7
$\begingroup$

The standard way to treat a tight binding method in a magnetic is to replace the hopping matrix element:

$t_{i,j}\rightarrow e^{i\int_i^j\mathbf{A(x)}.d\mathbf{x}}$

the so called "Peierls substitution", How is this justified?

The usual way people try to justify this is to perform a consistency check (i.e. to show it's gauge invariant). Is there a more rigorous way to drive this? What are it's limits of validity?

$\endgroup$
  • 1
    $\begingroup$ One answer to your question is that in a tight-binding model it actually does not matter how you choose these phases for each $t_{ij}$, as long as the product of the phases along a closed path in the lattice agrees with the Aharonov-Bohm phase set by the external magnetic flux. The individual phases do not have invariant meanings. Once the magnetic fluxes are fixed, the spectrum is determined. $\endgroup$ – Meng Cheng Apr 23 '15 at 16:41
  • $\begingroup$ @MengCheng, What you said is correct but is again a consistency check. I was wondering if there is anyway to actually drive this result. $\endgroup$ – Yahya Alavirad Apr 24 '15 at 1:17
  • 1
    $\begingroup$ This results from the modification of the definition of a Wannier state. For Wannier states in a magnetic field, you attach a vector potential related phase. You may read Wannier's original papers. $\endgroup$ – DKS Apr 25 '15 at 11:38
10
$\begingroup$

The Hamiltonian is given by \begin{equation} H=\frac{\mathbf{p}^2}{2m}+U\left(\mathbf{r}\right), \end{equation} where $U\left(\mathbf{r}\right)$ is the potential landscape due to the crystal lattice. The Bloch theorem asserts that the solution to the problem \begin{equation} H\Psi_{\mathbf{k}}=E\left(\mathbf{k}\right)\Psi_{\mathbf{k}}, \end{equation} is to be sought in the Bloch sum form \begin{equation} \Psi_{\mathbf{k}}=\frac{1}{\sqrt{N}}\sum_{\mathbf{R}}e^{i\mathbf{k}\cdot\mathbf{R}}\phi\left(\mathbf{r}-\mathbf{R}\right), \end{equation} where $N$ is the number of unit cells, and $\phi$ are atomic orbitals, also known as Wannier states. The corresponding eigenvalues $E\left(\mathbf{k}\right)$, which form bands depending on the crystal momentum $\mathbf{k}$, are obtained by calculating the matrix element $\langle\Psi_{\mathbf{k}}|H|\Psi_{\mathbf{k}}\rangle$

\begin{equation} \frac{1}{N}\sum_{\mathbf{R}\mathbf{R}^{\prime}}e^{i\mathbf{k}\left(\mathbf{R}^{\prime}-\mathbf{R}\right)} \int d\mathbf{r}\phi^*\left(\mathbf{r}-\mathbf{R}\right)H\phi\left(\mathbf{r}-\mathbf{R}^{\prime}\right) \end{equation} and ultimately depend on material-related hopping integrals $t_{12}=-\int d\mathbf{r}\phi^*\left(\mathbf{r}-\mathbf{R_1}\right)H\phi\left(\mathbf{r}-\mathbf{R_2}\right)$.

In the presence of the magnetic field the Hamiltonian changes to

\begin{equation} H=\frac{\left(\mathbf{p}-q\mathbf{A}\right)^2}{2m}+U\left(\mathbf{r}\right), \end{equation} where $q$ is the charge of the particle. The additional term introduces complications, and the original Bloch sum becomes inadequate. As it turns out, simply adding a phase term

\begin{equation} \Psi_{\mathbf{k}}=\frac{1}{\sqrt{N}}\sum_{\mathbf{R}}e^{i\left(\mathbf{k}\cdot\mathbf{R}+\frac{q}{\hbar}G_{\mathbf{R}}\right)}\phi\left(\mathbf{r}-\mathbf{R}\right), \end{equation} where

\begin{equation} G_{\mathbf{R}}=\int_{\mathbf{R}}^{\mathbf{r}}\mathbf{A}\cdot d\mathbf{l}, \end{equation} resolves the issue. The hopping matrix elements now read

