# Why is the number of allowed $\vec k$ states in the first Brillouin zone equal to the number of unit cells?

From my lecturer's notes I have that

The density of states $$g_{\bf k}$$ in reciprocal space for travelling waves in three dimensions is uniform in reciprocal space and is equal to $$g_{\bf k} = \frac{L_x}{2\pi}\times\frac{L_y}{2\pi}\times\frac{L_z}{2\pi}=\frac{V}{(2\pi)^3}$$ where $$V$$ is the total volume of the lattice in real space.

$$N_{\text{states}}=g_{\bf k}\int_{\text{BZ}}d^3{\bf k}=\frac{V}{(2\pi)^3}\times\frac{(2\pi)^3}{a^3}=N_{\text{unit cells}}=N_{\text{atoms}}\tag{1}$$ where $$a$$ is the interatomic lattice spacing and $$\int_{\text{BZ}}$$ denotes integration over the first Brillouin zone. The last equality holds as I am only considering one atom in each cell.

To put this question into context consider part of the derivation for the internal energy $$U$$ of a phonon band in the high-temperature limit:

Bose-Einstein statistics tells us that the number of phonons with energy $$\epsilon_{\bf k}$$ at temperature $$T$$ is given by $$n_B(\epsilon_{\bf k},T)=\frac{1}{e^{\beta \epsilon_{\bf k}}-1}$$ where $$\beta=\frac{1}{k_B T}$$

So $$U\approx\sum_{{\bf k} \, \in \, \text{BZ}}n_B(\epsilon_{\bf k},T) \epsilon_{\bf k}= \sum_{{\bf k} \, \in \, \text{BZ}}\frac{\epsilon_{\bf k}}{e^{\beta \epsilon_{\bf k}}-1}\longrightarrow \frac{V}{(2\pi)^3}\int_{\text{BZ}}\frac{1}{e^{\beta \epsilon_{\bf k}}-1}d^3{\bf k}\tag{2}$$

In the high-temperature limit $$k_B T \gg \epsilon_{\bf k}\implies\exp(\beta \epsilon_{\bf k})\approx 1+\beta \epsilon_{\bf k}\implies n_B(\epsilon_{\bf k},T)\approx\frac{k_B T}{\epsilon_{\bf k}}$$ Using the left hand side of $$(2)$$, $$U\approx\sum_{{\bf k} \, \in \, \text{BZ}}n_B(\epsilon_{\bf k},T) \epsilon_{\bf k}=k_B T\sum_{{\bf k} \, \in \, \text{BZ}}1=k_B T \times N_{\text{atoms}}$$ by virtue of $$(1)$$

So does this really mean that $$\sum_{{\bf k} \, \in \, \text{BZ}}1=N_{\text{atoms}}?$$

So let us look at some very simple real-space lattices and their corresponding reciprocal lattices to see if this is really true:

Well from the above images I can see that there is exactly one and only one reciprocal lattice vector in the 1st BZ.

Now here is the problem, $$\sum_{{\bf k} \, \in \, \text{BZ}}1\stackrel{?}{=}N_{\text{atoms}}$$ cannot be true as it is clearly not equal to the number of unit cells/states/atoms.

In fact the only way for that equation to hold is if it reads $$\sum_{{\bf k} \, \in \, \text{BZ}}1=1$$ then this way the number of atoms is guaranteed to be equal to the number of reciprocal lattice vectors in the 1st BZ (even if it is forever unity).

Clearly, I'm missing the point so if someone could kindly explain how $$\sum_{{\bf k} \, \in \, \text{BZ}}1\stackrel{?}{=}N_{\text{atoms}}$$ can be true it would be most appreciated.

Thank you.

• Please, add the source of the pictures. – Massimo Ortolano May 7 at 8:55

$$k$$ states in the Brillioun zone are not reciprocal lattice vectors. Reciprocal lattice vectors set the periodicity of k-space, i.e. the size of the Brillioun zine. They take you from one Brillioun zone to another. The (symmetry distinct) $$k$$ vectors defining states are points within the Brillioun zone.