1. Introduction
In our previous works [1, 2], we solved the two-dimensional strongly correlated problem of the electron system by the slave particle representation. A strongly correlated electron system is exactly mapped to a weak coupling slave particle system by exactly dealing with the local constraints which are either Dirac’s first-class ones when the system is in the atomic limit of the electrons [1] or the second-class ones in the mean field states [2]. The essential step for the exact mapping is taking a gauge fixing condition that is consistent with the local constraints. For the first-class constraint, the consistent gauge theory was established by Fradkin, Vilkovisky, and Batalin [3–5] half a century ago. We explained their theory in a comprehensible language for the condensed matter physicists [1, 2]. The gauge fixing term added to the Lagrangian of the gauge theory is in a Becchi–Rouet–Stora–Tyutin (BRST) [6–8] exact form, while the BRST charge acting on the physical state exactly enforces both the local constraints and the gauge fixing condition [9]. The first-class constraint’s problem has been completely solved so that we will not concern ourselves with it in this paper anymore.
There is no a systematic way to consistently solve a gauge theory with the second-class constraints. We have solved the mean field theory where the current constraint is the second-class one in the t–J model by considering the BRST symmetry [2]. In principle, we have obtained a consistent gauge theory with the second-class constraint. We found that the Lorenz gauge is a consistent one while the other familiar ones, such as the axial gauge or the Coulomb gauge, are not. However, when performing perturbation calculations, we encountered a problem: the free propagator of the gauge field is ill-defined because of the skew U(1) gauge symmetry of the Lorenz gauge fixing term. We modified the gauge fixing term so that the ill-defined problem of the free propagator was resolved while it remained BRST invariant. We then performed the perturbation calculation for the strange metal phase of the t–J model and obtained some results that are consistent with the experimental measurements on cuprates. However, we did not check whether the gauge fixing term we used was a BRST exact form or not. If this BRST invariant term is BRST-closed but not an exact form, the added term may not be in the same BRST cohomology class as the vanishing pure gauge field Lagrangian L0(aμ) = 0 before the gauge fixing. For example, we can add a Maxwell term plus the Lorenz gauge fixing term and the ghost term to L0(aμ) = 0, which is BRST closed but not exact. This obviously introduces additional dynamics to the system and changes the physics. On the other hand, although the free propagator of the gauge field under the Lorenz gauge is ill-defined, can we still perform the perturbation calculation in this case? In this paper, we will clarify the confusion and inconsistency in the previous work and to address the incompleteness therein. We attach an appendix where we show the BRST formalism of the t–J model on a lattice rather than the continuum limit (see appendix A ).
2. Brief review on the mean field theory of the t–J model
We consider the t–J model with the Hamiltonian on a square lattice,
$\begin{eqnarray}\begin{array}{r}{H}_{t-J}=-t\displaystyle \sum _{\langle ij\rangle ,\sigma }{c}_{i\sigma }^{\dagger }{c}_{j\sigma }+J\displaystyle \sum _{\langle ij\rangle }\left({{\boldsymbol{S}}}_{i}\cdot {{\boldsymbol{S}}}_{j}-\frac{1}{4}{n}_{i}{n}_{j}\right),\end{array}\end{eqnarray}$
where ciσ is the electron annihilation operator at a lattice site i with spin σ; ${S}_{i}^{a}=\frac{1}{2}{\sum }_{\sigma ,{\sigma }^{{\prime} }}{c}_{i\sigma }^{\dagger }{\sigma }_{\sigma {\sigma }^{{\prime} }}^{a}{c}_{i{\sigma }^{{\prime} }}$ are the spin operators, and σa (a = x, y, z) are Pauli matrices. The hopping amplitude t and the exchange amplitude J are fixed for the nearest neighbor sites. The constraint is that there is no double occupation at each lattice site, i.e. ${c}_{i}^{\dagger }{c}_{i}\leqslant 1$ for all i with a fixed total electron number.In the slave boson representation, the electron operator is decomposed into the fermionic spinon and bosonic holon, ${c}_{i\sigma }^{\dagger }={f}_{i\sigma }^{\dagger }{h}_{i}$, where ${f}_{i\sigma }^{\dagger }$ is the spinon creation operator and hi is the holon annihilation operator. This decomposition is valid when the local constraint is enforced for every site i by
$\begin{eqnarray}\begin{array}{r}{G}_{i}={h}_{i}^{\dagger }{h}_{i}+\displaystyle \sum _{\sigma }{f}_{i\sigma }^{\dagger }{f}_{i\sigma }-1=0.\end{array}\end{eqnarray}$
The mean field Hamiltonian of the t–J model in the slave boson representation reads [10, 11], $\begin{eqnarray}\begin{array}{rcl}{L}_{{\rm{M}}{\rm{F}}} & = & \displaystyle \frac{J}{4}\displaystyle \sum _{\langle ij\rangle }\left[\left|{\gamma }^{f}{| }^{2}+| {{\rm{\Delta }}}_{a}{| }^{2}-\displaystyle \sum _{\sigma }({\gamma }^{f\dagger }{{\rm{e}}}^{{\rm{i}}{a}_{ij}}{f}_{i\sigma }^{\dagger }{f}_{j\sigma }+{\rm{h}}.{\rm{c}}.\right)\right]\\ & & +\displaystyle \sum _{\langle ij\rangle }\displaystyle \frac{J}{4}[{{\rm{\Delta }}}_{a}{{\rm{e}}}^{{\rm{i}}{\phi }_{ij}}({f}_{i\uparrow }^{\dagger }{f}_{j\downarrow }^{\dagger }-{f}_{i\downarrow }^{\dagger }{f}_{j\uparrow }^{\dagger })+{\rm{h}}.{\rm{c}}.]\\ & & +\displaystyle \sum _{i}{h}_{i}^{\dagger }({\partial }_{\tau }-{\mu }_{h}){h}_{i}+\displaystyle \sum _{i\sigma }{f}_{i\sigma }^{\dagger }({\partial }_{\tau }-{\mu }_{f}){f}_{i\sigma }\\ & & -t\displaystyle \sum _{\langle ij\rangle }({{\rm{e}}}^{{\rm{i}}{a}_{ij}}({\gamma }^{f}{h}_{i}^{\dagger }{h}_{j}+{\gamma }^{h\dagger }{f}_{i\sigma }^{\dagger }{f}_{j\sigma })+{\rm{h}}.{\rm{c}}.)\\ & & +\displaystyle \sum _{i}ig{\lambda }_{i}{G}_{i},\end{array}\end{eqnarray}$
where Δa for a = x, y labels the pairing parameter in the a-link, and γh,f are the hopping parameters for the spinon and the holon. We choose Δa and γh,f as expectation values in the mean field approximation. The phase fields aij and Φij, which obey the periodic boundary condition, are quantum fluctuations introduced to compensate for the gauge symmetry breaking due to the mean field approximation. Besides the temporal gauge field λi, the mean field theory has a spatial gauge invariance under the transformations $({f}_{i\sigma }(\tau ),{h}_{i}(\tau ))\to {{\rm{e}}}^{{\rm{i}}{\theta }_{i}(\tau )}({f}_{i\sigma }(\tau ),{h}_{i}(\tau ))$, and aij → aij + θi − θj and Φij → Φij + θi + θj. The equation of motion of aij leads to the constraint of vanishing counterflow between the holon and spinon currents due to the mean field approximations [12], i.e. $\begin{eqnarray}\begin{array}{r}{J}_{ij}={J}_{ij}^{f}+{J}_{ij}^{h}=0.\end{array}\end{eqnarray}$
This is also a local constraint. We have shown that the variation of Φij does not result in new constraints [2]. Therefore, our problem is how to quantize the mean field theory with the constraints Gi = 0 and Jij = 0 under proper gauge fixing conditions.3. Brief review on our previous work [2]
We briefly review the main results obtained in [2]. In the continuum limit, the mean field Lagrangian (3 ) with a d-wave pairing is given by 6 ) can be obtained by a so-call off-shell BRST exact form (see e.g. [9]). Defining δB = εs, we introduce the off-shell BRST Lagrangian 7 ). The auxiliary field Π is the Nakanishi–Lautrup field. This off-shell gauge fixing term is explicitly BRST exact. The Euler–Lagrange equation of Π given Π = ξ−1(ζ∂τδλ + ∑b∂bδab) and substituting which into ${L}_{\mathrm{GFL}}^{\mathrm{off}}$, one recovers equation (6 ). However, it is easy to check that the free propagator of the gauge field is ill-defined because ${D}_{\mu \nu }^{(0)-1}(i\omega ,{\boldsymbol{q}})\propto {q}_{\mu }{q}_{\nu }$, so that $\det {D}^{(0)-1}=0$.
