Landau Levels

Purpose

A uniform magnetic field quantizes the transverse kinetic energy of a charged particle into equally spaced Landau levels. For a spinless particle of mass $m$ and signed charge $q$ in

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

the two-dimensional orbital energies are

$$E_n = \hbar\omega_c \left(n+\frac12\right), \qquad n=0,1,2,\ldots,$$

where

$$\omega_c = \frac{\lvert q\rvert B}{m}.$$

Each level has a macroscopic guiding-center degeneracy. This card collects the energy, length, state-counting, gauge, and spin-extension formulas. The full oscillator reduction is at Landau Levels, and the careful finite-area counting argument is at Degeneracy of Landau Levels.

At a glance

Task	Formula
Cyclotron frequency	$\omega_c=\lvert q\rvert B/m$
Magnetic length	$\ell_B=\sqrt{\hbar/(\lvert q\rvert B)}$
Spinless two-dimensional energy	$E_n=\hbar\omega_c(n+1/2)$
Three-dimensional energy	$E_{n,k_z}=\hbar\omega_c(n+1/2)+\hbar^2k_z^2/(2m)$
Degeneracy per area	$N_\Phi/A=1/(2\pi\ell_B^2)=\lvert q\rvert B/h$
Charge-$q$ flux quantum	$\Phi_0=h/\lvert q\rvert$
Cyclotron radius squared	$\langle\eta^2\rangle_n=(2n+1)\ell_B^2$
Filling factor	$\nu=N/N_\Phi=2\pi\ell_B^2n_{2\mathrm D}$

Magnetic scales

The cyclotron frequency and magnetic length are

$$\omega_c = \frac{\lvert q\rvert B}{m}, \qquad \ell_B = \sqrt{ \frac{\hbar}{\lvert q\rvert B} }.$$

They obey the useful identity

$$m\omega_c\ell_B^2 = \hbar.$$

The sign of $q$ controls the direction of cyclotron motion and the orientation of magnetic translation algebra, but not the positive energy spacing $\hbar\omega_c$. The magnetic length is not the classical orbit radius. It is the basic quantum width from which the level-dependent orbit scale is built.

For charge magnitude $\lvert q\rvert$, define

$$\Phi_0 = \frac{h}{\lvert q\rvert}.$$

For an electron this is $h/e$. It should not be confused with the superconducting flux quantum $h/(2e)$, which reflects charge-$2e$ pairs.

Hamiltonian and oscillator algebra

With minimal coupling,

$$H = \frac{\boldsymbol\pi^2}{2m}, \qquad \boldsymbol\pi = \mathbf p-q\mathbf A,$$

where $\nabla\times\mathbf A=\mathbf B$. In the transverse plane,

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

Let

$$s = \operatorname{sgn}(qB)$$

and define

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

Then

$$[a,a^\dagger]=1$$

and

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

Therefore

$$E_n = \hbar\omega_c \left(n+\frac12\right).$$

The oscillator number $n$ labels cyclotron excitation. It does not label the many distinct guiding-center states within the same Landau level.

Landau-gauge basis

Choose

$$\mathbf A = Bx\,\hat{\mathbf y}.$$

Then $p_y$ commutes with the Hamiltonian, and a convenient basis is

$$\psi_{n,k_y}(x,y) = \frac{e^{ik_yy}}{\sqrt{L_y}} \varphi_n(x-x_0),$$

where

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

The sign of $x_0$ therefore depends on the signed charge. The normalized oscillator function is

$$\varphi_n(\xi) = \frac{ H_n(\xi/\ell_B) }{ \pi^{1/4}\sqrt{2^n n!\ell_B} } \exp\!\left( -\frac{\xi^2}{2\ell_B^2} \right).$$

Substitution reduces the Hamiltonian to

$$H_\perp = \frac{p_x^2}{2m} + \frac12m\omega_c^2(x-x_0)^2.$$

The energy is independent of $k_y$. Different $k_y$ values shift the guiding center and generate the degeneracy. The canonical momentum label $k_y$ and the center coordinate $x_0$ depend on gauge; the level energy and total state count do not.

The symmetric gauge

$$\mathbf A = \frac12\mathbf B\times\mathbf r$$

organizes the same degenerate subspace by angular-momentum-like labels. Use Landau Gauge and Symmetric Gauge when translating between those bases.

