1
$\begingroup$

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 $$ H= H_0 + L_\mu^A A_\mu + 1/2 L_{\mu v}^{AA} A_\mu A_v +... $$ since the partial derivative is w res to $H$? or am I missing something since we're taking a derivative over $A$ instead of $r$?

$\endgroup$
1
  • 1
    $\begingroup$ Yes, it's a typo. And the notation is unclear. This is pre-print, so you could let the authors know, or if it got published somewhere you could find the actual journal article and see if it is fixed already. $\endgroup$
    – hft
    Commented May 19, 2022 at 3:03

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ yes as for the last paragraph, ty $\endgroup$
    – anon.jpg
    Commented May 20, 2022 at 21:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.