Landau Levels Revisited

Landau levels are usually first solved as a wave-mechanics problem: choose a gauge, reduce the Hamiltonian to a harmonic oscillator, and count the degenerate states. That derivation belongs to Landau Levels and Degeneracy of Landau Levels.

This page revisits the same system from the symmetry and gauge-geometry viewpoint. The central idea is that a charged particle in a uniform magnetic field has two independent noncommutative planes:

• the cyclotron plane, controlled by kinetic momentum and responsible for the energy ladder;
• the guiding-center plane, controlled by magnetic translations and responsible for degeneracy.

This split explains why the Landau spectrum is gauge invariant even though common wavefunctions and labels look very different in Landau and symmetric gauge.

Landau Levels in Solids uses this cyclotron-versus-guiding-center separation but owns the material-band specialization. Mass tensors, Zeeman and orbital shifts, valley multiplicity, nonparabolicity, and Dirac-band contrasts are therefore inputs and limits of that solids treatment, not extensions of the ideal geometry derived here.

Ideal Model

Consider a spinless nonrelativistic particle of charge $q$ and mass $m$ in a uniform magnetic field

$$\mathbf B = B\hat{\mathbf z}, \qquad B>0.$$

With $\Phi=0$, minimal coupling gives the transverse Hamiltonian

$$H = \frac{1}{2m} \left( \pi_x^2+\pi_y^2 \right), \qquad \boldsymbol\pi = \mathbf p-q\mathbf A.$$

The spectrum is

$$E_n = \hbar\omega_c \left( n+\frac12 \right), \qquad \omega_c = \frac{\lvert q\rvert B}{m}.$$

This page does not rederive the coordinate-space oscillator solution. Instead it explains why this oscillator structure is gauge independent.

Kinetic-Momentum Algebra

The kinetic momenta obey

$$[\pi_x,\pi_y] = i\hbar qB.$$

Define the magnetic length and sign

$$\ell_B = \sqrt{ \frac{\hbar}{\lvert q\rvert B} }, \qquad s = \operatorname{sgn}(qB).$$

Then

$$[\pi_x,\pi_y] = i s\frac{\hbar^2}{\ell_B^2}.$$

A gauge-independent cyclotron lowering operator is

$$a = \frac{\ell_B}{\sqrt2\hbar} \left( \pi_x+i s\pi_y \right), \qquad [a,a^\dagger]=1.$$

The Hamiltonian becomes

$$H = \hbar\omega_c \left( a^\dagger a+\frac12 \right).$$

This is the algebraic core of Landau quantization. No particular vector-potential gauge appears in the final ladder algebra.

Guiding-Center Coordinates

The kinetic momenta describe the cyclotron motion. The center of that cyclotron orbit is described by the guiding-center coordinates

$$X = x+\frac{\pi_y}{qB}, \qquad Y = y-\frac{\pi_x}{qB}.$$

They commute with the kinetic momenta:

$$[X,\pi_i]=[Y,\pi_i]=0,$$

and therefore commute with the Hamiltonian:

$$[X,H]=[Y,H]=0.$$

But the guiding-center coordinates do not commute with each other:

$$[X,Y] = -\frac{i\hbar}{qB} = -i\,s\,\ell_B^2.$$

This is the degeneracy algebra. Since $H$ only sees the cyclotron oscillator, shifting the guiding center costs no energy in the ideal infinite system. But $X$ and $Y$ cannot both be diagonalized. A Landau level is not degenerate because a gauge calculation accidentally left a free label; it is degenerate because the guiding-center plane is a quantum phase space.

Gauge Choices as Coordinate Choices

Landau gauge and symmetric gauge organize the same Hilbert space differently.

In Landau gauge,

$$\mathbf A_L = Bx\,\hat{\mathbf y},$$

the Hamiltonian commutes with $\hat p_y$. A state label $k_y$ fixes the oscillator center

$$x_0 = \frac{\hbar k_y}{qB}.$$

This $x_0$ is a guiding-center coordinate in disguise.

In symmetric gauge,

$$\mathbf A_S = \frac{B}{2} \left( -y\,\hat{\mathbf x} + x\,\hat{\mathbf y} \right),$$

rotational symmetry about the $z$ axis is manifest. States are naturally organized by angular labels and radial localization rather than by strip-like guiding-center positions.

The two descriptions are gauge related. The wavefunctions are not expected to be equal pointwise; they represent the same physical subspaces after the appropriate gauge phase and basis change.

Magnetic Translations

Ordinary translations are not the right symmetry operators in a magnetic field because a spatial shift generally changes the vector-potential representative. Magnetic translations combine a shift with the compensating phase needed for gauge covariance.

In the guiding-center convention above,

$$\mathsf T_B(\mathbf a) = \exp\left[ \frac{i}{\hbar} qB(a_xY-a_yX) \right]$$

shifts the guiding center by $\mathbf a$ and commutes with $H$.

