Spontaneous Mass Generation and Chiral Symmetry Breaking in a Lattice Nambu–Jona-Lasinio Model

Abstract

We study a lattice Nambu–Jona-Lasinio model with interacting staggered fermions in the Kogut–Susskind Hamiltonian formalism. The model has a discrete chiral symmetry but not the usual continuous chiral symmetry. In a strong coupling regime for the four-fermion interaction, we prove that the mass of the fermions is spontaneously generated at sufficiently low non-zero temperatures in the dimensions $\nu \ge 3$ of the model and zero temperature in $\nu \ge 2$. Due to the phase transition, the discrete chiral symmetry of the model is broken. Our analysis is based on the reflection positivity for fermions and the method of the infrared bound.

Appendices

A Chirality and Mass for the Staggered Fermions

Following [2], we review the chirality of the free fermion part of the Hamiltonian $H^{(\Lambda )}(m)$ of (2.2) in the case of the three dimension $\nu =3$. For a more careful mathematical discussion on the staggered fermions, we refer the reader to [31].

Let us consider the Hamiltonian of [2] with the mass term which is given by

$$\begin{aligned} H_\textrm{free}^{(\Lambda )}:= & {} i\kappa \sum _{x\in \Lambda \subset \mathbb {Z}^3} \Bigl \{[\psi ^\dagger (x)\psi (x+e_1)-\psi ^\dagger (x+e_1)\psi (x)]\nonumber \\{} & {} +\,i[\psi ^\dagger (x)\psi (x+e_2)+\psi ^\dagger (x+e_2)\psi (x)](-1)^{x^{(1)}+x^{(2)}}\nonumber \\{} & {} +\,[\psi ^\dagger (x)\psi (x+e_3)-\psi ^\dagger (x+e_3)\psi (x)](-1)^{x^{(1)}+x^{(2)}}\Bigr \}\nonumber \\{} & {} +\,m\sum _{x\in \Lambda \subset \mathbb {Z}^3}(-1)^{x^{(1)}+x^{(2)}+x^{(3)}}\psi ^\dagger (x)\psi (x). \end{aligned}$$

(A.1)

Here, we impose the anti-periodic boundary conditions,

$$\begin{aligned} \psi (L+1,x^{(2)},x^{(3)})= & {} -\psi (-L+1,x^{(2)},x^{(3)}),\\ \psi (x^{(1)},L+1,x^{(3)})= & {} -\psi (x^{(1)},-L+1,x^{(3)}), \end{aligned}$$

and

$$\begin{aligned} \psi (x^{(1)},x^{(2)},L+1)=-\psi (x^{(1)},x^{(2)},-L+1) \end{aligned}$$

for the fermion operator $\psi (x)$ at the right boundaries. Similarly,

$$\begin{aligned} \psi (-L,x^{(2)},x^{(3)})= & {} -\psi (L,x^{(2)},x^{(3)}),\\ \psi (x^{(1)},-L,x^{(3)})= & {} -\psi (x^{(1)},L,x^{(3)}), \end{aligned}$$

and

$$\begin{aligned} \psi (x^{(1)},x^{(2)},-L)=-\psi (x^{(1)},x^{(2)},L) \end{aligned}$$

for the left boundaries. These are equivalent to the anti-periodic boundary conditions for the Hamiltonian $H^{(\Lambda )}(m)$. These conditions are realized by the Fourier transform of $\psi (x)$,

$$\begin{aligned} \psi (x)=\frac{1}{\sqrt{\vert \Lambda \vert }}\sum _k e^{ikx}\hat{\psi }(k), \end{aligned}$$

(A.2)

with the momenta $k=(k^{(1)},k^{(2)},k^{(3)})$ which satisfy

$$\begin{aligned} \exp [ik^{(i)}\cdot 2L]=-1 \quad \text{ for } i=1,2,3. \end{aligned}$$

(A.3)

The second term in the summand of the first sum in the right-hand side of (A.1) is different from the corresponding term of the Hamiltonian $H^{(\Lambda )}(m)$ of (2.2). In order to show that the two Hamiltonians are unitary equivalent to each other, we introduce a unitary transformation,

$$\begin{aligned} U_\textrm{free}:=\prod _{x\;:\;x^{(2)}=\textrm{odd}}\exp [-(i\pi /2)\psi ^\dagger (x)\psi (x)]. \end{aligned}$$

(A.4)

Then, one has

$$\begin{aligned} U_\textrm{free}^\dagger \psi (x) U_\textrm{free}= {\left\{ \begin{array}{ll} -i\psi (x) &{} \text{ for } x^{(2)}=\textrm{odd};\\ \psi (x) &{} \text{ for } x^{(2)}=\textrm{even}. \end{array}\right. } \end{aligned}$$

(A.5)

