Quantum Hall effect

In two dimensions, when classical electrons are subjected to a magnetic field they follow circular cyclotron orbits. When the system is treated quantum mechanically, these orbits are quantized. To determine the values of the energy levels the Schrödinger equation must be solved.

Since the system is subjected to a magnetic field, it has to be introduced as an electromagnetic vector potential in the Schrödinger equation. The system considered is an electron gas that is free to move in the x and y directions, but is tightly confined in the z direction. Then, a magnetic field is applied in the z direction and according to the Landau gauge the electromagnetic vector potential is ${\displaystyle \mathbf {A} =(0,Bx,0)}$ and the scalar potential is ${\displaystyle \phi =0}$. Thus the Schrödinger equation for a particle of charge ${\displaystyle q}$ and effective mass ${\displaystyle m^{*}}$ in this system is:

${\displaystyle \left\{{\frac {1}{2m^{*}}}\left[\mathbf {p} -q\mathbf {A} \right]^{2}+V(z)\right\}\psi (x,y,z)=\varepsilon \psi (x,y,z)}$

where ${\displaystyle \mathbf {p} }$ is the canonical momentum, which is replaced by the operator ${\displaystyle -i\hbar \nabla }$ and ${\displaystyle \varepsilon }$ is the total energy.

To solve this equation it is possible to separate it into two equations since the magnetic field just affects the movement along x and y axes. The total energy becomes then, the sum of two contributions ${\displaystyle \varepsilon =\varepsilon _{z}+\varepsilon _{xy}}$. The corresponding equations in z axis is:

${\displaystyle \left[-{\frac {\hbar ^{2}}{2m^{*}}}{\partial ^{2} \over \partial z^{2}}+V(z)\right]u(z)=\varepsilon _{z}u(z)}$

To simplify things, the solution ${\displaystyle V(z)}$ is considered as an infinite well. Thus the solutions for the z direction are the energies ${\textstyle \varepsilon _{z}={\frac {n_{z}^{2}\pi ^{2}\hbar ^{2}}{2m^{*}L^{2}}}}$, ${\displaystyle n_{z}=1,2,3...}$ and the wavefunctions are sinusoidal. For the ${\displaystyle x}$ and ${\displaystyle y}$ directions, the solution of the Schrödinger equation can be chosen to be the product of a plane wave in ${\displaystyle y}$-direction with some unknown function of ${\displaystyle x}$, i.e., ${\displaystyle \psi _{xy}=u(x)e^{ik_{y}y}}$. This is because the vector potential does not depend on ${\displaystyle y}$ and the momentum operator ${\displaystyle {\hat {p}}_{y}}$ therefore commutes with the Hamiltonian. By substituting this Ansatz into the Schrödinger equation one gets the one-dimensional harmonic oscillator equation centered at ${\textstyle x_{k_{y}}={\frac {\hbar k_{y}}{eB}}}$.

${\displaystyle \left[-{\frac {\hbar ^{2}}{2m^{*}}}{\partial ^{2} \over \partial x^{2}}+{\frac {1}{2}}m^{*}\omega _{\rm {c}}^{2}(x-l_{B}^{2}k_{y})^{2}\right]u(x)=\varepsilon _{xy}u(x)}$

where ${\textstyle \omega _{\rm {c}}={\frac {eB}{m^{*}}}}$ is defined as the cyclotron frequency and ${\textstyle l_{B}^{2}={\frac {\hbar }{eB}}}$ the magnetic length. The energies are:

${\displaystyle \varepsilon _{xy}\equiv \varepsilon _{n_{x}}=\hbar \omega _{\rm {c}}\left(n_{x}+{\frac {1}{2}}\right)}$, ${\displaystyle n_{x}=1,2,3...}$

And the wavefunctions for the motion in the ${\displaystyle xy}$ plane are given by the product of a plane wave in ${\displaystyle y}$ and Hermite polynomials attenuated by the gaussian function in ${\displaystyle x}$, which are the wavefunctions of a harmonic oscillator.

From the expression for the Landau levels one notices that the energy depends only on ${\displaystyle n_{x}}$, not on ${\displaystyle k_{y}}$. States with the same ${\displaystyle n_{x}}$ but different ${\displaystyle k_{y}}$ are degenerate.

At zero field, the density of states per unit surface for the two-dimensional electron gas taking into account degeneration due to spin is independent of the energy

${\displaystyle n_{\rm {2D}}={\frac {m^{*}}{\pi \hbar ^{2}}}}$.

