_{X}\xi_{X^{\prime}})\,{\rm
tr}\,(\xi_{A}\xi_{X^{\prime}}\xi_{X})-{\rm
tr}\,(\xi_{A}\xi_{X}\xi_{X^{\prime}})\,{\rm
tr}\,(\xi_{A}\xi_{X}\xi_{X^{\prime}})\Bigr{)}\ .$

For $\xi_{A}=1$ the traces set $X^{\prime}=X$, and setting all masses $m_{X}$
equal (i.e., taking the continuum limit), the factor on the second line of
this result is equal to $9\times 4^{2}\times 16=144\times 16$ (a 4 for each
trace; the 16 comes from the $\sum_{X}$). With $n_{r}=1/4$, the total
prefactor becomes equal to one, which is what one would have obtained directly
in the continuum. For $a>0$, the taste pion $\pi_{X}$ runs around the loop,
and these loops are averaged over all tastes.

When $\xi_{A}\neq 1$, a little more work is needed. First, we have that

${\rm tr}\,(\xi_{A}\xi_{X}\xi_{X^{\prime}})={\rm
tr}\,((\xi_{A}\xi_{X}\xi_{X^{\prime}})^{T})={\rm
tr}\,(\xi_{X^{\prime}}^{T}\xi_{X}^{T}\xi_{A}^{T})={\rm
tr}\,(\xi_{A}\xi_{X^{\prime}}\xi_{X}^{\prime})^{*}\ ,$ (226)

and thus

${\rm tr}\,(\xi_{A}\xi_{X}\xi_{X^{\prime}})\,{\rm
tr}\,(\xi_{A}\xi_{X^{\prime}}\xi_{X})=16\delta_{X^{\prime},X+A}\ .$ (227)

This takes care of the mixed trace in Eq. (V.2). We still get the factor 16,
and the current inserts taste $A$ into the pion loop. For the unmixed trace,
we use that

$\xi_{A}\xi_{X}=(-1)^{|A||X|+A\cdot X}\xi_{X}\xi_{A}\ ,$ (228)

with $|X|=\sum_{\mu}X_{\mu}$ and $A\cdot X=\sum_{\mu}A_{\mu}X_{\mu}$. Thus

$({\rm tr}\,(\xi_{A}\xi_{X}\xi_{X^{\prime}}))^{2}=16(-1)^{|A||X|+A\cdot
X}\delta_{X^{\prime},X+A}\ .$ (229)

For $\xi_{A}=1$ the right-hand side is equal to $\delta_{X,X^{\prime}}$, as
before. For any $\xi_{A}\neq 1$, $(-1)^{|A||X|+A\cdot X}=+1$ for half of the
tastes $X$, and $-1$ for the other half. Thus, when all masses become
degenerate in the continuum limit, the trace term on the second line of Eq.
(V.2) becomes equal to 11 when $(-1)^{|A||X|+A\cdot X}=-1$, and 9 when
$(-1)^{|A||X|+A\cdot X}=+1$. The total prefactor thus becomes

$\frac{n_{r}^{2}}{144}\times 4^{2}\times 8\times(11+9)=\frac{10}{9}\times
16n_{r}^{2}\ ,$ (230)

which is $10/9$ times the continuum limit for $\xi_{A}=1$, as we expect from
Eq. (219). Here the factors 4 again come from the traces, and the factor
$8\times(11+9)$ from the $\sum_{X}$. Again, the current $j_{\mu}^{A}$ inserts
a taste $A$ into the pion loop. This verifies Eq. (219) at this order SChPT,
in the continuum limit. For $a>0$, the sums in Eq. (V.2) have to be carried
out to obtain the complete result in explicit form. The Kronecker delta in Eq.
(227) reduces the sum over $X$ and $X^{\prime}$ to a single sum.

##  VI  Staggered states and operators

In this chapter, we first consider the spectrum of QCD with staggered fermions
and general properties of staggered correlation functions, after which we
discuss the construction of operators that create and annihilate states in the
staggered Hilbert space. Most of the chapter will deal with mesons, but in the
final section we will briefly discuss aspects of baryonic states in QCD with
staggered fermions. The material of the first three sections can be found in
Refs. [27, 28, 29], while the material of the final section can be found in
Refs. [28, 30]. Another useful reference is Ref. [31].

###  VI.1  Staggered spectrum: states and operators

We will consider states in momentum space, because the irreducible
representations of a theory with staggered fermions live in momentum space.
Operators acting on the staggered Hilbert space will be denoted with a hat.