$\begin{eqnarray}\begin{array}{rcl}{L}_{\mathrm{MFC}} & = & \int {{\rm{d}}}^{2}r\left[\displaystyle \sum _{\sigma }{f}_{\sigma }^{\dagger }({\partial }_{\tau }-{\mu }_{f}-{\rm{i}}g\delta \lambda ){f}_{\sigma }\right.\\ & & \left.+{h}^{\dagger }({\partial }_{\tau }-{\mu }_{h}-{\rm{i}}g\delta \lambda )h\Space{0ex}{2.5ex}{0ex}\right]\\ & & -\int {{\rm{d}}}^{2}r\left[\displaystyle \frac{1}{2{m}_{f}}\displaystyle \sum _{\sigma ,a}{f}_{\sigma }^{\dagger }{(-{\rm{i}}{\partial }_{a}-g\delta {a}_{a})}^{2}{f}_{\sigma }\right.\\ & & -\left.\displaystyle \frac{1}{2{m}_{h}}\displaystyle \sum _{a}{h}^{\dagger }{(-{\rm{i}}{\partial }_{a}-g\delta {a}_{a})}^{2}h\right]\\ & & +\displaystyle \frac{1}{2}\int {{\rm{d}}}^{2}r\displaystyle \sum _{a}({{\rm{\Delta }}}_{a}{\partial }_{a}({{\rm{e}}}^{{\rm{i}}\phi /2}{f}_{\uparrow }^{\dagger }){\partial }_{a}({{\rm{e}}}^{{\rm{i}}\phi /2}{f}_{\downarrow }^{\dagger })+{\rm{h}}.{\rm{c}}.),\end{array}\end{eqnarray}$
where the uniform RVB state is taken in the mean field state of the gauge field due to time-reversal symmetry [13]. The pure gauge field Lagrangian is L0(aμ) = 0 where aμ = (aτ, ai) = (δλ, δai). To retain the constraints and remove the gauge redundancy, we tried the BRST invariant gauge fixing term as $\begin{eqnarray}\begin{array}{rcl}{L}_{\mathrm{GFL}} & = & -\int {{\rm{d}}}^{2}r\displaystyle \frac{1}{2\xi }{\left(\zeta {\partial }_{\tau }\delta \lambda +\displaystyle \sum _{a}{\partial }_{a}\delta {a}_{a}\right)}^{2}\\ & & +\int {{\rm{d}}}^{2}r\bar{u}\left(\zeta {\partial }_{\tau }^{2}+\displaystyle \sum _{a}{\partial }_{a}^{2}\right)u,\end{array}\end{eqnarray}$
where ξ is an arbitrary (gauge) parameter and ζ−1 is similar to the speed of light. This is called the Lorenz gauge fixing and is invariant under the BRST transformation $\begin{eqnarray}\begin{array}{rcl}{\delta }_{{\rm{B}}}{f}_{\sigma } & = & {\rm{i}}\epsilon gu{f}_{\sigma },{\delta }_{{\rm{B}}}h={\rm{i}}\epsilon guh,{\delta }_{{\rm{B}}}\phi =2{\rm{i}}\epsilon gu,\\ {\delta }_{{\rm{B}}}\delta \lambda & = & \epsilon {\partial }_{\tau }u,{\delta }_{{\rm{B}}}\delta {a}_{b}=\epsilon {\partial }_{b}u,{\delta }_{{\rm{B}}}u=0,\\ {\delta }_{{\rm{B}}}\bar{u} & = & \epsilon {\xi }^{-1}\left(\zeta {\partial }_{\tau }\delta \lambda +\displaystyle \sum _{b}{\partial }_{b}\delta {a}_{b}\right).\end{array}\end{eqnarray}$
The BRST transformation is nilpotent, i.e. ${\delta }_{{\rm{B}}}^{2}=0$. (${\delta }_{{\rm{B}}}^{2}\bar{u}=0$ due to the equation of motion for u. We here use the on-shell BRST transformation. Notice that equation ( $\begin{eqnarray}\begin{array}{l}{L}_{\mathrm{GFL}}^{\mathrm{off}}=s\left[\bar{u}\left(\displaystyle \frac{\xi }{2}{\rm{\Pi }}-\left(\zeta {\partial }_{\tau }\delta \lambda +\displaystyle \sum _{b}{\partial }_{b}\delta {a}_{b}\right)\right)\right],\end{array}\end{eqnarray}$
and the BRST transformations are modified as $\begin{eqnarray}\begin{array}{r}{\delta }_{{\rm{B}}}\bar{u}={\rm{\Pi }},\quad {\delta }_{{\rm{B}}}{\rm{\Pi }}=0,\end{array}\end{eqnarray}$
while other fields’ are the same as those in equation (To overcome this difficulty and get a well-defined free gauge propagator, we introduced another gauge fixing term 11 ) implies A/ζ2 = B/ζ = C, which reduces to the Lorenz gauge for C = 1/ξ. It is easy to verify that the free propagator of the gauge field is well-defined except when D = E = 0. In our previous work [2], we set B = 0 while keeping D, E ≠ 0 to decouple the temporal and spatial components for convenience. The perturbation calculations in [2] are all based on LMFC + LGF.
$\begin{eqnarray}\begin{array}{rcl}{L}_{{\rm{G}}{\rm{F}}} & = & \frac{A}{2}{({\partial }_{\tau }\delta \lambda )}^{2}+B\displaystyle \sum _{b}{\partial }_{\tau }\delta {a}_{b}{\partial }_{b}\delta \lambda +\frac{C}{2}{\left(\displaystyle \sum _{b}{\partial }_{b}\delta {a}_{b}\right)}^{2}\\ & & +\frac{D}{2}\displaystyle \sum _{b}{({\partial }_{\tau }\delta {a}_{b})}^{2}+\frac{E}{2}\displaystyle \sum _{b}{({\partial }_{b}\delta \lambda )}^{2}+\bar{u}Ku,\end{array}\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{r}A=C{\zeta }^{2},\quad C\zeta =B+D=B+E,\quad E=D,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{r}K=-C\xi \left(\zeta {\partial }_{\tau }^{2}+\displaystyle \sum _{b}{\partial }_{b}^{2}\right).\end{array}\end{eqnarray}$
This gauge fixing term is also the BRST invariant. Furthermore, when D = E = 0, equation (4. Flaws in the previous section and a new point of view
4.1. The BRST exact form of the gauge fixing term
We added a BRST invariant term equation (10 ) to L0(aμ) = 0 as the gauge fixing. However, we did not prove whether the added gauge fixing term is of the BRST exact form or not. According to the BRST cohomology, only when the added gauge fixing term is BRST exact does it not introduce extra dynamics to the system. For example, the Maxwell term is BRST invariant but not BRST exact. If we add this term with a Lorenz gauge fixing to L0(aμ) = 0, extra gauge field dynamics is introduced to the system.
Let us determine what kind of gauge fixing terms are BRST exact. We consider a general BRST transformation 16 ) can be obtained from a BRST exact off-shell Lagrangian
$\begin{eqnarray}\begin{array}{r}{\delta }_{{\rm{B}}}{a}_{\mu }=\epsilon {\partial }_{\mu }u,{\delta }_{{\rm{B}}}u=0,{\delta }_{{\rm{B}}}\bar{u}=\epsilon F({a}_{\mu }),\end{array}\end{eqnarray}$
with F(aμ) to be determined. Denoting δB = εs, a BRST exact form which is added to the Lagrangian is assumed to be proportional to $\begin{eqnarray}\begin{array}{rcl}s\left(\bar{u}\displaystyle \sum _{\mu \nu }{C}_{\mu \nu }{\partial }_{\mu }{a}_{\nu }\right) & = & F\displaystyle \sum _{\mu \nu }{C}_{\mu \nu }{\partial }_{\mu }{a}_{\nu }\\ & & -\bar{u}\displaystyle \sum _{\mu \nu }{C}_{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }u.\end{array}\end{eqnarray}$
The equation of motion for u is ∑μνCμν∂μ∂νu = 0. This leads to $\begin{eqnarray}\begin{array}{r}F({a}_{\mu })=\frac{1}{2\xi }\displaystyle \sum _{\mu ,\nu }{C}_{\mu \nu }{\partial }_{\mu }{a}_{\nu },\end{array}\end{eqnarray}$
due to the requirement of s2 = 0. Therefore, a general BRST exact term is given by $\begin{eqnarray}\begin{array}{rcl}{L}_{{\rm{G}}{\rm{G}}{\rm{F}}} & = & \frac{1}{2\xi }{\left(\displaystyle \sum _{\mu \nu }{C}_{\mu \nu }{\partial }_{\mu }{a}_{\nu }\right)}^{2}-\bar{u}\displaystyle \sum _{\mu \nu }{C}_{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }u.\end{array}\end{eqnarray}$
The Lorenz gauge is recovered by taking $\begin{eqnarray}\begin{array}{r}{C}_{\mu \nu }={\delta }_{\mu \nu }.\end{array}\end{eqnarray}$
As before, equation ( $\begin{eqnarray}\begin{array}{r}{L}_{{\rm{G}}{\rm{G}}{\rm{F}}}^{{\rm{o}}{\rm{f}}{\rm{f}}}=s\left[\bar{u}\left(\frac{\xi }{2}{\rm{\Pi }}-\displaystyle \sum _{\mu \nu }{C}_{\mu \nu }{\partial }_{\mu }{a}_{\nu }\right)\right].\end{array}\end{eqnarray}$
The inverse of the free gauge propagator corresponding to equation (16 ) can be read out, ${D}^{(0)-1}\propto {\tilde{k}}_{\mu }{\tilde{k}}_{\nu }$ for ${\tilde{k}}_{\mu }\,={\sum }_{\rho }{C}_{\mu \rho }{k}_{\rho }$. Therefore the free propagator of the gauge field D(0) now is ill-defined because $\det ({D}^{(0)-1})\propto \det ({\tilde{k}}_{\mu }{\tilde{k}}_{\nu })=0$.