As the field is turned on, the density of states collapses from the constant to a Dirac comb, a series of Dirac ${\displaystyle \delta }$ functions, corresponding to the Landau levels separated ${\displaystyle \Delta \varepsilon _{xy}=\hbar \omega _{\rm {c}}}$. At finite temperature, however, the Landau levels acquire a width ${\textstyle \Gamma ={\frac {\hbar }{\tau _{i}}}}$ being ${\displaystyle \tau _{i}}$ the time between scattering events. Commonly it is assumed that the precise shape of Landau levels is a Gaussian or Lorentzian profile.

Another feature is that the wave functions form parallel strips in the ${\displaystyle y}$-direction spaced equally along the ${\displaystyle x}$-axis, along the lines of ${\displaystyle \mathbf {A} }$. Since there is nothing special about any direction in the ${\displaystyle xy}$-plane if the vector potential was differently chosen one should find circular symmetry.

Given a sample of dimensions ${\displaystyle L_{x}\times L_{y}}$ and applying the periodic boundary conditions in the ${\displaystyle y}$-direction ${\textstyle k={\frac {2\pi }{L_{y}}}j}$ being ${\displaystyle j}$ an integer, one gets that each parabolic potential is placed at a value ${\displaystyle x_{k}=l_{B}^{2}k}$.

Parabolic potentials along the ${\displaystyle x}$-axis centered at ${\displaystyle x_{k}}$ with the 1st wave functions corresponding to an infinite well confinement in the ${\displaystyle z}$ direction. In the ${\displaystyle y}$-direction there are travelling plane waves.

The number of states for each Landau Level and ${\displaystyle k}$ can be calculated from the ratio between the total magnetic flux that passes through the sample and the magnetic flux corresponding to a state.

${\displaystyle N_{B}={\frac {\phi }{\phi _{0}}}={\frac {BA}{BL_{y}\Delta x_{k}}}={\frac {A}{2\pi l_{B}^{2}}}{\begin{array}{lcr}&l_{B}&\\&=&\\&&\end{array}}{\frac {AeB}{2\pi \hbar }}{\begin{array}{lcr}&\omega _{\rm {c}}&\\&=&\\&&\end{array}}{\frac {m^{*}\omega _{\rm {c}}A}{2\pi \hbar }}}$

Thus the density of states per unit surface is

${\displaystyle n_{B}={\frac {m^{*}\omega _{\rm {c}}}{2\pi \hbar }}}$.

Note the dependency of the density of states with the magnetic field. The larger the magnetic field is, the more states are in each Landau level. As a consequence, there is more confinement in the system since less energy levels are occupied.

Rewriting the last expression as ${\textstyle n_{B}={\frac {\hbar \omega _{\rm {c}}}{2}}{\frac {m^{*}}{\pi \hbar ^{2}}}}$ it is clear that each Landau level contains as many states as in a 2DEG in a ${\displaystyle \Delta \varepsilon =\hbar \omega _{\rm {c}}}$.

Given the fact that electrons are fermions, for each state available in the Landau levels it corresponds two electrons, one electron with each value for the spin ${\textstyle s=\pm {\frac {1}{2}}}$. However, if a large magnetic field is applied, the energies split into two levels due to the magnetic moment associated with the alignment of the spin with the magnetic field. The difference in the energies is ${\textstyle \Delta E=\pm {\frac {1}{2}}g\mu _{\rm {B}}B}$ being ${\displaystyle g}$ a factor which depends on the material (${\displaystyle g=2}$ for free electrons) and ${\displaystyle \mu _{\rm {B}}}$ the Bohr magneton. The sign ${\displaystyle +}$ is taken when the spin is parallel to the field and ${\displaystyle -}$ when it is antiparallel. This fact called spin splitting implies that the density of states for each level is reduced by a half. Note that ${\displaystyle \Delta E}$ is proportional to the magnetic field so, the larger the magnetic field is, the more relevant is the split.

Density of states in a magnetic field, neglecting spin splitting. (a)The states in each range ${\displaystyle \hbar \omega _{\rm {c}}}$ are squeezed into a ${\displaystyle \delta }$-function Landau level. (b) Landau levels have a non-zero width ${\displaystyle \Gamma }$ in a more realistic picture and overlap if ${\displaystyle \hbar \omega _{\rm {c}}<\Gamma }$. (c) The levels become distinct when ${\displaystyle \hbar \omega _{\rm {c}}>\Gamma }$.

