4.3.2 Landau Quantization#

Prompts

Why does a charged particle in a uniform magnetic field behave like a 2D harmonic oscillator? What plays the role of position and momentum?
Define the guiding-center coordinates \(\hat{X}, \hat{Y}\). Why do they not commute, and what is the magnetic length \(\ell_B\)?
Construct ladder operators \(\hat{a}\) from the kinetic momentum and \(\hat{b}\) from the guiding-center coordinates. Why does only \(\hat{a}^\dagger\hat{a}\) appear in the Hamiltonian?
Eigenstates are labeled \(\vert n,m\rangle\). What is the physical meaning of \(n\) and \(m\), and why is the energy independent of \(m\)?
Why does each Landau level have macroscopic degeneracy \(N_\Phi = BA/\Phi_0\)?

Lecture Notes#

Overview#

A charged particle in a uniform magnetic field becomes a 2D quantum harmonic oscillator: the kinetic energy is quantized into evenly-spaced Landau levels, and each level carries a macroscopic degeneracy controlled by the total magnetic flux through the sample. The cleanest route is the symmetric gauge, which makes a clean separation between the cyclotron motion (which sets the energy) and the guiding-center motion (which sets the degeneracy).

Hamiltonian in Symmetric Gauge#

Take a uniform field \(\boldsymbol{B} = B\hat{z}\) (with \(B > 0\), \(q > 0\)) and restrict to motion in the \(xy\)-plane. We work in the symmetric gauge

\[ \boldsymbol{A} = \frac{B}{2}\,(-y, x, 0), \]

so the Hamiltonian (§4.1.1 minimal coupling) takes the symmetric form

\[ \hat{H} = \frac{1}{2m}(\hat{\boldsymbol{p}} - q\boldsymbol{A})^{2} = \frac{\hat{\pi}_x^2 + \hat{\pi}_y^2}{2m}, \]

with kinetic momentum

\[\begin{split} \begin{split} \hat{\pi}_x &\;=\; \hat{p}_x + \frac{qB}{2}\,\hat{y},\\ \hat{\pi}_y &\;=\; \hat{p}_y - \frac{qB}{2}\,\hat{x}. \end{split} \end{split}\]

The dynamics is gauge-invariant; the kinetic momenta \(\hat{\pi}_{x,y}\) and the guiding-center coordinates introduced below are gauge-invariant operators. We choose symmetric gauge for algebraic transparency.

Two Commuting Canonical Pairs

Classically, a cyclotron orbit has two pieces: (i) the velocity (which rotates at \(\omega_c\) around the orbit) and (ii) the guiding center — the constant centre of the orbit. Quantum mechanics promotes both to operators, and they form two independent canonical pairs.

Guiding-center coordinates. Echoing the classical relation \(X = x + v_y/\omega_c\), \(Y = y - v_x/\omega_c\) (§4.3.1), define

\[\begin{split} \begin{split} \hat{X} &\;=\; \hat{x} + \frac{\hat{\pi}_y}{qB},\\ \hat{Y} &\;=\; \hat{y} - \frac{\hat{\pi}_x}{qB}. \end{split} \end{split}\]

Both are gauge-invariant: they involve only kinetic momenta and bare positions.

Canonical commutators in Landau problem

After defining the magnetic length \(\ell_B\), write all canonical commutators in that scale:

\[\begin{split} \begin{split} [\hat{\pi}_x, \hat{\pi}_y] &= \mathrm{i}\,\frac{\hbar^2}{\ell_B^2},\\ [\hat{X}, \hat{Y}] &= -\mathrm{i}\,\ell_B^2, \end{split} \end{split}\]

with all cross-commutators vanishing,

\[ [\hat{X}, \hat{\pi}_i] = [\hat{Y}, \hat{\pi}_i] = 0 \quad (i=x,y). \]

Magnetic length

\[ \boxed{\;\ell_B \;\equiv\; \sqrt{\frac{\hbar}{qB}}\;} \]

(eq-magnetic-length)

\(\ell_B\) is the intrinsic length scale of the Landau problem: it sets both cyclotron-orbit size and the area per guiding-center state.

The new feature: in zero magnetic field, position \((\hat{x},\hat{y})\) commutes and momentum \((\hat{p}_x,\hat{p}_y)\) commutes. In a uniform field, the kinetic momenta no longer commute — the cyclotron pair \((\hat{\pi}_x,\hat{\pi}_y)\) is canonically conjugate with scale \(\hbar^2/\ell_B^2\) — and the guiding-center coordinates no longer commute either, with their own uncertainty relation \(\Delta X\,\Delta Y \geq \ell_B^2/2\).

