# Calculating the nr. of phonon modes depending on atom nr. in a crystal and dimensions

I am trying to understand how we can determine the correct nr. of phonon modes of a crystal lattice, because I am trying to show that the volume of a state in reciprocal space is $$(\frac{2\pi}{L})^3$$. In order to do that, I need to accurately calculate the nr. of phonon modes within the first Brillouin zone. For simplicity both , the direct lattice and the reciprocal one are simple cubic.

Now when we consider a 1D lattice of N atoms, we say that we have N oscillating modes/phonon modes.

Now, if we consider a 2D lattice of N atoms, because of the fact that each atom has 2 degrees of freedom, we have 2N modes. But the problem is the following. In each dimension we have $$N^{\frac 1 2}$$ atoms. Going by the logic in the 1D case, we will have $$N^{\frac 1 2}$$ modes in each dimension, which means we must have $$N^{\frac 1 2} * N^{\frac 1 2}=N$$ modes in total. As you can see there is a problem. Initially I get 2N modes, and here I get N modes. And by not being able to calculate the correct nr. of modes within the first Brillouin zone, I cannot find the volume (in the general 3D case) of a state/mode.

Can someone help me understand how dimensionality and nr. of Atoms play a role in deciding the nr. of phonon modes?

• Back when you did the 1D case, there was only one way for the vibration, namely longitudinal. In 2D, each 1D strand can also vibrate transversely. Commented May 19, 2023 at 0:46

A crystal with $$N$$ unit cells supports exactly $$N$$ phonon modes. This is true in any dimension.

To solve Newton's second law for the motion of the atoms, one invokes a plane-wave ansatz

$$\sum_{\vec{\ell}}\exp[i(\vec{q}\cdot\vec{\ell}-\omega t)]$$

where $$\vec{\ell}$$ labels a unit cell and $$\vec{q}$$ is the wavevector of the vibration. One then demands the ansatz obey periodic boundary conditions:

$$\exp[i(\vec{q}\cdot\vec{\ell}-\omega t)] = \exp[i(\vec{q}\cdot(\vec{\ell}+N_i\vec{a}_i)-\omega t)]$$

where $$\vec{a}_i$$ is a primitive vector and $$N_i$$ is the number of unit cells in the direction $$i$$. It follows from the above that

$$\vec{q}\cdot N_i\vec{a}_i=2\pi n_i$$

where $$n_i$$ is an integer. Expanding $$\vec{q}$$ in the reciprocal basis:

$$\vec{q}=\sum_j q_j\vec{b}_j$$

it follows that

$$\left(\sum_j q_j\vec{b}_j\right)\cdot N_i\vec{a}_i = N_i\sum_jq_j\left(\vec{a}_i\cdot\vec{b}_j\right) = N_i\sum_jq_j\left(2\pi\delta_{ij}\right) = 2\pi q_iN_i$$

and thus

$$2\pi q_iN_i = 2\pi n_i \implies q_i=\frac{n_i}{N_i}$$

Each integer $$n_i$$ is restricted to a range $$n_i\in[0,N_i-1]$$. To see this, note that if $$n_i=N_i$$, then $$\vec{q}$$ reduces to a sum of reciprocal space vectors, which does not change the plane-wave ansatz. Similarly, if we decide to keep $$n_i$$, there is no new information contained in $$-n_i$$, since we can change the summation index in the ansatz from $$\vec{\ell}\rightarrow-\vec{\ell}$$. Thus the number of unique wavevectors (alternatively, phonon modes) is

$$\prod_i\mbox{# values }n_i\mbox{ can take} =\prod_iN_i =N$$