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:

Square lattice

Triangular lattice

Reciprocal space

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.

  • $\begingroup$ Please, add the source of the pictures. $\endgroup$ – 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.

Now working in real or reciprocal space is just a choice of basis, but the total number of states is just the dimension of your Hilbert space and so is independent of the choice of basis, so the number of states in k-space must be the same as the number in real space. In more physical terms, we can imagine putting one particle in each state and counting them. Since particle number is a physical observable it must be basis independent, but if all states are occupied then clearly we have just counted the number of states, so the number of states must also be basis independent.


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