Under this transformation, only the second term in the summand of the first sum in the right-hand side of (A.1) changes as follows:

$$\begin{aligned}{} & {} U_\textrm{free}^\dagger [\psi ^\dagger (x)\psi (x+e_2)+\psi ^\dagger (x+e_2)\psi (x)]U_\textrm{free}\nonumber \\{} & {} \quad ={\left\{ \begin{array}{ll} i\psi ^\dagger (x)\psi (x+e_2)-i\psi ^\dagger (x+e_2)\psi (x) &{} \text{ for } x^{(2)}=\textrm{odd};\\ -i\psi ^\dagger (x)\psi (x+e_2)+i\psi ^\dagger (x+e_2)\psi (x) &{} \text{ for } x^{(2)}=\textrm{even} \end{array}\right. }\nonumber \\{} & {} \quad =-i(-1)^{x^{(2)}}[\psi ^\dagger (x)\psi (x+e_2)-\psi ^\dagger (x+e_2)\psi (x)]. \end{aligned}$$

(A.6)

Therefore, the transformed Hamiltonian $U_\textrm{free}^\dagger H_\textrm{free}^{(\Lambda )}U_\textrm{free}$ has the desired form, which is equal to the free part of the Hamiltonian $H^{(\Lambda )}(m)$ of (2.2).

Since the Hamiltonian $H_\textrm{free}^{(\Lambda )}$ of (A.1) has the quadratic form in the fermion operators $\psi (x)$, it can be diagonalized in terms of fermion operators $\mathcal {A}_i$ as follows:

$$\begin{aligned} H_\textrm{free}^{(\Lambda )}=\sum _i E_i \mathcal {A}_i^\dagger \mathcal {A}_i, \end{aligned}$$

(A.7)

where $E_i$ are the energy eigenvalues. One has

$$\begin{aligned} {[}H_\textrm{free}^{(\Lambda )},\mathcal {A}_i^\dagger ]=E_i\mathcal {A}_i^\dagger \end{aligned}$$

(A.8)

by using the anticommutation relations for $\mathcal {A}_i$. In order to find the solution $\mathcal {A}^\dagger$ for the equation $[H_\textrm{free}^{(\Lambda )},\mathcal {A}^\dagger ]=E\mathcal {A}^\dagger$, we set

$$\begin{aligned} \mathcal {A}^\dagger =\sum _{x\in \Lambda } v(x)\psi ^\dagger (x), \end{aligned}$$

(A.9)

where v(x) is a function of the site $x\in \Lambda$. We choose the function v(x) so that v(x) satisfies the same anti-periodic boundary conditions as those of $\psi (x)$. Substituting this into $[H_\textrm{free}^{(\Lambda )},\mathcal {A}^\dagger ]=E\mathcal {A}^\dagger$, one has

$$\begin{aligned}{} & {} \sum _{x\in \Lambda } [H_\textrm{free}^{(\Lambda )},v(x)\psi ^\dagger (x)]\nonumber \\{} & {} \quad =i\kappa \sum _{x\in \Lambda } v(x)\Bigl \{[\psi ^\dagger (x-e_1)-\psi ^\dagger (x+e_1)] +i(-1)^{x^{(1)}+x^{(2)}}[-\psi ^\dagger (x-e_2)+\psi ^\dagger (x+e_2)]\nonumber \\{} & {} \qquad +(-1)^{x^{(1)}+x^{(2)}}[\psi ^\dagger (x-e_3)-\psi ^\dagger (x+e_3)]\Bigr \} +m\sum _{x\in \Lambda }v(x)(-1)^{x^{(1)}+x^{(2)}+x^{(3)}}\psi ^\dagger (x)\nonumber \\{} & {} \quad =E\sum _{x\in \Lambda }v(x)\psi ^\dagger (x). \end{aligned}$$

(A.10)

From the coefficients of $\psi ^\dagger (x)$, one obtains the eigenvalue equation,

$$\begin{aligned}{} & {} i\kappa \bigl \{[v(x+e_1)-v(x-e_1)]+i(-1)^{x^{(1)}+x^{(2)}}[v(x+e_2)-v(x-e_2)]\nonumber \\{} & {} \quad +(-1)^{x^{(1)}+x^{(2)}}[v(x+e_3)-v(x-e_3) \bigr \}+m(-1)^{x^{(1)}+x^{(2)}+x^{(3)}}v(x) =Ev(x),\nonumber \\ \end{aligned}$$

(A.11)

for the single particle wavefunction v(x).

For the function v(x), we introduce the Fourier transform,

$$\begin{aligned} v(x)=\frac{1}{\sqrt{\vert \Lambda \vert }}\sum _k e^{ikx}\hat{v}(k), \end{aligned}$$

(A.12)

with the momentum $k=(k^{(1)},k^{(2)},k^{(3)})\in (-\pi ,\pi ]^3$. Substituting this into the above Eq. (A.11), one has

$$\begin{aligned} -2\kappa \bigl [\sin k^{(1)}\hat{v}(k)+i\sin k^{(2)}\hat{v}(k+\pi _{12})+\sin k^{(3)}\hat{v}(k+\pi _{12})\bigr ] +m\hat{v}(k+\pi _{123})=E\hat{v}(k), \end{aligned}$$

