# Number of states in a given Landau level

For an electron in a uniform magnetic field, in free space, we seek to find the number of allowed states in a given rectangle $L_x L_y$ (for some fixed Landau level). In effect we are tiling 2-D space with these rectangles and finding out how many states each rectangle contains.

The number of allowed states is quantized, or made finite, by two conditions. First, the periodic boundary condition $e^{i k_y y/ \hbar} = 1$ on the y-dimension (note the gauge is $\vec{A}=$<$0$ $Bx$ $0$>, $B$ being the magnitude of the magnetic field pointing in the $+z$ direction thus y is the "not harmonic oscillator" dimension; and x is the dimension in which harmonic oscillation occurs) ensures that the y-dimension boundaries of the rectangles seam up nicely with each other.

Then there is the fact that the values of $k_y$ are also limited to those that do not, since the value of $k_y$ affects the center position of the harmonic oscillation that occurs in the $x$ dimension, limited to those that do not push the center of the harmonic oscillator out of the width $L_x$ of the rectangle.

If you work all this out, you get that the allowable $k_y$s are given by

$k_y = \dfrac{2 \pi N}{L_y}\hbar$

($N$ is an integer)

(We're now at the section "Landau levels" in https://en.wikipedia.org/wiki/Landau_quantization)

and

$0 \le N < \dfrac{m w_c L_x L_y}{2 \pi \hbar}$

This upper bound on $N$ they say equals:

(1) $\dfrac{B L_x L_y}{(hc/e)} = \dfrac{\Phi}{\Phi_0}$

where $\Phi_0 = h/2e$, the fundamental quantum of flux and $\Phi = B \cdot (Area) = B L_x L_y$, the flux through the rectangle $L_x L_y$.

The two sides of (1) are not the same, right?

## 1 Answer

I guess the $2$ depends on taking into account spin or not. Spin can be up or down, so just multiply the total number of allowed states by $2$.

In http://www.phys.ufl.edu/~pjh/teaching/phy4605/notes/landau.pdf on page 8 they conclude the allowed number of states for a given fixed Landau level, is $\dfrac{eB}{h}$. They do not mention spin.

So when Wikipedia says $\Phi_0 = \dfrac{h}{2e}$ is the fundamental quantum of flux, presumably they are taking into account spin (other places say $\dfrac{h}{e}$ is the flux quantum).

If you take into account spin, the final result is not $\dfrac{eB}{h}$ but $\dfrac{2eB}{h}$. For example

http://www2.physics.ox.ac.uk/sites/default/files/BandMT_12.pdf

section 12.2: "the factor $2$ takes into account spin .. the number of states per unit area per Landau level = $\dfrac{2eB}{h}$"