On the other hand, the off-shell version of equation (10 ) can be written as (up to a surface term)
$\begin{eqnarray}\begin{array}{rcl}{L}_{{\rm{G}}{\rm{F}}}^{{\rm{o}}{\rm{f}}{\rm{f}}} & = & \frac{1}{2}\left(A-\frac{{\zeta }^{2}}{\xi }\right){\left({\partial }_{\tau }\delta \lambda \right)}^{2}\\ & & +\left(B-\frac{\zeta }{\xi }\right)\left(\displaystyle \sum _{b}{\partial }_{b}\delta {a}_{b}\right)({\partial }_{\tau }\delta \lambda )\\ & & +\frac{1}{2}\left(C-\frac{1}{\xi }\right)\displaystyle \sum _{b}{({\partial }_{b}\delta {a}_{b})}^{2}\\ & & +\frac{D}{2}\displaystyle \sum _{b}{({\partial }_{\tau }\delta {a}_{b})}^{2}+\frac{E}{2}\displaystyle \sum _{b}{({\partial }_{b}\delta \lambda )}^{2}\\ & & +Cs\left[\bar{u}\left(\frac{\xi }{2}{\rm{\Pi }}-\left(\zeta {\partial }_{\tau }\delta \lambda +\displaystyle \sum _{b}{\partial }_{b}\delta {a}_{b}\right)\right)\right].\end{array}\end{eqnarray}$
The last term is a BRST exact form. The rest terms are actually gauge invariant and thus are not of the BRST exact form. Therefore, ${L}_{{\rm{G}}{\rm{F}}}^{{\rm{o}}{\rm{f}}{\rm{f}}}$ is not BRST exact. Obviously, when E = D = 0 and C = 1/ξ, ${L}_{{\rm{G}}{\rm{F}}}^{{\rm{o}}{\rm{f}}{\rm{f}}}$ recovers ${L}_{{\rm{G}}\mathrm{FL}}^{{\rm{o}}{\rm{f}}{\rm{f}}}$. Thus, the gauge invariant terms in ${L}_{{\rm{G}}{\rm{F}}}^{{\rm{o}}{\rm{f}}{\rm{f}}}$ can be thought as the regulator to ensure a well-defined full propagator of the gauge field.4.2. The exact local constraint problem
The BRST symmetry is a global symmetry; Noether’s theorem gives the conserved BRST charge Q which counts the ghost number of a state. If we denote the BRST exact term in equation (19 ) as s$\Psi$ which serves as the gauge fixing term and a small variation of such a gauge fixing condition as $s\tilde{\delta }{\rm{\Psi }}$, the variation of any physical matrix element vanishes [9], i.e.19 ), the on-shell BRST charge is given by [2]
$\begin{eqnarray}\begin{array}{rcl}\tilde{\delta }\langle \alpha | \beta \rangle & = & -\left\langle \alpha | \tilde{\delta }\displaystyle \int {\rm{d}}\tau L| \beta \right\rangle =-\langle \alpha | \{{Q}^{{\rm{o}}{\rm{f}}{\rm{f}}},\tilde{\delta }{\rm{\Psi }}\}| \beta \rangle \\ & = & 0.\end{array}\end{eqnarray}$
Since $\tilde{\delta }{\rm{\Psi }}$ is arbitrary, one must have $\begin{eqnarray}\begin{array}{r}\langle \alpha | {Q}^{{\rm{o}}{\rm{f}}{\rm{f}}}=0,\,{Q}^{{\rm{o}}{\rm{f}}{\rm{f}}}| \beta \rangle =0.\end{array}\end{eqnarray}$
Integrating over the auxiliary field, this proves the BRST charge free requirement of the physical states, $\begin{eqnarray*}Q| {\rm{p}}{\rm{h}}{\rm{y}}{\rm{s}}\rangle =0.\end{eqnarray*}$
For the theory with the gauge fixing term in equation ( $\begin{eqnarray}\begin{array}{rcl}Q & = & \displaystyle \int {{\rm{d}}}^{2}x\,\left({\rm{i}}gG+E\displaystyle \sum _{b}{\partial }_{b}^{2}\delta \lambda -D\displaystyle \sum _{b}{\partial }_{\tau }{\partial }_{b}\delta {a}_{b}\right)u\\ & & +\left[A{\partial }_{\tau }\delta \lambda +(B+D)\displaystyle \sum _{b}{\partial }_{b}\delta {a}_{b}\right]{\partial }_{\tau }u.\end{array}\end{eqnarray}$
The constraints read out from the BRST charge free of the physical states are given by $\begin{eqnarray}\begin{array}{rcl}\left({\rm{i}}gG+E\displaystyle \sum _{b}{\partial }_{b}^{2}\delta \lambda -D\displaystyle \sum _{b}{\partial }_{\tau }{\partial }_{b}\delta {a}_{b}\right) & = & 0,\\ \left[A{\partial }_{\tau }\delta \lambda +(B+D)\displaystyle \sum _{b}{\partial }_{b}\delta {a}_{b}\right] & = & 0.\end{array}\end{eqnarray}$
While the second equation is the gauge fixing condition, the first one does not exactly give the local constraint G = 0 unless both E and D are zero. In deriving the conserved charge from Noether’s theorem, the Euler–Lagrange equations of the fields are used. For example, the equations of motion of the gauge field δai gives $\begin{eqnarray}\begin{array}{r}{J}_{b}(\delta a)=B{\partial }_{\tau }{\partial }_{b}\delta \lambda +D{\partial }_{\tau }^{2}\delta {a}_{b}+C{\partial }_{b}\left(\displaystyle \sum _{c}{\partial }_{c}\delta {a}_{c}\right),\end{array}\end{eqnarray}$
which also does not yield the constraint Jb = 0.4.3. Completeness of consistent gauge theory
In light of the discussions in section 4 , our theory in [2] still has some problems: (1) Extra artificial gauge field dynamics was introduced. (2) Both the local number and current constraints are not exactly recovered. The above two problems can be solved by applying the Lorenz gauge in the D = E = 0 limit. On the one hand, the gauge fixing term is BRST exact so that no extra dynamics is introduced by hand; on the other hand, the BRST charge becomes 11 ). However, as we have shown that the free propagator of the gauge field is ill-defined for the Lorenz gauge. The perturbation calculation therefore seems to be invalid. Instead of equation (10 ), one may try to add a Maxwell term to the gauge field as the usual treatment in quantum electrodynamics where the propagator of the gauge field is well-defined. But, there will be extra contributions to the BRST charge (22 ) and the current (24 ) stemming from the Maxwell term (because the Maxwell term is not BRST-exact) unless the strength of the Maxwell term goes to zero, and the problem of ill-defined propagator remains.
$\begin{eqnarray}\begin{array}{r}Q=\displaystyle \int {{\rm{d}}}^{2}x\,\left({\rm{i}}gGu+B\left[\zeta {\partial }_{\tau }\delta \lambda +\displaystyle \sum _{b}{\partial }_{b}\delta {a}_{b}\right]{\partial }_{\tau }u\right).\end{array}\end{eqnarray}$
This gives the local constraint: G = 0 and the Lorenz gauge ζ∂τδλ + ∑b∂bδab = 0. The vanishing counterflow constraint becomes $\begin{eqnarray}\begin{array}{r}{J}_{b}(\delta a)=C{\partial }_{b}\left(\zeta {\partial }_{\tau }\delta \lambda +\displaystyle \sum _{c}{\partial }_{c}\delta {a}_{c}\right)=0,\end{array}\end{eqnarray}$
where the second equality comes from the Lorenz gauge because of equation (Can we have a solution to these problems? Let us recall the source of the ill-defined nature of the free propagator of the gauge field. As we have observed, the zero mode in the gauge field actually comes from the skew U(1) gauge symmetry of the Lorenz gauge fixing term [2], i.e. when 19 ), in order to keep the BRST symmetry of the theory, and take D = E → 0 limit after the calculations. In this case, if we denote the gauge field propagator as ${D}_{\mu \nu }^{{\prime} }$, then the free propagator ${D}_{\mu \nu }^{{}^{{\prime} }(0)}$ is invertible. According to Dyson’s equation, the full propagator is B ). We then have a well-defined DRPA. The same process also works for adding a Maxwell term and then taking the zero strength limit.