The group law is projective:

$$\mathsf T_B(\mathbf a) \mathsf T_B(\mathbf b) = \exp\left[ \frac{iqB}{\hbar} (a_xb_y-a_yb_x) \right] \mathsf T_B(\mathbf b) \mathsf T_B(\mathbf a).$$

The phase is the magnetic flux through the parallelogram in units of $\hbar/q$. Thus Landau-level degeneracy is tied directly to magnetic flux and to the same holonomy idea that appears in the Aharonov–Bohm effect.

One State per Flux Quantum

Because

$$[X,Y] = -i\,s\,\ell_B^2,$$

the guiding-center plane has a quantum cell area of order $2\pi\ell_B^2$. A large region of area $A$ supports approximately

$$N_\Phi = \frac{A}{2\pi\ell_B^2} = \frac{\lvert q\rvert BA}{h}$$

independent orbital states in each spinless Landau level.

Equivalently,

$$N_\Phi = \frac{\Phi}{\Phi_0}, \qquad \Phi = BA, \qquad \Phi_0 = \frac{h}{\lvert q\rvert}.$$

This is a flux count, not a special property of a rectangular Landau-gauge box. Boundaries and global boundary conditions decide the exact finite-size bookkeeping, but the bulk density is geometric.

Boundary and Torus Caveats

On the infinite plane, every Landau level has infinite degeneracy. In a finite sample, edges and confinement reorganize the states. In a rectangle with open boundaries, guiding centers near the edge turn into edge-sensitive states. On a torus, magnetic translations around the two cycles are consistent only when the total flux is quantized:

$$\frac{q\Phi_{\mathrm{tot}}}{\hbar} \in 2\pi\mathbb Z.$$

This condition is the global version of the magnetic-translation phase. It is also the entry point to quantum Hall topology, where filled Landau levels have quantized Hall response. The detailed many-body and response theory belongs to quantum matter; the present page only identifies the single-particle symmetry geometry.

What Perturbations Do

The ideal degeneracy is fragile to perturbations that depend on the guiding center. A scalar potential, boundary confinement, disorder, or interactions can split or broaden a Landau level. The cyclotron gap, however, is controlled by $\hbar\omega_c$ as long as the uniform-field approximation remains meaningful.

This is why one distinguishes:

• Landau quantization, the cyclotron oscillator energy ladder;
• Landau degeneracy, the guiding-center degeneracy inside a level;
• quantum Hall physics, the many-body and topological response of filled or partially filled levels.

Those are connected, but they are not the same claim.

Common Mistakes

• Treating the Landau-gauge label $k_y$ as a gauge-invariant physical momentum.
• Thinking Landau and symmetric gauge describe different spectra because their wavefunctions look different.
• Forgetting that kinetic momentum components fail to commute in a magnetic field.
• Counting degeneracy without specifying area, boundary conditions, or spin/internal factors.
• Assuming the ideal degeneracy survives arbitrary scalar potentials or edges unchanged.
• Calling every Landau-level fact “topological” without distinguishing algebraic degeneracy, flux quantization, and response topology.

References

• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
• R. E. Prange and S. M. Girvin, eds., The Quantum Hall Effect, 2nd ed., Springer, 1990.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• D. Tong, Lectures on the Quantum Hall Effect, 2016.
• J. K. Jain, Composite Fermions, Cambridge University Press, 2007.

Exercises

1. Verify the cyclotron ladder algebra.

Solution

Let

$$a = \frac{\ell_B}{\sqrt2\hbar} \left( \pi_x+i s\pi_y \right), \qquad a^\dagger = \frac{\ell_B}{\sqrt2\hbar} \left( \pi_x-i s\pi_y \right).$$

Using $[\pi_x,\pi_y]=is\hbar^2/\ell_B^2$,

$$\begin{aligned} [a,a^\dagger] &= \frac{\ell_B^2}{2\hbar^2} [\pi_x+i s\pi_y,\pi_x-i s\pi_y] \\ &= \frac{\ell_B^2}{2\hbar^2} \left( -2is[\pi_x,\pi_y] \right) \\ &= \frac{\ell_B^2}{2\hbar^2} \left( 2\frac{\hbar^2}{\ell_B^2} \right) = 1. \end{aligned}$$

2. Show that $X=x+\pi_y/(qB)$ commutes with $\pi_x$.

Solution

Use $[x,\pi_x]=i\hbar$ and $[\pi_y,\pi_x]=-i\hbar qB$:

$$\begin{aligned} [X,\pi_x] &= [x,\pi_x] + \frac{1}{qB}[\pi_y,\pi_x] \\ &= i\hbar - i\hbar = 0. \end{aligned}$$

The other guiding-center commutators with $\pi_i$ work similarly.

3. Estimate the number of spinless Landau orbitals in a disk of area $A$.

Solution

The bulk degeneracy is one orbital per area $2\pi\ell_B^2$, so

$$N_\Phi \approx \frac{A}{2\pi\ell_B^2} = \frac{\lvert q\rvert BA}{h}.$$

For a finite disk this count has edge and integrality corrections, but the large-area density is fixed by magnetic flux.