# Peierls substitution vs minimal coupling

In the presence of vector potential (let's assume it's uniform),

a tight-binding Hamiltonian will be changed according to the Peierls substitution:

$$t_{ij}c_i^{\dagger}c_j \to t_{ij}e^{iqA|i-j|}c_i^{\dagger}c_j$$

when transformed to Bloch basis, it becomes: $$\hbar k\to \hbar k-qA$$

Which is the same as minimal coupling.

Are these two approaches just the same thing?

• They are exactly the same. The minimal coupling $-i\partial_x\rightarrow -i\partial_x-qA$ is basically the continuum limit of the Peierls substitution of a tight-binding model. Feb 4, 2015 at 6:00

TL;DR: No, Peierls substitution is an approximation, resembling exact minimal coupling (the fine difference is between crystal quasimomentum and true momentum.)

Minimal coupling refers to the coupling term that we have in an exact Hamiltonian. E.g., if we consider an electron in a periodic potential, we can write its Hamiltonian as $$H=\frac{\mathbf{p}^2}{2m}+U(\mathbf{r})\longrightarrow H=\frac{1}{2m}\left(\mathbf{p}-\frac{e}{c}\mathbf{A}(\mathbf{r})\right)^2+U(\mathbf{r})$$ As we know, the solutions of the crystal Hamiltonian are Bloch states, i.e., we have the sent of eigenfunctions $$\left[\frac{\mathbf{p}^2}{2m}+U(\mathbf{r})\right]\varphi_{n,\mathbf{k}}(\mathbf{r})=\epsilon_{n}(\mathbf{k})\varphi_{n,\mathbf{k}}(\mathbf{r}),$$ where $$n$$ is the band index and $$\mathbf{k}$$ spans the first Brillouin zone.

One often considers this Hamiltonian in the limit of small $$|\mathbf{k}|$$, i.e., the wave function slowly varying over the scale of hundreds or thousands of atoms - an approximation often used in discussing semiconductor nanostructures. We then would write the Hamiltonian for the band of interest (e.g., the conduction band, $$n=c$$) as $$H_c=\epsilon_c\left(\frac{\mathbf{p}}{\hbar}\right)=\epsilon_c(-i\nabla_\mathbf{r}),$$ where we have replaced the quasimomentum $$\hbar\mathbf{k}$$ by true momentum $$\mathbf{p}$$. Of course, if we calculate the matrix elements of momentum in the full Bloch basis, it is not identical with quasimomentum, but in the one-band approximation and in adiabatic limit, the substitution works fine (sometimes this is referred to as envelope function approximation).

One then inserts the magnetic field using Peierls substitution as $$H_c=\epsilon_c\left(\frac{\mathbf{p}}{\hbar}\right)=\epsilon_c(-i\nabla_\mathbf{r})\longrightarrow \epsilon_c\left(-i\nabla_\mathbf{r}-\frac{e}{\hbar c}\mathbf{A}(\mathbf{r})\right)$$

As correctly pointed out by @TavernSa there is an extensive discussion devoted to this subject in volume 9 of Landau&Lifshitz Statistical Physics. Part 2., which has solid-state physics hidden in it as one of the chapters.

The tight-binding approximation presented in the OP takes a somewhat different path - first limiting the discussion to only one band (one atomic orbital), then including the magnetic field, and only then diagonalizing the Hamiltonian. However, the final result would be the same.

Under the substitution ℏk→ℏk−qA

$$\langle p{\mid}x\rangle =\langle 0{\mid} a_{p} a_{x}^{+}{\mid}0\rangle =\exp(-ipx/ h)$$

will become

$$\langle p{\mid}x\rangle =\langle 0{\mid} a_{p} a_{x}^{+}{\mid}0\rangle =\exp(-i(p-qA)x/ h)$$

effectively, the change in operator:

$$a_{p} a_{x}^{+} \rightarrow a_{p} a_{x}^{+} e^{iqAx/h}$$

Then it looks as if:

$$a_{x}^{+} \rightarrow a_{x}^{+} e^{iqAx/h}$$

$$a_{x} \rightarrow a_{x} e^{-iqAx/h}$$

Actually, this is just $$\Phi \rightarrow e^{iqAx /h} \Phi$$ for the substitution solution of Schrodinger equation.

Peierls substitution is just an assumption. It first appears in the paper of Hofstadter butterfly. The magnetic field should be small so that it could be valid. In old times (Luttinger,Kohn), people try to justify it but they don't have a yes or no answer. You can find the history in books (Landau for example).

But people today tend to take it for granted and consider it as "exact". Basically, you can always use cold atom physics to realize anything.....