Appendix A. Derivation of RG equations for the model Hamiltonian
We consider the bosonised model Hamiltonian with \(g_u=0\) after rescaling the fields,
$$\begin{aligned} \phi ^{\prime } = \frac{\phi }{\sqrt{K}} \end{aligned}$$
and
$$\begin{aligned} \theta ^{\prime } = \sqrt{K}\theta . \end{aligned}$$
Now the model Hamiltonian can be rewritten as
$$\begin{aligned} H= & {} \frac{v}{2} [ {({\partial _x \phi ^{\prime } })}^2 + {({\partial _x \theta ^{\prime } })}^2 ] - \mu \sqrt{\frac{K}{\pi }} \partial _x \phi ^{\prime } \nonumber \\&+ \frac{B}{\pi } \cos (\sqrt{4 \pi K} \phi ^{\prime }) - \frac{\Delta }{\pi } \cos \left( \sqrt{\frac{4 \pi }{K} \theta ^{\prime }}\right) , \end{aligned}$$
(A.1)
where \(\theta (x)\) and \(\phi (x)\) are the dual fields and K is the Luttinger liquid parameter of the system. Writing the Lagrangian using the Hamilton’s equations,
$$\begin{aligned} \partial _{x} \theta ^{\prime } = -\frac{1}{v} \partial _{t} \phi ^{\prime } \end{aligned}$$
and
$$\begin{aligned} \partial _{x}\phi ^{\prime } = -\frac{1}{v} \partial _{t} \theta ^{\prime } \end{aligned}$$
leads to
$$\begin{aligned} {\mathcal {L}}_0^{(\phi )}&= \Pi _{\phi ^{\prime }} \partial _t \phi ^{\prime }-H_0^{\prime } = \frac{1}{2}[v^{-1}(\partial _t \phi ^{\prime })^2-v(\partial _x \phi ^{\prime })^2],\nonumber \\ {\mathcal {L}}_0^{(\theta )}&=\Pi _{\theta ^{\prime }} \partial _t \theta ^{\prime }-H_0^{\prime }= \frac{1}{2}[v^{-1}(\partial _t \theta ^{\prime })^2-v(\partial _x \theta ^{\prime })^2].\nonumber \\ \end{aligned}$$
(A.2)
Thus, the complete form of \({\mathcal {L}}_0= {\mathcal {L}}_0^{(\phi )}+{\mathcal {L}}_0^{(\theta )}\) in terms of imaginary time, i.e, \(\tau =it\), can be written as
$$\begin{aligned} {\mathcal {L}}_0&= -\frac{1}{4} [v^{-1}(\partial _{\tau } \phi ^{\prime })^2 {+} v^{-1}(\partial _{\tau } \theta ^{\prime })^2 {+} v(\partial _x \phi ^{\prime })^2 \\&\qquad {+} v(\partial _x \theta ^{\prime })^2]. \end{aligned}$$
Lagrangian of the interaction terms \({\mathcal {L}}_{\mathrm {int}}=-H_{\mathrm {int}}\) are
$$\begin{aligned} {\mathcal {L}}_{\mathrm {int}}&= -\dfrac{i\mu }{v} \sqrt{\dfrac{K}{\pi }} \partial _{\tau } \theta ^{\prime } - \dfrac{B}{\pi }\cos ( \sqrt{4 \pi K} \phi ^{\prime })\nonumber \\&\quad + \dfrac{\Delta }{\pi } \cos \left( \sqrt{\dfrac{4 \pi }{K}} \theta ^{\prime } \right) . \end{aligned}$$
(A.3)
Now we write the partition function \({\mathcal {Z}}\) as
$$\begin{aligned} {\mathcal {Z}}= \int {\mathcal {D}}\phi {\mathcal {D}}\theta \mathrm {e}^{-S_E[\phi ,\theta ]}, \end{aligned}$$
(A.4)
where
$$\begin{aligned} S_E= \int \mathrm {d}\tau \,\mathrm {d}x {\mathcal {L}} = \int \mathrm {d}\tau \, \mathrm {d}x ({\mathcal {L}}_0+{\mathcal {L}}_{\mathrm {int}}) \end{aligned}$$
is the Euclidean action. Thus, we have
$$\begin{aligned} {\mathcal {Z}}= & {} \int {\mathcal {D}}\phi {\mathcal {D}}\theta \, \exp \left[ - \int _{-\Lambda }^{\Lambda } \frac{\mathrm {d}\omega }{2\pi }|\omega | \left( \frac{|\phi (\omega )|^2}{2K}\right. \right. \nonumber \\&+\left. \left. \frac{K|\theta (\omega )|^2}{2}\right) + \int \mathrm {d}\tau \left( {\mathcal {L}}_{\mathrm {int}}\right) \right] . \end{aligned}$$
(A.5)