(A.13)

where we have written

$$\begin{aligned} \pi _{12}:=(\pi ,\pi ,0) \quad \text{ and }\quad \pi _{123}:=(\pi ,\pi ,\pi ). \end{aligned}$$

We also write

$$\begin{aligned} \pi _1:=(\pi ,0,0), \ \pi _2:=(0,\pi ,0), \ \pi _3:=(0,0,\pi ) \end{aligned}$$

and

$$\begin{aligned} \pi _{13}:=(\pi ,0,\pi ), \ \pi _{23}:=(0,\pi ,\pi ). \end{aligned}$$

When $m=0$, these momenta as well as $k=0$ yield the zero energy $E=0$. Following [2], we define

$$\begin{aligned} \begin{pmatrix} \hat{v}_\textrm{u}^{(1)}(k)\\ \hat{v}_\textrm{u}^{(2)}(k)\\ \hat{v}_\textrm{u}^{(3)}(k)\\ \hat{v}_\textrm{u}^{(4)}(k) \end{pmatrix} :=U_\textrm{u} \begin{pmatrix} \hat{v}(k)\\ \hat{v}(k+\pi _3)\\ \hat{v}(k+\pi _{12})\\ \hat{v}(k+\pi _{123})\end{pmatrix} \end{aligned}$$

(A.14)

for the u-quark, where we have written

$$\begin{aligned} U_\textrm{u}:=\frac{1}{2}\begin{pmatrix} 1 &{} 1 &{} 1 &{} 1 \\ 1 &{} -1 &{} -1 &{} 1 \\ 1 &{} -1 &{} 1 &{} -1 \\ 1 &{} 1 &{} -1 &{} -1 \end{pmatrix}. \end{aligned}$$

(A.15)

Clearly, the decomposition in the above Eq. (A.14) is justified only for a small k. However, one can expect that the contributions for large k do not affect the continuum limit. In fact, in the sence of the norm for the resolvent of the Dirac operator, the statement was proved by [31].

Similarly, for the d-quark,

$$\begin{aligned} \begin{pmatrix} \hat{v}_\textrm{d}^{(1)}(k)\\ \hat{v}_\textrm{d}^{(2)}(k)\\ \hat{v}_\textrm{d}^{(3)}(k)\\ \hat{v}_\textrm{d}^{(4)}(k) \end{pmatrix} :=U_\textrm{d} \begin{pmatrix} \hat{v}(k+\pi _1)\\ \hat{v}(k+\pi _2) \\ \hat{v}(k+\pi _{23})\\ \hat{v}(k+\pi _{13}) \end{pmatrix}, \end{aligned}$$

(A.16)

where

$$\begin{aligned} U_\textrm{d}:=\frac{1}{2}\begin{pmatrix} -1 &{} 1 &{} -1 &{} 1 \\ -1 &{} -1 &{} -1 &{} -1 \\ 1 &{} -1 &{} -1 &{} 1 \\ 1 &{} 1 &{} -1 &{} -1 \end{pmatrix}. \end{aligned}$$

(A.17)

We write

$$\begin{aligned} \hat{v}_\textrm{f}(k):=\begin{pmatrix} \hat{v}_\textrm{f}^{(1)}(k) \\ \hat{v}_\textrm{f}^{(2)}(k) \\ \hat{v}_\textrm{f}^{(3)}(k) \\ \hat{v}_\textrm{f}^{(4)}(k) \end{pmatrix} \end{aligned}$$

for $\textrm{f}=\textrm{u,d}$. We also write

$$\begin{aligned} \sigma ^{(1)}:= & {} \begin{pmatrix} 0 &{} 1 \\ 1 &{} 0 \end{pmatrix},\quad \sigma ^{(2)}:=\begin{pmatrix} 0 &{} -i \\ i &{} 0 \end{pmatrix},\quad \sigma ^{(3)}:=\begin{pmatrix} 1 &{} 0 \\ 0 &{} -1 \end{pmatrix},\\ \sin k:= & {} (\sin k^{(1)},\sin k^{(2)},\sin k^{(3)}), \end{aligned}$$

and

$$\begin{aligned} \sin k\cdot \sigma :=\sin k^{(1)}\sigma ^{(1)}+\sin k^{(2)}\sigma ^{(2)}+\sin k^{(3)}\sigma ^{(3)}. \end{aligned}$$

By using these expressions and the eigenvalue equation (A.13), one has

$$\begin{aligned} \left[ -2\kappa \begin{pmatrix} 0 &{} \sin k\cdot \sigma \\ \sin k\cdot \sigma &{} 0 \end{pmatrix} +m\gamma _0 \right] \hat{v}_\textrm{f}(k)=E\hat{v}_\textrm{f}(k) \end{aligned}$$

(A.18)

for low momenta k, where we have written

$$\begin{aligned} \gamma _0:=\begin{pmatrix} 1 &{} 0 &{} 0 &{} 0\\ 0 &{} 1 &{} 0 &{} 0\\ 0 &{} 0 &{} -1 &{} 0 \\ 0 &{} 0 &{} 0 &{} -1 \end{pmatrix}. \end{aligned}$$

This is nothing but the desired Dirac equation for small momenta k.

