# Why this is the density of points in $k$-space?

I'm reading a solid state physics book and there's something which is confusing me, related to the free electron gas.

After solving Schrodinger's equation with $V = 0$ and with periodic boundary conditions, one finds out that the allowed values of the components of $\mathbf{k}$ are:

$$k_x = \dfrac{2n_x\pi}{L}, \quad k_y=\dfrac{2n_y \pi}{L}, \quad k_z = \dfrac{2n_z\pi}{L}.$$

In the book I'm reading the author says that it follows from this that: there is one allowed wavevector - that is, one distinct triplet of quantum numbers $k_x,k_y,k_z$ - for the volume element $(2\pi/L)^3$ of $\mathbf{k}$ space.

After that he says that this implies that in the sphere of radius $k_F$ the total number of states is

$$2 \dfrac{4\pi k_F^3/3}{(2\pi/L)^3}=\dfrac{V}{3\pi^2}k_F^3 = N,$$

where the factor $2$ comes from spin.

Now, why is that the case? Why it follows from the possible values of $k_x,k_y,k_z$ that density of points in $k$-space? I really can't understand this properly.

Consider $k_x$, $k_y$, and $k_z$ defining the three orthogonal axes of a three dimensional space. This is what he calls $\boldsymbol{k}$ space.

The allowed values of $\boldsymbol{k}$, that is $k_i = 2 n_i \pi / L$, are represented by evenly spaced points in this $\boldsymbol{k}$ space. Each allowed point is separated from its closest neighbours by a distance $2\pi/L$ along each axis. To see this, just compute the separation between points with consecutive integers $n_i$ along each axis:

$$\frac{2(n+1)\pi}{L} - \frac{2n\pi}{L} = \frac{2\pi}{L}$$

Therefore there is a $\boldsymbol{k}$ space volume of $(2\pi/L)^3$ for each allowed point.

You could imagine each allowed point being the centre of a cube with side length $2\pi/L$. All these cubes would exactly fill the space.

Then he postulates that allowed states are occupied for $|\boldsymbol{k}| \leq k_F$. The $\boldsymbol{k}$ space volume of all such states is just the volume of a sphere with radius $k_F$ (assuming that $k_F \gg 2\pi/L$), that is, $4\pi k_F^3/3$. The total number of states is the total volume divided by the volume of each state, multiplied by $2$ for spin, which is your final formula.

We have the allowed discrete values of electron wave vector $\mathbf{k}=(2\pi/L)(\mathbf{e}_xn_x+\mathbf{e}_yn_y+\mathbf{e}_yn_y)$ in the box, where $n_x,n_y,n_z=0,\pm1,\pm2,\ldots$

Then we place noninteracting electrons in this box at zero temperature. The first two electrons will occupy the lowest energy state $\mathbf{k}=0$ with opposite spin proejctions. The third, fourth and subsequent electrons will occupy available states with smallest $|\mathbf{k}|$ in order to minimize the total energy of the system $$E=\sum_{\mathbf{k}}\frac{\hbar^2\mathbf{k}^2}{2m}.$$ Finally, large number $N$ of electrons will occupy $N/2$ points in $\mathbf{k}$-space, two electrons with opposite spins at each momentum. These points are located on cubic "lattice" (or mesh) of the period $2\pi/L$ and situated as close as possible to the origin in order to minimize the energy. If $N\gg1$ and we can neglect the discreteness of the mesh, these points reside approximately inside a sphere of some radius $k_\mathrm{F}$, (see the picture, source).

So, if you want to find how many $\mathbf{k}$-points fit inside the sphere, you need to divide its volume $(4/3)\pi k_\mathrm{F}^3$ in $\mathbf{k}$-space on the elementary volume $(2\pi/L)^3$ shared by each point: $$N/2=\frac{(4/3)\pi k_\mathrm{F}^3}{(2\pi/L)^3}.$$

• Thanks for the answer. I understand the point, just the one thing that I'm still not getting, and is probably the simpler of them all. Why is "$(2\pi/L)^3$ the elementary volume shared by each point"? I really couldn't get this yet. Thanks for the help!
– Gold
Nov 23 '16 at 1:35
• $\mathbf{k}$-space can be divided into cubes of the size $2\pi/L$ is such a way that there is exactly one $\mathbf{k}$-point in each cube. For example, see this picture for two-dimensional case: wiki.fysik.dtu.dk/ase/_images/periodic-images-1.svg Distance between neighboring points is $2\pi/L$, and each point is inside its own square of the same size $2\pi/L$. Nov 23 '16 at 9:17
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