$\begin{eqnarray*}{a}_{\mu }\to {a}_{\mu }+{\epsilon }_{\mu \nu }{\partial }_{\nu }\theta ,\end{eqnarray*}$
the Lorenz gauge fixing condition is invariant, $\begin{eqnarray*}{\partial }_{\mu }{a}_{\mu }\to {\partial }_{\mu }({a}_{\mu }+{\epsilon }_{\mu \nu }{\partial }_{\nu }\theta )={\partial }_{\mu }{a}_{\mu }.\end{eqnarray*}$
This leads to the situation where the inverse of the free propagator of the gauge field, ${D}_{\mu \nu }^{(0)}\propto {q}_{\mu }{q}_{\nu }$ has a vanishing determinant. However, this skew U(1) gauge invariance holds only for the gauge fixing term. The whole Lagrangian LMFC + LGFL is not skew gauge invariant because the minimal coupling between the gauge field and the matter field breaks this symmetry. Thus, the full propagator Dμν is well-defined. To avoid the divergence of ${D}_{\mu \nu }^{(0)}$, we use the Lagrangian LGF, i.e. equation ( $\begin{eqnarray}\begin{array}{r}{D}_{\mu \nu }^{{\prime} }={[{({D}^{{}^{{\prime} }(0)-1}-{{\rm{\Pi }}{}^{{\prime} }}^{* })}^{-1}]}_{\mu \nu },\end{array}\end{eqnarray}$
where ${{\rm{\Pi }}{}^{{\prime} }}^{* }$ is the vacuum polarization of the holon h and the spinon fσ. When D, E → 0, we have the well-defined $\begin{eqnarray}\begin{array}{r}{D}_{\mu \nu }={[{({D}^{(0)-1}-{{\rm{\Pi }}}^{* })}^{-1}]}_{\mu \nu }.\end{array}\end{eqnarray}$
For example, if we take the random phase approximation (RPA) for the propagator, one has $\begin{eqnarray}\begin{array}{r}{D}_{\mu \nu }^{{}^{{\prime} }{\rm{R}}{\rm{P}}{\rm{A}}}={[{({D}^{{}^{{\prime} }(0)-1}-{{\rm{\Pi }}}^{{}^{{\prime} }(0)* })}^{-1}]}_{\mu \nu },\end{array}\end{eqnarray}$
where ${{\rm{\Pi }}}^{{}^{{\prime} }(0)* }={{\rm{\Pi }}}^{(0)* }$ does not have the contribution from the gauge field propagator. And ${D}^{{}^{{\prime} }{\rm{R}}{\rm{P}}{\rm{A}}}\to {D}^{{\rm{R}}{\rm{P}}{\rm{A}}}$ when D, E → 0 (see appendix We can replace D(0) with DRPA in all calculations. In this way, we are able to perform a perturbative calculation for the consistent gauge theory in the slave boson representation of the t–J model in the strong coupling limit.
5. Conclusions
We now complete the construction of the consistent gauge theory for the U(1) slave particle representation of the t–J model. We demonstrated that the gauge fixing condition is BRST-exact and show that local constraints are exactly preserved in the Lorenz gauge. The difficulty of ill-defined free gauge propagator was also resolved. This work thus establishes a concrete framework for systematic perturbation theory. The further tasks will be to perform the hard work of the perturbation calculations for the physical observables. For example, the first task is to re-calculate the key observables in the strange metal phase from [2] using the new, consistent formalism presented here. We provide a consistent formalism that duals a strongly correlated system into a controllable weakly coupled gauge theory with constraints, this may pave a way towards a deeper understanding of the physics in other strongly correlated systems, such as the overdoped regime of the superconducting phase and the pseudo gap regime of cuprates, Mott physics, and spin liquids.
Appendix A BRST symmetry on lattice
For the completeness of this paper, we write some results of the lattice version of consistent gauge theory with the BRST symmetry in this appendix.
For the t–J model, the partition function on lattice reads
$\begin{eqnarray}\begin{array}{r}Z=\displaystyle \int DfD{f}^{\dagger }DbD{b}^{* }D\lambda D\chi D{\rm{\Delta }}\exp \left(-{\displaystyle \int }_{0}^{\beta }{\rm{d}}\tau {L}_{1}\right),\end{array}\end{eqnarray}$
with $\begin{eqnarray}\begin{array}{rcl}{L}_{1} & = & \tilde{J}\displaystyle \sum _{\langle ij\rangle }(| {\chi }_{ij}{| }^{2}+| {{\rm{\Delta }}}_{ij}{| }^{2})+\displaystyle \sum _{i,\sigma }{f}_{i\sigma }^{\dagger }({\partial }_{\tau }-{\mu }_{f}){f}_{i\sigma }\\ & & +\displaystyle \sum _{i}{b}_{i}^{* }({\partial }_{\tau }-{\mu }_{b}){b}_{i}-\displaystyle \sum _{ij}{t}_{ij}{b}_{i}{b}_{j}^{* }{f}_{i\sigma }^{\dagger }{f}_{j\sigma }\\ & & -{\rm{i}}\displaystyle \sum _{i}{\lambda }_{i}({f}_{i\sigma }^{\dagger }{f}_{i\sigma }+{b}_{i}^{* }{b}_{i}-1)\\ & & -\tilde{J}\left[\displaystyle \sum _{\langle ij\rangle }{\chi }_{ij}^{* }\left(\displaystyle \sum _{\sigma }{f}_{i\sigma }^{\dagger }{f}_{j\sigma }+{\rm{c}}.{\rm{c}}.\right)\right]\\ & & +\tilde{J}\left[\displaystyle \sum _{\langle ij\rangle }{{\rm{\Delta }}}_{ij}({f}_{i\uparrow }^{\dagger }{f}_{j\downarrow }^{\dagger }-{f}_{i\downarrow }^{\dagger }{f}_{j\uparrow }^{\dagger })+{\rm{c}}.{\rm{c}}.\right],\end{array}\end{eqnarray}$
with $\begin{eqnarray}\tilde{J}=J/4,\,{\chi }_{ij}=\displaystyle \sum _{\sigma }({f}_{i\sigma }^{\dagger }{f}_{j\sigma }),\,{{\rm{\Delta }}}_{ij}=({f}_{i\uparrow }{f}_{j\downarrow }-{f}_{i\downarrow }{f}_{j\uparrow }).\end{eqnarray}$
There is a gauge invariance, $\begin{eqnarray}\begin{array}{rcl}{f}_{i\sigma } & \to & {{\rm{e}}}^{{\rm{i}}{\theta }_{i}}{f}_{i\sigma },\quad {b}_{i}\to {{\rm{e}}}^{{\rm{i}}{\theta }_{i}}{b}_{i},\quad {\chi }_{ij}\to {{\rm{e}}}^{-{\rm{i}}{\theta }_{i}}{\chi }_{ij}{{\rm{e}}}^{{\rm{i}}{\theta }_{j}},\\ {{\rm{\Delta }}}_{ij} & \to & {{\rm{e}}}^{{\rm{i}}{\theta }_{i}}{{\rm{\Delta }}}_{ij}{{\rm{e}}}^{{\rm{i}}{\theta }_{j}},\quad {\lambda }_{i}\to {\lambda }_{i}+{\partial }_{\tau }{\theta }_{i}.\end{array}\end{eqnarray}$
In the mean field theory, one chooses ${\chi }_{ij}={\sum }_{\sigma }\langle {f}_{i\sigma }^{\dagger }{f}_{j\sigma }\rangle $ and Δij = ⟨fi↑fj↓ − fi↓fj↑⟩. In the uniform RVB mean field theory, one assume $\begin{eqnarray}\begin{array}{r}{\chi }_{ij}=\chi ={\rm{r}}{\rm{e}}{\rm{a}}{\rm{l}}.\end{array}\end{eqnarray}$