Now we divide the fields into slow and fast modes and integrate out the fast modes. The field \(\phi (\tau )=\phi _s(\tau )+\phi _f(\tau )\) and \(\theta (\tau )=\theta _s(\tau )+\theta _f(\tau )\), where
$$\begin{aligned}&\phi _{s}(\tau )=\int _{-\Lambda /b}^{\Lambda /b} \frac{\mathrm {d}\omega }{2\pi } \mathrm {e}^{-i\omega \tau }\phi (\omega ), \;\;\;\;\;\\&\phi _f(\tau )=\int _{\Lambda /b<|\omega _n|<\Lambda } \frac{\mathrm {d}\omega }{2\pi } \mathrm {e}^{-i\omega \tau }\phi (\omega )\\&\theta _{s}(\tau )=\int _{-\Lambda /b}^{\Lambda /b} \frac{\mathrm {d}\omega }{2\pi } \mathrm {e}^{-i\omega \tau }\theta (\omega ), \;\;\;\;\;\\&\theta _f(\tau )=\int _{\Lambda /b<|\omega _n|<\Lambda } \frac{\mathrm {d}\omega }{2\pi } \mathrm {e}^{-i\omega \tau }\theta (\omega ). \end{aligned}$$
Thus, \({\mathcal {Z}}\) can be written as
$$\begin{aligned} {\mathcal {Z}}= & {} \int {\mathcal {D}}\phi _s {\mathcal {D}}\phi _f {\mathcal {D}}\theta _s {\mathcal {D}}\theta _f\nonumber \\&\times \mathrm {e}^{-S_s(\phi _s,\theta _s)} \mathrm {e}^{-S_f(\phi _f,\theta _f)} \mathrm {e}^{-S_{int}(\phi ,\theta )}. \end{aligned}$$
(A.6)
Now the effective action can be written as
$$\begin{aligned} S_{\mathrm {eff}}(\phi _s,\theta _s)= & {} S_s(\phi _s,\theta _s) \nonumber \\ {\!}{\!}&{\!}{\!}- \ln \langle \mathrm {e}^{-S_{\mathrm {int}}(\phi ,\theta )} \rangle _f. \end{aligned}$$
(A.7)
Writing the cumulant expansion up to the second order, we have
$$\begin{aligned} S_{\mathrm {eff}}(\phi _s,\theta _s){\!}{\!}{\!}{\!}= & {} {\!}{\!}{\!}{\!}S_s(\phi _s,\theta _s)+ \left\langle S_{\mathrm {int}}(\phi ,\theta ) \right\rangle _f \nonumber \\&\qquad -&\frac{1}{2} ( \langle S^2_{\mathrm {int}}(\phi ,\theta )\rangle _f - \left\langle S_{\mathrm {int}}(\phi ,\theta ) \right\rangle ^2_f) . \end{aligned}$$
(A.8)
At first, we calculate the first-order cumulant expansion
$$\begin{aligned} \left\langle S_{\mathrm {int}}(\phi ,\theta ) \right\rangle _f {\!}{\!}{\!}{\!}{\!}{\!}{\!}= & {} {\!}{\!}{\!}{\!}{\!}{\!}{\!} \int \mathrm {d}\tau \left\langle \frac{i\mu }{v\sqrt{\pi }} \partial _{\tau } \theta \right\rangle _f \nonumber \\&\qquad +&\int \mathrm {d}\tau \frac{B}{\pi }\langle \cos (\sqrt{4\pi }\phi (\tau ))\rangle _f \nonumber \\&\qquad -&\int \mathrm {d}\tau \frac{\Delta }{\pi }\langle \cos (\sqrt{4\pi }\theta (\tau ))\rangle _f. \end{aligned}$$
(A.9)
The second term can be written as
$$\begin{aligned} \begin{aligned}&\langle S_B(\phi _s,\phi _f)\rangle \\&\quad = \int \mathrm {d}\tau \left( \frac{B}{\pi }\right) \int {\mathcal {D}}\phi _f \mathrm {}^{-S_f[\phi _f]} \cos (\sqrt{4\pi }\phi (\tau )) \\&\quad =\frac{B}{2\pi } \int \mathrm {d}\tau \left\{ \mathrm {e}^{i\sqrt{4\pi }\phi _s(\tau )} \mathrm {e}^{-\int _f \frac{\mathrm {d}\omega }{2\pi } \frac{4\pi }{2} \frac{K}{|\omega |}} + \mathrm {H.c.} \right\} \\&\quad = b^{-K} \int \mathrm {d}\tau \left( \frac{B}{\pi } \right) \cos [\sqrt{4\pi }\phi _s(\tau )]. \end{aligned} \end{aligned}$$
Similarly, for coupling \(\Delta \) we have
$$\begin{aligned} \begin{aligned} \langle S_\Delta (\theta _s,\theta _f)\rangle&= b^{-{1}/{K}}\int \mathrm {d}\tau \left( \frac{\Delta }{\pi } \right) \cos [\sqrt{4\pi }\theta _s(\tau )]. \end{aligned}\nonumber \\ \end{aligned}$$
