In this paper, on page 3, the authors go from the tight binding model w the Peierls substitution $$ H = \sum_{i,j} \sum_{a,b} t_{a,b} \exp\left(i \int_{\textbf{R}_{j,b}}^{\textbf{R}_{i,a}} dr'_{\mu} A_{\mu} (\textbf{r}',t ) \right) c_{i,a}^\dagger c_{j,b} $$ to $$ H= H_0 + \sum_{i,j} \sum_{a,b} t_{a,b} (L_\mu^A A_\mu + 1/2 L_{\mu v}^{AA} A_\mu A_v +..) c_{i,a}^\dagger c_{j,b} $$ where $H_0 $ is $H$ without the exponential, and then define $ L_{\mu}^A = ( \partial_{A_\mu} H)|_{A=0} $ and so on.

However, shouldn't this second line just be $$ H= H_0 + L_\mu^A A_\mu + 1/2 L_{\mu v}^{AA} A_\mu A_v +... $$

Yes, their expression is wrong, for the reason you already know. This is probably just a typo.

The expansion is obtained (as they say in the paper) by using the Taylor series expansion of the exponential: $$ H = \sum_{i,j} \sum_{a,b} t_{a,b} \left(1 + i \int_{\textbf{R}_{j,b}}^{\textbf{R}_{i,a}} dr'_{\mu} A_{\mu} (\textbf{r}',t ) - \frac{1}{2}\int_{\textbf{R}_{j,b}}^{\textbf{R}_{i,a}} dr'_{\mu} A_{\mu} (\textbf{r}',t ) \int_{\textbf{R}_{j,b}}^{\textbf{R}_{i,a}} dr''_{\mu} A_{\mu} (\textbf{r}'',t ) +\ldots \right) c_{i,a}^\dagger c_{j,b} $$

Your expression also looks questionable to me, because the notation is not super clear. You are dotting $L_\mu$ with $A_\mu$, but there is no free $A_\mu$ so I suppose your expression probably should have (or implicitly has) a convolution as well as a dot product. I suppose this is the case, since I also suppose the $\partial_{A_\mu}$ means a functional derivative. And so, it must be convolved with the $A$ field to make sense.

In this paper, on page 3, the authors go from the tight binding model w the Peierls substitution $$ H = \sum_{i,j} \sum_{a,b} t_{a,b} \exp\left(i \int_{\textbf{R}_{j,b}}^{\textbf{R}_{i,a}} dr'_{\mu} A_{\mu} (\textbf{r}',t ) \right) c_{i,a}^\dagger c_{j,b} $$

I note here, for completeness, that in their expression $t$ also depends on $i$ and $j$, and they use the notation $t_{ab}(i,j)$.

to $$ H= H_0 + \sum_{i,j} \sum_{a,b} t_{a,b} (L_\mu^A A_\mu + 1/2 L_{\mu v}^{AA} A_\mu A_v +..) c_{i,a}^\dagger c_{j,b} $$ where $H_0 $ is $H$ without the exponential, and then define $ L_{\mu}^A = ( \partial_{A_\mu} H)|_{A=0} $ and so on.

However, shouldn't this second line just be $$ H= H_0 + L_\mu^A A_\mu + 1/2 L_{\mu v}^{AA} A_\mu A_v +... $$

Yes, their expression is wrong, for the reason you already know. This is probably just a typo.

The expansion is obtained (as they say in the paper) by using the Taylor series expansion of the exponential: $$ H = \sum_{i,j} \sum_{a,b} t_{a,b}(i,j) \left(1 + i \int_{\textbf{R}_{j,b}}^{\textbf{R}_{i,a}} dr'_{\mu} A_{\mu} (\textbf{r}',t ) - \frac{1}{2}\int_{\textbf{R}_{j,b}}^{\textbf{R}_{i,a}} dr'_{\mu} A_{\mu} (\textbf{r}',t ) \int_{\textbf{R}_{j,b}}^{\textbf{R}_{i,a}} dr''_{\mu} A_{\mu} (\textbf{r}'',t ) +\ldots \right) c_{i,a}^\dagger c_{j,b} $$

Your expression is more correct, but also looks a little questionable to me, probably because the notation is not super clear. You are dotting $L_\mu$ with $A_\mu$, but there is no free $A_\mu(\vec r,t)$ so I suppose your expression probably should have (or implicitly has) a convolution as well as a dot product. I suppose this is the case, since I also suppose the $\partial_{A_\mu}$ means a functional derivative. And so, it must be convolved with the $A$ field to make sense.

