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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?

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  • $\begingroup$ 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. $\endgroup$ May 19, 2023 at 0:46

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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 $

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