## 3 Perturbative Approach to Quenched Disorder
### Superfluid Response and Quenched Disorder  
Now, what can we say about the disorder influence on the superfluid motion of the system? Actually, within the Landau-Khalatnikov two-fluid formulation [58; 59], the calculation proceeds in a similar fashion, provided that the theory is modified accordingly to the following points [50; 52; 54]:  
* within the Landau two-fluid model, we assume that the normal part of the system is in motion with velocity \(\mathbf{v}\), so \(\partial_{\tau}\to\partial_{\tau}-i\mathbf{v}\cdot\
abla\);
* Superfluid and normal parts do not exchange momentum. This means that a superflow can be imposed by means of phase twist of the field \(\psi(\mathbf{r},\tau)\), i.e., \(\psi\to\exp\big{(}im\mathbf{v}_{s}\cdot\mathbf{r}/\hbar\big{)}\psi\) with \(\mathbf{v}_{s}\) the superfluid velocity [60; 61];
* as a consequence of the previous points, we can redefine the chemical potential as \[\mu_{\rm eff}=\mu-\frac{1}{2}m\mathbf{v}_{s}\cdot(\mathbf{v}_{s}-2\mathbf{v} )\.\] (44)  
The resulting Gaussian partition function (or grand potential) is formally similar to the one in Equation (27), with the replacement \(\mu\to\mu_{\rm eff}\) and the (inverse) Gaussian propagator given by  
\[\mathcal{G}^{-1}(\mathbf{q},\omega)= \begin{pmatrix}-i\hbar\omega_{n}+\epsilon _{q}-\mu+2g\psi_{0}^{2}+\hbar\mathbf{q}\cdot(\mathbf{v}-\mathbf{v}_{s})&g \psi_{0}^{2}\\ g\psi_{0}^{2}&+i\hbar\omega_{n}+\epsilon_{q}-\mu+2g\psi_{0}^{2}-\hbar\mathbf{q }\cdot(\mathbf{v}-\mathbf{v}_{s})\end{pmatrix}  \tag{45}\]  
whose determinant is given by  
\[\det\mathcal{G}^{-1}(\mathbf{q},\omega_{n})=E_{q}^{2}-\big{[}i\hbar\omega_{n} -\hbar\cdot\mathbf{q}\cdot(\mathbf{v}-\mathbf{v}_{s})\big{]}^{2}\, \tag{46}\]  
where \(E_{q}\) as in Equation (29) with \(\mu\to\mu_{\rm eff}\). By following this prescription, the _phase-twisted_ grand potential can be derived from Equation (33), reading  
\[\langle\Omega_{\rm dis}(\mu_{\rm eff},\mathbf{v})\rangle_{\rm th+dis}=-\gamma^{ 2}\psi_{0}^{2}\sum_{\mathbf{q}}\frac{\epsilon_{q}-\mu_{\rm eff}+g\psi_{0}^{2}} {E_{q}^{2}-\big{[}\hbar\mathbf{q}\cdot(\mathbf{v}-\mathbf{v}_{s})\big{]}^{2} }. \tag{47}\]  
We are interested in the linear-response regime, where the density current  
\[\mathbf{g}=-\frac{1}{L^{d}}\frac{\Omega(\mu_{\rm eff}(\mathbf{v}),\mathbf{v}) }{\partial\mathbf{v}} \tag{48}\]is taken up to the first order in \((\mathbf{v}-\mathbf{v}_{s})\). Therefore, we begin with an expansion of Equation (47) up to the quadratic order in \(\hbar\mathbf{q}\cdot(\mathbf{v}-\mathbf{v}_{s})\). For symmetry reasons, the linear term of this expansion is identically zero, such that we have to deal with the following equation:  
\[\frac{\langle\Omega_{\text{dis}}(\mu_{\text{eff}},\mathbf{v})\rangle_{\text{ th+dis}}}{L^{d}}=-\frac{\gamma^{2}\psi_{0}^{2}}{(2\pi)^{d}}\int d^{d}\mathbf{q} \bigg{[}\frac{\epsilon_{q}-\mu_{\text{eff}}+g\psi_{0}^{2}}{E_{q}^{2}}+\frac{ \hbar^{2}q^{2}}{d}\bigg{(}\frac{\epsilon_{q}-\mu_{\text{eff}}+g\psi_{0}^{2}}{ E_{q}^{4}}\bigg{)}(\mathbf{v}-\mathbf{v}_{s})^{2}\bigg{]}\, \tag{49}\]  
where we have made use of the identity  
\[\int d^{d}\mathbf{q}\,g_{i}q_{j}f(|\mathbf{q}|)=d^{-1}\delta_{ij}\int d^{d} \mathbf{q}\,|\mathbf{q}|^{2}f(|\mathbf{q}|). \tag{50}\]  
Since the first term in the equation above depends only on \(\mu_{\text{eff}}\) and it holds \(\frac{\partial\mu_{\text{eff}}}{\partial\mathbf{v}}=m\mathbf{v}_{s}\), it will be the disorder contribution to the \(n\mathbf{v}_{s}\) term in \(\mathbf{g}\). More precisely, by considering also the thermal depletion due to the quasiparticle excitations, Equation (48) reads  
\[\mathbf{g}=n\mathbf{v}_{s}+\frac{1}{d}\int\frac{d^{d}\mathbf{q}}{(2\pi)^{d}} \bigg{[}\frac{\beta\hbar^{2}q^{2}}{m}\frac{e^{\beta E_{q}}}{(e^{\beta E_{q}}- 1)^{2}}+4\gamma^{2}\psi_{0}^{2}\frac{\epsilon_{q}(\epsilon_{q}-\mu+g\psi_{0}^ {2})}{E_{q}^{4}}\bigg{]}(\mathbf{v}-\mathbf{v}_{s}). \tag{51}\]  
At this point, we finally replace Equation (18) for the broken-symmetry case in the equation above.