Phonon - Operator Formalism

Operator Formalism

The phonon Hamiltonian is given by

In terms of the operators, these are given by

Here, in expressing the Hamiltonian (quantum mechanics) in operator formalism, we have not taken into account the term, since if we take an infinite lattice or, for that matter a continuum, the terms will add up giving an infinity. Hence, it is "renormalized" by putting the factor of to 0 arguing that the difference in energy is what we measure and not the absolute value of it. Hence, the factor is absent in the operator formalised expression for the Hamiltonian.
The ground state also called the "vacuum state" is the state composed of no phonons. Hence, the energy of the ground state is 0. When, a system is in state, we say there are phonons of type . The are called the occupation number of the phonons. Energy of a single phonon of type being, the total energy of a general phonon system is given by . In other words, the phonons are non-interacting. The action of creation and annihilation operators are given by

$a^{\dagger}_{\alpha}|n_{1}...n_{\alpha -1}n_{\alpha}n_{\alpha +1}...\rangle = \sqrt{n_{\alpha} +1}|n_{1}...,n_{\alpha -1}, n_{\alpha}+1, n_{\alpha+1}...\rangle$

and,

i.e. creates a phonon of type while annihilates. Hence, they are respectively the creation and annihilation operator for phonons. Analogous to the Quantum harmonic oscillator case, we can define particle number operator as . The number operator commutes with a string of products of the creation and annihilation operators if, the number of 's are equal to number of 's.
Phonons are bosons since, i.e. they are symmetric under exchange.

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