(A.10)
Thus, the first-order cumulant expansions can be rewritten as
$$\begin{aligned} \begin{aligned} \left\langle S_{\mathrm {int}}(\phi ,\theta ) \right\rangle _f&=b^{-K}\int \mathrm {d}\tau \left( \frac{B}{\pi } \cos (\sqrt{4\pi }\phi _s(\tau ))\right) \\&\quad - b^{-{1}/{K}}\int \mathrm {d}\tau \left( \frac{\Delta }{\pi }\cos (\sqrt{4\pi }\theta _s(\tau ))\right) . \end{aligned}\nonumber \\ \end{aligned}$$
(A.11)
Now we calculate the second-order cumulant expansion which has the following terms:
$$\begin{aligned}&-\frac{1}{2}(\langle S_{\mathrm {int}}^2\rangle -\left\langle S_{\mathrm {int}}\right\rangle ^2) = -\frac{1}{2} \int \mathrm {d}\tau \, \mathrm {d}\tau ^{\prime } \nonumber \\&\times \left( -\frac{\mu ^2}{v^2\pi } \left\langle \partial _{\tau }\phi (\tau )\partial _{\tau ^{\prime }}\phi (\tau ^{\prime })\right\rangle - \left\langle \partial _{\tau }\phi (\tau ) \right\rangle \left\langle \partial _{\tau ^{\prime }}\phi (\tau ^{\prime })\right\rangle \right) \nonumber \\&-\frac{1}{2} \int \mathrm {d}\tau \,\mathrm {d}\tau ^{\prime }\left( \frac{B^2}{\pi ^2} \langle \cos (\sqrt{4\pi }\phi (\tau )) \cos (\sqrt{4\pi }\phi (\tau ^{\prime }))\rangle \right. \nonumber \\&\left. - \langle \cos (\sqrt{4\pi }\phi (\tau )) \rangle \langle \cos (\sqrt{4\pi }\phi (\tau ^{\prime }))\rangle \right) \nonumber \\&-\frac{1}{2} \int \mathrm {d}\tau \,\mathrm {d}\tau ^{\prime } \left( \frac{\Delta ^2}{\pi ^2} \langle \cos (\sqrt{4\pi }\theta (\tau )) \cos (\sqrt{4\pi }\theta (\tau ^{\prime }))\rangle \right. \nonumber \\&\left. - \langle \cos (\sqrt{4\pi }\theta (\tau )) \rangle \langle \cos (\sqrt{4\pi }\theta (\tau ^{\prime }))\rangle \right) \nonumber \\&-\frac{1}{2} \int \mathrm {d}\tau \, \mathrm {d}\tau ^{\prime }\left( \frac{i\mu B}{v\sqrt{\pi }\pi } \langle \partial _{\tau }\phi (\tau )\cos (\sqrt{4\pi }\phi (\tau ^{\prime }))\rangle \right. \nonumber \\&\times \left. \left\langle \partial _{\tau }\phi (\tau )\right\rangle - \langle \cos (\sqrt{4\pi }\phi (\tau ^{\prime })) \rangle \right) \nonumber \\&-\frac{1}{2} \int \mathrm {d}\tau \, \mathrm {d}\tau ^{\prime }\bigg (- \frac{ i\mu \Delta }{v\sqrt{\pi }\pi } \langle \partial _{\tau }\phi (\tau ) \cos (\sqrt{4\pi }\theta (\tau ^{\prime }))\rangle \nonumber \\&- \left\langle \partial _{\tau }\phi (\tau )\right\rangle \langle \cos (\sqrt{4\pi }\theta (\tau ^{\prime }))\rangle \bigg )\nonumber \\&-\frac{1}{2} \int \mathrm {d}\tau \,\mathrm {d}\tau ^{\prime }\bigg ( \frac{Bi\mu }{\pi v\sqrt{\pi }} \langle \cos (\sqrt{4\pi }\phi (\tau )) \partial _{\tau ^{\prime }}\phi (\tau ^{\prime })\rangle \nonumber \\&- \langle \cos (\sqrt{4\pi }\phi (\tau )) \rangle \left\langle \partial _{\tau ^{\prime }}\phi (\tau ^{\prime })\right\rangle \bigg )\nonumber \\&{-}\frac{1}{2} \int \mathrm {d}\tau \mathrm {d}\tau ^{\prime }\!