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## Main Question or Discussion Point

I have been reading about Landau levels for a two-dimensional system of charged particles in a perpendicular magnetic field and I have trouble understanding why there is degeneracy in the system. Let me provide some background to my problem.

In the presence of a magnetic field, the momentum of a charge particle changes from ##p_{i}## to ##\pi_{i}\equiv p_{i}+eA_{i}##, where ##e## is the charge of the particle.

It is also possible to define another kind of momentum ##\tilde{\pi}_{i}=p_{i}-eA_{i}##.

For a two-dimensional system with a magnetic field pointing in the ##z##-direction, we can work in the symmetric gauge

$${\bf{A}}=\left(-\frac{yB}{2},\frac{xB}{2},0\right),$$

and define two sets of creation and annihilation operators ##a, a^{\dagger}## and ##b, b^{\dagger}## such that a generic state of the system is given by

$$|n,m\rangle=\frac{a^{\dagger n}b^{\dagger}n}{\sqrt{n!m!}}|0,0\rangle,$$

where ##|0,0\rangle## is the unique ground state annihilated by both ##a## and ##b##.

Now, ##|0,0\rangle## is the unique ground state annihilated by both ##a## and ##b## since ##\pi_{i}## and ##\tilde{\pi}_{i}## commute in the complex plane defined by ##(z=x-iy,\bar{z}=x+iy)## and commuting observables have a common basis of eigenfunctions.

Does this not mean that the quantum numbers labelled by ##n## and ##m## give the same wavefunctions?

Why does the energy of the state depend only on ##n##, but not on ##m##?

In the presence of a magnetic field, the momentum of a charge particle changes from ##p_{i}## to ##\pi_{i}\equiv p_{i}+eA_{i}##, where ##e## is the charge of the particle.

It is also possible to define another kind of momentum ##\tilde{\pi}_{i}=p_{i}-eA_{i}##.

For a two-dimensional system with a magnetic field pointing in the ##z##-direction, we can work in the symmetric gauge

$${\bf{A}}=\left(-\frac{yB}{2},\frac{xB}{2},0\right),$$

and define two sets of creation and annihilation operators ##a, a^{\dagger}## and ##b, b^{\dagger}## such that a generic state of the system is given by

$$|n,m\rangle=\frac{a^{\dagger n}b^{\dagger}n}{\sqrt{n!m!}}|0,0\rangle,$$

where ##|0,0\rangle## is the unique ground state annihilated by both ##a## and ##b##.

Now, ##|0,0\rangle## is the unique ground state annihilated by both ##a## and ##b## since ##\pi_{i}## and ##\tilde{\pi}_{i}## commute in the complex plane defined by ##(z=x-iy,\bar{z}=x+iy)## and commuting observables have a common basis of eigenfunctions.

Does this not mean that the quantum numbers labelled by ##n## and ##m## give the same wavefunctions?

Why does the energy of the state depend only on ##n##, but not on ##m##?