Spectral index

Landau quantization in quantum mechanics is the quantization of the cyclotron orbits of charged particles in magnetic fields. As a result, the charged particles can only occupy orbits with discrete energy values, called Landau levels. The Landau levels are degenerate, with the number of electrons per level directly proportional to the strength of the applied magnetic field. Landau quantization is directly responsible for oscillations in electronic properties of materials as a function of the applied magnetic field. It is named after the Soviet physicist Lev Landau.

Derivation

Consider a two-dimensional system of non-interacting particles with charge $q$ and spin $S$ confined to an area $A=L_{x}L_{y}$ in the x-y plane. Apply a uniform magnetic field $\mathbf {B} ={\begin{pmatrix}0\\0\\B\end{pmatrix}}$ along the z-axis. Using CGS units, the Hamiltonian of this system is

${\hat {H}}={\frac {1}{2m}}({\hat {\mathbf {p} }}-q{\hat {\mathbf {A} }}/c)^{2}.$

Here, ${\hat {\mathbf {p} }}$ is the canonical momentum operator and ${\hat {\mathbf {A} }}$ is the electromagnetic vector potential, which is related to the magnetic field by

$\mathbf {B} =\mathbf {\nabla } \times \mathbf {A} .\,$

There is some freedom in the choice of vector potential for a given magnetic field. However, the Hamiltonian is gauge invariant, which means that adding the gradient of a scalar field to A changes the overall phase of the wave function by an amount corresponding to the scalar field. Physical properties are not influenced by the specific choice of gauge. For simplicity in calculation, choose the Landau gauge, which is

${\hat {\mathbf {A} }}={\begin{pmatrix}0\\Bx\\0\end{pmatrix}}.$

where $B=|\mathbf {B} |$ and ${\hat {x}}$ is the x component of the position operator. In this gauge the Hamiltonian is

${\hat {H}}={\frac {{\hat {p}}_{x}^{2}}{2m}}+{\frac {1}{2m}}\left({\hat {p}}_{y}-{\frac {qB{\hat {x}}}{c}}\right)^{2}.$

The operator ${\hat {p}}_{y}$ commutes with this Hamiltonian since the operator ${\hat {y}}$ is absent due to the choice of gauge. Then the operator ${\hat {p}}_{y}$ can be replaced by its eigenvalue $\hbar k_{y}$. The Hamiltonian can also be written more simply by noting that the cyclotron frequency is $\omega _{c}=qB/mc$, giving

${\hat {H}}={\frac {{\hat {p}}_{x}^{2}}{2m}}+{\frac {1}{2}}m\omega _{c}^{2}\left({\hat {x}}-{\frac {\hbar k_{y}}{m\omega _{c}}}\right)^{2}.$

This is exactly the Hamiltonian for the quantum harmonic oscillator, except shifted in coordinate space by $x_{0}={\frac {\hbar k_{y}}{m\omega _{c}}}$.

To find the energies, note that translating the harmonic oscillator potential left or right does not change the energies. The energies of this system are identical to those of the quantum harmonic oscillator:

$E_{n}=\hbar \omega _{c}\left(n+{\frac {1}{2}}\right),\quad n\geq 0.\,$

The energy does not depend on the quantum number $k_{y}$, so there will be degeneracies.

For the wave functions, recall that ${\hat {p}}_{y}$ commutes with the Hamiltonian. Then the wave function factors into a product of momentum eigenstates in the $y$ direction and harmonic oscillator eigenstates $|\phi _{n}\rangle$ shifted by an amount $x_{0}$ in the $x$ direction:

$\Psi (x,y)=e^{ik_{y}y}\phi _{n}(x-x_{0}).\,$

In sum, the state of the electron is characterized by two quantum numbers, $n$ and $k_{y}$.

Landau levels

Each set of wave functions with the same value of $n$ is called a Landau level. Effects of Landau levels are only observed when the mean thermal energy is smaller than the energy level separation, $kT\ll \hbar \omega _{c}$, meaning low temperatures and strong magnetic fields.

Each Landau level is degenerate due to the second quantum number $k_{y}$. If periodic boundary conditions are assumed, $k_{y}$ can take the values $k_{y}={\frac {2\pi N}{L_{y}}}$, where $N$ is an integer. The allowed values of $N$ are further restricted by the condition that the center of the oscillator $x_{0}$ must physically lie within the system, $0\leq x_{0}<L_{x}$. This gives the following range for $N$:

$0\leq N<{\frac {m\omega _{c}L_{x}L_{y}}{2\pi \hbar }}.$

For particles with charge $q=Ze$, the upper bound on $N$ can be simply written as a ratio of fluxes:

${\frac {ZBL_{x}L_{y}}{(hc/e)}}=Z{\frac {\Phi }{\Phi _{0}}},$

where $\Phi _{0}=hc/e$ is the fundamental quantum of flux and $\Phi =BA$ is the flux through the system (with area $A=L_{x}L_{y}$). Thus for particles with spin $S$, the maximum number $D$ of particles per Landau level is

$D=Z(2S+1){\frac {\Phi }{\Phi _{0}}}.$

The above gives only a rough idea of the effects of finite-size geometry. Strictly speaking, using the standard solution of the harmonic oscillator is only valid for systems unbounded in the x-direction (infinite strips). If the size $L_{x}$ is finite, boundary conditions in that direction give rise to non-standard quantization conditions on the magnetic field, involving (in principle) both solutions to the Hermite equation. The filling of these levels with many electrons is still an active area of research.

The apparent "oscillator center" $x_{0}$ is in this sense spurious, as the system has no reference point on the x-axis. It is, however, an indication of the very real issue of translational symmetry breaking: orbits in a magnetic field are circles, so how to choose their centers? Related issues on a lattice have also been discussed at length.

Generally, Landau levels are observed in electronic systems, where $Z=1$ and $S=1/2$. As the magnetic field is increased, more and more electrons can fit into a given Landau level. The occupation of the highest Landau level ranges from completely full to entirely empty, leading to oscillations in various electronic properties (see De Haas-van Alphen effect and Shubnikov-De Haas effect).

If Zeeman splitting is included, each Landau level splits into a pair, one for spin up electrons and the other for spin down electrons. Then the occupation of each spin Landau level is just the ratio of fluxes $D=\Phi /\Phi _{0}$. Zeeman splitting will have a significant effect on the Landau levels because their energy scales are the same, $2\mu _{B}B=\hbar \omega$. However, the Fermi energy and ground state energy stay roughly the same in a system with many filled levels since pairs of split energy levels cancel each other out when summed.

Discussion

This derivation treats x and y as slightly asymmetric. However, because of the symmetry of the system, there is no physical quantity which differentiates these coordinates. The same result could have been obtained with an appropriate exchange of x and y.

Additionally, the above derivation assumed an electron confined in the z-direction, which is a relevant experimental situation — found in two-dimensional electron gases, for instance. This assumption is not essential for the obtained results. If electrons are free to move along the z direction, the wave function acquires an additional multiplicative term $e^{ik_{z}z}$; the energy corresponding to this free motion, ${\frac {\hbar ^{2}k_{z}^{2}}{2m}}$, has to be added to E. This term fills the separation in energy of the different Landau levels, blurring the effect of the quantization. In any case, the motion in the x-y-plane, perpendicular to the magnetic field, is quantized.

See also

• De Haas–van Alphen effect
• Shubnikov–De Haas effect
• Quantum Hall effect
• Laughlin wavefunction
• Coulomb potential between two current loops embedded in a magnetic field

Further reading

• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534