Chapter 4 Hopping expansion Cross-check

The partition function for one complex field in the $g=0$ theory is

$$Z=\int d\mu(\phi) \,e^{-S[\phi]}\\ S= -\kappa\sum_{x}\left[ \phi^\dagger(x) \sum_\mu\phi({x+\mu}) +h.c. \right]\,,$$

The field has the costrain $\phi^\dagger \phi=1$ so $\phi \in U(1)$ and $d\mu$ is the Haar measure over $U(1)$.

4.0.1 Haar measure

The Haar measure must satisfy $$\int d\mu(\phi)=1\\ d\mu(\phi')= d\mu(\phi) \quad\mbox{with}\quad \phi'=e^{i\delta}\phi$$ these properties are statisfied by $d\mu=d\theta/2\pi$ with $\phi=e^{i\theta}$ $$\int_0^{2\pi} \frac{d\theta}{2\pi}=1\\ \frac{d\theta'}{2\pi}=\frac{d\theta}{2\pi} \quad\mbox{with}\quad \phi'=e^{i\delta}\phi\,.$$ With these definitions we can compute $$\int d\mu(\phi) \phi =\int_0^{2\pi} \frac{d\theta}{2\pi} e^{i\theta}=\oint_{\circlearrowleft} \frac{dz}{2\pi (iz)}z =0 \,,\quad\quad z=e^{i\theta}$$ and similar for $\phi^\dagger$ $$\int d\mu(\phi) \phi^\dagger =\int_0^{2\pi} \frac{d\theta}{2\pi} e^{-i\theta}=\oint_{\circlearrowright} \frac{dz}{2\pi (-iz)}z=0 \,.$$ For all the other power of $\phi$ we observe that there will not be poles in the integrand and thus the integral will be zero $$\int d\mu(\phi) \phi^n=0 \quad \mbox{for}\quad n>0\,.$$

4.0.2 Partition function for small hopping parameter $\kappa$

Expanding the partition function around $\kappa=0$ we obtain

$$Z=\int \prod_x \frac{d\theta(x)}{2\pi} \left[1+\kappa\sum_{x}\left[ \phi^\dagger(x) \sum_\mu\phi({x+\mu}) +h.c. \right]\right]+O(\kappa^2)=1+O(\kappa^2)\,,$$ we have use the fact the in the $O(\kappa)$ term there is at least one integral of the form $\int d\theta(x) \phi(x)=0$. Notice that up to order $O(\kappa^2)$ the partition function is already normalized to 1.

4.0.3 Energy

We define the Energy observable as $$E=\frac{1}{2V}\sum_{x}\left[ \phi^\dagger(x) \sum_\mu\phi({x+\mu}) +h.c. \right]=\frac{V}{2\kappa}S\,.$$ It’s expectation value is

$$\langle E\rangle=\int \prod_x \frac{d\theta(x)}{2\pi} \left[1+\kappa\sum_{x}\left[ \phi^\dagger(x) \sum_\mu\phi({x+\mu}) +h.c. \right]\right]\frac{1}{2V}\sum_{x'}\left[ \phi^\dagger(x') \sum_\mu'\phi({x'+\mu'}) +h.c. \right]+O(\kappa^2)\,.$$ The term $O(\kappa^0)$ is zero since there always be an integral of $\int d\theta(x) \phi(x)=0$. In the term $O(\kappa)$ the only non zero contribution comes when the fileds in the Energy observable cancel the fields in the action, this happen only if $x=x'$ and $\mu=\mu'$ giving a contribution $8V$

• $2$ for $\phi^\dagger(x)\phi(x+\mu) (\phi(x)\phi^\dagger(x+\mu))+\phi(x)\phi^\dagger(x+\mu) (\phi(x)^\dagger\phi(x+\mu))$;
• $4$ sum over $\mu$;
• $V$ sum over $x$.

thus toghether with the overall factor $\kappa/2V$ we get $$\langle E\rangle=4\kappa +O(\kappa^2)$$

4.0.4 Numerical test

## Warning: `gather_()` was deprecated in tidyr 1.2.0.
## Please use `gather()` instead.
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_lifecycle_warnings()` to see where this warning was generated.