Guiding-center and cyclotron coordinates

The position separates into guiding-center and cyclotron pieces:

$$\mathbf r = \mathbf R+\boldsymbol\eta,$$

with

$$\begin{aligned} R_x &= x+\frac{\pi_y}{qB}, & R_y &= y-\frac{\pi_x}{qB}, \\ \eta_x &= -\frac{\pi_y}{qB}, & \eta_y &= \frac{\pi_x}{qB}. \end{aligned}$$

These sectors commute with one another:

$$[R_i,\eta_j]=0.$$

Within each sector,

$$[R_x,R_y] = -is\ell_B^2, \qquad [\eta_x,\eta_y] = is\ell_B^2.$$

The Hamiltonian depends only on $\boldsymbol\eta$; the independent guiding center $\mathbf R$ labels states at fixed energy. This is the gauge-independent content behind the Landau-gauge parameter $k_y$.

The root-mean-square cyclotron coordinate is

$$\sqrt{\langle\eta^2\rangle_n} = \sqrt{2n+1}\,\ell_B.$$

Thus the lowest Landau level has a finite transverse quantum extent even though $n=0$.

Orbital degeneracy and flux counting

Consider a rectangle $L_x\times L_y$ and impose periodic boundary conditions along $y$. Then

$$k_y = \frac{2\pi j}{L_y}, \qquad j\in\mathbb Z.$$

Adjacent guiding centers are separated in magnitude by

$$\lvert\Delta x_0\rvert = \frac{2\pi\ell_B^2}{L_y}.$$

The number whose centers fit across the bulk width $L_x$ is

$$N_\Phi = \frac{L_xL_y}{2\pi\ell_B^2} = \frac{\lvert q\rvert BA}{h} = \frac{BA}{\Phi_0}.$$

Hence the orbital degeneracy per area is

$$\frac{N_\Phi}{A} = \frac{1}{2\pi\ell_B^2} = \frac{\lvert q\rvert B}{h}.$$

This counts one spinless orbital per charge-$q$ flux quantum. On a torus, consistent magnetic boundary conditions quantize the total flux and make $N_\Phi$ an integer. In a finite sample with physical edges, edge states and boundary details modify the literal center-counting picture. The bulk density of states remains the central result.

Do not multiply this count by two unless an unresolved twofold spin degeneracy is actually present.

Three-dimensional spectrum

For free motion parallel to the field,

$$H = H_\perp+\frac{p_z^2}{2m}.$$

The spectrum is

$$E_{n,k_z} = \hbar\omega_c \left(n+\frac12\right) + \frac{\hbar^2k_z^2}{2m}.$$

The magnetic field discretizes transverse kinetic energy but leaves a continuous longitudinal dispersion on infinite space. A finite length or an additional potential can quantize $k_z$ separately.

Adding spin carefully

The orbital formulas above are spinless. If the particle has magnetic moment

$$\boldsymbol\mu_s = g\frac{q}{2m}\mathbf S,$$

the Zeeman Hamiltonian is

$$H_Z = -\boldsymbol\mu_s\cdot\mathbf B.$$

For $S_z\lvert s,m_s\rangle=\hbar m_s\lvert s,m_s\rangle$,

$$\Delta E_Z = -g\frac{qB}{2m}\hbar m_s.$$

The combined ideal energy is therefore

$$E_{n,m_s} = \hbar\omega_c \left(n+\frac12\right) - g\frac{qB}{2m}\hbar m_s.$$

For a free electron, $q=-e$ and $\Delta E_Z=g\mu_BBm_s$, with $\mu_B=e\hbar/(2m_e)$. In solids, the effective mass controlling $\omega_c$ and the effective $g$ factor can differ strongly from their vacuum values. State those parameters before comparing orbital and Zeeman gaps.

Relativistic particles, graphene-like Dirac bands, and particles with strong spin-orbit coupling have different Landau spectra; they are not obtained by blindly appending this Zeeman term.

Filling factor

For $N$ spinless particles in area $A$, define the number density $n_{2\mathrm D}=N/A$. The orbital filling factor is

$$\nu = \frac{N}{N_\Phi} = 2\pi\ell_B^2n_{2\mathrm D} = \frac{n_{2\mathrm D}h}{\lvert q\rvert B}.$$

In a clean noninteracting picture, integer $\nu$ means an integer number of spin-resolved orbital Landau levels is filled. Quantized Hall plateaus require the response, disorder localization, and edge structure developed at the Integer Quantum Hall Effect canonical page; flux counting alone is not a derivation of the plateau physics.