Derivation: Commutators

Kinetic momenta. Using \([\hat{p}_i, f(\hat{\boldsymbol{x}})] = -\mathrm{i}\hbar\,\partial_i f\) — so \([\hat{p}_x, A_y] = -\mathrm{i}\hbar\,\partial_x A_y\) and \([A_x, \hat{p}_y] = -[\hat{p}_y, A_x] = +\mathrm{i}\hbar\,\partial_y A_x\) — expand bilinearly and substitute:

\[\begin{split} \begin{split} [\hat{\pi}_x, \hat{\pi}_y] &= [\hat{p}_x - qA_x,\,\hat{p}_y - qA_y]\\ &= [\hat{p}_x, \hat{p}_y] - q[\hat{p}_x, A_y] - q[A_x, \hat{p}_y] + q^2[A_x, A_y]\\ &= 0 - q(-\mathrm{i}\hbar\,\partial_x A_y) - q(+\mathrm{i}\hbar\,\partial_y A_x) + 0\\ &= \mathrm{i}q\hbar(\partial_x A_y - \partial_y A_x)\\ &= \mathrm{i}q\hbar B = \mathrm{i}\,\frac{\hbar^2}{\ell_B^2}, \end{split} \end{split}\]

where the last line uses \((\nabla\times\boldsymbol{A})_z = \partial_x A_y - \partial_y A_x = B\) and \(\ell_B^2 = \hbar/(qB)\) from eq-magnetic-length.

Guiding center. With \(\hat{X} = \hat{x} + \hat{\pi}_y/(qB)\) and \(\hat{Y} = \hat{y} - \hat{\pi}_x/(qB)\), the four sub-commutators are \([\hat{x},\hat{y}]=0\), \([\hat{x},\hat{\pi}_x] = \mathrm{i}\hbar\), \([\hat{\pi}_y,\hat{y}] = -\mathrm{i}\hbar\), and \([\hat{\pi}_y,\hat{\pi}_x] = -\mathrm{i}q\hbar B\) (the last by antisymmetry of the kinetic-momentum result above). Expand bilinearly and substitute:

\[\begin{split} \begin{split} [\hat{X},\hat{Y}] &= [\hat{x},\hat{y}] - \frac{[\hat{x},\hat{\pi}_x]}{qB} + \frac{[\hat{\pi}_y,\hat{y}]}{qB} - \frac{[\hat{\pi}_y,\hat{\pi}_x]}{(qB)^2}\\ &= 0 - \frac{\mathrm{i}\hbar}{qB} + \frac{-\mathrm{i}\hbar}{qB} - \frac{-\mathrm{i}q\hbar B}{(qB)^2}\\ &= -\frac{\mathrm{i}\hbar}{qB} - \frac{\mathrm{i}\hbar}{qB} + \frac{\mathrm{i}\hbar}{qB}\\ &= -\frac{\mathrm{i}\hbar}{qB} = -\mathrm{i}\,\ell_B^2, \end{split} \end{split}\]

where the last equality uses \(\ell_B^2 = \hbar/(qB)\) from eq-magnetic-length.

Cross-commutators. \([\hat{X},\hat{\pi}_x] = [\hat{x},\hat{\pi}_x] + [\hat{\pi}_y,\hat{\pi}_x]/(qB) = \mathrm{i}\hbar + (-\mathrm{i}q\hbar B)/(qB) = \mathrm{i}\hbar - \mathrm{i}\hbar = 0\), and similarly for the other three combinations.

Ladder Operators#

Each canonical pair admits a single set of ladder operators.

Cyclotron and guiding-center ladders

Cyclotron ladder (from \((\hat{\pi}_x, \hat{\pi}_y)\)):

\[\begin{split} \begin{split} \hat{a} &\;=\; \frac{\ell_B}{\sqrt{2}\,\hbar}\left(\hat{\pi}_x + \mathrm{i}\hat{\pi}_y\right),\\ [\hat{a},\hat{a}^\dagger] &= 1. \end{split} \end{split}\]

Guiding-center ladder (from \((\hat{X},\hat{Y})\)):

\[\begin{split} \begin{split} \hat{b} &\;=\; \frac{\hat{X} - \mathrm{i}\hat{Y}}{\sqrt{2}\,\ell_B},\\ [\hat{b},\hat{b}^\dagger] &= 1. \end{split} \end{split}\]

The two sets commute: \([\hat{a},\hat{b}] = [\hat{a},\hat{b}^\dagger] = 0\).

