Quantum Hall Effect

  • A quantum state in two dimensions

  • 2D electron gas in strong perpendicular magnetic fields.

  • Can exist for bosons too!

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Landau Levels

  • Single particle Hamiltonian with vector potential

$$ H = -\frac{1}{2m}\left(\nabla -i q \mathbf{A}\right)^2, \label{Landau} $$

  • For perpendicular field

$$ \partial_x A_y - \partial_y A_x = B. $$

  • We choose symmetric gauge

$$ A_x = -\frac{1}{2} B y,\quad A_y = \frac{1}{2} B x. $$

$H$ in complex coordinates

$$ \begin{align} z = x + i y \quad \bar z = x - iy,\\ \partial_z = \frac{1}{2}\left(\partial_x - i\partial_y\right) \quad \partial_{\bar z} = \frac{1}{2}\left(\partial_x + i\partial_y\right).\\ H = -\frac{2}{m}\left(\partial_z -\frac{qB \bar z}{4}\right)\left(\partial_{\bar z} +\frac{qB z}{4}\right) + \frac{\omega_c}{2} \end{align} $$

  • $\omega_c = \frac{qB}{m}$ is the cyclotron frequency

  • States that satisfy $$ \left(\partial_{\bar z} +\frac{qB z}{4}\right)\psi(z,\bar z) = 0 $$ have energy $\omega_c/2$ and belong to the lowest Landau level (LLL).

Show that these states have the form

$$ \psi(z,\bar z) = f(z) \exp\left(-\frac{qB}{4}\left|z\right|^2\right), $$

where $f(z)$ is an arbitrary analytic function.

  • $f(z)=a_0 + a_1z + a_2 z^2+\ldots$ for arbitrary $a_n$

  • Landau levels highly degenerate

Notation

  • Write LLL wavefunction as $f(z)$, leaving Gaussian in inner product

$$ \bra{f_1}f_2\rangle = \int \frac{d^2z}{2\pi} \overline{f_1(z)}f_2(z) \exp\left(-\left|z\right|^2/2\right), $$

  • Units with magnetic length $\ell \equiv (qB)^{-1/2}=1$

  • A possible orthonormal basis is $f_n(z) = \frac{z^n}{\sqrt{2^n n!}}$

Show this

Filled LLL of Fermions

  • Lift degeneracy by adding harmonic potential $$ V_\text{harm}(x,y) = \frac{v}{2}\left(x^2+y^2\right) = \frac{v}{2}\left|z\right|^2. \label{many_HarmonicRound} $$

  • Problem: $V_\text{harm}(z,\bar z)f(z)$ is not in LLL

  • Consider only the action of $V$ in the LLL subspace.

By considering matrix elements $\bra{f_1}V\ket{f_2}$ between LLL states, show (by integrating by parts) that it is possible to replace any occurrence of $\bar z$ in $V$ with $2\partial_z$ acting on the analytic part of the wavefunction.

Note that the order is important: all the $\partial_z$ must stand to the left of the $z$, Thus $$ V_\text{harm}\longrightarrow v\partial_z z = v\left(1+z \partial_z\right). \label{many_HarmonicProject} $$

  • Applied to $f_n(z)$, $V$ just counts the degree: $V_\text{harm} f_n = v(1+n)f_n$.

  • Fill states $n=0,1,\ldots N-1$

  • As for 1D fermions, single particle states are powers $z^n$.

  • Identical arguments imply

$$ \Psi(z_1,\ldots, z_N) = \prod_{j<k}^N (z_j-z_k) \exp\left(-\frac{1}{4}\sum_{j=1}^N\left|z_j\right|^2\right) \label{many_nu1} $$

Show that the density is $$ \rho_1(z,\bar z) = \frac{e^{-|z|^2/2}}{2\pi}\sum_{n=0}^{N-1} \frac{\left|z\right|^{2n}}{2^n n!} = \frac{1}{2\pi} \frac{\Gamma(N,|z|^2/2)}{(N-1)!}. \label{many_LLLdensity} $$ $\Gamma(s,x) = \int^\infty_x t^{s-1}e^{-t}dt$ is incomplete gamma fn.

drawing

  • At small $\left|z\right|$, approximate by infinite sum, and $\rho_1\to \frac{1}{2\pi}$

  • The density is fixed at this value until we reach $\sim\sqrt{2N}$, then falls to zero on scale of $\ell_B$.

  • The filled LLL is described by a circular droplet of fixed density $\rho_1 = 1/(2\pi)$: one state per flux quantum $\Phi_0=2\pi/q$

  • Changing confining potential causes the droplet to deform, but density to remain constant

The Laughlin Wavefunction

  • In 1983 Robert Laughlin proposed the generalization $$ \Psi_m(z_1,\ldots, z_N) = \prod_{j<k}^N (z_j-z_k)^{m} \exp\left(-\frac{1}{4}\sum_{j=1}^N\left|z_j\right|^2\right). \label{many_nu} $$

  • $m$ odd / even for fermions / bosons. $m\neq 1$ not a product state!

