A method often used to couple a lattice tight binding model to a magnetic field is the Peierls substitution, whereby one changes all hopping elements (schematically) as $t_{ij}\mapsto t_{ij}\exp(\mathrm{i}Q\int A\,\mathrm{d}l)$. I have looked in a number of places, and there seems to be no consistent agreement about whether the sign in the exponent is to be positive or negative. The Wikipedia page gives three arguments for the method; they seem equally plausible to me, yet give opposite signs. I have reproduced (and slightly altered) two of these arguments below ($\hbar=1$). Which conclusion is correct and why?
Method 1:
The hopping matrix element $t_{ij}$ is proportional to the amplitude $\langle \mathbf{r}_i, t_i|\mathbf{r}_j, t_j\rangle$, which can be calculated for a free particle as $$ \langle \mathbf{r}_i, t_i|\mathbf{r}_j, t_j\rangle =\int_{\mathbf{r}(t_i)}^{\mathbf{r}(t_j)}\mathcal{D}\mathbf{r}(t)\,\exp\left(\mathrm{i}\mathcal{S}_0\right), $$ where $\mathcal{S}_0=\int L_0\,\mathrm{d}t$ is the action. If the particle has charge $Q$ then the effect of introducing a magnetic field is to change $L_0\mapsto L_0+Q \mathbf{v}\cdot\mathbf{A}$, so the action changes as \begin{align} \mathcal{S}_0=\int_{t_i}^{t_j}L_0\,\mathrm{d}t\mapsto& \int_{t_i}^{t_j}L_0+Q\mathbf{v}\cdot\mathbf{A}\,\mathrm{d}t\\ &=\int_{t_i}^{t_j}L_0\,\mathrm{d}t +Q\int_{t_i}^{t_j}\mathbf{A}\cdot\frac{\mathrm{d}\mathbf{r}}{\mathrm{d} t}\,\mathrm{d}t \\ &= \mathcal{S}_0+Q\int_{\mathbf{r}_i}^{\mathbf{r}_j}\mathbf{A}\cdot\mathrm{d}\mathbf{r} \end{align} and the amplitude changes to \begin{align} \langle \mathbf{r}_i, t_i|\mathbf{r}_j, t_j\rangle\mapsto &\int_{\mathbf{r}(t_i)}^{\mathbf{r}(t_j)}\mathcal{D}\mathbf{r}(t)\,\exp\left(\mathrm{i}\mathcal{S}_0+\mathrm{i}Q\int_{\mathbf{r}_i}^{\mathbf{r}_j}\mathbf{A}\cdot\mathrm{d}\mathbf{r}\right)\\ &\approx \exp\left(\mathrm{i}Q\int_{\mathbf{r}_i}^{\mathbf{r}_j}\mathbf{A}\cdot\mathrm{d}\mathbf{r}\right)\int_{\mathbf{r}(t_i)}^{\mathbf{r}(t_j)}\mathcal{D}\mathbf{r}(t)\,\exp\left(\mathrm{i}\mathcal{S}_0\right)\\ &=\exp\left(\mathrm{i}Q\int_{\mathbf{r}_i}^{\mathbf{r}_j}\mathbf{A}\cdot\mathrm{d}\mathbf{r}\right)\langle \mathbf{r}_i, t_i|\mathbf{r}_j, t_j\rangle \end{align} where in the second line we have made the (saddle point) approximation that $\mathbf{A}$ varies little over paths which contribute the most to the amplitude. Thus, the hopping element must change as $$ t_{ij}\propto \langle \mathbf{r}_i, t_i|\mathbf{r}_j, t_j\rangle\mapsto \exp\left(\mathrm{i}Q\int_{\mathbf{r}_i}^{\mathbf{r}_j}\mathbf{A}\cdot\mathrm{d}\mathbf{r}\right)\langle \mathbf{r}_i, t_i|\mathbf{r}_j, t_j\rangle\propto \exp\left(+\mathrm{i}Q\int_{\mathbf{r}_i}^{\mathbf{r}_j}\mathbf{A}\cdot\mathrm{d}\mathbf{r}\right) t_{ij} $$
On the other hand, we can make an argument about the required substitution by considering a lattice model
Method 2:
Consider the tight binding Hamiltonian $$ H_0 = -\sum_{ij} t_{ij} c_{i}^{\dagger} c_j \sim -\sum_{ij}t_{ij}|\phi_{\mathbf{R}_i}\rangle\langle\phi_{\mathbf{R}_j}| $$ where $t_{ij} = t^*_{ji}$. We can rewrite the state $|\phi_{\mathbf{R}_i}\rangle$ using the translation operator as $$ |\phi_{\mathbf{R}_i}\rangle = \mathrm{e}^{-\mathrm{i}(\mathbf{R}_i-\mathbf{R}_j)\cdot\mathbf{p}}|\phi_{\mathbf{R}_j}\rangle $$ Under minimal coupling, we change $\mathbf{p}\mapsto \mathbf{p}-Q\mathbf{A}$, which means that \begin{align} \mathrm{e}^{-\mathrm{i}(\mathbf{R}_i-\mathbf{R}_j)\cdot\mathbf{p}}|\phi_{\mathbf{R}_j}\rangle\mapsto &\mathrm{e}^{-\mathrm{i}(\mathbf{R}_i-\mathbf{R}_j)\cdot(\mathbf{p}-Q\mathbf{A})}|\phi_{\mathbf{R}_j}\rangle\\ &\approx\mathrm{e}^{\mathrm{i}(\mathbf{R}_i-\mathbf{R}_j)\cdot Q\mathbf{A}}\mathrm{e}^{-\mathrm{i}(\mathbf{R}_i-\mathbf{R}_j)\cdot\mathbf{p}}|\phi_{\mathbf{R}_j}\rangle \\ &= \mathrm{e}^{\mathrm{i}(\mathbf{R}_i-\mathbf{R}_j)\cdot Q\mathbf{A}}|\phi_{\mathbf{R}_i}\rangle \end{align} where we have neglected higher order terms. The effect of this on the Hamiltonian is $$ H_0\mapsto- \sum_{ij}t_{ij} \mathrm{e}^{\mathrm{i}(\mathbf{R}_i-\mathbf{R}_j)\cdot Q\mathbf{A}}|\phi_{\mathbf{R}_i}\rangle\langle\phi_{\mathbf{R}_j}| $$ which is equivalent to $$ t_{ij}\mapsto\exp\left(\mathrm{i}(\mathbf{R}_i-\mathbf{R}_j)\cdot Q\mathbf{A}\right) t_{ij}\approx \exp\left(-\mathrm{i}Q \int_{\mathbf{R}_i}^{\mathbf{R}_j} \mathbf{A}\cdot\mathrm{d}\mathbf{r}\right) t_{ij} $$