In order to discuss the chirality in the case with $m=0$, we define

$$\begin{aligned} \gamma _5:=\begin{pmatrix} 0 &{} 0 &{} 1 &{} 0 \\ 0 &{} 0 &{} 0 &{} 1 \\ 1 &{} 0 &{} 0 &{} 0 \\ 0 &{} 1 &{} 0 &{} 0 \end{pmatrix}. \end{aligned}$$

(A.19)

Then, one has

$$\begin{aligned} \Bigl [\gamma _5,\begin{pmatrix} 0 &{} \sigma ^{(i)} \\ \sigma ^{(i)} &{} 0 \end{pmatrix}\Bigr ]=0. \end{aligned}$$

(A.20)

for all the indices $i=1,2,3$. This implies that $\gamma _5$ is the conserved quantity, and its eigenvalues take $\pm 1$, i.e., the chirality. Since $\gamma _5$ does not commute with $\gamma _0$, it is not conserved in the case with $m\ne 0$. Actually, one has the following important relation:

$$\begin{aligned} \gamma _5 \gamma _0 \gamma _5=-\gamma _0. \end{aligned}$$

(A.21)

As mentioned above, we have ignored the contributions from large momenta k. Then, in the case of $m=0$, the chirality is conserved. This procedure is justified in the sense of the norm for the resolvent of the Dirac operator [31]. Therefore, we can expect that the free fermion Hamiltonian is chiral invariant in the continuum limit with $m=0$, although we will not give the precise proof in the present paper.

B Chiral Invariance of the Interaction Hamiltonian Under a Low-Energy Approximation

In this appendix, we show that the interaction Hamiltonian of $H^{(\Lambda )}(m)$ of (2.2) is invariant under a certain chiral transformation under a low-energy approximation. The chiral transformation is discrete in the sense of [2]. As mentioned at the end of the preceding section, we have ignored the contributions from large momenta k in order to show the chiral invariance of the free fermion Hamiltonian with the vanishing mass parameter, $m=0$. For the purpose of the present section, we will also use this approximation. We hope that the following argument for the continuum limit will be justified in future studies. In passing, we stress that our main results for the phase transition are still mathematically rigorous, because we have proved the particle-hole symmetry breaking in the main part of the present paper.

To begin with, we introduce the following Fourier transform:

$$\begin{aligned} \psi (x)=\frac{1}{\sqrt{\vert \Lambda \vert }}e^{i\pi x^{(2)}}\sum _k e^{ikx}\hat{\psi }(k). \end{aligned}$$

(B.1)

We consider first the free part of the Hamiltonian. Substituting the Fourier transform (B.1) into the expression (A.1) of the Hamiltonian $H_\textrm{free}^{(\Lambda )}$, one has

$$\begin{aligned} H_\textrm{free}^{(\Lambda )}= & {} \sum _k \Bigl \{-2\kappa [\sin k^{(1)}\hat{\psi }^\dagger (k)\hat{\psi }(k) +i\sin k^{(2)}\hat{\psi }^\dagger (k)\hat{\psi }(k+\pi _{12})\nonumber \\{} & {} +\sin k^{(3)}\hat{\psi }^\dagger (k)\hat{\psi }(k+\pi _{12})]+m\hat{\psi }^\dagger (k)\hat{\psi }(k+\pi _{123})\Bigr \}\nonumber \\= & {} H_\textrm{free,u}^{(\Lambda )}+H_\textrm{free,d}^{(\Lambda )}, \end{aligned}$$

(B.2)

where we have written

$$\begin{aligned} H_\textrm{free,u}^{(\Lambda )}= & {} \sum _k \Bigl \{-\kappa [\sin k^{(1)}\hat{\psi }^\dagger (k)\hat{\psi }(k) +i\sin k^{(2)}\hat{\psi }^\dagger (k)\hat{\psi }(k+\pi _{12})\nonumber \\{} & {} +\sin k^{(3)}\hat{\psi }^\dagger (k)\hat{\psi }(k+\pi _{12})] +\frac{m}{2}\hat{\psi }^\dagger (k)\hat{\psi }(k+\pi _{123})\Bigl \} \end{aligned}$$

(B.3)

and

$$\begin{aligned} H_\textrm{free,d}^{(\Lambda )}= & {} \sum _k \Bigl \{-\kappa [-\sin k^{(1)}\hat{\psi }^\dagger (k+\pi _1)\hat{\psi }(k+\pi _1) +i\sin k^{(2)}\hat{\psi }^\dagger (k+\pi _1)\hat{\psi }(k+\pi _{2})\nonumber \\{} & {} +\sin k^{(3)}\hat{\psi }^\dagger (k+\pi _1)\hat{\psi }(k+\pi _{2})] +\frac{m}{2}\hat{\psi }^\dagger (k+\pi _1)\hat{\psi }(k+\pi _{23})\Bigl \}. \end{aligned}$$

(B.4)

