# Line integral in Peierls substitution

I'm trying to understand the reasoning behind Peierls substitution. The final result seems to be simply replacing the hopping elements

$$t_{ij} \to t_{ij} e^{i \frac{q}{\hbar} \int_i^j \vec{A} \cdot d \vec{r}}$$

The line integral in the exponent is taken along the straight line connecting the two sites. Aside from having the shortest euclidean distance, I do not see what is so special about this path? From my understanding, the difference of the line integrals along two different paths is proportional to the magnetic flux enclosed by the two paths.

$$\oint \vec{A} \cdot d\vec{r} = \iint \vec{B} \cdot d\vec{S}$$

So it seems to me the choice of path is actually quite important.

In the tight-binding method, the finite difference approximation $$\psi'(x) \approx \frac{\psi(x+a/2) - \psi(x-a/2)}{a}$$ is made, which means that the momentum operator can be approximated in the following way. $$p \approx \frac{2 \hbar}{a} \sin\left(\frac{a}{2\hbar} p \right) + \mathcal{O}(a^2)$$

Using the double-angle formula for the cosine, the tight-binding Hamiltonian for a nearest-neighbor hopping on a cubic lattice is recovered. $$H = \sum_{\mu}^{x,y,z} \frac{p_\mu^2}{2m} \approx \sum_{\mu} \frac{\hbar^2}{m a^2} \left[1-\cos\left(\frac{a}{\hbar} p_\mu \right)\right]$$

This is perhaps more clear when we identify the hopping constant as $$t = \frac{\hbar^2}{2ma^2}$$ and $$k_\mu = \frac{p_\mu}{\hbar}$$. Usually, the constant identity term is also neglected for simplicity. Then, $$H = -2t \sum_{\mu} \cos k_\mu a$$

The electromagnetic interaction in the continuum model can be incorporated by the minimal coupling substitution $$\vec{p} \to \vec{p} - q\vec{A}$$, so that the single particle Hamiltonian for a charged particle reads $$H = \sum_{\mu} \frac{(p_\mu - q A_\mu)^2}{2m} + q \phi$$

The claim now is that the line integral form of the Peirels substitution (with the straight line path) can be obtained by just performing the minimal coupling substitution in momentum space. But first, some mathematical results need to be stated first. In the theory of first-order differential equations, the integrating factor method is often used to factorize the linear differential operator $$e^{-f(x)} \partial_x e^{f(x)} = f'(x) + \partial_x$$

which allows us to take the exponential easily $$e^{-f(x)} e^{a \partial_x} e^{f(x)} = e^{a (f'(x) + \partial_x)}$$

But the understanding of the operator identity is that there is always a test function on the right. $$e^{a (f'(x) + \partial_x)} \psi(x) = e^{\int_x^{x+a} f'(t) d t} \psi(x+a)$$

And the form of the Peirels substitution is apparent, with the integral occurring in the exponent. Therefore, the acting of the Hamiltonian on a wavefunction is $$H \psi(\vec{x}) = -t \sum_{\mu} e^{- i \frac{q}{\hbar} \int_{\vec{x}}^{\vec{x}+\vec{a}_\mu} A_\mu d x_\mu} \psi(\vec{x}+\vec{a}_\mu) + e^{-i \frac{q}{\hbar} \int_{\vec{x}}^{\vec{x}-\vec{a}_\mu} A_\mu d x_\mu} \psi(\vec{x}-\vec{a}_\mu) \\ = -t \sum_{\mu} e^{- i \frac{q}{\hbar} \int_{\vec{x}}^{\vec{x}+\vec{a}_\mu} \vec{A} \cdot d \vec{r}} \psi(\vec{x}+\vec{a}_\mu) + e^{-i \frac{q}{\hbar} \int_{\vec{x}}^{\vec{x}-\vec{a}_\mu} \vec{A} \cdot d \vec{r}} \psi(\vec{x}-\vec{a}_\mu)$$

$$\vec{a}_\mu = a \hat{e}_\mu$$ is a shorthand, and the integral as seen should be taken along the straight line path. Under a gauge transformation $$\vec{A} \to \vec{A}' = \vec{A} + \nabla \Lambda$$, it can be shown that the transformed Hamiltonian $$H'$$ satisfies $$H'(e^{iq\Lambda/\hbar} \psi) = e^{iq\Lambda/\hbar} H \psi$$