Calculation workflow

1. Record the signed charge $q$ but use $\lvert q\rvert$ in $\omega_c$ and $\ell_B$.
2. Decide whether the model is two- or three-dimensional and whether spin is included.
3. Compute $\omega_c$, $\ell_B$, and the spinless orbital energies.
4. Use $A/(2\pi\ell_B^2)$ only after stating the sample geometry and boundary approximation.
5. Keep gauge-dependent labels such as $k_y$ separate from gauge-invariant observables.
6. Add Zeeman, confinement, disorder, lattice, or interaction terms only as explicit extensions of the ideal model.

Common mistakes

• Using $qB/m$ as a positive cyclotron frequency for a negatively charged particle.
• Losing the signed $q$ in the Landau-gauge center $x_0=\hbar k_y/(qB)$.
• Calling $k_y$ an ordinary gauge-invariant mechanical momentum.
• Confusing $\ell_B$ with the level-dependent radius $\sqrt{2n+1}\,\ell_B$.
• Counting $A/(2\pi\ell_B^2)$ as an exact finite-edge degeneracy without specifying boundary conditions.
• Adding an automatic factor of two for spin after using a spinless formula.
• Forgetting the free $k_z$ term in three dimensions.
• Treating different-looking Landau- and symmetric-gauge wavefunctions as different spectra.
• Assuming a uniform magnetic field confines the guiding center in the bulk.
• Applying the quadratic-band result to relativistic or Dirac particles.

Canonical links

• Particle in a Uniform Magnetic Field introduces the Lorentz-force and kinetic-momentum structure.
• Minimal Coupling in Wave Mechanics fixes the sign convention in $\mathbf p-q\mathbf A$.
• Degeneracy of Landau Levels gives the boundary-aware flux-counting derivation.
• Magnetic Translations develops the noncommuting guiding-center symmetry.
• Landau Levels Revisited reorganizes the problem algebraically.
• Charged Harmonic Oscillator in a Magnetic Field shows how parabolic confinement lifts the ideal guiding-center degeneracy.
• Exactly Solvable Landau-Level System summarizes the model, observables, and failure modes.

References

• L. D. Landau, “Diamagnetismus der Metalle,” Zeitschrift fur Physik 64, 629-637 (1930).
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. E. Prange and S. M. Girvin, eds., The Quantum Hall Effect, 2nd ed., Springer, 1990.

Exercises

1. Show that the kinetic-momentum ladder operators satisfy $[a,a^\dagger]=1$.

Solution

Using $[\pi_x,\pi_y]=i\hbar qB=i\hbar s\lvert q\rvert B$,

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

The final equality uses $\ell_B^2=\hbar/(\lvert q\rvert B)$.

2. Derive the bulk degeneracy of one spinless Landau level in a rectangle.

Solution

Periodic boundary conditions along $y$ give

$$k_y=\frac{2\pi j}{L_y}.$$

Because $x_0=\hbar k_y/(qB)$, adjacent centers have magnitude spacing

$$\lvert\Delta x_0\rvert = \frac{2\pi\hbar}{\lvert q\rvert BL_y} = \frac{2\pi\ell_B^2}{L_y}.$$

The number of bulk centers in width $L_x$ is therefore

$$N_\Phi = \frac{L_x}{\lvert\Delta x_0\rvert} = \frac{L_xL_y}{2\pi\ell_B^2} = \frac{\lvert q\rvert BA}{h}.$$

3. A three-dimensional spinless particle is in the state $n=2$ with longitudinal wave number $k_z$. Find its energy and transverse root-mean-square cyclotron radius.

Solution

The energy is

$$E_{2,k_z} = \frac{5}{2}\hbar\omega_c + \frac{\hbar^2k_z^2}{2m}.$$

The cyclotron-coordinate expectation value is

$$\langle\eta^2\rangle_2 = (2\cdot2+1)\ell_B^2 = 5\ell_B^2,$$

so

$$\sqrt{\langle\eta^2\rangle_2} = \sqrt5\,\ell_B.$$

4. A spinless two-dimensional gas has number density $n_{2\mathrm D}$. Find the field at which the orbital filling factor is $\nu=1$.

Solution

Use

$$\nu = \frac{n_{2\mathrm D}h}{\lvert q\rvert B}.$$

Setting $\nu=1$ gives

$$B = \frac{n_{2\mathrm D}h}{\lvert q\rvert}.$$

This is a flux-counting result. By itself it does not establish a quantized Hall plateau, which also depends on the spectrum, occupied states, response, and localization physics.