Here we neglect the Δ field and consider the χ and λ fields. There are amplitude and phase fluctuations of the χ field, but the amplitude fluctuations are massive and do not play an important role in the low-energy limit. Furthermore, the mean field ground state we consider is the uniform RVB state, which is topologically trivial. Then, the ghost zero modes are ignored. Therefore the relevant Lagrangian to start with is $\begin{eqnarray}\begin{array}{rcl}{L}_{0} & = & \displaystyle \sum _{i\sigma }{f}_{i\sigma }^{\dagger }({\partial }_{\tau }-{\mu }_{f}+{\rm{i}}{a}_{0}({r}_{i})){f}_{i\sigma }\\ & & +\displaystyle \sum _{i}{b}_{i}^{* }({\partial }_{\tau }-{\mu }_{b}+{\rm{i}}{a}_{0}({r}_{i})){b}_{i}\\ & & -\tilde{J}\chi \displaystyle \sum _{\langle ij\rangle ,\sigma }({{\rm{e}}}^{{\rm{i}}{a}_{ij}}{f}_{i\sigma }^{\dagger }{f}_{j\sigma }+{\rm{h}}.{\rm{c}}.)\\ & & -t\chi \displaystyle \sum _{\langle ij\rangle }({{\rm{e}}}^{{\rm{i}}{a}_{ij}}{b}_{i}^{* }{b}_{j}+{\rm{h}}.{\rm{c}}.),\end{array}\end{eqnarray}$
with $\begin{eqnarray}\begin{array}{r}{a}_{ij}\to {a}_{ij}+{\theta }_{i}-{\theta }_{j},\quad {a}_{0}(i)\to {a}_{0}(i)+{\partial }_{\tau }{\theta }_{i}(\tau ).\end{array}\end{eqnarray}$
The equation of motion of aij leads to the constraint of vanishing counterflow between the holon and spinon currents, i.e. $\begin{eqnarray}\begin{array}{r}{J}_{ij}={J}_{ij}^{f}+{J}_{ij}^{h}=0.\end{array}\end{eqnarray}$
This is also a local constraint. Note that the gauge field appears in the expression of the spinon and holon currents. And the constraint holds for non-vanishing gauge configurations. The variation of Φij does not result in new constraints (Higgs mechanism).A.1. Gauge fixing term equation (10 ) on lattice
We abbreviate δλ and δaa as λ and aa for convenience. The gauge fixing terms (10 ) are
$\begin{eqnarray}\begin{array}{rcl}{L}_{{\rm{G}}{\rm{F}}} & = & \frac{A}{2}{({\partial }_{\tau }{a}_{\tau })}^{2}+\displaystyle \sum _{a}\frac{D}{2}{({\partial }_{\tau }{a}_{a})}^{2}\\ & & +\displaystyle \sum _{a}\frac{E}{2}{({\partial }_{a}{a}_{\tau })}^{2}+\displaystyle \sum _{ab}\frac{C}{2}({\partial }_{a}{a}_{a})({\partial }_{b}{a}_{b})\\ & & +\displaystyle \sum _{a}B({\partial }_{\tau }{a}_{a})({\partial }_{a}{a}_{\tau })+{L}_{{\rm{g}}{\rm{h}}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{r}{L}_{{\rm{g}}{\rm{h}}}=\bar{u}Ku,\end{array}\end{eqnarray}$
where μ = τ, a, a, b = x, y for two spatial dimensions.The lattice version for aτ(r, τ) is just λi(τ) for i, the two-dimensional lattice index. aa(r) comes from the ${U}_{ij}={{\rm{e}}}^{{\rm{i}}g{a}_{ij}}$. The spatial derivative is given by $\frac{\partial f}{\partial {x}_{\hat{\delta }}}\to {{\rm{\Delta }}}_{i,\hat{\delta }}f={f}_{i+\hat{\delta }}-{f}_{i}$. ${{\rm{\Delta }}}_{i,\hat{\delta }}^{+}$ means $\hat{\delta }=\hat{x},\hat{y}$ in the forward direction. Thus,
$\begin{eqnarray}\begin{array}{r}{\partial }_{a}{a}_{0}\to {\lambda }_{i+\hat{\delta }}-{\lambda }_{i}.\end{array}\end{eqnarray}$
On the other hand, ${a}_{ij}=({{\boldsymbol{r}}}_{j}-{{\boldsymbol{r}}}_{i})\cdot {\boldsymbol{a}}(\frac{{{\boldsymbol{r}}}_{i}+{{\boldsymbol{r}}}_{j}}{2})$.In the usual treatment of the BRST fermionic ghost, ui lives on lattice sites.
$\begin{eqnarray}\begin{array}{r}\displaystyle \int {{\rm{d}}}^{2}r\bar{u}\left(\zeta {\partial }_{\tau }^{2}+\displaystyle \sum _{a}{\partial }_{a}^{2}\right)u\to \displaystyle \sum _{i}{\bar{u}}_{i}({\partial }_{\tau }^{2}+{{\rm{\Delta }}}^{2}){u}_{i},\end{array}\end{eqnarray}$
where Δ2 is the Laplacian on lattice in two-dimensions, ${{\rm{\Delta }}}^{2}{u}_{i}={\sum }_{{{\boldsymbol{e}}}_{j}}({u}_{i+{{\boldsymbol{e}}}_{j}}+{u}_{i-{{\boldsymbol{e}}}_{j}}-2{u}_{i})$, and ej is the unit vector.A.2. Checking lattice BRST symmetry
The Fourier transformation is defined as ${f}_{i}\,={\sum }_{n}\int {{\rm{d}}}^{2}k{f}_{{\omega }_{n},k}{{\rm{e}}}^{{\omega }_{n}\tau +{\rm{i}}\overrightarrow{k}\cdot \overrightarrow{r}}$, and ${a}_{i,i+{{\boldsymbol{e}}}_{j}}=\int {{\rm{d}}}^{2}k{a}_{k}{{\rm{e}}}^{{\rm{i}}k(i+i+{{\boldsymbol{e}}}_{j})/2}$, where we assume all the gauge fields are real. The Lagrangian we are now considering is
$\begin{eqnarray}\begin{array}{r}L={L}_{0}+{L}_{{\rm{g}}{\rm{f}}}+{L}_{{\rm{g}}{\rm{h}}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{L}_{0} & = & \displaystyle \sum _{i\sigma }{f}_{i\sigma }^{\dagger }({\partial }_{\tau }-{\mu }_{f}+{\rm{i}}{\lambda }_{i}){f}_{i\sigma }\\ & & +\displaystyle \sum _{i}{b}_{i}^{* }({\partial }_{\tau }-{\mu }_{b}+{\rm{i}}{\lambda }_{i}){b}_{i}\\ & & -\tilde{J}\chi \displaystyle \sum _{\langle ij\rangle ,\sigma }({{\rm{e}}}^{{\rm{i}}{a}_{ij}}{f}_{i\sigma }^{\dagger }{f}_{j\sigma }+{\rm{h}}.{\rm{c}}.)\\ & & -t\chi \displaystyle \sum _{\langle ij\rangle }({{\rm{e}}}^{{\rm{i}}{a}_{ij}}{b}_{i}^{* }{b}_{j}+{\rm{h}}.{\rm{c}}.)\\ & = & \displaystyle \sum _{i\sigma }{f}_{i\sigma }^{\dagger }({\partial }_{\tau }-{\mu }_{f}+{\rm{i}}{\lambda }_{i}){f}_{i\sigma }\\ & & +\displaystyle \sum _{i}{b}_{i}^{* }({\partial }_{\tau }-{\mu }_{b}+{\rm{i}}{\lambda }_{i}){b}_{i}\\ & & -\tilde{J}\chi \displaystyle \sum _{\langle ij\rangle ,\sigma }\left(\left(1+{\rm{i}}{a}_{ij}-\displaystyle \frac{{a}_{ij}^{2}}{2}+{ \mathcal O }({a}^{3})\right){f}_{i\sigma }^{\dagger }{f}_{j\sigma }+{\rm{h}}.{\rm{c}}.\right)\\ & & -t\chi \displaystyle \sum _{\langle ij\rangle }\left(\left(1+{\rm{i}}{a}_{ij}-\displaystyle \frac{{a}_{ij}^{2}}{2}+{ \mathcal O }({a}^{3})\right){b}_{i}^{* }{b}_{j}+{\rm{h}}.{\rm{c}}.\right)\\ & = & \int {{\rm{d}}}^{2}{\boldsymbol{k}}{{\rm{d}}}^{2}{\boldsymbol{p}}{{\rm{d}}}^{2}{\boldsymbol{q}}\left(\displaystyle \sum _{\sigma }(({\omega }_{n}-{\mu }_{f}){f}_{k\sigma }^{\dagger }{f}_{k\sigma }\right.