This Letter presents variational ground-state and excited-state wave functions which describe the condensation of a two-dimensional electron gas into a new state of matter.

$m=2$ boson state

  • With the repulsive potential $$ H_{\text{int}} = g\sum_{j<k}\delta(\br_j-\br_k),\qquad g>0 \label{many_delta} $$ $m=2$ state has zero interaction energy

  • Any state with zero interaction energy must have $\Psi_2(z_1,\ldots, z_N)$ as a factor

  • If wavefunction has higher degree, then in the presence of potential $V_\text{harm}$ it will have higher energy, so $\Psi_2$ is ground state

Particle distribution

drawing

  • Monte Carlo samples from uncorrelated (uniform) distrubution vs. the Laughlin probability distribution $\lvert\Psi_3(z_1,\ldots z_N)\rvert^2$.

  • Note uniformity of particle distribution, in contrast with the sample of uncorrelated particles on the left

The Plasma Analogy

  • Probability distribution $\lvert\Psi_m(z_1,\ldots, z_N)\rvert^2$ of particles interpreted as Boltzmann distribution of particles in a classical 2D plasma

$$ \nabla^2 V = -q\delta(\br). $$

  • In 2D the solution describing a point charge is $$ V_\text{point}(\br) = -\frac{q}{2\pi}\log\left|\br\right|, $$
  • Constant background charge density $-\rho_0$ gives rise to a potential $$ V_\text{bg}(\br) = \frac{\rho_0}{4} \left|\br\right|^2. $$

  • System of identical charges in a background charge has overall electrostatic energy $$ V(\br_1,\ldots,\br_N) = -\frac{q^2}{2\pi} \sum_{j<k}\log\left|\br_j-\br_k\right| + \frac{q\rho_0}{4}\sum_j \left|\br_j\right|^2. $$

  • At finite temperature (unnormalized) probability of finding particles at locations $\br_1, \ldots, \br_N$ is $$ \exp[-\beta V(\brN)] = \left|\Psi_m(\brN)\right|^2, $$ where we set $\beta q^2/(2\pi) = 2m$ and $\beta q\rho_0 = 2$.

  • These are not physical electrostatic fields!

$N\to\infty$ limit

  • Introduce a continuum charge density $\rho(\br)$ and write the energy as a functional $$ \begin{align} \beta V[\rho] = -m\int d^2\br\, d^2\br'\rho(\br)\log\left|\br-\br'\right|\rho(\br') + \frac{1}{2}\int d^2\br\left|\br\right|^2\rho(\br). \end{align} $$

Show that minimizing the energy with respect to $\rho(\br)$ – corresponding to finding the most likely configuration – leads to the condition $$ -2m\int d^2\br’\log\left|\br-\br’\right|\rho(\br’) + \frac{1}{2}\left|\br\right|^2 = 0. $$

Show that applying $\nabla^2$ to both sides gives $$ \rho(\br) = \frac{1}{2\pi m}. $$

  • Conclusion: density is fixed at $1/m$ of value for $m=1$ case

  • Result applies where density is non-zero, so we get a uniform droplet of radius $\sqrt{2Nm}$

  • $1/m$ is called the filling fraction of the state.

Excited states: fractional charge

  • Simplest excited state is quasihole $$ \begin{align} \Psi_\text{hole}(z_1,\ldots, z_N|Z) &= \left(\prod_j (Z-z_j)\right)\Psi_m(z_1,\ldots, z_N)\\ V(\brN)&=-\frac{q^2}{2\pi m}\sum_j \log\left|\br_j-\mathbf{R}\right|-\frac{q^2}{2\pi} \sum_{j<k}\log\left|\br_j-\br_k\right|\\ &+ \frac{\rho q_0}{4}\sum_j \left|\br_j\right|^2. \end{align} $$

  • Charge $q/m$ at point $\mathbf{R} = (X, Y)$, where $Z=X+iY$

  • Plasma will screen this charge, leaving a ‘hole’ in density distribution of charge $-q/m$

Hole state

  • 20 holes with charge $-20/m$ for clarity

Fractional Statistics

  • Two quasihole wavefunction

$$ \Psi_\text{2 hole}(z_1,\ldots, z_N|Z_1,Z_2) = \left(\prod_j (Z_1-z_j)(Z_2-z_j)\right)\Psi_m(z_1,\ldots, z_N). $$

  • $\vert\Psi_\text{2 hole}(z_1,\ldots, z_N\vert Z_1,Z_2)\rvert^2$ is plasma with two $1/m$ charges at positions $\mathbf{R}_{1,2}$

  • Each surrounded by depleted density of $-1/m$ of a particle

  • (See notes) Quasiholes are anyons, particles with fractional statistics!