We write

$$\begin{aligned} \tilde{\Psi }_\textrm{u}(k):= \begin{pmatrix} \hat{\psi }(k) \\ \hat{\psi }(k+\pi _3)\\ \hat{\psi }(k+\pi _{12})\\ \hat{\psi }(k+\pi _{123}) \end{pmatrix}. \end{aligned}$$

(B.5)

Then, one has

$$\begin{aligned} H_\textrm{free,u}^{(\Lambda )}= & {} \frac{1}{4}\sum _k \tilde{\Psi }_\textrm{u}^\dagger (k)\Bigl \{ -\kappa \Bigl [\sin k^{(1)} \begin{pmatrix} 1 &{} 0 \\ 0 &{} -1 \end{pmatrix} +\sin k^{(2)} \begin{pmatrix} 0 &{} i \\ -i &{} 0 \end{pmatrix} +\sin k^{(3)} \begin{pmatrix} 0 &{} \sigma ^{(3)} \\ \sigma ^{(3)} &{} 0 \end{pmatrix} \Bigr ]\nonumber \\{} & {} +\frac{m}{2}\begin{pmatrix} 0 &{} \sigma ^{(2)}\\ \sigma ^{(2)} &{} 0 \end{pmatrix} \Bigr \}\tilde{\Psi }_\textrm{u}(k). \end{aligned}$$

(B.6)

Further, we write

$$\begin{aligned} \hat{\Psi }_\textrm{u}(k):=\begin{pmatrix} \hat{\Psi }_\textrm{u}^{(1)}(k) \\ \hat{\Psi }_\textrm{u}^{(2)}(k) \\ \hat{\Psi }_\textrm{u}^{(3)}(k) \\ \hat{\Psi }_\textrm{u}^{(4)}(k)\end{pmatrix} :=U_\textrm{u}\tilde{\Psi }_\textrm{u}(k) \end{aligned}$$

(B.7)

in terms of $U_\textrm{u}$ of (A.15). Then, the above Hamiltonian has the desired form,

$$\begin{aligned} H_\textrm{free,u}^{(\Lambda )}=\frac{1}{4}\sum _k \hat{\Psi }_\textrm{u}^\dagger (k)\Bigl [ -\kappa \begin{pmatrix} 0 &{} \sin k\cdot \sigma \\ \sin k\cdot \sigma &{} 0 \end{pmatrix} +\frac{m}{2}\gamma _0\Bigr ]\hat{\Psi }_\textrm{u}(k). \end{aligned}$$

(B.8)

However, the momenta k should be restricted to small values so that the components of the operators $\hat{\Psi }_\textrm{u}(k)$ are independent of each other.

Similarly,

$$\begin{aligned} \tilde{\Psi }_\textrm{d}(k):=\begin{pmatrix} \hat{\psi }(k+\pi _1) \\ \hat{\psi }(k+\pi _2) \\ \hat{\psi }(k+\pi _{23}) \\ \hat{\psi }(k+\pi _{13}) \end{pmatrix}. \end{aligned}$$

(B.9)

Then,

$$\begin{aligned} H_\textrm{free,d}^{(\Lambda )}= & {} \frac{1}{4}\sum _k \tilde{\Psi }_\textrm{d}^\dagger (k)\Bigl \{ -\kappa \Bigl [\sin k^{(1)} \begin{pmatrix} -\sigma ^{(3)} &{} 0 \\ 0 &{} \sigma ^{(3)} \end{pmatrix} +\sin k^{(2)} \begin{pmatrix} -\sigma ^{(2)} &{} 0 \\ 0 &{} \sigma ^{(2)} \end{pmatrix}\nonumber \\{} & {} +\sin k^{(3)} \begin{pmatrix} \sigma ^{(1)} &{} 0 \\ 0 &{} -\sigma ^{(1)} \end{pmatrix} \Bigr ] +\frac{m}{2}\begin{pmatrix} 0 &{} 1 \\ 1 &{} 0 \end{pmatrix} \Bigr \}\tilde{\Psi }_\textrm{d}(k). \end{aligned}$$

(B.10)

Further, we write

$$\begin{aligned} \hat{\Psi }_\textrm{d}(k):=\begin{pmatrix} \hat{\Psi }_\textrm{d}^{(1)}(k) \\ \hat{\Psi }_\textrm{d}^{(2)}(k) \\ \hat{\Psi }_\textrm{d}^{(3)}(k) \\ \hat{\Psi }_\textrm{d}^{(4)}(k)\end{pmatrix} :=U_\textrm{d}\tilde{\Psi }_\textrm{d}(k). \end{aligned}$$

(B.11)

These yield the same form of the Hamiltonian,

$$\begin{aligned} H_\textrm{free,d}^{(\Lambda )}=\frac{1}{4}\sum _k \hat{\Psi }_\textrm{d}^\dagger (k)\Bigl [ -\kappa \begin{pmatrix} 0 &{} \sin k\cdot \sigma \\ \sin k\cdot \sigma &{} 0 \end{pmatrix} +\frac{m}{2}\gamma _0\Bigr ]\hat{\Psi }_\textrm{d}(k). \end{aligned}$$