\bigg (\! {-}\frac{B\Delta }{\pi ^2} \langle \cos (\sqrt{4\pi }\phi (\tau )) \cos (\sqrt{4\pi }\theta (\tau ^{\prime }))\rangle \nonumber \\&- \langle \cos (\sqrt{4\pi }\phi (\tau )) \rangle \langle \cos (\sqrt{4\pi }\theta (\tau ^{\prime }))\rangle \bigg )\nonumber \\&-\frac{1}{2} \int \mathrm {d}\tau \,\mathrm {d}\tau ^{\prime }\bigg (- \frac{\Delta i \mu }{\pi v\sqrt{\pi }} \langle \cos (\sqrt{4\pi }\theta (\tau )) \partial _{\tau ^{\prime }}\phi (\tau ^{\prime })\rangle \nonumber \\&- \langle \cos (\sqrt{4\pi }\theta (\tau )) \rangle \left\langle \partial _{\tau ^{\prime }}\phi (\tau ^{\prime })\right\rangle \bigg )\nonumber \\&{-}\frac{1}{2} \int \mathrm {d}\tau \mathrm {d}\tau ^{\prime }\!\bigg (\! {-}\frac{\Delta B}{\pi ^2} \langle \cos (\sqrt{4\pi }\theta (\tau )) \cos (\sqrt{4\pi }\phi (\tau ^{\prime }))\rangle \nonumber \\&- \langle \cos (\sqrt{4\pi }\theta (\tau )) \rangle \langle \cos (\sqrt{4\pi }\phi (\tau ^{\prime }))\rangle \bigg ). \end{aligned}$$
(A.12)
Now we calculate each term separately. \(B^2\) term can be written as
$$\begin{aligned}&-\frac{1}{2} \int \mathrm {d}\tau \, \mathrm {d}\tau ^{\prime }\bigg ( \frac{B^2}{\pi ^2} \langle \cos (\sqrt{4\pi }\phi (\tau )) \cos (\sqrt{4\pi }\phi (\tau ^{\prime }))\rangle \nonumber \\&\qquad - \langle \cos (\sqrt{4\pi }\phi (\tau )) \rangle \langle \cos (\sqrt{4\pi }\phi (\tau ^{\prime }))\rangle \bigg )\nonumber \\&\quad = -\frac{B^2}{2\pi ^2} \int \mathrm {d}\tau \, \mathrm {d}\tau ^{\prime } [ \cos \sqrt{4\pi }[\phi _s(\tau )+\phi _s(\tau ^{\prime })] \nonumber \\&\qquad \times ( \mathrm {e}^{-2\pi \langle ( \phi _f(\tau ) + \phi _f(\tau ^{\prime }))^2 \rangle } - \mathrm {e}^{-2\pi [\langle \phi _f^2(\tau ) \rangle + \langle \phi _f^2(\tau ^{\prime }) \rangle ]} )]\nonumber \\&\qquad \quad -\frac{B^2}{2\pi ^2} \int \mathrm {d}\tau \, \mathrm {d}\tau ^{\prime } [ \cos \sqrt{4\pi }[\phi _s(\tau )-\phi _s(\tau ^{\prime })] \nonumber \\&\qquad \times ( \mathrm {e}^{-2\pi \left\langle \left( \phi _f(\tau ) - \phi _f(\tau ^{\prime })\right) ^2 \right\rangle } - \mathrm {e}^{-2\pi [\left\langle \phi _f^2(\tau ) \right\rangle + \left\langle \phi _f^2(\tau ^{\prime }) \right\rangle ]} ) ] .\nonumber \\ \end{aligned}$$
(A.13)
The correlation function can be calculated as
$$\begin{aligned}&\mathrm {e}^{-2\pi \big \langle \left( \phi _f(\tau ) {+} \phi _f(\tau ^{\prime })\right) ^2 \big \rangle } {=} b^{-4K},\quad \\&\mathrm {e}^{-2\pi \big \langle \left( \phi _f(\tau ) {-} \phi _f(\tau ^{\prime })\right) ^2 \big \rangle } {=} 1 \end{aligned}$$
and
$$\begin{aligned} \mathrm {e}^{-2\pi [\big \langle \phi _f^2(\tau ) \big \rangle + \big \langle \phi _f^2(\tau ^{\prime }) \big \rangle ]}=b^{-2K}. \end{aligned}$$
Thus, by setting \(\tau ^{\prime }\rightarrow \tau \) we have
$$\begin{aligned}&-\frac{1}{2} \int \mathrm {d}\tau \, \mathrm {d}\tau ^{\prime }\bigg ( \frac{B^2}{\pi ^2} \langle \cos (\sqrt{4\pi }\phi (\tau )) \cos (\sqrt{4\pi }\phi (\tau ^{\prime }))\rangle \nonumber \\&\qquad - \langle \cos (\sqrt{4\pi }\phi (\tau )) \rangle \langle \cos (\sqrt{4\pi }\phi (\tau ^{\prime }))\rangle \bigg ) \nonumber \\&= -\frac{B^2}{2\pi ^2} \int \mathrm {d}\tau \bigg [ \cos [2\sqrt{4\pi }\phi _s(\tau )] \nonumber \\&\qquad \times ( b^{-4K} - b^{-2K} ) + \left[ 1-\frac{1}{2}(\partial _{\tau }\phi _s)^2\right] (1- b^{-2K} ) \bigg ] .\nonumber \\ \end{aligned}$$
(A.14)
Here we have