In order to get the number of occupied Landau levels, one defines the so-called filling factor ${\displaystyle \nu }$ as the ratio between the density of states in a 2DEG and the density of states in the Landau levels.

${\displaystyle \nu ={\frac {n_{\rm {2D}}}{n_{B}}}={\frac {hn_{\rm {2D}}}{eB}}}$

In general the filling factor ${\displaystyle \nu }$ is not an integer. It happens to be an integer when there is an exact number of filled Landau levels. Instead, it becomes a non-integer when the top level is not fully occupied. In actual experiments, one varies the magnetic field and fixes electron density (and not the Fermi energy!) or varies the electron density and fixes the magnetic field. Both cases correspond to a continuous variation of the filling factor ${\displaystyle \nu }$ and one cannot expect ${\displaystyle \nu }$ to be an integer. Since ${\displaystyle n_{B}\propto B}$, by increasing the magnetic field, the Landau levels move up in energy and the number of states in each level grow, so fewer electrons occupy the top level until it becomes empty. If the magnetic field keeps increasing, eventually, all electrons will be in the lowest Landau level (${\displaystyle \nu <1}$) and this is called the magnetic quantum limit.

Occupation of Landau levels in a magnetic field neglecting the spin splitting, showing how the Fermi level moves to maintain a constant density of electrons. The fields are in the ratio ${\displaystyle 2:3:4}$ and give ${\displaystyle \nu =4,{\frac {8}{3}}}$ and ${\displaystyle 2}$.

It is possible to relate the filling factor to the resistivity and hence, to the conductivity of the system. When ${\displaystyle \nu }$ is an integer, the Fermi energy lies in between Landau levels where there are no states available for carriers, so the conductivity becomes zero (it is considered that the magnetic field is big enough so that there is no overlap between Landau levels, otherwise there would be few electrons and the conductivity would be approximately ${\displaystyle 0}$). Consequently, the resistivity becomes zero too (At very high magnetic fields it is proven that longitudinal conductivity and resistivity are proportional).[21]

With the conductivity ${\displaystyle \sigma =\rho ^{-1}}$ one finds

${\displaystyle \sigma ={\frac {1}{\det \rho }}{\begin{pmatrix}\rho _{yy}&-\rho _{xy}\\-\rho _{yx}&\rho _{xx}\end{pmatrix}}\;.}$

If the longitudinal resistivity is zero and transversal is finite, then ${\displaystyle \det \rho \neq 0}$. Thus both, the longitudinal conductivity and resistivity become zero.

Instead, when ${\displaystyle \nu }$ is a half-integer, the Fermi energy is located at the peak of the density distribution of some Landau Level. This means that the conductivity will have a maximum .

This distribution of minimums and maximums corresponds to ¨quantum oscillations¨ called Shubnikov–de Haas oscillations which become more relevant as the magnetic field increases. Obviously, the height of the peaks are larger as the magnetic field increases since the density of states increases with the field, so there are more carriers which contribute to the resistivity. It is interesting to notice that if the magnetic field is very small, the longitudinal resistivity is a constant which means that the classical result is reached.

Longitudinal and transverse (Hall) resistivity, ${\displaystyle \rho _{xx}}$ and ${\displaystyle \rho _{xy}}$, of a two-dimensional electron gas as a function of magnetic field. Both vertical axes were divided by the quantum unit of conductance ${\displaystyle e^{2}/h}$ (units are misleading). The filling factor ${\displaystyle \nu }$ is displayed for the last 4 plateaus.

From the classical relation of the transverse resistivity ${\textstyle \rho _{xy}={\frac {B}{en_{\rm {2D}}}}}$ and substituting ${\textstyle n_{\rm {2D}}=\nu {\frac {eB}{h}}}$ one finds out the quantization of the transverse resistivity and conductivity:

${\displaystyle \rho _{xy}={\frac {h}{\nu e^{2}}}\Rightarrow \sigma =\nu {\frac {e^{2}}{h}}}$

One concludes then, that the transverse resistivity is a multiple of the inverse of the so-called conductance quantum ${\displaystyle e^{2}/h}$ if the filling factor is an integer. In experiments, however, plateaus are observed for whole plateaus of filling values ${\displaystyle \nu }$, which indicates that there are in fact electron states between the Landau levels. These states are localized in, for example, impurities of the material where they are trapped in orbits so they can not contribute to the conductivity. That is why the resistivity remains constant in between Landau levels. Again if the magnetic field decreases, one gets the classical result in which the resistivity is proportional to the magnetic field.