\begin{align} \langle\Psi_{\mathbf{k}}|H|\Psi_{\mathbf{k}}\rangle=&\frac{1}{N}\sum_{\mathbf{R}\mathbf{R}^{\prime}}e^{i\mathbf{k}\left(\mathbf{R}^{\prime}-\mathbf{R}\right)}\int d\mathbf{r}e^{-i\frac{q}{\hbar}G_{\mathbf{R}}}\phi^*\left(\mathbf{r}-\mathbf{R}\right)\left[\frac{\left(\mathbf{p}-q\mathbf{A}\right)^2}{2m}+U\left(\mathbf{r}\right)\right]e^{i\frac{q}{\hbar}G_{\mathbf{R}^{\prime}}}\phi\left(\mathbf{r}-\mathbf{R}^{\prime}\right)\nonumber\\ =&\frac{1}{N}\sum_{\mathbf{R}\mathbf{R}^{\prime}}e^{i\mathbf{k}\left(\mathbf{R}^{\prime}-\mathbf{R}\right)}e^{i\frac{q}{\hbar}\int_{\mathbf{R}^{\prime}}^{\mathbf{R}}\mathbf{A}\cdot d\mathbf{l}}\nonumber\\&\times\int d\mathbf{r}e^{i\frac{q}{\hbar}\Phi\left(\mathbf{r}\right)}\phi^*\left(\mathbf{r}-\mathbf{R}\right)\left[\frac{\left(\mathbf{p}-q\mathbf{A}+q\nabla G_{\mathbf{R}^{\prime}}\right)^2}{2m}+U\left(\mathbf{r}\right)\right]\phi\left(\mathbf{r}-\mathbf{R}^{\prime}\right)\nonumber\\ =&\frac{1}{N}\sum_{\mathbf{R}\mathbf{R}^{\prime}}e^{i\mathbf{k}\left(\mathbf{R}^{\prime}-\mathbf{R}\right)}e^{i\frac{q}{\hbar}\int_{\mathbf{R}^{\prime}}^{\mathbf{R}}\mathbf{A}\cdot d\mathbf{l}}\int d\mathbf{r}\phi^*\left(\mathbf{r}-\mathbf{R}\right)\left[\frac{\mathbf{p}^2}{2m}+U\left(\mathbf{r}\right)\right]\phi\left(\mathbf{r}-\mathbf{R}^{\prime}\right). \end{align} The relation $\nabla G_{\mathbf{R}^{\prime}}=\mathbf{A}$ holds for the tight binding condition and in the case when the magnetic field is invariant at the scale of the crystal lattice. On the other hand, the flux $\Phi\left(\mathbf{r}\right)=\oint_{\mathbf{R}^{\prime}\rightarrow\mathbf{r}\rightarrow\mathbf{R}}\mathbf{A}\cdot d\mathbf{l}$ is larger when the integrand $\mathbf{r}$ is further from the two vectors $\mathbf{R}$ and $\mathbf{R}^{\prime}$, where the atomic orbitals Wannier states are effectively zero, while the flux is vanishing where the hopping integral is nonzero. Having these two things in mind helps explain the transition from the second to the third line.

Now it becomes obvious that the matrix elements are the same as in the case without magnetic field, apart from the phase factor picked up, which is denoted the Peierls phase. This is tremendously convenient, since then we get to use the same material parameters regardless of the magnetic field value, and the corresponding phase is computationally trivial to take into account. For electrons it amounts to replacing the hopping term $t_{ij}$ with $t_{ij}e^{i\frac{e}{\hbar}\int_i^j\mathbf{A}\cdot d\mathbf{l}}$. Finally, note that a beautiful and elucidating explanation for this phase can also be found in Feynman's Lectures (Vol. III, Chapter 21).

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Wannier states are not atomic orbitals. $\endgroup$ – DKS Jun 17 '15 at 12:35
  • $\begingroup$ @L.Su you are right, I have edited my answer to reflect that. $\endgroup$ – mgphys Jun 18 '15 at 8:20
  • $\begingroup$ @mgphys If we use the convention for Wannier state like this, the Wannier functions at different sites will not be orthogonal, right? How do we explain this? $\endgroup$ – Chuan Chen Oct 16 '18 at 6:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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