$$\begin{aligned}&\frac{\mathrm {d}}{\mathrm {d}\tau }\cos [\sqrt{4\pi }(\phi _s(\tau )-\phi _s(\tau ^{\prime }))]\nonumber \\&\quad = -\sin [\sqrt{4\pi }(\phi _s(\tau )-\phi _s(\tau ^{\prime }))] \left( -\frac{\partial \phi _s}{\partial \tau ^{\prime }}\right) . \end{aligned}$$
(A.15)
$$\begin{aligned}&\frac{\mathrm {d}^2}{\mathrm {d}\tau ^{\prime 2}}\cos [\sqrt{4\pi }(\phi _s(\tau )-\phi _s(\tau ^{\prime }))]\nonumber \\&\quad = - \cos [\sqrt{4\pi }(\phi _s(\tau )-\phi _s(\tau ^{\prime }))] \left( -\frac{\partial \phi _s}{\partial \tau ^{\prime }}\right) ^2 \nonumber \\&\qquad + \sin [\sqrt{4\pi }(\phi _s(\tau )-\phi _s(\tau ^{\prime }))] \left( -\frac{\partial ^2 \phi _s}{\partial \tau ^{\prime 2}}\right) ^2. \end{aligned}$$
(A.16)
As \(\tau \rightarrow \tau ^{\prime }\)
$$\begin{aligned} \rightarrow \left( \frac{\partial \phi _s}{\partial \tau ^{\prime }}\right) ^2. \end{aligned}$$
(A.17)
Thus, we have \(\cos \sqrt{4\pi }[\phi _s(\tau )-\phi _s(\tau ^{\prime })]= 1-\frac{1}{2}(\partial _{\tau }\phi _s)^2\). The first part of the second term is field-independent. This term can be neglected and we remain with
$$\begin{aligned}&-\frac{1}{2} \int \mathrm {d}\tau \, \mathrm {d}\tau ^{\prime }\bigg ( \frac{B^2}{\pi ^2} \langle \cos (\sqrt{4\pi }\phi (\tau )) \cos (\sqrt{4\pi }\phi (\tau ^{\prime }))\rangle \nonumber \\&\qquad \qquad \qquad \quad - \langle \cos (\sqrt{4\pi }\phi (\tau )) \rangle \langle \cos (\sqrt{4\pi }\phi (\tau ^{\prime }))\rangle \bigg ) \nonumber \\&\quad = \frac{B^2}{4\pi ^2}(1- b^{-2K})\int \mathrm {d}\tau (\partial _{\tau }\phi _s)^2 \nonumber \\&\qquad - \frac{B^2}{2\pi ^2}(b^{-4K}- b^{-2K})\int \mathrm {d}\tau \cos [\sqrt{16\pi }\phi _s(\tau )].\nonumber \\ \end{aligned}$$
(A.18)
Similarly, one can also follow the same procedure for \(\Delta ^2\) term.
$$\begin{aligned}&-\frac{1}{2} \int \mathrm {d}\tau \, \mathrm {d}\tau ^{\prime } \bigg ( \frac{\Delta ^2}{\pi ^2} \langle \cos (\sqrt{4\pi }\theta (\tau )) \cos (\sqrt{4\pi }\theta (\tau ^{\prime }))\rangle \nonumber \\&\quad - \langle \cos (\sqrt{4\pi }\theta (\tau )) \rangle \langle \cos (\sqrt{4\pi }\theta (\tau ^{\prime }))\rangle \bigg )\nonumber \\&\qquad = \frac{\Delta ^2}{4\pi ^2}(1- b^{-{2}/{K}})\int \mathrm {d}\tau (\partial _{\tau }\theta _s)^2. \end{aligned}$$
(A.19)
Now we calculate \(\Delta i\mu \) term,
$$\begin{aligned}&-\frac{1}{2} \int \mathrm {d}\tau \, \mathrm {d}\tau ^{\prime }\bigg (- \frac{\Delta i \mu }{\pi v\sqrt{\pi }} \langle \cos (\sqrt{4\pi }\theta (\tau )) \partial _{\tau ^{\prime }}\theta (\tau ^{\prime })\rangle \nonumber \\&\quad - \langle \cos (\sqrt{4\pi }\theta (\tau )) \rangle \left\langle \partial _{\tau ^{\prime }}\theta (\tau ^{\prime })\right\rangle \bigg )\nonumber \\&= \frac{\Delta i \mu }{2\pi v\sqrt{\pi }}\int \mathrm {d}\tau \, \mathrm {d}\tau ^{\prime }[ \langle \cos [\sqrt{4\pi }\theta (\tau )](\partial _{\tau ^{\prime }}\theta _s(\tau ^{\prime })\nonumber \\&\quad +\partial _{\tau ^{\prime }}\theta _f(\tau ^{\prime }))\rangle ]\nonumber \\&\quad -\frac{\Delta i \mu }{2\pi v\sqrt{\pi }}\int \mathrm {d}\tau \, \mathrm {d}\tau ^{\prime }[ \langle \cos [\sqrt{4\pi }\theta (\tau )] \rangle \langle \partial _{\tau ^{\prime }}\theta _s(\tau ^{\prime })\nonumber \\&\quad +\partial _{\tau ^{\prime }}\theta _f(\tau ^{\prime }) \rangle ] \nonumber \\&=\frac{\Delta i \mu }{2\pi v\sqrt{\pi }}\int \mathrm {d}\tau \,\mathrm {d}\tau ^{\prime }[\langle \cos [\sqrt{4\pi }\theta (\tau )]\partial _{\tau ^{\prime }}\theta _f(\tau ^{\prime })\rangle ] .\nonumber \\ \end{aligned}$$