Thus, $\hat{S}_{\mu}$ is the shift operator acting on the Hilbert space.
Defining $\hat{T}_{\mu}=\hat{S}_{\mu}^{2}$, $\hat{T}_{\mu}$ is a normal
translation over two lattice spacings in the $\mu$ direction. In momentum
space, the eigenvalues are

$\hat{T}_{k}:\quad e^{2ip_{k}}\ ,\qquad\hat{T}_{4}:\quad e^{-2E}\ ,$ (231)

where $p_{k}$ with $-\pi/2<p_{k}\leq\pi/2$ are the components of the physical
spatial momentum of a state, and $E$ is the energy. The operator $\hat{T}_{4}$
can be taken as the transfer matrix of the theory. With Eq. (231) we can
define the operators $\hat{T}_{\mu}^{-{1\over 2}}$ as having eigenvalues
$e^{-ip_{k}}$ and $e^{E}$, and thus we can define

$\hat{\Xi}_{\mu}=\hat{S}_{\mu}\hat{T}_{\mu}^{-{1\over 2}}\ ,$ (232)

with

$\hat{\Xi}_{\mu}^{2}=1\ .$ (233)

We will consider states with $\vec{p}=0$ (except in Sec. VI.3). Such states
transform in representations of the geometrical restframe group ($GRF$)

$GRF=G(\Xi_{\mu},R^{(k\ell)},I_{s})\ ,$ (234)

where the spatial rotations $R^{(k\ell)}=R^{(k\ell)}(\pi/2)$ generate the
little group of $\vec{p}=0$ states, and we recall that
$I_{s}=I_{1}I_{2}I_{3}=\Gamma_{4}\Xi_{4}$ in the defining representation. The
notation $G(X)$ denotes the finite group generated by the elements $X$. Since
$\Xi_{4}\in GRF$, also parity $P=I_{s}\Xi_{4}=\Gamma_{4}\in GRF$, and, with
$P$ commuting with all other elements of the group,

$GRF=G(\Xi_{\mu},R^{(k\ell)})\times\\{1,\ P\\}\ .$ (235)

A maximally commuting set of elements of $GRF$ is the set $\\{\Xi_{4},\
\Xi_{1}\Xi_{2},\ R^{(12)},\ P\\}$.

We will label states in the Hilbert space by the eigenvalues $E$, the
eigenvalue $\sigma$ of $P$, and the eigenvalue $\sigma_{t}$ of $\Xi_{4}$:

$\displaystyle\hat{S}_{4}|Er\sigma;\sigma_{t}\rangle$ $\displaystyle=$
$\displaystyle\sigma_{t}e^{-E}|Er\sigma;\sigma_{t}\rangle\ ,$ (236)
$\displaystyle\Xi_{4}|Er\sigma;\sigma_{t}\rangle$ $\displaystyle=$
$\displaystyle\sigma_{t}|Er\sigma;\sigma_{t}\rangle\ ,$ $\displaystyle
P|Er\sigma;\sigma_{t}\rangle$ $\displaystyle=$
$\displaystyle\sigma|Er\sigma;\sigma_{t}\rangle\ ,$ $\displaystyle
I_{s}|Er\sigma;\sigma_{t}\rangle$ $\displaystyle=$
$\displaystyle\sigma\sigma_{t}|Er\sigma;\sigma_{t}\rangle\equiv\sigma_{s}|Er\sigma;\sigma_{t}\rangle\
.$

The label $r$ denotes an irrep of $G(\Xi_{\mu},R^{(k\ell)})$; within such an
irrep, we can label the state with the eigenvalue $\sigma_{t}$ of $\Xi_{4}$.
The final equation defines the parity $\sigma_{s}$ of $I_{s}$.

Next, we introduce local operators $\hat{\phi}$ and $\hat{\overline{\phi}}$ at
fixed time $t$, which thus transform in representations of a smaller group,
the geometric time-slice group

$GTS=G(\Xi_{m},R^{(k\ell)},I_{s})\ ,$ (237)

which does not contain $\Xi_{4}$. We define time-dependent operators (i.e., we
use the Heisenberg picture) through

$\hat{\phi}(t)=\hat{S}_{4}^{-t}\hat{\phi}\hat{S}_{4}^{t}\
,\qquad\hat{\overline{\phi}}(t)=\hat{S}_{4}^{-t}\hat{\overline{\phi}}\hat{S}_{4}^{t}\
.$ (238)