(B.12)

Now let us consider the interaction term of the Hamiltonian $H^{(\Lambda )}(m)$ of (2.2). From the expression (B.1), one has

$$\begin{aligned}{} & {} \sum _{x\in \Lambda } \psi ^\dagger (x)\psi (x)\psi ^\dagger (x+e_\mu )\psi (x+e_\mu )\nonumber \\{} & {} \quad =\frac{1}{\vert \Lambda \vert ^2}\sum _{x\in \Lambda }\sum _{k_1,k_2,k_3,k_4}e^{-ik_1x}e^{ik_2x}e^{-ik_3(x+e_\mu )}e^{ik_4(x+e_\mu )} \hat{\psi }^\dagger (k_1)\hat{\psi }(k_2)\hat{\psi }^\dagger (k_3)\hat{\psi }(k_4)\nonumber \\{} & {} \quad =\frac{1}{\vert \Lambda \vert }\sum _{k_1,k_2,k_3,k_4}\delta _{k_1+k_3,k_2+k_4}e^{i(k_4-k_3)e_\mu } \hat{\psi }^\dagger (k_1)\hat{\psi }(k_2)\hat{\psi }^\dagger (k_3)\hat{\psi }(k_4). \end{aligned}$$

(B.13)

In order to treat the region of the low energies, we consider the corresponding momenta,

$$\begin{aligned} k_i=p_i+K_i \ \ \text{ for } i=1,2,3,4, \end{aligned}$$

where $p_i$ are small momenta, and $K_i$ are given by

$$\begin{aligned} K_i=(K_i^{(1)},K_i^{(2)},K_i^{(3)}) \end{aligned}$$

(B.14)

with $K_i^{(j)}\in \{0,\pi \}$ for $i=1,2,3,4$ and $j=1,2,3$. The momentum conservation of the Kronecker delta implies

$$\begin{aligned} p_1+p_3=p_2+p_4 \quad \text{ and }\quad K_1+K_3=K_2+K_4. \end{aligned}$$

(B.15)

Let $K=(K^{(1)},K^{(2)},K^{(3)})$ for $K^{(i)}\in \{0,\pi \}$, $i=1,2,3$. We define a signature $\textrm{sgn}(K)$ for K as follows:

$$\begin{aligned} \textrm{sgn}(K):=(-1)^{\iota (K^{(1)})+\iota (K^{(2)})+\iota (K^{(3)})}, \end{aligned}$$

(B.16)

where

$$\begin{aligned} \iota (K^{(i)}):={\left\{ \begin{array}{ll} 1, &{} \text{ if } \ K^{(i)}=\pi ;\\ 0, &{} \text{ if } \ K^{(i)}=0 \end{array}\right. } \end{aligned}$$

(B.17)

for $i=1,2,3$. Then, the second relation of the momentum conservation law (B.15) implies

$$\begin{aligned} \textrm{sgn}(K_1)\textrm{sgn}(K_2)\textrm{sgn}(K_3)\textrm{sgn}(K_4)=+1. \end{aligned}$$

(B.18)

Having these observations in mind, let us consider the chiral transformation $\gamma _5$. From (A.15), (A.19), (B.5) and (B.7), we have

$$\begin{aligned} U_\textrm{u}^\dagger \gamma _5 U_\textrm{u}\tilde{\Psi }_\textrm{u}= & {} \begin{pmatrix} 1 &{} 0 &{} 0 &{} 0 \\ 0 &{} -1 &{} 0 &{} 0 \\ 0 &{} 0 &{} 1 &{} 0 \\ 0 &{} 0 &{} 0 &{} -1 \end{pmatrix} \begin{pmatrix} \hat{\psi }(k) \\ \hat{\psi }(k+\pi _3)\\ \hat{\psi }(k+\pi _{12})\\ \hat{\psi }(k+\pi _{123}) \end{pmatrix}\nonumber \\= & {} \begin{pmatrix} \hat{\psi }(k) \\ -\hat{\psi }(k+\pi _3)\\ \hat{\psi }(k+\pi _{12})\\ -\hat{\psi }(k+\pi _{123}) \end{pmatrix} =\begin{pmatrix} (\gamma _5\hat{\psi })(k) \\ (\gamma _5\hat{\psi })(k+\pi _3)\\ (\gamma _5\hat{\psi })(k+\pi _{12})\\ (\gamma _5\hat{\psi })(k+\pi _{123}) \end{pmatrix}, \end{aligned}$$

(B.19)

where we have written

$$\begin{aligned} (\gamma _5\hat{\psi })(k+K)=\textrm{sgn}(K)\hat{\psi }(k+K) \end{aligned}$$

(B.20)