(A.20)
The correlation function \(\langle \cos [\sqrt{4\pi }\theta (\tau )]\partial _{\tau ^{\prime }}\theta _f(\tau ^{\prime })\rangle \) can be calculated as
$$\begin{aligned}&\langle \cos [\sqrt{4\pi }\theta (\tau )]\partial _{\tau ^{\prime }}\theta _f(\tau ^{\prime })\rangle \nonumber \\&\quad = \lim _{\epsilon \rightarrow 0} \frac{1}{2i\epsilon \sqrt{\pi }}\partial _{\tau ^{\prime }} \langle \mathrm {e}^{2i\epsilon \sqrt{\pi }\theta _f(\tau ^{\prime })}\cos [\sqrt{4\pi }\theta (\tau )]\rangle \nonumber \\&\quad = -2\sqrt{\pi } \sin [\sqrt{4\pi }\theta _s(\tau )]\partial _{\tau ^{\prime }}\langle \theta _f(\tau ^{\prime })\theta _f(\tau ) \rangle \mathrm {e}^{-2\pi \big \langle \theta ^2_f(\tau ) \big \rangle }.\nonumber \\ \end{aligned}$$
(A.21)
Thus, we have
$$\begin{aligned} \begin{aligned}&=-\frac{\Delta i \mu }{\pi v}\int \mathrm {d}\tau \,\mathrm {d}\tau ^{\prime } \sin [\sqrt{4\pi }\theta _s(\tau )]\partial _{\tau ^{\prime }}\langle \theta _f(\tau ^{\prime })\theta _f(\tau ) \rangle \\&\quad \times \mathrm {e}^{-2\pi \big \langle \theta ^2_f(\tau ) \big \rangle }\\&=-\frac{\Delta i \mu }{\pi v}\int \mathrm {d}\tau \sin [\sqrt{4\pi }\theta _s(\tau )]\left( -\frac{1}{2\pi K}\ln b\right) \\&\quad \times \mathrm {e}^{-2\pi (-\frac{1}{2\pi K}\ln b)}\\&=\frac{\Delta i \mu }{2\pi ^2 v}\int \mathrm {d}\tau \sin [\sqrt{4\pi }\theta _s(\tau )](\mathrm {e}^{\frac{1}{ K}\ln b}-1)( \mathrm {e}^{-\frac{1}{ K}\ln b})\\&=-\frac{ \Delta \mu }{2\pi ^2 v}(1-b^{-{1}/{K}})\int \mathrm {d}\tau \cos [\sqrt{4\pi }\theta _s(\tau )]. \end{aligned} \end{aligned}$$
Thus, the combined \(\Delta i\mu \) and \(i\mu \Delta \) terms is
$$\begin{aligned}&-\frac{1}{2} \int \mathrm {d}\tau \, \mathrm {d}\tau ^{\prime }\bigg (- \frac{\Delta i \mu }{\pi v\sqrt{\pi }} \langle \cos (\sqrt{4\pi }\theta (\tau )) \partial _{\tau ^{\prime }}\theta (\tau ^{\prime })\rangle \nonumber \\&\quad - \langle \cos (\sqrt{4\pi }\theta (\tau )) \rangle \left\langle \partial _{\tau ^{\prime }}\theta (\tau ^{\prime })\right\rangle \bigg )\nonumber \\&\quad =-\frac{ \Delta \mu }{\pi ^2 v}(1-b^{-{1}/{K}})\int \mathrm {d}\tau \cos [\sqrt{4\pi }\theta _s(\tau )]. \end{aligned}$$
(A.22)
In the case of \(Bi\mu \), the correlation function \(\langle \phi _f(\tau ) \theta _f(\tau ^{\prime }) \rangle \) is
$$\begin{aligned}&\langle \phi _f(\tau ) \theta _f(\tau ^{\prime }) \rangle \nonumber \\&\quad = \langle (\phi _{R\uparrow }+\phi _{L\downarrow })(\phi _{R\uparrow }^{\prime }-\phi _{L\downarrow }^{\prime }) \rangle \nonumber \\&\quad {=} \langle \phi _{R\uparrow }\phi _{R\uparrow }^{\prime } {-} \phi _{R\uparrow }\phi _{L\downarrow }^{\prime } {+} \phi _{L\downarrow } \phi _{R\uparrow }^{\prime } {-} \phi _{L\downarrow }\phi _{L\downarrow }^{\prime } \rangle {=}0. \end{aligned}$$
(A.23)
Thus, the combined \(Bi\mu \) and \(i\mu B\) terms are equal to zero. Now, we calculate the term \(B\Delta \),