There is a $U(1)$ charge operator $\hat{Q}$ (quark number) with

$[\hat{Q},\hat{\phi}]=q\hat{\phi}\
,\qquad[\hat{Q},\hat{\overline{\phi}}]=-q\hat{\overline{\phi}}\ ,$ (239)

with $q$ the charge of $\hat{\phi}$ (we can take $q=1$ for the operator
$\hat{\chi}$), and a charge conjugation operator $\hat{C}_{0}$ with

$\hat{C}_{0}^{-1}\hat{\phi}\,\hat{C}_{0}=\tau\hat{\overline{\phi}}\
,\qquad\hat{C}_{0}^{-1}\hat{\overline{\phi}}\,\hat{C}_{0}=\overline{\tau}\hat{\phi}\
.$ (240)

The phases $\tau$ and $\overline{\tau}$ depend on convention, but the phase
$\sigma_{c}=\tau\overline{\tau}$ defined by

$\hat{C}_{0}^{-2}\hat{\phi}\,\hat{C}_{0}^{2}=\sigma_{c}\hat{\phi}$ (241)

is not. Using the definition of $C_{0}$ in Sec. II.4 we find that
$\sigma_{c}=-1$ for the staggered quark field $\chi$, and thus that it is
equal to $+1$ for mesonic operators, and $-1$ for baryonic operators. The
fields $\hat{\phi}$ and $\hat{\overline{\phi}}$ transform in a representation
$r$ of $GTS$, which is isomorphic to $G(\Xi_{\mu},R^{(k\ell)})$ and can thus
be taken to be the same irrep as in Eq. (236).

We note that $GRF=GTS\times\\{1,\ P\\}$, but the fields $\hat{\phi}$ and
$\hat{\overline{\phi}}$ do not have $P$ quantum numbers, since the definition
of $P$ involves a translation in time. We can consider operators $\phi$ and
$\overline{\phi}$ with a well-defined value $\sigma_{s}$ of $I_{s}$:

$\hat{I}_{s}^{-1}\hat{\phi}\hat{I}_{s}=\sigma_{s}\hat{\phi}\
,\qquad\hat{I}_{s}^{-1}\hat{\overline{\phi}}\hat{I}_{s}=\sigma_{s}\hat{\overline{\phi}}\
.$ (242)

($\hat{\phi}$ and $\hat{\overline{\phi}}$ have the same $I_{s}$ quantum number
because $\hat{I}_{s}$ commutes with $\hat{C}_{0}$.) If $\hat{\phi}$,
$\hat{\overline{\phi}}$ have the $I_{s}$ quantum number $\sigma_{s}$, we have
that

$\langle
Er\sigma;\sigma_{t}|\hat{\overline{\phi}}|0\rangle=0\qquad\mbox{unless}\
\sigma_{s}=\sigma\sigma_{t}\ ,$ (243)

because $\hat{I}_{s}=\hat{P}\hat{\Xi}_{4}$.

An example of $\hat{\phi}$ is

$\sum_{\vec{m}}\hat{\chi}(\vec{x}+2\vec{m})\ ,$ (244)

which transforms in an 8-dimensional irrep of $GTS$. This irrep decomposes as

${\bf 8}\to A_{1}^{+}+A_{1}^{-}+F_{1}^{+}+F_{1}^{-}\ ,$ (245)

where $A_{1}$ is the trivial irrep of the cubic group and $F_{1}$ the vector
irrep. The superscript denotes the value of $I_{s}$. These irreps of the cubic
group can be projected out as [28]

$\displaystyle A_{1}^{+}:\qquad$ $\displaystyle\sum_{\vec{x}\ {\rm
even}}\hat{\chi}(\vec{x})\ ,$ $\displaystyle A_{1}^{-}:\qquad$
$\displaystyle\sum_{\vec{x}\ {\rm even}}\hat{\chi}(\vec{x}+1+2+3)\ ,$
$\displaystyle F_{1}^{-}:\qquad$ $\displaystyle\sum_{\vec{x}\ {\rm
even}}\left(\begin{array}[]{c}\hat{\chi}(\vec{x}+1)\\\
\hat{\chi}(\vec{x}+2)\\\ \hat{\chi}(\vec{x}+3)\end{array}\right)\ ,$ (250)
$\displaystyle F_{1}^{+}:\qquad$ $\displaystyle\sum_{\vec{x}\ {\rm
even}}\left(\begin{array}[]{c}\hat{\chi}(\vec{x}+2+3)\\\
-\hat{\chi}(\vec{x}+1+3)\\\ \hat{\chi}(\vec{x}+1+2)\end{array}\right)\ .$
(254)