Similarly,

$$\begin{aligned} U_\textrm{d}^\dagger \gamma _5 U_\textrm{d}\tilde{\Psi }_\textrm{d}= \begin{pmatrix} -\hat{\psi }(k+\pi _1) \\ -\hat{\psi }(k+\pi _2) \\ \hat{\psi }(k+\pi _{23}) \\ \hat{\psi }(k+\pi _{13}) \end{pmatrix} = \begin{pmatrix} (\gamma _5\hat{\psi })(k+\pi _1) \\ (\gamma _5\hat{\psi })(k+\pi _2) \\ (\gamma _5\hat{\psi })(k+\pi _{23}) \\ (\gamma _5\hat{\psi })(k+\pi _{13}) \end{pmatrix}. \end{aligned}$$

(B.21)

The operator in the right-hand side of the interaction Hamiltonian (B.13) is transformed under the transformation $\gamma _5$ as follows:

$$\begin{aligned}{} & {} \hat{\psi }^\dagger (p_1+K_1)\hat{\psi }(p_2+K_2)\hat{\psi }^\dagger (p_3+K_3)\hat{\psi }(p_4+K_4)\nonumber \\{} & {} \quad \rightarrow (\gamma _5\hat{\psi })^\dagger (p_1+K_1)(\gamma _5\hat{\psi })(p_2+K_2)(\gamma _5\hat{\psi })^\dagger (p_3+K_3) (\gamma _5\hat{\psi })(p_4+K_4)\nonumber \\{} & {} \quad =\textrm{sgn}(K_1)\textrm{sgn}(K_2)\textrm{sgn}(K_3)\textrm{sgn}(K_4) \hat{\psi }^\dagger (p_1+K_1)\hat{\psi }(p_2+K_2)\hat{\psi }^\dagger (p_3+K_3)\hat{\psi }(p_4+K_4)\nonumber \\{} & {} \quad =\hat{\psi }^\dagger (p_1+K_1)\hat{\psi }(p_2+K_2)\hat{\psi }^\dagger (p_3+K_3)\hat{\psi }(p_4+K_4), \end{aligned}$$

(B.22)

where we have used (B.18) and (B.20). Thus, the interaction Hamiltonian is invariant under the discrete chiral transformation $\gamma _5$ when the energies are restricted to the low values. This implies that, under the same restriction, the total Hamiltonian without the mass term is invariant under the discrete chiral transformation $\gamma _5$.

C Chiral Symmetry Breaking in an Infinite-Volume and Continuum Limit

In this appendix, we discuss the infinite-volume and continuum limit. For the present Hamiltonian, we set $m=0$, i.e., we consider the case without the mass term in the Hamiltonian. We also assume that the contributions from the high energies can be neglected in the continuum limit. Namely, we will continue to use the low-energy approximation. Therefore, the discussion here is fairly formal, but we believe it will be helpful to the reader for comprehending the physics in the infinite-volume and continuum limit. In particular, the form (C.10) of an interaction below is standard, although one might think that interactions generally exhibit a very complicated form in the representation by using the staggered fermions.

1.1 C.1 Chiral symmetry breaking

We write $\langle \cdots \rangle$ for the expectation value in the infinite-volume and continuum limit. Let us deal with the case that the expectation value $\langle \cdots \rangle$ is invariant under the discrete chiral transformation $\gamma _5$. Consider the two-point correlation function,

$$\begin{aligned} \langle \hat{\psi }^\dagger (p_1+K_1)\hat{\psi }(p_2+K_2)\rangle . \end{aligned}$$

(C.1)

From the argument of the preceding Appendix B, we obtain the following result: When $\textrm{sgn}(K_1)\textrm{sgn}(K_2)=-1$, this expectation value is vanishing under the above assumption. Clearly, generic n-point correlation functions have the same property as well.

Similarly, from the relation (A.21), one can show that the expectation value of the mass term of the Hamiltonian is also vanishing under the same assumption on the expectation value. Conversely, if the expectation value of the mass term shows a non-vanishing value, then the chiral symmetry must be broken.

1.2 C.2 The interaction Hamiltonian in an infinite-volume and continuum limit

If one expresses the interactions in the right-hand of (B.13) in terms of the operators of the u- and d- quarks, then the expressions become very complicated. The reason is that the interaction processes allow the momentum transfer given by (B.14). In order to clarify this point, we consider more general interactions which have the following form:

$$\begin{aligned} \sum _{x,y\in \Lambda }\psi ^\dagger (x)\psi (x)G^{(\Lambda )}(x-y)(-1)^{x^{(1)}+x^{(2)}+x^{(3)}+y^{(1)}+y^{(2)}+y^{(3)}} \psi ^\dagger (y)\psi (y), \end{aligned}$$

(C.2)

where the function $G^{(\Lambda )}(x)$ of the site $x\in \Lambda$ is given by

$$\begin{aligned} G^{(\Lambda )}(x):=\frac{1}{\vert \Lambda \vert }\sum _k \hat{G}(k)e^{ikx} \end{aligned}$$

(C.3)

with a real-valued function $\hat{G}(k)$ of the momentum k. Here, we require the condition $\textrm{supp}\; \hat{G}(k)\subset [-\varepsilon _0,\varepsilon _0]^3$ with a small positive $\varepsilon _0$. This condition forbids the momentum transfer given by (B.14) in the processes of the interaction. Moreover, because of the anti-periodic boundary condition (A.3) for the momentum k, we restrict the argument $x-y$ of the function $G^{(\Lambda )}(x-y)$ in (C.2) into the region $\vert x^{(i)}-y^{(i)}\vert \le L$ for $i=1,2,3$ by relying on the periodic boundary condition of the lattice $\Lambda$.

