Quantizing Wave Vectors in 2D Electron Gas: Periodic vs Hard Wall Boundary Conditions?

I'm studying the behavior of a two-dimensional electron gas in a magnetic field, focusing on the quantization of wave vectors and the resulting energy levels, specifically Landau levels. However, I've come across a point of confusion regarding the application of boundary conditions and the consequent quantization of the wave vector component $$k_y$$.

In the absence of a magnetic field, I understand that a particle in a box would have its wave vector quantized in units of $$\pi/L$$ due to the hard wall boundary conditions, where the wave function must be zero at the walls.

Conversely, in the presence of a magnetic field, as in the case of Landau levels, I've seen in most textbooks that the wave vector component $$k_y$$ quantized in units of $$2\pi/L$$ instead, implying periodic boundary conditions.$$\psi(x,y)=\psi(x,y+L_y)$$

Could someone help clarify the following points?

1. Why is $$k_y$$ quantized as $$2\pi n/L_y$$ under periodic boundary conditions for a 2DEG in a magnetic field, rather than $$\pi n/L_y$$ as one might expect from a "box" analogy?
2. How does "having a finite size $$L_y$$" reconcile with the use of periodic boundary conditions in this context?
3. Is there a physical intuition or a solid-state physics convention that dictates the choice of periodic boundary conditions over hard wall conditions when discussing Landau levels and the quantization of $$k_y$$?

Any insights or explanations would be greatly appreciated, especially those that can bridge my understanding of quantum mechanics in confined systems with the behaviors observed in solid-state physics.

• This perhaps: en.wikipedia.org/wiki/Jellium. Commented Jan 18 at 0:35
• Commented Jan 18 at 7:52

1. Why is $$k_y$$ quantized as $$2πn/L_y$$ under periodic boundary conditions for a 2DEG in a magnetic field, rather than $$πn/L_y$$ as one might expect from a "box" analogy?

For a system of finite size containing a thermodynamically large number of particles, it should not matter precisely what boundary conditions are used at the edges.

There is a particular name for this Periodic Boundary Condition: https://en.wikipedia.org/wiki/Born%E2%80%93von_Karman_boundary_condition

BvK PBC is very useful particularly for your question

1. Is there a physical intuition or a solid-state physics convention that dictates the choice of periodic boundary conditions over hard wall conditions when discussing Landau levels and the quantization of $$k_y$$?

Because in solid state physics we are often interested in conduction, which hard wall standing wave boundary conditions cannot treat at all, whereas BvK PBC allows us to pretend to treat it. It is also almost always the case that BvK PBC makes for perfectly unperturbed energy levels (of equal spacing, say), where the perturbation I am talking about comes from boundary conditions. Remember that then you were using standing wave boundary conditions, the normal mode decomposition gives almost analytic energy eigenvalues, with small deviations at the end of the spectrum. BvK PBC ensures that there will be no small deviations at all, and the normal mode decomposition is trivial to state.

1. How does "having a finite size $$L_y$$" reconcile with the use of periodic boundary conditions in this context?

This is actually irreconcilable. If $$L_y$$ is small, then there is no way that BvK PBC can be used. You would actually have to consider the boundary conditions exactly.

This is because the use of BvK PBC is to imagine that you have an infinitely large system, from which you are mentally chopping out a box of width $$L_y$$, and hoping that such a boundary-less mental copy is a good approximation of the actual system you want to study. When $$L_y$$ is actually small, then you cannot assert that this approximation is tolerable.

For a system of finite size containing a thermodynamically large number of particles, it should not matter precisely what boundary conditions are used at the edges. The quantitative behavior should be dominated by what happens in the bulk interior of the system, just because that is where almost all the matter is located. (Note that the dimensionality doesn't really matter here; the key points can be seen by talking about just a one-dimensional system.) With this in mind, we may come to the tentative conclusion that the precise boundary conditions we choose should not matter very much—and while this isn't absolutely always true, we can see how it works for the comparison to periodic to hard wall boundary conditions.

The key observation is that the microscopic energy levels are not exactly the same between the two boundary conditions, but the differences are very small—and thus irrelevant for the behavior of macroscopic systems. Consider the allowed energy levels in the two cases. With the hard wall boundaries, there is one state for each wave vector $$k=\frac{n\pi}{L}$$, with energy $$\frac{\hbar^{2}k^{2}}{2m}$$. In contrast, with the periodic boundary conditions, the allowed wave vectors are, as noted in the question $$k=\frac{2n\pi}{L}$$. However—and this is the key point—there are two states with this wave vector, with (orthogonal) wave functions proportional to $$\sin ky$$ and $$\cos ky$$.

The upshot is that the density of states is effectively the same in either case; the number of states in a range $$\Delta k$$ (where $$\Delta k\gg L^{-1}$$, but still microscopic) is $$\rho_{k}\,\Delta k=\frac{L}{\pi}\Delta k$$. Half of the states have slightly different energies between the two systems of boundary conditions, but these energy differences approach zero as $$L$$ becomes large. So the boundary conditions don't make a difference for the macroscopic properties of the system, as we are free to use whichever is more convenient mathematically. (In most cases, the periodic boundary conditions are just easier to work with, but there are occasional exceptions to this rule.)