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I am considering a tight-binding model in a magnetic field, and studying the Peierls substitution

$$t_{jk} \to t_{jk} e^{i\frac{q}{\hbar}∫_j^k \vec{A} \cdot d \vec{r}}$$

In some papers, such as this one, the author assumed periodic boundary conditions, and he chose the Landau gauge to represent the magnetic field. My concern is regarding gauge invariance when periodic boundary condition is used.

Let's look at the Landau gauge specifically. We choose

\begin{align*} \vec{A} &= B_0 x \hat{y} \\ \phi &= 0 \end{align*}

This corresponds to a uniform magnetic field $$\vec{B} = \nabla \times \vec{A} = B_0 \hat{z}$$

I'm worried as $\vec{A}$ is not periodic in the $x$ direction. So to be safe, let's only consider periodic boundary conditions in the $y$-direction. Our choice of Landau gauge happens to have periodicity in the $y$-direction. So I am happy that I could apply the Peierls substitution to my tight-binding Hamiltonian. The result is that matrix elements corresponding to edges along the $y$ direction will acquire a phase.

Here is the part that confuses me. Suppose I have a friend who disagrees with me about the choice of origin. He claims that the vector potential should be $$\vec{A'} = B_0 (x-x_0) \hat{y}$$

where $x_0$ is a constant. This can also be understood as a gauge transformation. By taking the curl, we obtain the same magnetic field as before. However, the Hamiltonian obtained using Peirels substitution with $\vec{A'}$ is different from that obtained using Peierls substitution with $\vec{A}$ (at least they do not have the same eigenvalues anymore in general). Whose Hamiltonian is the correct one?

I guess what I'm really asking is how to deal with gauge invariance when there is periodic boundary conditions involved. I have thought of using polar coordinates $(r, \theta)$ such that periodic in $\theta$ but not in $r$. But before that, maybe there is a better way to address this issue, or that I might just have a misunderstanding somewhere. Any help would be much appreciated.

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However, the Hamiltonian obtained using Peirels substitution with A′ is different from that obtained using Peierls substitution with A (at least they do not have the same eigenvalues anymore in general). Whose Hamiltonian is the correct one?

They are both correct. The Hamiltonian is not gauge invariant. It is different in different gauges.

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  • $\begingroup$ Thanks for the confirmation hft! Without the periodic boundary condition, I think that a gauge transformation will only cause a similarity transformation to the Hamiltonian, but the important physical quantities, such as the spectrum, remain the same. However, in this case, we can have non-equivalent Hamiltonians describing the same physical system? Do you know how to resolve this? Or is there a reason why this is not really an issue? I understand that Peierls substitution is after all just an approximation. $\endgroup$
    – Bio
    Nov 3, 2021 at 2:30
  • $\begingroup$ I think the resolution comes from the fact that the wave function also changes when the gauge changes. The wave function in the new gauge will have a phase factor of $e^{iqB_0x_0y/\hbar}$ relative to the other gauge. This should cause the physical observables to be gauge independent. See also, this answer: physics.stackexchange.com/questions/439550/… $\endgroup$
    – hft
    Nov 3, 2021 at 6:13

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