The appearance of the irreps $A_{1}$ and $F_{1}$ for a fermionic field can be
understood as follows. In the rest frame, the relevant continuum group is
$SU(2)_{\rm spin}\times SU(2)_{\rm taste}$. The group $G(R^{(k\ell)})$ is a
discrete subgroup of the diagonal $SU(2)$ subgroup of the continuum symmetry
group, and the relevant decomposition is, for the groups $SU(2)_{\rm
spin}\times SU(2)_{\rm taste}\supset SU(2)_{\rm diag}\supset G(R^{(k\ell)})$,

$\left({\scriptstyle{{1\over 2}}}_{\rm spin},\ {\scriptstyle{{1\over 2}}}_{\rm
taste}\right)\to{\bf 0}+{\bf 1}\to A_{1}+F_{1}\ .$ (255)

Now, let us consider a correlation function of the form

$C(t)=\langle 0|\hat{\phi}(t_{1})\hat{\overline{\phi}}(t_{2})|0\rangle\ ,$
(256)

with $t=t_{1}-t_{2}>0$. Using Eqs. (236), (238) and (243), this can be written
as (note that $\hat{S}_{4}|0\rangle=|0\rangle$)

$\displaystyle C(t)$ $\displaystyle=$ $\displaystyle\sum_{E\sigma}\langle
0|\hat{\phi}(t_{1})|Er\sigma;\sigma_{t}\rangle\langle
Er\sigma;\sigma_{t}|\hat{\overline{\phi}}(t_{2})|0\rangle$ $\displaystyle=$
$\displaystyle\sum_{E\sigma}\langle
0|\hat{\phi}\hat{S}_{4}^{t_{1}}|Er\sigma;\sigma_{t}\rangle\langle
Er\sigma;\sigma_{t}|\hat{S}_{4}^{-t_{2}}\hat{\overline{\phi}}|0\rangle$
$\displaystyle=$ $\displaystyle\sum_{E\sigma}\sigma_{t}^{t}e^{-Et}\langle
0|\hat{\phi}|Er\sigma;\sigma_{t}\rangle\langle
Er\sigma;\sigma_{t}|\hat{\overline{\phi}}|0\rangle$ $\displaystyle=$
$\displaystyle\sum_{E_{+}}|\langle
0|\hat{\phi}|E_{+}r;\sigma_{t}=\sigma_{s}\rangle|^{2}\sigma_{s}^{t}e^{-E_{+}t}$
$\displaystyle+\sum_{E_{-}}|\langle
0|\hat{\phi}|E_{-}r;\sigma_{t}=-\sigma_{s}\rangle|^{2}(-\sigma_{s})^{t}e^{-E_{-}t}\
,$

where in the first term the sum is over states with positive parity
$\sigma=\sigma_{t}\sigma_{s}$ and energies $E_{+}$, and in the second term the
sum is over states with negative parity and energies $E_{-}$. The subscript on
$E_{\pm}$ indicates the parity $\sigma$. If, for example, the operator
$\hat{\phi}$ has $I_{s}$ parity $\sigma_{s}=+1$, the negative $P$ parity
states yield a rapidly oscillating contribution to $C(t)$. The correlator gets
contributions from both parities because $P$ is not well-defined for operators
localized in time.

The result (VI.1) is valid when the time extent of the lattice is infinite.
If, instead, it is $2T$ (it is convenient to take the lattice even in all
directions for staggered fermions), then, instead of considering $\langle
0|\hat{\phi}(t_{1})\hat{\overline{\phi}}(t_{2})|0\rangle$, we should consider
${\rm
Tr}\,(\hat{S}_{4}^{2T-t}\hat{\phi}\hat{S}_{4}^{t}\hat{\overline{\phi}})$, with
${\rm Tr}\,$ the trace in Hilbert space. Let us instead derive the correction
to Eq. (VI.1) more intuitively.