The result from the fit in the region $\kappa<$ 0.06 gives

$\langle E \rangle [\kappa<$ 0.06 $] =$ -0.002840827 + 4.11504 $\kappa$

summary(fit)

## 
## Call:
## lm(formula = E_0 ~ k0, data = df)
## 
## Residuals:
##        Min         1Q     Median         3Q        Max 
## -3.250e-04 -2.384e-04 -4.208e-05  1.687e-04  8.339e-04 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -0.0028408  0.0001359   -20.9   <2e-16 ***
## k0           4.1150405  0.0035760  1150.8   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.0002694 on 59 degrees of freedom
## Multiple R-squared:      1,  Adjusted R-squared:      1 
## F-statistic: 1.324e+06 on 1 and 59 DF,  p-value: < 2.2e-16

We repeat for smaller values of $\kappa$

for (treshold in c(0.05,0.04,0.03)  ){
  fit<- lm(E_0~k0 , data = df[which(df$k0<treshold),])
  # chi2 and error?
  a<-coef(fit)[["(Intercept)"]]
  b<-coef(fit)[["k0"]]
  cat("$\\langle E \\rangle [\\kappa<$", treshold,"$] =$",a ,"+",b, "$\\kappa$\n\n")
}

$\langle E \rangle [\kappa<$ 0.05 $] =$ -0.00207288 + 4.090886 $\kappa$

$\langle E \rangle [\kappa<$ 0.04 $] =$ -0.001394341 + 4.068108 $\kappa$

$\langle E \rangle [\kappa<$ 0.03 $] =$ -0.0009570297 + 4.05229 $\kappa$

4.0.5 g term

We can do a similar computation setting $\kappa_0=\kappa_1=0$ and expanding for small $g$. We compute the expectation value of $$E_g=\frac{1}{2V}\sum_{x}\left[ \phi_1^\dagger(x) \phi_0^3({x}) +h.c. \right]\,.$$ $$\langle E_g\rangle=\int \prod_x \frac{d\theta(x)}{2\pi} \left[1-\frac{g}{2}\sum_{x}\left[ \phi_1^\dagger(x) \phi_0^3({x}) +h.c. \right]\right]\frac{1}{2V}\sum_{x'}\left[ \phi_1^\dagger(x')\phi_0^3(x') +h.c. \right]+O(g^2)=-\frac{g}{2}\,.$$

## 
## Call:
## lm(formula = E_0 ~ g, data = df)
## 
## Residuals:
##        Min         1Q     Median         3Q        Max 
## -0.0009660 -0.0001208  0.0001143  0.0002147  0.0006750 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -0.0002917  0.0001338   -2.18   0.0499 *  
## g           -0.4912214  0.0010625 -462.33   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.0004177 on 12 degrees of freedom
## Multiple R-squared:  0.9999, Adjusted R-squared:  0.9999 
## F-statistic: 2.137e+05 on 1 and 12 DF,  p-value: < 2.2e-16

4.0.6 Numerical integration g term

Setting $\kappa_0=\kappa_1=0$ we can numerically solve the integral

$$\langle E_g\rangle=\frac{\int \prod_x \frac{d\theta_1(x) d\theta_2(x) }{(2\pi)^2} e^{-S} E_g}{\int \prod_x \frac{d\theta_1(x) d\theta_2(x) }{(2\pi)^2} e^{-S}}=\frac{\int_0^{2\pi} \frac{d\theta_1 d\theta_2 }{(2\pi)^2} e^{-g \mbox{Re}[\phi_1\phi_0^3] } \mbox{Re}[\phi_1\phi_0^3] }{\int_0^{2\pi}\frac{d\theta_1 d\theta_2 }{(2\pi)^2} e^{-g \mbox{Re}[\phi_1\phi_0^3] } }$$

Using Mathematica we get $$\langle E_g\rangle(g=3)=-0.809985$$

while our numerical simulation gives on an $4^4$ lattice

df<- read.table("../check_g/E_T4_L4_msq010000000.000000_msq110000000.000000_l00.000000_l10.000000_mu0.000000_g3.000000_rep0")
m<-mean(df[,3])
err<- sd(df[,3])
cat("$$\\overline{ E_g} (g=3)=",mean_print(m,err),"$$\n")

$$\overline{ E_g} (g=3)= -0.810(17)$$

#probably the sd can not compute the variance because the vectori is too long
#below we try with binning")
bins<-1000
df_b<-rep(0,bins-1)
len<-length(df[,3])
for (i in c(1:bins-1)){
  df_b[i]<- mean(df[c(1:len/bins)+i *len/bins ,3])
}
m<-mean(df_b)
err<- sd(df_b)
cat("$$\\overline{ E_g} (g=3)=",mean_print(m,err),"$$\n")

$$\overline{ E_g} (g=3)= -0.8099(17)$$