I have always thought that a gauge transformation of a quantum Hamiltonian $H(\Psi,\Psi^{\dagger},A)$ ($A$ is the vector potential and $\Psi$ a matter field) is given by:
$$\Psi(r) \rightarrow \Psi(r) e^{i \Lambda(r)}; \quad A(r) \rightarrow A(r) -\frac{1}{e} \nabla \Lambda(r)$$
i.e. we transform both the light and the matter parts.
As an example consider ($c = 1$) a free minimally-coupled Hamiltonian:
$$ H = \int d^3r \Psi^{\dagger}(r) \left(\frac{(p-eA(r))^2}{2m}\right) \Psi(r)$$
$H$ is invariant under this gauge transformation, so obviously manifests gauge invariance.
Paper https://arxiv.org/abs/2009.11088 adopts this reasoning (Eqs. 11,12).
However I have recently come across what seems like a different and confusing definition of gauge transform in a different paper https://arxiv.org/abs/2005.08994 where they only transform the matter part of the Hamiltonian (paragraph right below Eq. 2).
To be more concrete, they consider a tight-binding Hamiltonian $H_0$:
$$H_{0}=-\sum_{j, \delta} \sum_{\alpha \beta} t_{\alpha \beta}^{\delta} c_{\boldsymbol{R}_{j, \alpha}}^{\dagger} c_{\boldsymbol{R}_{\boldsymbol{j}, \beta}+\boldsymbol{\delta}}$$
which can be minimally-coupled via a Peierls substitution $t_{\alpha \beta}^{\delta} \rightarrow t_{\alpha \beta}^{\delta} e^{-i e \lambda / c}$ with $\lambda=\int_{\boldsymbol{R}_{j, \alpha}}^{\boldsymbol{R}_{\boldsymbol{j}, \beta}+\boldsymbol{\delta}} d \boldsymbol{r} \cdot \hat{\boldsymbol{A}}(\boldsymbol{r})$.
They then proceed to demostrate gauge invariance by considering a uniform and classical $\mathbf{\hat{A}} = \mathbf{A_0}$. In this case the Peierls phase factor clearly becomes $\lambda=\boldsymbol{A}_{0} \cdot\left(\boldsymbol{R}_{\boldsymbol{j}, \beta}+\boldsymbol{\delta}-\boldsymbol{R}_{\boldsymbol{j}, \alpha}\right)$.
Now here is what I am confused about: they state that $\lambda$ can now be absorbed by a "gauge transform" $c_{\boldsymbol{R}_{j, \alpha}} \rightarrow e^{i e \boldsymbol{A}_{0} \cdot \boldsymbol{R}_{\boldsymbol{j}, \alpha} / c} c_{\boldsymbol{R}_{\boldsymbol{j}, \alpha}}$. Obviously this operation absorbs the phase factor, but I would not call it a "gauge transform", because they have not transformed the vector potential.
I have also come across some lecture notes https://topocondmat.org/w2_majorana/Peierls.html which also claim the same thing as the aforementioned paper.
So what is the correct definition of a gauge transformation in this case? Is it :
$$\Psi(r) \rightarrow \Psi(r) e^{i \Lambda(r)}; \quad A(r) \rightarrow A(r) -\frac{1}{e} \nabla \Lambda(r)$$
or just one of these two?