\\ & & +{\rm{i}}{\lambda }_{p}{f}_{k+p,\sigma }^{\dagger }{f}_{k\sigma })+(({\nu }_{n}-{\mu }_{b}){b}_{k}^{* }{b}_{k}+{\rm{i}}{\lambda }_{p}{b}_{k+p}^{\dagger }{b}_{k})\\ & & -{\tilde{J}}_{\chi }\left(2\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}\cos ({{\boldsymbol{k}}\cdot {\boldsymbol{e}}}_{j}){f}_{k\sigma }^{\dagger }{f}_{k\sigma }\right.\\ & & +2\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}\sin \left(\left(\displaystyle \frac{{\boldsymbol{k}}}{2}+{\boldsymbol{p}}\right)\cdot {{\boldsymbol{e}}}_{j}\right){a}_{k}{f}_{k+p,\sigma }^{\dagger }{f}_{p\sigma }\\ & & \left.-\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}\cos \left(\left(\displaystyle \frac{{\boldsymbol{k}}}{2}+\displaystyle \frac{{\boldsymbol{p}}}{2}+{\boldsymbol{q}}\right)\cdot {{\boldsymbol{e}}}_{j}\right){a}_{k}{a}_{p}{f}_{k+p+q,\sigma }^{\dagger }{f}_{q\sigma }\right)\\ & & -{t}_{\chi }\left(2\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}\cos ({{\boldsymbol{k}}\cdot {\boldsymbol{e}}}_{j}){b}_{k}^{* }{b}_{k}\right.\\ & & +2\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}\sin \left(\left(\displaystyle \frac{{\boldsymbol{k}}}{2}+{\boldsymbol{p}}\right)\cdot {{\boldsymbol{e}}}_{j}\right){a}_{k}{b}_{k+p}^{* }{b}_{p}\\ & & -\left.\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}\cos \left(\left(\displaystyle \frac{{\boldsymbol{k}}}{2}+\displaystyle \frac{{\boldsymbol{p}}}{2}+{\boldsymbol{q}}\right)\cdot {{\boldsymbol{e}}}_{j}\right){a}_{k}{a}_{p}{b}_{k+p+q}^{* }{b}_{q}\right),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{L}_{{\rm{g}}{\rm{f}}} & = & \displaystyle \sum _{i}\left(\displaystyle \frac{A}{2}{({\partial }_{\tau }{\lambda }_{i})}^{2}+\displaystyle \frac{D}{2}\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{({\partial }_{\tau }{a}_{i,i+{{\boldsymbol{e}}}_{j}})}^{2}\right.\\ & & +\displaystyle \frac{E}{2}\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{({\lambda }_{i+{{\boldsymbol{e}}}_{j}}-{\lambda }_{i})}^{2}+\displaystyle \frac{C}{2}\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{({a}_{i,i+{{\boldsymbol{e}}}_{j}}-{a}_{i-{{\boldsymbol{e}}}_{j},i})}^{2}\\ & & +\left.B\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}({\partial }_{\tau }{a}_{i,i+{{\boldsymbol{e}}}_{j}}({\lambda }_{i+{{\boldsymbol{e}}}_{j}}-{\lambda }_{i}))\right)\\ & = & \displaystyle \sum _{i}\left(\displaystyle \frac{A}{2}{({\partial }_{\tau }{\lambda }_{i})}^{2}+\displaystyle \frac{D}{2}\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{({\partial }_{\tau }{a}_{i,i+{{\boldsymbol{e}}}_{j}})}^{2}\right.\\ & & +\displaystyle \frac{E}{2}\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{({{\rm{\Delta }}}_{i,{{\boldsymbol{e}}}_{j}}^{+}{\lambda }_{i})}^{2}+\displaystyle \frac{C}{2}\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{({{\rm{\Delta }}}_{i,{{\boldsymbol{e}}}_{j}}^{-}{a}_{i,i+{{\boldsymbol{e}}}_{j}})}^{2}\\ & & \left.+B\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}({\partial }_{\tau }{a}_{i,i+{{\boldsymbol{e}}}_{j}}{{\rm{\Delta }}}_{i,{{\boldsymbol{e}}}_{j}}^{+}{\lambda }_{i})\right)\\ & = & \int {{\rm{d}}}^{2}k\left(\displaystyle \frac{A{\nu }_{n}^{2}{\lambda }_{k}{\lambda }_{-k}}{2}+\displaystyle \frac{D}{2}({\nu }_{n}^{2}{a}_{k}{a}_{-k})\right.\\ & & +\displaystyle \frac{E}{2}\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}(2-2\cos ({{\boldsymbol{k}}\cdot {\boldsymbol{e}}}_{j})){\lambda }_{k}{\lambda }_{-k}\\ & & +\displaystyle \frac{B}{2}({\nu }_{n}(\exp ({\rm{i}}({{\boldsymbol{k}}\cdot {\boldsymbol{e}}}_{j}))-1){a}_{k}{\lambda }_{-k}+{\rm{h}}.{\rm{c}}.)\\ & & +\displaystyle \frac{C}{2}({a}_{-{k}_{x}},{a}_{-{k}_{y}})\\ & & \times \left[\begin{array}{cc}2-2\cos ({{\boldsymbol{k}}\cdot {\boldsymbol{e}}}_{x}) & 4\sin \left(\displaystyle \frac{{{\boldsymbol{k}}\cdot {\boldsymbol{e}}}_{x}}{2}\right)\sin \left(\displaystyle \frac{{{\boldsymbol{k}}\cdot {\boldsymbol{e}}}_{y}}{2}\right)\\ 4\sin \left(\displaystyle \frac{{{\boldsymbol{k}}\cdot {\boldsymbol{e}}}_{x}}{2}\right)\sin \left(\displaystyle \frac{{{\boldsymbol{k}}\cdot {\boldsymbol{e}}}_{y}}{2}\right) & 2-2\cos ({{\boldsymbol{k}}\cdot {\boldsymbol{e}}}_{y})\end{array}\right]\\ & & \left.\times \left(\begin{array}{l}{a}_{{k}_{x}}\\ {a}_{{k}_{y}}\end{array}\right)\right)\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{L}_{{\rm{g}}{\rm{h}}} & = & -C\displaystyle \sum _{i}{\bar{u}}_{i}(\zeta {\partial }_{\tau }^{2}+{{\rm{\Delta }}}^{2}){u}_{i}\\ & = & C\displaystyle \int {{\rm{d}}}^{2}k\left(\zeta {\omega }_{n}^{2}+2\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}(\cos ({{\boldsymbol{k}}\cdot {\boldsymbol{e}}}_{j})-1)\right){\bar{u}}_{k}{u}_{k},\end{array}\end{eqnarray}$
where ${\lambda }_{-k}={\lambda }_{k}^{\dagger }$, and ${a}_{-k}={a}_{k}^{\dagger }$.The BRST transformations are:
$\begin{eqnarray}\begin{array}{rcl}{\delta }_{{\rm{B}}}{f}_{i,\sigma } & = & -{\rm{i}}\epsilon g{u}_{i}{f}_{i,\sigma },{\delta }_{{\rm{B}}}{h}_{i}=-{\rm{i}}\epsilon g{u}_{i}{h}_{i},\\ {\delta }_{{\rm{B}}}{\lambda }_{i} & = & \epsilon {\partial }_{\tau }{u}_{i},\\ {\delta }_{{\rm{B}}}{a}_{i,i+{{\boldsymbol{e}}}_{j}} & = & \epsilon {{\rm{\Delta }}}_{i,{{\boldsymbol{e}}}_{j}}^{+}{u}_{i}=\epsilon ({u}_{i+{{\boldsymbol{e}}}_{j}}-{u}_{i}),{\delta }_{{\rm{B}}}{u}_{i}=0,\\ {\delta }_{{\rm{B}}}{\bar{u}}_{i} & = & \frac{1}{\xi }\left(\zeta {\partial }_{\tau }{\lambda }_{i}+\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{{\rm{\Delta }}}_{i,{{\boldsymbol{e}}}_{j}}^{-}{a}_{i,i+{{\boldsymbol{e}}}_{j}}\right)\\ & = & \frac{1}{\xi }\left(\zeta {\partial }_{\tau }{\lambda }_{i}+\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}({a}_{i,i+{{\boldsymbol{e}}}_{j}}-{a}_{i-{{\boldsymbol{e}}}_{j},i})\right).\end{array}\end{eqnarray}$
