Quantum Harmonic Oscillator - Harmonic Oscillators Lattice: Phonons

Harmonic Oscillators Lattice: Phonons

We can extend the notion of harmonic oscillator to a one lattice of many particles. Consider a one-dimensional quantum mechanical harmonic chain of N identical atoms. This is the simplest quantum mechanical model of a lattice, and we will see how phonons arise from it. The formalism that we will develop for this model is readily generalizable to two and three dimensions.

As in the previous section, we denote the positions of the masses by, as measured from their equilibrium positions (i.e. if particle is at its equilibrium position.) In two or more dimensions, the are vector quantities. The Hamiltonian for this system is

where is the mass of each atom (assuming is equal for all), and and are the position and momentum operators for the th atom and the sum is made over the nearest neighboors (nn). However, as one expect, in a lattice could also appear waves that could behave as particles. For the treatment of waves, is custom to treat with the fourier space which uses normal modes of the wavevector as variables instead coordinates of particles. The number of normal modes are the same to the particles however the fourier space are very useful given the periodicity of the system.

We introduce, then, a set of "normal coordinates", defined as the discrete Fourier transforms of the 's and "conjugate momenta" defined as the Fourier transforms of the 's:

$Q_k = {1\over\sqrt{N}} \sum_{l} e^{ikal} x_l$

$\Pi_{k} = {1\over\sqrt{N}} \sum_{l} e^{-ikal} p_l.$

The quantity will turn out to be the wave number of the phonon, i.e. divided by the wavelength. It takes on quantized values, because the number of atoms is finite.

This choice remains the desired conmutation relations in either real space or wave vector space

$\begin{align} \left&=i\hbar\delta_{l,m} \\ \left &={1\over N} \sum_{l,m} e^{ikal} e^{-ik'am} \\ &= {i \hbar\over N} \sum_{m} e^{iam\left(k-k'\right)} = i\hbar\delta_{k,k'} \\ \left &= \left = 0 \end{align}$

From the general result

$\begin{align} \sum_{l}x_l x_{l+m}&={1\over N}\sum_{kk'}Q_k Q_k'\sum_{l} e^{ial\left(k+k'\right)}e^{iamk'}= \sum_{k}Q_k Q_{-k}e^{iamk} \\ \sum_{l}{p_l}^2 &= \sum_{k}\Pi_k \Pi_{-k} \end{align}$

It is easy to show that the potential energy term is

${1\over 2} m \omega^2 \sum_{j} (x_j - x_{j+1})^2= {1\over 2}\omega^2\sum_{k}Q_k Q_{-k}(2-e^{ika}-e^{-ika})= {1\over 2} \sum_{k}{\omega_k}^2Q_k Q_{-k}$

where

The Hamiltonian may be written in wave vector space as

$\mathbf{H} = {1\over {2m}}\sum_k \left( { \Pi_k\Pi_{-k} } + m^2 \omega_k^2 Q_k Q_{-k} \right)$

Notice that the couplings between the position variables have been transformed away; if the 's and 's were hermitian(which they are not), the transformed Hamiltonian would describe uncoupled harmonic oscillators.

The form of the quantization depends on the choice of boundary conditions; for simplicity, we impose periodic boundary conditions, defining the th atom as equivalent to the first atom. Physically, this corresponds to joining the chain at its ends. The resulting quantization is

$k=k_n = {2n\pi \over Na} \quad \hbox{for}\ n = 0, \pm1, \pm2, ..., \pm {N \over 2}.\$

The upper bound to comes from the minimum wavelength, which is twice the lattice spacing, as discussed above.

The harmonic oscillator eigenvalues or energy levels for the mode are :

If we ignore the zero-point energy then the levels are evenly spaced at :

So an exact amount of energy must be supplied to the harmonic oscillator lattice to push it to the next energy level. In comparison to the photon case when the electromagnetic field is quantised, the quantum of vibrational energy is called a phonon.

All quantum systems show wave-like and particle-like properties. The particle-like properties of the phonon are best understood using the methods of second quantization and operator techniques described later.

Read more about this topic: Quantum Harmonic Oscillator

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