next up previous contents
Next: Impurities Up: Introducing the quantum Hall Previous: Uniform states   Contents

Integer quantum Hall effect

To better understand the quantum Hall effect, we need to take in to account the fact that the world is quantum. Hence we write down a Hamiltonian for a system of electrons in $ 2+1$ dimensions living in a magnetic field,

$\displaystyle H=\sum_{i} \frac{1}{2m} \big(\vec{p}_{i} - \vec{A}(\vec{x}_{i}) \big)^{2}$ (2.8)

and try to solve Schrödinger's equation. As you probably have noticed, we neglected interactions between electrons. As you will see this approach will be fruitful for a class of quantum Hall effects, namely the integer quantum Hall effects, but we have thrown away to much to understand the Hall effect for fractional filling.

While the particles are non-interacting, the operators $ \vec{p}_{i}$ en $ \vec{x}_{i}$ commute for different $ i$, hence we can assume that the wave function is a product of single particle wave functions. The assumption of non-interacting electrons thus reduces the problem to a one particle problem. We will now drop the particle label, $ i$, from the operators for clarity. The magnetic field is pointed perpendicular to the plane. A gauge field configuration consistent with this condition is, $ (0,-Bx,0)$. Inserting this in the one particle Hamiltonian yields,

$\displaystyle H= \frac{1}{2m} \Big( p_{x}^{2} + \big( p_{y} -Bx \big)^{2} \Big).$ (2.9)

The operator $ p_{y}$ commutes with this Hamiltonian, so we immediately know the wave function is of the form,

$\displaystyle \phi(x,y)=e^{iky}v(x).$ (2.10)

By plugging this ansatz in the Schrödinger equation, we find an equation for $ v(x)$.

$\displaystyle \frac{1}{2m} \Big( p_{x}^{2} + \big( k - Bx \big)^{2} \Big) v(x) = E v(x).$ (2.11)

This equation is simply the Schrödinger equation for a harmonic oscillator with its potential shifted by $ x_{0} \equiv \frac{k}{B}$. We can write our Hamiltonian in terms of creation and annihilation operators, which we define in the usual fashion,

$\displaystyle H= \omega_{c} \big( A^{\dagger}A + \frac{1}{2} \big),$ (2.12)
$\displaystyle A^{(\dagger)}=\sqrt{\frac{1}{2}}(x-x_{0}) \pm i \sqrt{\frac{1}{2}} p_{x},$ (2.13)

and define the ground state as the state annihilated by $ A$. In the equation for the Hamiltonian appears, $ \omega_{c}=\frac{eB}{mc}$, which is the cyclotron frequency. Furthermore we found it convenient to make $ x$ dimensionless by absorbing a factor $ \sqrt{B}$ or in conventional units $ \sqrt{\frac{eB}{c\hbar}}$. This (inverse) length scale is know as the magnetic length. From the requirement that the ground state is annihilated by $ A$, we find a differential equation which we can solve to yield the ground state wave function. Properly normalized it reads,

$\displaystyle \phi_{00}(x)=\left(\frac{B}{\pi}\right)^{\frac{1}{4}} e^{-\frac{1}{2}(x-x_{0})^{2}}e^{iky}.$ (2.14)

Higher states are then created by applying $ \frac{1}{n!} (A^{\dagger})^{n}$ on this state. It is easy to see this operation yields a polynomial ($ P_{n}$) in front of the exponential. If we multiply this new state with $ e^{+\frac{1}{2}(x-x_{0})^2}$ from the left (and drop some constants) we yield a generating equation for this polynomial.

$\displaystyle P_{n}(x)=e^{\frac{1}{2}x^{2}}(A^{\dagger})^{n}e^{-\frac{1}{2}x^{2}}=e^{\frac{1}{2}x^{2}}(x-\partial_{x})^{n}e^{-\frac{1}{2}x^{2}}.$ (2.15)

This equation is precisely the generating equation for the Hermite polynomials. The properly normalized wave functions then read,

$\displaystyle \phi_{mk}(x)=\left(\frac{\sqrt{B}}{2^{m}m!\sqrt{\pi}}\right)^{\frac{1}{2}} H_{m}( x-x_{0}) e^{-\frac{1}{2}(x-x_{0})^{2}}e^{iky}.$ (2.16)

The energy of such states are easily determined,

$\displaystyle H\phi_{mk}= \omega_{c}(m+\frac{1}{2}) \phi_{mk}, \quad m \in \mathbbm{Z}.$ (2.17)

Note that for each state we can still choose $ k$ freely, hence we have degenerate energy levels. If we assume our space is finite, for definiteness we choose a rectangle of dimensions $ L_{x} \times L_{y}$, the number of states in a level is finite and can be calculated. First of all the eigenvalues of $ p_{y}$ are quantized, $ k=\frac{2\pi}{L_{y}}n$. For the center of the wave functions ($ x_{0}$) (2.16) to lay on the disk, we have to impose $ 0<x_{0}<L_{x}$. This constrains $ n$ to the domain $ 0<n<\frac{BL_{x}L_{y}}{2\pi
}$. Hence the maximum number of states in a Landau level is $ \frac{BL_{x}L_{y}}{2\pi}$.

We have now solved the one-particle problem, we know all states and their quantum numbers. Let us now go back to the original many-body problem. We already concluded that the wave function is the product of one-particle wave functions, but one more essential ingredient is needed to write down the total wave function. This ingredient is the fact that electrons are fermions. Fermions obey the Pauli-principle, that means that their wave function has to be anti-symmetric. Now we can write down the total wave function, which is the product of one-particle wave functions, but anti-symmetrized in the particle coordinates.

next up previous contents
Next: Impurities Up: Introducing the quantum Hall Previous: Uniform states   Contents
Tim Dijkstra 2002-05-08