Let us check $\begin{eqnarray}\begin{array}{rcl}{\delta }_{{\rm{B}}}{L}_{{\rm{g}}{\rm{h}}} & = & -C\xi \displaystyle \sum _{i}{\delta }_{{\rm{B}}}{\bar{u}}_{i}(\zeta {\partial }_{\tau }^{2}+{{\rm{\Delta }}}^{2}){u}_{i}\\ & = & -\epsilon C\displaystyle \sum _{i}\left(\zeta {\partial }_{\tau }{\lambda }_{i}+\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{{\rm{\Delta }}}_{i,{{\boldsymbol{e}}}_{j}}^{-}{a}_{i,i+{{\boldsymbol{e}}}_{j}}\right)\\ & & \times (\zeta {\partial }_{\tau }^{2}+{{\rm{\Delta }}}^{2}){u}_{i},\\ & = & -\epsilon C\displaystyle \sum _{i}\left({\zeta }^{2}{\partial }_{\tau }{\lambda }_{i}{\partial }_{\tau }^{2}{u}_{i}+\zeta {\partial }_{\tau }{\lambda }_{i}{{\rm{\Delta }}}^{2}{u}_{i}\right.\\ & & +\zeta \left(\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{{\rm{\Delta }}}_{i,{{\boldsymbol{e}}}_{j}}^{-}{a}_{i,i+{{\boldsymbol{e}}}_{j}}\right){\partial }_{\tau }^{2}{u}_{i}\\ & & \left.+\left(\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{{\rm{\Delta }}}_{i,{{\boldsymbol{e}}}_{j}}^{-}{a}_{i,{{\boldsymbol{e}}}_{j}}\right){{\rm{\Delta }}}^{2}{u}_{i}\right),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{\delta }_{{\rm{B}}}\left(\displaystyle \sum _{i}\frac{A}{2}{({\partial }_{\tau }{\lambda }_{i})}^{2}\right) & = & \displaystyle \sum _{i}A({\partial }_{\tau }{\lambda }_{i}){\partial }_{\tau }({\delta }_{{\rm{B}}}{\lambda }_{i})\\ & = & \epsilon \displaystyle \sum _{i}A({\partial }_{\tau }{\lambda }_{i})({\partial }_{\tau }^{2}{u}_{i}),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{\delta }_{{\rm{B}}}\displaystyle \sum _{i}\left(\frac{D}{2}\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{({\partial }_{\tau }{a}_{i,i+{{\boldsymbol{e}}}_{j}})}^{2}\right)\\ =\,D\displaystyle \sum _{i}\left(\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{\partial }_{\tau }{a}_{i,i+{{\boldsymbol{e}}}_{j}}{\partial }_{\tau }{\delta }_{{\rm{B}}}{a}_{i,i+{{\boldsymbol{e}}}_{j}}\right)\\ =\,\epsilon D\displaystyle \sum _{i}\left(\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{\partial }_{\tau }{a}_{i,i+{{\boldsymbol{e}}}_{j}}{\partial }_{\tau }({u}_{i+{{\boldsymbol{e}}}_{j}}-{u}_{i})\right)\\ =\,\epsilon D\displaystyle \sum _{i}\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}({a}_{i,i+{{\boldsymbol{e}}}_{j}}-{a}_{i-{{\boldsymbol{e}}}_{j},i}){\partial }_{\tau }^{2}{u}_{i}\\ =\,\epsilon D\displaystyle \sum _{i}\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{{\rm{\Delta }}}_{i,{{\boldsymbol{e}}}_{j}}^{-}{a}_{i,i+{{\boldsymbol{e}}}_{j}}{\partial }_{\tau }^{2}{u}_{i},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{\delta }_{{\rm{B}}}\displaystyle \sum _{i}\left(\frac{E}{2}\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{({\lambda }_{i+{{\boldsymbol{e}}}_{j}}-{\lambda }_{i})}^{2}\right)\\ =\,\displaystyle \sum _{i}\left(E\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}({\lambda }_{i+{{\boldsymbol{e}}}_{j}}-{\lambda }_{i})({\delta }_{{\rm{B}}}{\lambda }_{i+{{\boldsymbol{e}}}_{j}}-{\delta }_{{\rm{B}}}{\lambda }_{i})\right)\\ =\,\epsilon \displaystyle \sum _{i}\left(E\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}({\lambda }_{i+{{\boldsymbol{e}}}_{j}}-{\lambda }_{i}){\partial }_{\tau }({u}_{i+{{\boldsymbol{e}}}_{j}}-{u}_{i})\right)\\ =\,\epsilon \displaystyle \sum _{i}E{\partial }_{\tau }{\lambda }_{i}{{\rm{\Delta }}}^{2}{u}_{i},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{\delta }_{{\rm{B}}}\left(\displaystyle \sum _{i}\frac{C}{2}\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{({{\rm{\Delta }}}_{i,{{\boldsymbol{e}}}_{j}}^{-}{a}_{i,i+{{\boldsymbol{e}}}_{j}})}^{2}\right)\\ =\,C\displaystyle \sum _{i}\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}({{\rm{\Delta }}}_{i,{{\boldsymbol{e}}}_{j}}^{-}{a}_{i,i+{{\boldsymbol{e}}}_{j}}){\delta }_{{\rm{B}}}({a}_{i,i+{{\boldsymbol{e}}}_{j}}-{a}_{i-{{\boldsymbol{e}}}_{j},i}),\\ =\,\epsilon C\displaystyle \sum _{i}\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}({{\rm{\Delta }}}_{i,{{\boldsymbol{e}}}_{j}}^{-}{a}_{i,i+{{\boldsymbol{e}}}_{j}})\\ \quad \times (({u}_{i+{{\boldsymbol{e}}}_{j}}-{u}_{i})-({u}_{i}-{u}_{i-{{\boldsymbol{e}}}_{j}}))\\ =\,\epsilon \displaystyle \sum _{i}C\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}({{\rm{\Delta }}}_{i,{{\boldsymbol{e}}}_{j}}^{-}{a}_{i,i+{{\boldsymbol{e}}}_{j}}){{\rm{\Delta }}}^{2}{u}_{i},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{\delta }_{{\rm{B}}}\left(\displaystyle \sum _{i}B\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}({\partial }_{\tau }{a}_{i,i+{{\boldsymbol{e}}}_{j}}({\lambda }_{i+{{\boldsymbol{e}}}_{j}}-{\lambda }_{i}))\right)\\ =\,\displaystyle \sum _{i}B\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}(({\partial }_{\tau }{\delta }_{{\rm{B}}}{a}_{i,i+{{\boldsymbol{e}}}_{j}})({\lambda }_{i+{{\boldsymbol{e}}}_{j}}-{\lambda }_{i})\\ \quad +({\partial }_{\tau }{a}_{i,i+{{\boldsymbol{e}}}_{j}}{\delta }_{{\rm{B}}}({\lambda }_{i+{{\boldsymbol{e}}}_{j}}-{\lambda }_{i})))\\ =\,\epsilon \displaystyle \sum _{i}\left(B{\partial }_{\tau }{\lambda }_{i}{{\rm{\Delta }}}^{2}{u}_{i}+B\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{{\rm{\Delta }}}_{i,{{\boldsymbol{e}}}_{j}}^{-}{a}_{i,i+{{\boldsymbol{e}}}_{j}}{\partial }_{\tau }^{2}{u}_{i}\right),\end{array}\end{eqnarray}$
up to some surface terms. To cancel these terms with the ghost part, the following condition should be satisfied, $\begin{eqnarray}\begin{array}{r}A=C{\zeta }^{2},\quad C\zeta =B+D=B+E,\quad E=D,\end{array}\end{eqnarray}$
and the BRST symmetry is preserved on the lattice.The equations of motion for λi, ${a}_{i,i+{{\boldsymbol{e}}}_{j}}$, and ui are 25 ) and (26 ) in the main text] and induce the lattice versions of Gauss and current constraints.