$$\begin{aligned}&{-}\frac{1}{2} \int \mathrm {d}\tau \mathrm {d}\tau ^{\prime }\!\bigg (\! {-}\frac{B\Delta }{\pi ^2} \langle \cos (\sqrt{4\pi }\phi (\tau )) \cos (\sqrt{4\pi }\theta (\tau ^{\prime }))\rangle \nonumber \\&\qquad - \langle \cos (\sqrt{4\pi }\phi (\tau )) \rangle \langle \cos (\sqrt{4\pi }\theta (\tau ^{\prime }))\rangle \bigg ) \nonumber \\&\quad =\frac{B\Delta }{4\pi ^2} \int \mathrm {d}\tau \, \mathrm {d}\tau ^{\prime } [ \cos \sqrt{4\pi }[\phi _s(\tau )+\theta _s(\tau ^{\prime })] \nonumber \\&\qquad \times ( \mathrm {e}^{-2\pi \langle \left( \phi _f(\tau ) + \theta _f(\tau ^{\prime })\right) ^2 \rangle } - \mathrm {e}^{-2\pi [\langle \phi _f^2(\tau ) \rangle + \langle \theta _f^2(\tau ^{\prime }) \rangle ]} ) ]\nonumber \\&\qquad + \frac{B\Delta }{4\pi ^2} \int \mathrm {d}\tau \, \mathrm {d}\tau ^{\prime } [\cos \sqrt{4\pi }[\phi _s(\tau )-\theta _s(\tau ^{\prime })] \nonumber \\&\qquad \times ( \mathrm {e}^{-2\pi \langle \left( \phi _f(\tau ) - \theta _f(\tau ^{\prime })\right) ^2 \rangle } - \mathrm {e}^{-2\pi [\langle \phi _f^2(\tau ) \rangle + \langle \theta _f^2(\tau ^{\prime }) \rangle ]} ) ] .\nonumber \\ \end{aligned}$$
(A.24)
Here the correlation function,
$$\begin{aligned}&\mathrm {e}^{-2\pi \langle \left( \phi _f(\tau ) \pm \theta _f(\tau ^{\prime })\right) ^2 \rangle } \\&\quad = \mathrm {e}^{-2\pi \langle \phi _f^2(\tau ) \rangle + \langle \theta _f^2(\tau ^{\prime })\rangle \pm 2\langle \phi _f(\tau )\theta _f(\tau ^{\prime })\rangle }. \end{aligned}$$
We know that \(\langle \phi _f(\tau )\theta _f(\tau ^{\prime })\rangle =0\). Thus, we have
$$\begin{aligned} \mathrm {e}^{-2\pi \big \langle \left( \phi _f(\tau ) \pm \theta _f(\tau ^{\prime })\right) ^2 \big \rangle } = \mathrm {e}^{-2\pi [\big \langle \phi _f^2(\tau ) \big \rangle + \big \langle \theta _f^2(\tau ^{\prime }) \big \rangle ]}. \end{aligned}$$
These two exponentials cancel each other making the whole term zero. Thus, the first- and second-order cumulant expansions can be rewritten as
$$\begin{aligned}&\left\langle S_{\mathrm {int}}(\phi ,\theta ) \right\rangle _f\nonumber \\&\quad =b^{-K}\int \mathrm {d}\tau \left( \frac{B}{\pi } \cos (\sqrt{4\pi }\phi _s(\tau ))\right) \nonumber \\&\qquad - b^{-{1}/{K}}\int \mathrm {d}\tau \left( \frac{\Delta }{\pi }\cos (\sqrt{4\pi }\theta _s(\tau ))\right) , \end{aligned}$$
(A.25)
$$\begin{aligned}&-\frac{1}{2}(\langle S_{\mathrm {in}}^2\rangle -\left\langle S_{\mathrm {int}}\right\rangle ^2)\nonumber \\&\quad = \frac{B^2}{4\pi ^2}(1- b^{-2K})\int \mathrm {d}\tau (\partial _{\tau }\phi _s)^2 \nonumber \\&\qquad - \frac{B^2}{2\pi ^2}(b^{-4K}- b^{-2K})\int \mathrm {d}\tau \cos [\sqrt{16\pi }\phi _s(\tau )] \nonumber \\&\qquad + \frac{\Delta ^2}{4\pi ^2}(1- b^{-{2}/{K}})\int \mathrm {d}\tau (\partial _{\tau }\theta _s)^2 \nonumber \\&\qquad -\frac{ \Delta \mu }{\pi ^2 v}(1-b^{-{1}/{K}})\int \mathrm {d}\tau \cos [\sqrt{4\pi }\theta _s(\tau )]. \end{aligned}$$
(A.26)
Now we rescale the first- and second-order cumulant expansions by replacing \(\tau =b\tau ^{\prime }\), \(\omega ={\omega ^{\prime }}/{b}\), \(\phi _s(\tau )=\phi ^{\prime }(\tau ^{\prime })\) and \(\phi (\omega )=b\phi ^{\prime }(\omega ^{\prime })\).