Key result: Hamiltonian uses only cyclotron ladder

(156)#\[ \boxed{\;\hat{H} \;=\; \hbar\omega_c\!\left(\hat{a}^\dagger\hat{a} + \tfrac{1}{2}\right),\quad \omega_c = qB/m.\;} \]

So \(\hat{a}^\dagger\) raises the energy by \(\hbar\omega_c\) (cyclotron excitation), while \(\hat{b}^\dagger\) changes only the guiding-center state and leaves energy unchanged.

Eigenstates and Their Physical Meaning#

Joint eigenstates of \(\hat{a}^\dagger\hat{a}\) and \(\hat{b}^\dagger\hat{b}\) are labeled by two quantum numbers,

\[\begin{split} \begin{split} \hat{a}^\dagger\hat{a}\,\vert n, m\rangle &= n\,\vert n, m\rangle,\\ \hat{b}^\dagger\hat{b}\,\vert n, m\rangle &= m\,\vert n, m\rangle,\\ n,m &= 0, 1, 2, \ldots, \end{split} \end{split}\]

constructed from the joint vacuum \(\vert 0, 0\rangle\) (annihilated by both \(\hat{a}\) and \(\hat{b}\)) by

\[ \vert n, m\rangle \;=\; \frac{(\hat{a}^\dagger)^n\,(\hat{b}^\dagger)^m}{\sqrt{n!\,m!}}\,\vert 0, 0\rangle. \]

The energy is

\[ E_{n,m} = E_n = \hbar\omega_c\left(n + \tfrac{1}{2}\right), \]

independent of \(m\).

Physical meaning of the quantum numbers.

\(n\) — the Landau-level index — quantizes the cyclotron motion. It is the kinetic-energy quantum number: each unit of \(n\) adds \(\hbar\omega_c\) of orbital energy.
\(m\) — the guiding-center quantum number — labels the position of the orbit center. It does not affect the energy, so all \(\vert n, m\rangle\) at fixed \(n\) form one degenerate Landau level.

The two-tower structure has a classical echo: cyclotron motion (rotation around the center) and guiding-center motion (location of the center) are independent classical degrees of freedom. Quantum mechanics quantizes the first as energy and the second as a positional index, with the noncommutativity \([\hat{X},\hat{Y}] = -\mathrm{i}\ell_B^2\) giving the guiding-center index its discreteness.

Macroscopic Degeneracy: State Counting#

For an infinite system the \(b\)-tower is infinite; for a finite sample of area \(A\) the degeneracy is finite. The commutator \([\hat{X},\hat{Y}] = -\mathrm{i}\ell_B^2\) and uncertainty relation \(\Delta X\,\Delta Y \ge \ell_B^2/2\) explain why guiding centers are quantized rather than continuously localizable.

A clean finite-size count is as follows. Consider a disk of radius \(R_{\max}\) (area \(A=\pi R_{\max}^2\)). From \(\hat{b}=(\hat{X}-\mathrm{i}\hat{Y})/(\sqrt{2}\,\ell_B)\), we get

\[ \hat{R}^2:=\hat{X}^2+\hat{Y}^2 = \ell_B^2(2\hat{b}^\dagger\hat{b}+1). \]

For a state with guiding-center quantum number \(m\) (i.e. \(\hat{b}^\dagger\hat{b}=m\)), this gives a typical guiding-center radius \(R^2 = \ell_B^2 (2m+1)\). Requiring the orbit center to lie inside the sample (\(R\lesssim R_{\max}\)) gives \(m_{\max}\simeq R_{\max}^2/(2\ell_B^2)\). Therefore the number of allowed \(m\) values at fixed \(n\) is

Landau-level degeneracy

(157)#\[ \boxed{\;N_{\mathrm{states}}^{\mathrm{(LL)}} \;=\; \frac{R_{\max}^2}{2\ell_B^2} \;=\; \frac{\Phi_{\mathrm{total}}}{\Phi_0}\;\equiv\; N_\Phi\;} \]

Here \(N_{\mathrm{states}}^{\mathrm{(LL)}}\) means: number of electron states available in one Landau level. The final equality shows this number happens to equal the number \(N_\Phi\) of flux quanta through the sample. For a \(1\,\mathrm{mm}^2\) sample at \(B = 1\,\mathrm{T}\), \(N_\Phi \sim 10^{10}\).

