Critical field-theory action of the quantum rotor model with long-range Interactions I am currently reading papers on the field theoretical description of phase transitions of the quantum rotor model for systems with algebraically decaying long-range interactions $J_{ij}\propto\frac{1}{|r_{ij}|^\alpha}$.

*

*Dutta et al. (2001): https://journals.aps.org/prb/abstract/10.1103/PhysRevB.64.184106

*Defenu et al. (2017): https://journals.aps.org/prb/abstract/10.1103/PhysRevB.96.104432
The authors use a modified version of the $\phi^4$ short-range interacting quantum rotor action (discussed e.g. in the books by Kleinert and Sachdev) and call it the "long-range quantum rotor action".
$$ S_{\text{LR,Quantum Rotor}} = \frac{1}{2} \int \frac{d^dq}{(2\pi)^d} \int \frac{d\omega}{2\pi }\left[\tilde g \omega^2+r+aq^\sigma+bq^2\right]\phi_{q}(i\omega)\phi_{-q}(-i\omega) \nonumber + u \int \frac{d\omega_1}{2\pi}...\frac{d\omega_4}{2\pi}\int \frac{d^dq_1}{(2\pi)^d}...\frac{d^dq_4}{(2\pi)^d} \delta^d(q_1+...+q_4)\delta(\omega_1+...+\omega_4) \nonumber [\phi_{q_1}(i\omega_1)\phi_{q_2}(i\omega_2)][\phi_{q_3}(i\omega_3)\phi_{q_4}(i\omega_4)]
$$
With $\sigma=\alpha-d$ being a 'dimension corrected' decay exponent of the coupling. The autors do not explain in detail or reference the derivation of this action. I am especially interested in the Origin of the $q^\sigma$. As this is the only part that differs from the short range action.
My questions

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*My question would be if somebody could explain the steps to formally derive this action from the Hamiltonian ( e.g. for the Ising case $H=-J\sum_{ij}\frac{1}{|r_{ij}|^\alpha}\sigma_i^z\sigma_j^z+h\sum_i\sigma_i^x$ ) expecially the $q^\sigma$ ?

*Is there a simple motivation for the $q^\sigma$ term ?

*Is there a reference explaining the issue ?

I would be thankful for any kind of input regarding this topic.
 A: It comes from the Fourier transform of the long-range term. In real-space (and let's take imaginary time), one would expect the critical field theory to be described by the action
$$
\mathcal{S} = \int d \tau \int dx \, \left[ \frac{1}{2} \left( \partial_{\tau} \phi_{\alpha} \right)^2 + \frac{1}{2} \left( \nabla \phi_{\alpha} \right)^2 + \frac{s}{2} \phi_{\alpha}^2 + \frac{u}{4!} \phi_{\alpha}^4 \right] \\
 - \, a  \int d \tau \int dx \, dx' \, \frac{\phi_{\alpha}(x) \phi_{\alpha}(x')}{|x - x'|^{d + \sigma}}.
$$
That is, the only difference with the short-range models studied in Kleinert and Sachdev is the addition of the last term with a long-range decay of $\alpha = d + \sigma$. (One typically considers $\sigma > 0$ so that energy is extensive in the thermodynamic limit.) It should be pretty intuitive where such a term comes from - the order parameter, which is the rotor or Ising variable in the microscopic model, is described by the field $\phi$ in the field theory. So if the microscopic model has a term $\sum_{i,j} \sigma_i \sigma_j/|i - j|^{\alpha}$, one expects a corresponding term in the continuum field theory.
Some usual manipulations show that the Fourier transform of this term can be written
$$
- a \int \frac{d \omega}{2 \pi} \int \frac{d^d q}{(2 \pi)^d} f(q) \phi_{q}(i \omega) \phi_{-q}(- i \omega),
$$
where
$$
f(q) = \int d^d x \frac{e^{i q \cdot x}}{|x|^{d + \sigma}}.
$$
(Let me know if you need any steps filled in here and I can edit my answer.) This can also be simplified as
$$
f(q) = C_d \int_0^{\infty} dx \frac{J_0(|q| x)}{x^{1 + \sigma}} "=" C_d |q|^{\sigma} \int_0^{\infty} dx \frac{J_0(x)}{x^{1 + \sigma}}.
$$
Here, $C_d$ is some constant which depends on $d$ that I didn't bother figuring out exactly (it can be absorbed into the definition of $a$ anyways), and $J_{\alpha}$ is the Bessel function of the first kind.
Of course, a problem with this manipulation is that the integral actually diverges for the physical case $\sigma > 0$. This corresponds to the portion of the action where the denominator $|x - x'|^{d + \sigma}$ goes to zero, so this is a UV divergence in our QFT, which we expect to occur anyways. One could imagine regularizing the above integral at small $x$, or alternatively, analytically continuing the result for $\sigma<0$ to positive $\sigma$ (in which case you can just read off the Fourier transform from this table, formula 502). In any case, one can conclude that the new term added looks like
$$
- a' \int \frac{d \omega}{2 \pi} \int \frac{d^d q}{(2 \pi)^d} |q|^{\sigma} \phi_{q}(i \omega) \phi_{-q}(- i \omega),
$$
for some constant $a'$.
