Consider a system of non-interacting particles with charge and spin confined to an area in the plane. Apply a uniform magnetic field \mathbf{B} = \begin{pmatrix}0\\0\\B\end{pmatrix} along the -axis. In
SI units, the
Hamiltonian of this system (here, the effects of spin are neglected) is \hat{H} = \frac{1}{2m} \left(\hat{\mathbf{p}} - q\hat{\mathbf{A}}\right)^2. Here, \hat{\mathbf{p}} is the
canonical momentum operator and \hat{\mathbf{A}} is the
operator for the
electromagnetic vector potential \mathbf{A} (in
position space \hat{\mathbf{A}} =\mathbf{A}). The vector potential is related to the
magnetic field by \mathbf{B}=\mathbf{\nabla}\times \mathbf{A}. There is some gauge freedom in the choice of vector potential for a given magnetic field. The Hamiltonian is
gauge invariant, which means that adding the gradient of a
scalar field to changes the overall phase of the
wave function by an amount corresponding to the scalar field. But physical properties are not influenced by the specific choice of gauge.
In the Landau gauge From the possible solutions for
A, a
gauge fixing introduced by Lev Landau is often used for charged particles in a constant magnetic field. When \mathbf{B} = \begin{pmatrix} 0 \\ 0 \\ B \end{pmatrix} then \mathbf{A} = \begin{pmatrix} 0 \\ B\cdot x \\ 0 \end{pmatrix} is a possible solution in the Landau gauge (not to be mixed up with the
Landau R_\xi gauge). In this gauge, the Hamiltonian is \hat{H} = \frac{\hat{p}_x^2}{2m} + \frac{1}{2m} \left(\hat{p}_y - qB\hat{x}\right)^2 + \frac{\hat{p}_z^2}{2m}. The operator \hat{p}_y commutes with this Hamiltonian, since the operator \hat y is absent for this choice of gauge. Thus the operator \hat{p}_y can be replaced by its eigenvalue \hbar k_y. Since \hat{z} does not appear in the Hamiltonian and only the z-momentum appears in the
kinetic energy, this motion along the z-direction is a free motion. The Hamiltonian can also be written more simply by noting that the
cyclotron frequency is \omega_c=qB/m, giving \hat{H} = \frac{\hat{p}_x^2}{2m} + \frac{1}{2} m \omega_{\rm c}^2 \left( \hat{x} - \frac{\hbar k_y}{m \omega_{\rm c}} \right)^2 + \frac{\hat{p}_z^2}{2m}. This is exactly the Hamiltonian for the
quantum harmonic oscillator, except with the minimum of the potential shifted in coordinate space by x_0=\hbar k_y/m\omega_c. To find the energies, note that translating the harmonic oscillator potential does not affect the energies. The energies of this system are thus identical to those of the standard
quantum harmonic oscillator, E_n=\hbar\omega_{\rm c}\left(n+\frac{1}{2}\right) + \frac{p_z^2}{2m},\quad n\geq 0. The energy does not depend on the quantum number k_y, so there will be a finite number of degeneracies (If the particle is placed in an unconfined space, this degeneracy will correspond to a continuous sequence of p_y). The value of p_z is continuous if the particle is unconfined in the z-direction and discrete if the particle is bounded in the z-direction also. Each set of wave functions with the same value of n is called a
Landau level. 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,z) = e^{i(k_y y+k_z z)} \phi_n(x-x_0) where k_z = p_z / \hbar and \phi_n(x-x_0) is the
n-th state for the quantum harmonic oscilator. In sum, the state of the electron is characterized by the
quantum numbers, n, k_y and k_z.
In the symmetric gauge The derivation treated and
y as asymmetric. However, by the symmetry of the system, there is no
physical quantity which distinguishes these coordinates. The same result could have been obtained with an appropriate interchange of and . A more adequate choice of gauge, is the symmetric gauge, which refers to the choice \hat{\mathbf{A}} =\frac{1}{2} \mathbf{B}\times \hat{\mathbf{r}} = \frac{1}{2}\begin{pmatrix} -By\\ Bx \\0 \end{pmatrix}. In terms of dimensionless lengths and energies, the Hamiltonian can be expressed as \hat{H} = \frac{1}{2} \left[\left(-i\frac{\partial}{\partial x} + \frac{y}{2}\right)^2 + \left(-i \frac{\partial}{\partial y} - \frac{x}{2}\right)^2 \right] The correct units can be restored by introducing factors of q, \hbar, \mathbf{B} and m . Consider operators \begin{align} \hat{a} &= \frac{1}{\sqrt{2}} \left[\left(\frac{x}{2} + \frac{\partial}{\partial x}\right) -i \left(\frac{y}{2} + \frac{\partial}{\partial y}\right)\right] \\ \hat{a}^{\dagger} &= \frac{1}{\sqrt{2}} \left[\left(\frac{x}{2} - \frac{\partial}{\partial x}\right) +i \left(\frac{y}{2} - \frac{\partial}{\partial y}\right)\right] \\ \hat{b} &= \frac{1}{\sqrt{2}} \left[\left(\frac{x}{2} + \frac{\partial}{\partial x}\right) +i \left(\frac{y}{2} + \frac{\partial}{\partial y}\right)\right] \\ \hat{b}^{\dagger} &= \frac{1}{\sqrt{2}} \left[\left(\frac{x}{2} - \frac{\partial}{\partial x}\right) -i \left(\frac{y}{2} - \frac{\partial}{\partial y}\right)\right] \end{align} These operators follow certain commutation relations [\hat{a}, \hat{a}^{\dagger}] = [\hat{b},\hat{b}^{\dagger}] = 1. In terms of above operators the Hamiltonian can be written as \hat{H} = \hbar\omega_{\rm c}\left(\hat{a}^{\dagger}\hat{a} + \frac{1}{2}\right), where we reintroduced the units back. The Landau level index n is the
eigenvalue of the operator \hat{N}=\hat{a}^{\dagger}\hat{a}. The application of \hat{b}^{\dagger} increases m_z by one unit while preserving n, whereas \hat{a}^{\dagger} application simultaneously increase n and decreases m_z by one unit. The analogy to
quantum harmonic oscillator provides solutions \hat{H} |n,m_z\rangle = E_n |n,m_z\rangle, where E_n = \hbar\omega_{\rm c}\left(n + \frac{1}{2}\right) and |n,m_z\rangle = \frac{(\hat{b}^{\dagger})^{m_z+n}}{\sqrt{(m_z+n)!}} \frac{(\hat{a}^{\dagger})^{n}}{\sqrt{n!}}|0,0\rangle. One may verify that the above states correspond to choosing wavefunctions proportional to \psi_{n,m_z}(x, y) = \left( \frac{\partial}{\partial w} - \frac{\bar{w}}{4} \right)^n w^{n + m_z} e^{-|w|^2 / 4} where w = x - i y. In particular, the lowest Landau level n = 0 consists of arbitrary analytic functions multiplying a Gaussian, \psi(x,y) = f(w) e^{-|w|^2/4}. ==Degeneracy of the Landau levels==