We start by considering a 1D chain of atoms. We assume that the atoms interact with their neighbours via a harmonic potential (as in lecture 6 - Bonds and spectra), which implies that they are connected by simple linear springs.
If atom $n$ has displacement $u_n$, then Newton's law gives us
@@ -50,30 +52,30 @@ and if we have a system of size $L = Na$ with periodic boundary conditions, we a
A similar problem: we consider an LCAO wave function $\psi = \sum_n c_n \phi_n$. We need to solve the Schrödinger equation
Formulating the equation of motion for electrons is similar: we consider an LCAO wave function $|\psi \rangle = \sum_n c_n |\phi_n\rangle$ and assume a nearest neighbour hopping $t$ and on-site energy $E_0$. We thus need to solve the Schrödinger equation
$$
E c_n = E_0 c_n + t c_{n+1} + t c_{n-1}
E c_n = E_0 c_n + t c_{n+1} + t c_{n-1}.
$$
The periodic boundary conditions mean $c_N = c_0$.
The periodic boundary conditions imply $c_N = c_0$.
### Key idea
### Key idea for solving these equations
Our task is to find all normal modes and to take $L → ∞$.
All atoms are similar $⇒$ the solution should be similar for all atoms $⇒$
For phonons:
$$u_n = e^{i \omega t - i k x} u_0,$$
All atoms are similar $⇒$ the solution should be similar for all atoms. We therefore attempt the following solution (ansatz)
for phonons:
$$u_n = Ae^{i \omega t - i k x_n},$$
and for electrons:
$$c_n = e^{i E t/\hbar - i k x} c_0.$$
(Here we wrote the time-dependent solution of the Schrödinger equation to emphasize the similarity between two systems.)
$$c_n = Be^{i E t/\hbar - i k x_n},$$
where $x_n=na$ and where we wrote the time-dependent solution of the Schrödinger equation to emphasize the similarity between the two systems.
Periodicity requires $e^{i k L} = 1$ $⇒$ $k = 2\pi p/L$, with $p\in \mathbb{Z}$.
As usual, periodicity quantizes k-space by requiring $e^{i k L} = 1$ $⇒$ $k = 2\pi p/L$, with $p\in \mathbb{Z}$.
In that case $c_n/c_0 = \exp(2 \pi n p a/L) = \exp(2 \pi n p/N)$, and changing $p→p+N$ corresponds to exactly the same solution, and we have $N$ different solutions in total.
As such, $c_n/A = \exp(2 \pi n p a/L) = \exp(2 \pi n p/N)$, and we see that changing $p→p+N$ corresponds to exactly the same solution, and that we have $N$ different solutions in total.
We can see that the solutions with different $k$ (or different $p$) are identical by plotting them:
We see that solutions with $k$ differing by an integer multiple of $2\pi/a$ are identical by plotting them:
Let's count how many different solutions we expect to find. We have a system with $N$ degrees of freedom (either $u_n$ or $c_n$), and therefore we need $N$ normal modes (or eigenstates).
How many different solutions did we expect to find? We have a system with $N$ degrees of freedom (either $u_n$ or $c_n$), and therefore we need $N$ normal modes (or eigenstates).
Because we proposed $N$ different planewave solutions, if we find an energy or frequency for each solution, we have fully solved the problem!
Because we proposed an ansatz with $N$ different plane-wave solutions, if we find an energy or frequency for each solution, we have fully solved the problem!