For the nearest-neighbor interaction given by (B.13), the function $\hat{G}(k)$ is given by

$$\begin{aligned} \hat{G}(k)=\cos k^{(1)} +\cos k^{(2)} +\cos k^{(3)}. \end{aligned}$$

(C.4)

Clearly, this does not satisfy the above condition $\textrm{supp}\; \hat{G}(k)\subset [-\varepsilon _0,\varepsilon _0]^3$.

By using the Fourier transform (C.3) of $G^{(\Lambda )}(x)$, we have

$$\begin{aligned}{} & {} \sum _{x,y\in \Lambda }\psi ^\dagger (x)\psi (x)G^{(\Lambda )}(x-y)(-1)^{x^{(1)}+x^{(2)}+x^{(3)}+y^{(1)}+y^{(2)}+y^{(3)}} \psi ^\dagger (y)\psi (y)\nonumber \\{} & {} \quad =\frac{1}{\vert \Lambda \vert }\sum _k\hat{G}(k)\sum _x\psi ^\dagger (x)\psi (x)e^{ikx}(-1)^{x^{(1)}+x^{(2)}+x^{(3)}}\nonumber \\{} & {} \qquad \times \sum _y \psi ^\dagger (y)\psi (y)e^{-iky}(-1)^{y^{(1)}+y^{(2)}+y^{(3)}}. \end{aligned}$$

(C.5)

Further, by using the expression (B.1) of $\psi (x)$, we have

$$\begin{aligned} \sum _x\psi ^\dagger (x)\psi (x)e^{ikx}(-1)^{x^{(1)}+x^{(2)}+x^{(3)}}= & {} \frac{1}{8}\sum _p\bigl [ \hat{\Psi }_\textrm{u}^\dagger (k+p)\gamma _0\hat{\Psi }_\textrm{u}(p) +\hat{\Psi }_\textrm{d}^\dagger (k+p)\gamma _0\hat{\Psi }_\textrm{d}(p)\bigr ]\nonumber \\= & {} \frac{1}{8}\sum _p\hat{\Psi }(k+p)\gamma _0\hat{\Psi }(p) \end{aligned}$$

(C.6)

in the same way that we treated the mass term of the Hamiltonian, where we have written

$$\begin{aligned} \hat{\Psi }(k):=\begin{pmatrix} \hat{\Psi }_\textrm{u}(k) \\ \hat{\Psi }_\textrm{d}(k) \end{pmatrix}. \end{aligned}$$

(C.7)

Therefore, by substituting this expression into the right-hand side in the above equation (C.5), the right-hand side is written

$$\begin{aligned} \frac{1}{\vert \Lambda \vert }\sum _{k,p,q}\hat{G}(k)\hat{\Psi }^\dagger (k+p)\gamma _0\hat{\Psi }(p)\hat{\Psi }^\dagger (q)\gamma _0\hat{\Psi }(q+k), \end{aligned}$$

(C.8)

where we have dropped the prefactor. Therefore, the formal infinite-volume and continuum limit is given by

$$\begin{aligned} \int _{\mathbb {R}^3} dx^{(1)}dx^{(2)}dx^{(3)} \int _{\mathbb {R}^3} dy^{(1)}dy^{(2)}dy^{(3)}\;\Psi ^\dagger (x)\gamma _0\Psi (x)G(x-y)\Psi ^\dagger (y)\gamma _0\Psi (y), \end{aligned}$$

(C.9)

where $\Psi (x)$ is the inverse Fourier transform of $\hat{\Psi }(k)$, and the function G(x) is the corresponding limit of $G^{(\Lambda )}(x)$.

When we take the continuum limit, the region $[-\varepsilon _0,\varepsilon _0]^3$ of the support condition $\textrm{supp}\; \hat{G}(k)\subset [-\varepsilon _0,\varepsilon _0]^3$ for the function $\hat{G}(k)$ is replaced by $[-\varepsilon _0/a,\varepsilon _0/a]^3$ with a small lattice constant $a>0$. Clearly, one can take $\varepsilon _0$ and a so that $\varepsilon _0/a$ takes a large value. This implies that one can take the function $\hat{G}(k)$ so that it has a large support. Therefore, the function G(x) in the real space can be taken to approach to the delta function in a sequence. Namely, $G_n(x)\rightarrow \delta (x)$ as $n\rightarrow \infty$. In this formal limit, the above interaction has the form,

$$\begin{aligned} \int _{\mathbb {R}^3} dx^{(1)}dx^{(2)}dx^{(3)}\;[\Psi ^\dagger (x)\gamma _0\Psi (x)]^2. \end{aligned}$$

(C.10)

This is a desired form of the interaction.