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Suppose I have a 2D square lattice in the xy-plane, and I apply a uniform magnetic field in the z-direction. To simplify calculations, I would like to assume periodic boundary conditions in both the x and y directions for my system, so that I can perform some kind of Fourier transform. This was done in the following paper Energy Levels and Wave Functions of Bloch Electrons in Rational and Irrational Magnetic Fields. However, is this actually allowed?

My concern is this: When we identify the opposite edges of a rectangle as identical, we end up with a torus. The surface of a torus defines an "inside" and an "outside". The magnetic field being uniform and perpendicular to the torus' surface means that all the field lines travel from inside to outside or vice versa. But by Gauss' law, the magnetic flux of an enclosed surface is always zero. There is a contradiction somewhere.

So the main question is if periodic boundary conditions are allowed in this kind of system with a magnetic field? If, yes, it would be even better if you can further justify it in the context of the paper above.

Urb
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Bio
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  • You only get a torus after you embed the 2-D plane in a 3-D space. And only if you do it in a particular class of ways. And assuming you have the correct orientions on the periodic boundary conditions. It is possible, for example, to simply assume the boundary conditions without an embedding. You could, for example, presume this is simply a small portion of a much larger system, and treat the boundaries as being "more of the same." This arises in many systems and conditions. I do it in reactor physics sometimes. It arises in some kinds of cosmology. – Dan Feb 10 '22 at 19:34
  • It is not uncommon to consider geometries with magnetic fields that can only be generated with monopoles. To my understanding, Laughlin does it here: https://journals.aps.org/prb/pdf/10.1103/PhysRevB.23.5632 – Rishi Feb 10 '22 at 20:32
  • I found this interesting lecture (https://youtu.be/U4zZhQz1Xqc "MIT 8.06 Quantum Physics III, Spring 2018 - L14.2 Quantization of the magnetic field on a torus"), see also: https://physics.stackexchange.com/q/406672/226902. I am trying to collect useful references here: https://physics.stackexchange.com/q/729770/226902 – Quillo Feb 06 '23 at 11:04

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The same issue arises for any closed, oriented surface, when you try to put a uniform magnetic field through the surface. Strictly speaking Gauss's law does not allow any non-zero total flux through a closed surface. However, at least in theory we can have magnetic monopoles. If that's allowed, then nonzero total flux through a closed surface becomes possible, but there is still a condition: the flux must be an integer multiple of flux quanta ($\hbar c/e$, where $e$ is the fundamental charge of the system). The quantization is necessary to make sure that quantum mechanical wavefunctions are single-valued. One can make very general mathematical argument about why it has to be the case on any closed surface, but we can also do things explicitly on a torus.

Below I set $\hbar=e=c=1$. Suppose we have a $L_x\times L_y$ torus (so identify $x\sim x+L_x, y\sim y+L_y$), and let's choose a Landau gauge:

$$ A_x(x,y)=0, A_y(x,y)=Bx $$

This is what one would do naively on an infinite plane. Now we have a problem: the periodic boundary condition is imposed, but $A_y$ is not periodic in $x$ ($\mathbf{A}$ has no $y$ dependence so the periodicity along $y$ is fine). In particular, $A_y(x+L_x,y)$ and $A_y(x,y)$ differ by $BL_x$. To rescue this, we should remember that vector potential has gauge ambiguity. Namely, if we change $\mathbf{A}$ by a gauge transformation, the result is physically equivalent. So if we can find a gauge transformation $\mathbf{A}\rightarrow \mathbf{A}+\nabla \chi$ to "patch" the difference between $A_y(x+L_x,y)$ and $A_y(x,y)$, then everything is fine. The gauge transformation we need is one satisfies $\partial_y \chi = BL_x$, so $\chi=BL_x y$.

We are not quite done yet. Gauge transformation changes the wavefunction by $\psi\rightarrow e^{i\chi}\psi$. Since $\psi$ is single-valued, the gauge transformation should not spoil the periodicity. Thus $e^{iB L_xy}$ should be periodic in $y$: $$ e^{iBL_x (y+L_y)}=e^{iBL_xy}, $$ which means $BL_xL_y$ must be $2\pi n$ where $n \in\mathbb{Z}$.

TheQuantumMan
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Meng Cheng
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  • $\nabla \chi = B L_x \hat{y}$, which is a constant vector field. How does it patch the difference between $A_y(x+L_x, y)$ and $A_y(x, y)$ if it is added at both points? – Bio Feb 12 '22 at 15:44
  • @MengCheng "Gauss's law does not allow any non-zero total flux through a closed surface." - this sounds to me completely wrong, why do you say that? For one Gauss law is about electric field not magnetic, but anyway, for a constant field on a torus $$\nabla H =0$$, and $$\nabla x H = 0 $$ so everything is ok! – Kphysics Feb 11 '24 at 11:23
  • @Kphysics Well $\nabla \cdot \mathbf{B}=0$ is also known as the Gauss's law for magnetism, which just says there is no monopole. It is not OK to just have some arbitrary constant field on a torus, that is the point I wanted to explain. Read about it before commenting. – Meng Cheng Feb 12 '24 at 19:12
  • @MengCheng Your wording can imply that if there were no monopoles, then there could be no flux on a torus. And here we have precisely the counterexample: $\nabla \cdot B = 0$ everywhere, but there is flux through a closed surface. For the monopole $\nabla \cdot B = \delta(x=0)$. – Kphysics Feb 13 '24 at 00:28
  • Also, Wikipedia says: "Gauss's theorem... is a theorem relating the flux of a vector field through a closed surface to the divergence of the field in the volume enclosed. " Here on the torus the surface is closed but it does not enclose a volume, so no theorem! – Kphysics Feb 13 '24 at 00:38
  • Another way to understand Meng Cheng's argument is not to appeal to Gauss law but rather $B = \nabla\times A$. If you impose that $A$ is globally defined on your domain, and if the domain is closed, then you necessarily have a zero total flux. This is stronger than Gauss law whenever the surface is not simply connected (like the torus), it's the distinction between closed and exact forms. In particular your counterexample does not apply anymore. – LPZ Mar 28 '24 at 17:42
  • Classically, $A$ need only be locally defined so there is no issue. In QM, $A$ needs to be globally defined. To get around this issue, you allow it to be multivalued such that any two ranches are related by a gauge transformation. – LPZ Mar 28 '24 at 17:43