4
$\begingroup$

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 $$ t_{ij}\mapsto\exp\left(\mathrm{i}(\mathbf{R}_i-\mathbf{R}_j)\cdot Q\mathbf{A}\right) t_{ij}\approx \exp\left(-\mathrm{i}Q \int_{\mathbf{R}_i}^{\mathbf{R}_j} \mathbf{A}\cdot\mathrm{d}\mathbf{r}\right) t_{ij} $$

$\endgroup$
4
  • $\begingroup$ The "right" sign depends on your convention for the sign of $Q$. Unfortunately, sources are split about 50/50 on this because the charge of the electron is inconveniently negative. So sometimes, $Q$ stands for the charge of the electron (negative), and other times $Q$ stands for the absolute value of the charge of the electron (positive). As long as you're consistent, there's no problem. $\endgroup$
    – knzhou
    Sep 15, 2023 at 23:46
  • $\begingroup$ @knzhou Correct me if I'm wrong, but a Lagrangian $L=\frac{1}{2}m(\dot{x})^2+Q \dot{x}A$ corresponds to a Hamiltonian $H=(p-QA)^2/(2m)$. Hence, the exponent of the translation operator should be $\propto (p-QA)$, with the relative sign between $A,p$ being the same as in the Hamiltonian. $\endgroup$ Sep 16, 2023 at 15:09
  • $\begingroup$ @TheQuantumMan Whoops, I read this too quickly! I think the first argument shows the additional phase you pick up given a vector potential, while the second argument shows how to define transition elements to get rid of that additional phase. Those are opposites, so we expect them to differ by a sign. The real problem is that the "blah $\to$ blah" notation in physics is horribly ambiguous. $\endgroup$
    – knzhou
    Sep 16, 2023 at 19:32
  • $\begingroup$ @knzhou maybe I'm missing something here, but in the 2nd argument, he just uses minimal substitution. If I understand correctly, what you say is that in the 2nd argument, he uses a gauge transformation to gauge away the additional phase, but why is this the case? I don't see it. $\endgroup$ Sep 17, 2023 at 13:53

0

Your Answer

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