# Degeneracy in Landau Levels

A subsection from "Landau Levels" from pg 21 from Lectures on Quantum Hall effect by David Tong.

He shows and derives the energy of a charged particle in a planar surface under the action of a magnetic field with the help of QHO $$a$$ and $$a\dagger$$ operators and goes on to show that there is a degeneracy since one degree of freedom has been lost in the process. All of this is followed by the following subsection where he illustrates the degenerate Landau levels.

The first line of the last para, "The degeneracy (1.21) is very very large".

So my question is how is this is the required degeneracy? To me it looks as if a small region was chosen and the number of encapsulated momentum states were simply counted, to my understanding (1.21) can be called the degeneracy only when the assumed region ($$L_{x} X L_{y}$$) encloses exactly all the states with a definite energy (one of the spectrum of energies $$E_{n}=(n+1/2)*\hbar*\omega$$). Clearly to me that should be the case, but I am not able to see as to why the small region chosen by Tong here should contain exactly all the states with one definite energy level, or maybe I am just misunderstanding something. Help please.

In reality, no system is infinite. Tong is simply taking a finite system of size $$L_x\times L_y$$ as to quantize the momenta. This is the whole system under consideration, not only a small cut-out. All states are counted.
The result is that the degeneracy scales with $$L_xL_y$$. You may now choose the system size to be anything you want.
• When you say "degeneracy scales with $L_{x}L_{y}$", you mean the number of degenerate states is proportional to the area being observed, right? But I still fail to see why/how all those different momentum states from (1.21) have the same n i.e., what is the argument for them to belong to the same Landau level? Commented Aug 4, 2019 at 17:42
• Yes, exactly. The energy of every state only depends on $n$. Furthermore, there are no constraints on $n$ or $k_y$, so for any given $n$ all states with quantum numbers $n$ and an arbitrary $k_y$ have the same energy. Commented Aug 5, 2019 at 13:39
• Yes I see but my doubt still remains. What I am asking is how those states in (1.21) share the same $n$ value? He hasn't mentioned why the states calculated have the same $n$ value i.e., the same energy level, has he not? Commented Aug 5, 2019 at 17:17
• @Hskritna The states $\left|n,k\right>$ have energy $E_n$ independent of $k$. So for each value of $k$, we get a different degenerate state. If we count all the possible values of $k$, we get the degeneracy.