$$\begin{aligned}&\left\langle S_{\mathrm {int}}(\phi ,\theta ) \right\rangle _f=b^{2-K}\int \mathrm {d}\tau ^{\prime } \left( \dfrac{B}{\pi } \cos (\sqrt{4\pi }\phi ^{\prime }(\tau ^{\prime }))\right) \nonumber \\&\quad - b^{2-\frac{1}{K}}\int \mathrm {d}\tau ^{\prime } \left( \dfrac{\Delta }{\pi }\cos (\sqrt{4\pi }\theta ^{\prime }(\tau ^{\prime }))\right) , \end{aligned}$$
(A.27)
$$\begin{aligned}&-\frac{1}{2}(\langle S_{\mathrm {int}}^2\rangle -\langle S_{\mathrm {int}}\rangle ^2)\nonumber \\&\quad = \frac{B^2}{4\pi ^2}(b^2- b^{2-2K})\int \mathrm {d}\tau ^{\prime } (\partial _{\tau ^{\prime }}\phi ^{\prime })^2 \nonumber \\&\qquad - \frac{B^2}{2\pi ^2}(b^{2-4K}- b^{2-2K})\int \mathrm {d}\tau ^{\prime } \cos [\sqrt{16\pi }\phi ^{\prime }(\tau ^{\prime })] \nonumber \\&\qquad + \frac{\Delta ^2}{4\pi ^2}(b^2- b^{2-\frac{2}{K}})\int \mathrm {d}\tau ^{\prime } (\partial _{\tau ^{\prime }}\theta ^{\prime })^2 \nonumber \\&\qquad -\frac{ \Delta \mu }{\pi ^2 v}(b^2-b^{2-\frac{1}{K}})\int \mathrm {d}\tau ^{\prime } \cos [\sqrt{4\pi }\theta ^{\prime }(\tau ^{\prime })]. \end{aligned}$$
(A.28)
Comparison of B terms gives
$$\begin{aligned} B^{\prime }=Bb^{2-K}. \end{aligned}$$
We put \(b=\mathrm {e}^{\mathrm {d}l}\) and expand the exponential up to second term, i.e., \(\mathrm {e}^{\mathrm {d}l}=1+\mathrm {d}l\). Then,
$$\begin{aligned} \begin{aligned} B^{\prime }&=B[1+(2-K)\mathrm {d}l],\\&=B+(2-K)B\mathrm {d}l. \end{aligned} \end{aligned}$$
We define \(B^{\prime }-B=\mathrm {d}B\). Thus, we have
$$\begin{aligned} \boxed {\frac{\mathrm {d}B}{\mathrm {d}l}=(2-K)B}. \end{aligned}$$
(A.29)
Comparison of \(\Delta \) terms gives
$$\begin{aligned} \begin{aligned} \Delta ^{\prime }&=\Delta ( b^{2-\frac{1}{K}})-\frac{ \Delta \mu }{\pi v}(b^2-b^{2-\frac{1}{K}})\\&=\Delta ( \mathrm {e}^{(2-\frac{1}{K})\mathrm {d}l})-\frac{ \Delta \mu }{\pi v}(\mathrm {e}^{2\mathrm {d}l}-\mathrm {e}^{(2-\frac{1}{K})\mathrm {d}l})\\&=\Delta + \left( 2-\frac{1}{K}\right) \Delta \mathrm {d}l - \frac{ \Delta \mu }{\pi vK}\mathrm {d}l. \end{aligned} \end{aligned}$$
Thus, the equation in differential form is
$$\begin{aligned} \boxed {\frac{\mathrm {d}\Delta }{\mathrm {d}l}=\left[ 2-\frac{1}{K}\left( 1+\frac{\mu }{v\pi }\right) \right] . } \end{aligned}$$
(A.30)
Comparison of K terms for \(\phi \) field gives
$$\begin{aligned} \begin{aligned} \frac{1}{K^{\prime }}&=\frac{1}{K}\left[ 1+\frac{B^2}{4\pi ^2}(b^2- b^{2-2K}) \right] \\&=\frac{1}{K}\left[ 1+\frac{B^2}{4\pi ^2}(\mathrm {e}^{2dl}- \mathrm {e}^{(2-2K)\mathrm {d}l})\right] \\&=\frac{1}{K} + \frac{B^2}{2\pi ^2}\mathrm {d}l. \end{aligned} \end{aligned}$$
Differential form can obtained as
$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}l}\left( \frac{1}{K} \right) = \frac{B^2}{2\pi ^2} . \end{aligned}$$
$$\begin{aligned} \frac{\mathrm {d}K}{\mathrm {d}l}=-\frac{B^2K^2}{2\pi ^2} . \end{aligned}$$
(A.31)
Similarly, for \(\theta \) field,
$$\begin{aligned} \begin{aligned} K^{\prime }&=K+\frac{\Delta ^2K}{4\pi ^2}(b^2- b^{2-{2}/{K}})\\&=K+\frac{\Delta ^2K}{4\pi ^2}(\mathrm{e}^{2dl}- \mathrm {e}^{(2-\frac{2}{K})dl})\\&=K+\frac{\Delta ^2}{2\pi ^2}\mathrm {d}l. \end{aligned} \end{aligned}$$
The differential form is
$$\begin{aligned} \frac{\mathrm {d}K}{\mathrm {d}l}=\frac{\Delta ^2}{2\pi ^2}. \end{aligned}$$
(A.32)
Thus, the complete form of the differential equation for K is
$$\begin{aligned} \boxed {\frac{\mathrm {d}K}{\mathrm {d}l}=\frac{1}{2\pi ^2}(\Delta ^2-B^2K^2)}. \end{aligned}$$
(A.33)
Thus, we obtain the whole set of RG equations,
$$\begin{aligned}&\frac{\mathrm {d}B}{\mathrm {d}l}=(2-K)B.\nonumber \\&\frac{\mathrm {d}\Delta }{\mathrm {d}l}=\left[ 2-\frac{1}{K}\left( 1+\frac{\mu }{v\pi }\right) \right] .\nonumber \\&\frac{\mathrm {d}K}{\mathrm {d}l}=\frac{1}{2\pi ^2}(\Delta ^2-B^2K^2). \end{aligned}$$
(A.34)