This is the structural fact that drives the integer (and fractional) quantum Hall effects: discrete energies \(E_n = (n+\tfrac{1}{2})\hbar\omega_c\) separated by gaps \(\hbar\omega_c\), with each level containing a macroscopic number \(N_\Phi\) (i.e. the number of available single-particle states in one level). As the magnetic field is swept, the filling factor \(\nu = N_e/N_\Phi\) changes continuously, but transport responds discretely — the topic of §4.3.3.

Discussion: counting Landau-level degeneracy

(a) For a 2D electron system with \(A = 1\,\mathrm{mm}^2\) at \(B = 5\,\mathrm{T}\) containing \(N_e = 10^{11}\) electrons, what is the filling factor \(\nu = N_e/N_\Phi\) (electrons divided by available states per Landau level)? How many Landau levels are partially occupied?

(b) Why does the system become incompressible — unable to absorb additional electrons at fixed Fermi energy — at integer \(\nu\)?

Poll: Landau-level energy spacing

The energy spacing between consecutive Landau levels is \(\Delta E = \hbar\omega_c = \hbar qB/m\). Why does it scale linearly with \(B\)?

(A) The zero-point energy of a 2D harmonic oscillator scales as \(\hbar\omega_c \propto B\).

(B) The density of states grows with \(B\), so spacings widen.

(D) Both (A) and (C).

Summary#

Symmetric gauge cleanly separates the dynamics into cyclotron motion \((\hat{\pi}_x, \hat{\pi}_y)\) and guiding-center motion \((\hat{X}, \hat{Y})\) — two commuting canonical pairs with \([\hat{\pi}_x, \hat{\pi}_y] = \mathrm{i}q\hbar B\) and \([\hat{X}, \hat{Y}] = -\mathrm{i}\ell_B^2\).
The magnetic length \(\ell_B = \sqrt{\hbar/(qB)}\) is the only intrinsic length scale.
Ladder operators \(\hat{a}\) (cyclotron) and \(\hat{b}\) (guiding-center) commute; the Hamiltonian is \(\hat{H} = \hbar\omega_c(\hat{a}^\dagger\hat{a} + \tfrac{1}{2})\).
States \(\vert n, m\rangle\): \(n\) = Landau-level index (energy), \(m\) = guiding-center index (degeneracy). Energy \(E_n = (n+\tfrac{1}{2})\hbar\omega_c\) is \(m\)-independent.
Each Landau level has \(N_\Phi\) available states, with \(N_\Phi = BA/\Phi_0\) (numerically equal to flux-quanta count).

Homework#

1. Lowest Landau level wavefunctions. In symmetric gauge \(\boldsymbol{A} = \frac{B}{2}(-y, x, 0)\), the lowest Landau level (LLL) states are

\[ \psi_m(\boldsymbol{r}) \propto z^m\,\mathrm{e}^{-\vert z\vert^2/(4\ell_B^2)}, \qquad z = x + \mathrm{i}y, \qquad m = 0,1,2,\ldots \]

(a) Verify by direct computation that \(\hat{a}\,\psi_m = 0\), where \(\hat{a} = (\ell_B/(\sqrt{2}\,\hbar))(\hat{\pi}_x + \mathrm{i}\hat{\pi}_y)\).

(b) Compute \(\langle\hat{r}^2\rangle\) for \(\psi_m\) and show it grows linearly with \(m\). Interpret as the squared distance of the guiding center from the origin.

(c) Show that \(\psi_m\) is an eigenstate of \(\hat{L}_z\) with eigenvalue \(m\hbar\). Connect \(m\) to the guiding-center quantum number from the lecture.

2. Cyclotron resonance. A circularly polarized electric field \(\boldsymbol{E}(t) = E_0(\cos\omega t,\, -\sin\omega t,\, 0)\) couples to a Landau-quantized particle of charge \(q\) via \(\hat{V}(t) = q\boldsymbol{E}(t)\cdot\hat{\boldsymbol{r}}\).

(a) Express \(\hat{x}\) and \(\hat{y}\) in terms of the cyclotron and guiding-center ladder operators \(\hat{a}, \hat{a}^\dagger, \hat{b}, \hat{b}^\dagger\). Show that the \(\hat{b}, \hat{b}^\dagger\) pieces leave the Landau-level index \(n\) unchanged, so \(\hat{V}(t)\) couples \(n\) only to \(n\pm 1\).