$\begin{eqnarray}\begin{array}{l}\lambda :A{\partial }_{\tau }^{2}{\lambda }_{i}+B\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{\partial }_{\tau }{{\rm{\Delta }}}_{i,{{\boldsymbol{e}}}_{j}}^{+}{a}_{i,i+{{\boldsymbol{e}}}_{j}}+E{{\rm{\Delta }}}^{2}{\lambda }_{i}\\ \quad =\,{\rm{i}}\left(\displaystyle \sum _{\sigma }{f}_{i\sigma }^{\dagger }{f}_{i\sigma }+{b}_{i}^{* }{b}_{i}-1\right).\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{i,i+{{\boldsymbol{e}}}_{j}}:\quad D{\partial }_{\tau }^{2}{a}_{i,i+{{\boldsymbol{e}}}_{j}}+C{{\rm{\Delta }}}^{2}{a}_{i,i+{{\boldsymbol{e}}}_{j}}+B{\partial }_{\tau }{{\rm{\Delta }}}_{i,{{\boldsymbol{e}}}_{j}}^{+}{\lambda }_{i}\\ \quad =\,{J}_{i,i+{{\boldsymbol{e}}}_{j}},\end{array}\end{eqnarray}$
$\begin{eqnarray}{\bar{u}}_{i}:(\zeta {\partial }_{\tau }^{2}+{{\rm{\Delta }}}^{2}){u}_{i}=0.\end{eqnarray}$
For the Noether current, $\begin{eqnarray}\begin{array}{rcl}{\delta }_{{\rm{B}}}S & = & \displaystyle \sum _{i}\displaystyle \int {\rm{d}}\tau {\partial }_{\mu }\left(\frac{\partial { \mathcal L }}{\partial {\partial }_{\mu }{{\rm{\Phi }}}_{i}}{\delta }_{{\rm{B}}}{{\rm{\Phi }}}_{i}\right)\\ & = & \epsilon \displaystyle \sum _{i}\displaystyle \int {\rm{d}}\tau {\partial }_{\mu }{K}^{\mu }.\end{array}\end{eqnarray}$
The BRST charge: $\begin{eqnarray}\begin{array}{rcl}Q & = & \displaystyle \sum _{i}\left(\displaystyle \frac{\partial { \mathcal L }}{\partial {\partial }_{\tau }{{\rm{\Phi }}}_{i}}{\delta }_{{\rm{B}}}{{\rm{\Phi }}}_{i}-{K}^{\tau }\right)\\ & = & \displaystyle \sum _{i}\left(-{\rm{i}}g\left(\displaystyle \sum _{\sigma }{f}_{i\sigma }^{\dagger }{f}_{i\sigma }+{b}_{i}^{* }{b}_{i}\right){u}_{i}\right.\\ & & +\left(A{\partial }_{\tau }{\lambda }_{i}+(B+D)\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{{\rm{\Delta }}}_{{{\boldsymbol{e}}}_{j}}^{-}{a}_{i,i+{{\boldsymbol{e}}}_{j}}\right){\partial }_{\tau }{u}_{i}\\ & & +\left.\left(-D\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{\partial }_{\tau }{{\rm{\Delta }}}_{{{\boldsymbol{e}}}_{j}}^{-}{a}_{i,i+{{\boldsymbol{e}}}_{j}}+E{{\rm{\Delta }}}^{2}\lambda \right){u}_{i}\right).\end{array}\end{eqnarray}$
In the E, D → 0 limit, the BRST charge Q and the current become $\begin{eqnarray}\begin{array}{rcl}Q{| }_{E,D\to 0} & = & \displaystyle \sum _{i}\left(-{\rm{i}}g\left(\displaystyle \sum _{\sigma }{f}_{i\sigma }^{\dagger }{f}_{i\sigma }+{b}_{i}^{* }{b}_{i}\right){u}_{i}\right.\\ & & \left.+\left(A{\partial }_{\tau }{\lambda }_{i}+B\displaystyle \sum _{{{\boldsymbol{e}}}_{j}}{{\rm{\Delta }}}_{{{\boldsymbol{e}}}_{j}}^{-}{a}_{i,i+{{\boldsymbol{e}}}_{j}}\right){\partial }_{\tau }{u}_{i}\right),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{r}{J}_{i,i+{{\boldsymbol{e}}}_{j}}{| }_{E,D\to 0}=C{{\rm{\Delta }}}^{2}{a}_{i,i+{{\boldsymbol{e}}}_{j}}+B{\partial }_{\tau }{{\rm{\Delta }}}_{i,{{\boldsymbol{e}}}_{j}}^{+}{\lambda }_{i},\end{array}\end{eqnarray}$
which are the same as the direct lattice version of the continuum limit [see equations (Appendix B An example of det((DμνRPA)−1) ≠ 0 in the limit of D = E → 0
In the main text we argue that the skew U(1) symmetry is broken in the presence of the matter field. To construct an explicit example of Dμν, for simplicity, we consider the gauge field is coupled to a relativistic Dirac field. Then due to the gauge symmetry, the vacuum polarization takes the following form,
$\begin{eqnarray}\begin{array}{r}{{\rm{\Pi }}}_{\mu \nu }={{\rm{\Pi }}}_{{k}^{2}}({k}^{2}{g}_{\mu \nu }-{k}_{\mu }{k}_{\nu }),\end{array}\end{eqnarray}$
where ${{\rm{\Pi }}}_{{k}^{2}}$ is a function of momentum k2. For the gauge field part, $\begin{eqnarray}\begin{array}{rcl}{L}_{{\rm{G}}{\rm{F}}} & = & \frac{A}{2}{({\partial }_{\tau }\delta \lambda )}^{2}+B\displaystyle \sum _{b}{\partial }_{\tau }\delta {a}_{b}{\partial }_{b}\delta \lambda \\ & & +\frac{C}{2}{\left(\displaystyle \sum _{b}{\partial }_{b}\delta {a}_{b}\right)}^{2}+\frac{D}{2}\displaystyle \sum _{b}{({\partial }_{\tau }\delta {a}_{b})}^{2}\\ & & +\frac{E}{2}\displaystyle \sum _{b}{({\partial }_{b}\delta \lambda )}^{2},\end{array}\end{eqnarray}$
where τ = −it. By substituting (i∂t, i∂r) = (ω, − k), and choosing the metric gμν = (1, − 1, − 1, − 1), the corresponding propagator ${D}_{\mu \nu }^{(0)}$ satisfies $\begin{eqnarray}\begin{array}{l}{({D}_{\mu \nu }^{(0)})}^{-1}\\ =\,\left[\begin{array}{ccc}-\frac{A}{2}{\omega }^{2}+\frac{E}{2}({k}_{1}^{2}+{k}_{2}^{2}) & -{\rm{i}}\frac{B}{2}\omega {k}_{1} & -{\rm{i}}\frac{B}{2}\omega {k}_{2}\\ -{\rm{i}}\frac{B}{2}\omega {k}_{1} & \frac{C}{2}{k}_{1}^{2}-\frac{D}{2}{\omega }^{2} & \frac{C}{2}{k}_{1}{k}_{2}\\ -{\rm{i}}\frac{B}{2}\omega {k}_{2} & \frac{C}{2}{k}_{1}{k}_{2} & \frac{C}{2}{k}_{2}^{2}-\frac{D}{2}{\omega }^{2}\end{array}\right],\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\det ({(\displaystyle {D}_{\mu \nu }^{(0)})}^{-1})=-\frac{1}{8}D{\omega }^{2}((A{\omega }^{2}(D{\omega }^{2}-C({k}_{1}^{2}+{k}_{2}^{2})))\\ \quad +({k}_{1}^{2}+{k}_{2}^{2})({B}^{2}{\omega }^{2}-DE{\omega }^{2}+CE({k}_{1}^{2}+{k}_{2}^{2}))).\end{array}\end{eqnarray}$
Here $\det ({({D}_{\mu \nu }^{(0)})}^{-1})=0$ in the limit D = E → 0. As discussed in the main text, after coupling the gauge field to the matter field, the inverse of the full propagator reads. $\begin{eqnarray}\begin{array}{r}{({D}_{\mu \nu }^{{\prime} })}^{-1}={({D}_{\mu \nu }^{(0)})}^{-1}-{{\rm{\Pi }}}_{\mu \nu },\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}\det ({({D}_{\mu \nu }^{{\prime} })}^{-1}) & = & -\frac{1}{8}(D{\omega }^{2}+2{{\rm{\Pi }}}_{{k}^{2}}(-{\omega }^{2}+{k}_{1}^{2}+{k}_{2}^{2}))\\ & & \times (({k}_{1}^{2}+{k}_{2}^{2})({\omega }^{2}({B}^{2}-4{\rm{i}}B{{\rm{\Pi }}}_{{k}^{2}}\\ & & +2(E-D){{\rm{\Pi }}}_{{k}^{2}}-DE)\\ & & +(CE+2C{{\rm{\Pi }}}_{{k}^{2}})({k}_{1}^{2}+{k}_{2}^{2}))\\ & & +A{\omega }^{2}((D-2{{\rm{\Pi }}}_{{k}^{2}}){\omega }^{2}-C({k}_{1}^{2}+{k}_{2}^{2}))).\end{array}\end{eqnarray}$
Denote ${({D}_{\mu \nu })}^{-1}$ and ${{\rm{\Pi }}}_{{k}^{2}}^{* }$ for the limit D = E → 0, $\begin{eqnarray}\begin{array}{rcl}\det ({({D}_{\mu \nu })}^{-1}) & = & -\frac{1}{4}{{\rm{\Pi }}}_{{k}^{2}}^{* }({\omega }^{2}-{k}_{1}^{2}-{k}_{2}^{2})\\ & & \times (A{\omega }^{2}(2{\omega }^{2}{{\rm{\Pi }}}_{{k}^{2}}^{* }+C({k}_{1}^{2}+{k}_{2}^{2}))\\ & & -({k}_{1}^{2}+{k}_{2}^{2})({B}^{2}{\omega }^{2}-4{\rm{i}}B{\omega }^{2}{{\rm{\Pi }}}_{{k}^{2}}^{* }\\ & & +2C{{\rm{\Pi }}}_{{k}^{2}}^{* }({k}_{1}^{2}+{k}_{2}^{2}))).\end{array}\end{eqnarray}$
By choosing RPA, ${{\rm{\Pi }}}_{{k}^{2}}^{* }\to {{\rm{\Pi }}}_{{k}^{2}}^{(0)* }$ which does not contain the contribution from the gauge field propagator, and $\det ({({D}_{\mu \nu }^{{\rm{R}}{\rm{P}}{\rm{A}}})}^{-1})\ne 0$.