In this paper, on page 3, the authors go from the tight binding model w the Peierls substitution $$ H = \sum_{i,j} \sum_{a,b} t_{a,b} \exp\left(i \int_{\textbf{R}_{j,b}}^{\textbf{R}_{i,a}} dr'_{\mu} A_{\mu} (\textbf{r}',t ) \right) c_{i,a}^\dagger c_{j,b} $$ to $$ H= H_0 + \sum_{i,j} \sum_{a,b} t_{a,b} (L_\mu^A A_\mu + 1/2 L_{\mu v}^{AA} A_\mu A_v +..) c_{i,a}^\dagger c_{j,b} $$ where $H_0 $ is $H$ without the exponential, and then define $ L_{\mu}^A = ( \partial_{A_\mu} H)|_{A=0} $ and so on.

However, shouldn't this second line just be $$ H= H_0 + L_\mu^A A_\mu + 1/2 L_{\mu v}^{AA} A_\mu A_v +... $$

Both expressions look wrong to me. Theirs for the reason you already know, yours because you still have to integrate over the dummy position $\vec r'$ in $A(\vec r',t)$. So unless there is some implicit convolution in your notation, yours looks wrong too.

The expansion is obtained by using the Taylor series expansion of the exponential: $$ H = \sum_{i,j} \sum_{a,b} t_{a,b} \left(1 + i \int_{\textbf{R}_{j,b}}^{\textbf{R}_{i,a}} dr'_{\mu} A_{\mu} (\textbf{r}',t ) - \frac{1}{2}\int_{\textbf{R}_{j,b}}^{\textbf{R}_{i,a}} dr'_{\mu} A_{\mu} (\textbf{r}',t ) \int_{\textbf{R}_{j,b}}^{\textbf{R}_{i,a}} dr''_{\mu} A_{\mu} (\textbf{r}'',t ) +\ldots \right) c_{i,a}^\dagger c_{j,b} $$

In this paper, on page 3, the authors go from the tight binding model w the Peierls substitution $$ H = \sum_{i,j} \sum_{a,b} t_{a,b} \exp\left(i \int_{\textbf{R}_{j,b}}^{\textbf{R}_{i,a}} dr'_{\mu} A_{\mu} (\textbf{r}',t ) \right) c_{i,a}^\dagger c_{j,b} $$ to $$ H= H_0 + \sum_{i,j} \sum_{a,b} t_{a,b} (L_\mu^A A_\mu + 1/2 L_{\mu v}^{AA} A_\mu A_v +..) c_{i,a}^\dagger c_{j,b} $$ where $H_0 $ is $H$ without the exponential, and then define $ L_{\mu}^A = ( \partial_{A_\mu} H)|_{A=0} $ and so on.

However, shouldn't this second line just be $$ H= H_0 + L_\mu^A A_\mu + 1/2 L_{\mu v}^{AA} A_\mu A_v +... $$

Yes, their expression is wrong, for the reason you already know. This is probably just a typo.

The expansion is obtained (as they say in the paper) by using the Taylor series expansion of the exponential: $$ H = \sum_{i,j} \sum_{a,b} t_{a,b} \left(1 + i \int_{\textbf{R}_{j,b}}^{\textbf{R}_{i,a}} dr'_{\mu} A_{\mu} (\textbf{r}',t ) - \frac{1}{2}\int_{\textbf{R}_{j,b}}^{\textbf{R}_{i,a}} dr'_{\mu} A_{\mu} (\textbf{r}',t ) \int_{\textbf{R}_{j,b}}^{\textbf{R}_{i,a}} dr''_{\mu} A_{\mu} (\textbf{r}'',t ) +\ldots \right) c_{i,a}^\dagger c_{j,b} $$

Your expression also looks questionable to me, because the notation is not super clear. You are dotting $L_\mu$ with $A_\mu$, but there is no free $A_\mu$ so I suppose your expression probably should have (or implicitly has) a convolution as well as a dot product. I suppose this is the case, since I also suppose the $\partial_{A_\mu}$ means a functional derivative. And so, it must be convolved with the $A$ field to make sense.