(b) Compute \(\vert\langle n+1\vert\hat{V}\vert n\rangle\vert^2\) at resonance (\(\omega = \omega_c\)). Show that it grows with \(n\).

(c) The absorption peaks sharply at \(\omega = \omega_c\) regardless of which Landau level is initially occupied. Explain how this cyclotron-resonance peak is used experimentally to measure effective mass.

3. Degeneracy and filling factor. Each Landau level has degeneracy \(N_\Phi = BA/\Phi_0\) on a sample of area \(A\).

(a) Show that doubling \(B\) doubles \(N_\Phi\) and halves \(\ell_B^2\). Interpret each effect physically (more states per level; tighter wavefunction).

(b) For a sample with fixed electron density \(n_e = N_e/A\), define the filling factor \(\nu = N_e/N_\Phi = n_e h/(qB)\). Why does \(\nu\) decrease as \(B\) increases?

(c) At \(\nu = 1\), one full Landau level is occupied. Adding one more electron costs the gap \(\hbar\omega_c\). Why does this gap make the system incompressible and produce a Hall-resistance plateau (anticipating §4.3.3)?

4. Gauge equivalence. The Landau gauge \(\boldsymbol{A}_L = (0, Bx, 0)\) and the symmetric gauge \(\boldsymbol{A}_S = \tfrac{B}{2}(-y, x, 0)\) both produce \(\boldsymbol{B} = B\hat{z}\).

(a) Find the gauge function \(\alpha(x, y)\) relating them, \(\boldsymbol{A}_S = \boldsymbol{A}_L + \nabla\alpha\).

(b) Eigenstates in the Landau gauge are labeled by \(\hat{p}_y\) (continuous translation in \(y\)); eigenstates in the symmetric gauge are labeled by \(\hat{L}_z\) (rotation about origin). Both label the same Landau-level degeneracy \(N_\Phi\). Explain how two different conserved quantities can label the same set of states.

(c) The combination \(\hat{X}^2 + \hat{Y}^2 = \ell_B^2(2\hat{b}^\dagger\hat{b} + 1)\) measures the squared distance of the guiding center from the origin. Use this to argue that the symmetric-gauge \(\hat{b}^\dagger\hat{b}\) quantum number is essentially \(\hat{L}_z/\hbar\) in the lowest Landau level.

5. Numerical scales. Use \(\ell_B = \sqrt{\hbar/(qB)} \approx 25.66\,\mathrm{nm}\cdot(1\,\mathrm{T}/B)^{1/2}\) for an electron.

(a) Compute \(\ell_B\) at \(B = 1\,\mathrm{T}\) and at \(B = 10\,\mathrm{T}\).

(b) Compute \(\hbar\omega_c\) at \(B = 10\,\mathrm{T}\) for an electron, and convert to kelvin via \(k_B T\). Comment on whether the QHE should be observable at \(T = 4\,\mathrm{K}\).

(c) For a \(1\,\mathrm{mm}^2\) sample at \(B = 10\,\mathrm{T}\), compute \(N_\Phi\) and compare to typical 2D electron densities \(n_e \sim 10^{11}\,\mathrm{cm}^{-2}\) to estimate the filling factor.

6. Landau levels in graphene. Specialize to electron carriers (\(q = -e\) with \(e > 0\)) throughout this problem. Graphene’s low-energy electrons obey a Dirac-like equation with linear dispersion \(E = \hbar v_F k\), leading to Landau levels \(E_n = \mathrm{sgn}(n)\,v_F\sqrt{2q\hbar\vert n\vert B}\) for \(n = 0, \pm 1, \pm 2, \ldots\).

(a) Show the level spacing is not uniform: \(E_1 - E_0 \neq E_2 - E_1\). For \(B = 10\,\mathrm{T}\) and \(v_F = 10^6\,\mathrm{m/s}\), compute \(E_1\) in meV.

(b) The \(n = 0\) level sits at exactly zero energy. Explain why this leads to a half-integer quantum Hall effect \(\sigma_{xy} = (n + 1/2)\cdot 4e^2/h\) (the factor 4 accounts for spin and valley).

(c) Compare \(E_1\) in graphene to \(\hbar\omega_c\) in GaAs at the same field. Why is the QHE observable at room temperature in graphene but requires millikelvins in GaAs?