If a particle can travel from $t_{2}$ to $t_{1}$, an anti-particle can travel
from $t_{1}$ to $t_{2}^{\prime}=t_{2}+2T$. This leads to the replacement
$e^{-Et}\to e^{-E(t_{2}^{\prime}-t_{1})}=e^{-E(2T-t)}$. With anti-
periodic/periodic boundary conditions for the staggered quarks, we pick up a
phase factor $\kappa=+1$, respectively, $\kappa=-1$ for fermionic operators,
while always $\kappa=+1$ for bosonic operators. Furthermore,

$\langle 0|\hat{\phi}|Er\sigma;\sigma_{t}\rangle\langle
Er\sigma;\sigma_{t}|\hat{\overline{\phi}}|0\rangle=\sigma_{c}\langle
0|\hat{C}_{0}^{-1}\hat{\overline{\phi}}\hat{C}_{0}|Er\sigma;\sigma_{t}\rangle\langle
Er\sigma;\sigma_{t}|\hat{C}_{0}^{-1}\hat{\phi}\hat{C}_{0}|0\rangle\ .$ (258)

Using invariance under charge-conjugation, Eq. (VI.1) gets replaced by

$\displaystyle C(t)$ $\displaystyle=$ $\displaystyle\sum_{E_{+}}|\langle
0|\hat{\phi}|E_{+}r;\sigma_{t}=\sigma_{s}\rangle|^{2}\sigma_{s}^{t}\left(e^{-E_{+}t}+\sigma_{c}\kappa\,e^{-(2T-t)E_{+}}\right)$
$\displaystyle+\sum_{E_{-}}|\langle
0|\hat{\phi}|E_{-}r;\sigma_{t}=-\sigma_{s}\rangle|^{2}(-\sigma_{s})^{t}\left(e^{-E_{-}t}+\sigma_{c}\kappa\,e^{-(2T-t)E_{-}}\right)+\dots\
.$

Contributions proportional to $|\langle n|\hat{\phi}|m\rangle|^{2}$ with both
$|n\rangle$ and $|m\rangle$ not equal to $|0\rangle$ are not shown. Note that
if $\langle 0|\hat{\phi}|0\rangle\neq 0$, we need to subtract $\langle
0|\hat{\phi}|0\rangle\langle 0|\hat{\overline{\phi}}|0\rangle$ to get the
connected part of $C(t)$.

###  VI.2  Mesons

In this subsection, we will specialize to meson operators, in particular,
operators bilinear in $\chi$ and $\bar{\chi}$ [27]. If $X_{\mu}=D(\Xi_{\mu})$
is the representation of $\Xi_{\mu}$ on a mesonic operator, they commute,
because

$D(\Xi_{\mu})D(\Xi_{\nu})=e^{iq\pi}D(\Xi_{\nu})D(\Xi_{\mu})\ ,$ (259)

and $q=0$ for mesonic operators. (This can be easily checked on the explicit
operators we will construct below.) Defining

$\tilde{X}_{m}=X_{1}X_{2}X_{3}X_{m}\ ,$ (260)

the restframe symmetry group for mesonic states is

$(RF)_{\rm mesons}=G(\tilde{X}_{m},R^{(k\ell)})\times\\{1,\
X_{4}\\}\times\\{1,\ X_{1}X_{2}X_{3}\\}\times\\{1,\ P\\}\times\\{1,\ C_{0}\\}\
.$ (261)

Mesonic states can thus be labeled by

$|E,\overline{r}^{\sigma_{t}\sigma_{123}},\sigma,\tau_{0}\rangle\ ,$ (262)

with $\overline{r}$ an irrep of $G(\tilde{X}_{m},R^{(k\ell)})$, and
$\sigma_{t}$, $\sigma_{123}$, $\sigma$ and $\tau_{0}$ the $X_{4}$, $X_{123}$,
$P$ and $C_{0}$ parities, respectively.

Defining171717After constructing meson operators, they can be made gauge
covariant by inserting averages over the shortest Wilson lines between $\chi$
and $\bar{\chi}$.

$D_{k}\chi(x)={1\over 2}\left(\chi(x+k)+\chi(x-k)\right)\ ,$ (263)

we define classes of bilinear operators (dropping hats on $\chi$ and
$\bar{\chi}$) on a fixed time slice $t$:

$\displaystyle M_{0}$ $\displaystyle=$ $\displaystyle\sum_{\vec{x}\ {\rm
even}}\bar{\chi}(x)\chi(x)\ ,$ (264) $\displaystyle M_{1}$ $\displaystyle=$
$\displaystyle\sum_{\vec{x}\ {\rm even}}\bar{\chi}(x)D_{1}\chi(x)\ ,$
$\displaystyle M_{12}$ $\displaystyle=$ $\displaystyle\sum_{\vec{x}\ {\rm
even}}\bar{\chi}(x)D_{1}D_{2}\chi(x)\ ,$ $\displaystyle M_{123}$
$\displaystyle=$ $\displaystyle\sum_{\vec{x}\ {\rm
even}}\bar{\chi}(x)D_{1}D_{2}D_{3}\chi(x)\ ,$

where all members of each class are obtained by taking the operator explicitly