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===Interpretation of phonons using second quantization techniques=== The above-derived Hamiltonian may look like a classical Hamiltonian function, but if it is interpreted as an [[Operator (physics)|operator]], then it describes a [[quantum field theory]] of non-interacting [[boson]]s.<ref name=girvinYang /> The [[second quantization]] technique, similar to the [[ladder operator]] method used for [[Quantum harmonic oscillator#Ladder operator method|quantum harmonic oscillators]], is a means of extracting energy [[eigenvalues]] without directly solving the differential equations. Given the Hamiltonian, <math>\mathcal{H}</math>, as well as the conjugate position, <math>Q_k</math>, and conjugate momentum <math>\Pi_{k}</math> defined in the quantum treatment section above, we can define [[creation and annihilation operators]]:<ref name=ashcrroftMermin>{{cite book |last1=Ashcroft |first1=Neil W. |last2=Mermin |first2=N. David |title=Solid State Physics |date=1976 |publisher=Saunders College Publishing |isbn=0-03-083993-9 |pages=780–783}}</ref> :<math>b_k=\sqrt\frac{m\omega_k}{2\hbar}\left(Q_k+\frac{i}{m\omega_k}\Pi_{-k}\right)</math> and <math>{b_k}^\dagger=\sqrt\frac{m\omega_k}{2\hbar}\left(Q_{-k}-\frac{i}{m\omega_k}\Pi_{k}\right)</math> The following commutators can be easily obtained by substituting in the [[canonical commutation relation]]: :<math>\left[b_k , {b_{k'}}^\dagger \right] = \delta_{k,k'} ,\quad \Big[b_k , b_{k'} \Big] = \left[{b_k}^\dagger , {b_{k'}}^\dagger \right] = 0</math> Using this, the operators ''b<sub>k</sub>''<sup>†</sup> and ''b<sub>k</sub>'' can be inverted to redefine the conjugate position and momentum as: :<math>Q_k=\sqrt{\frac{\hbar}{2m\omega_k}}\left({b_k}^\dagger+b_{-k}\right)</math> and <math>\Pi_k=i\sqrt{\frac{\hbar m\omega_k}{2}}\left({b_k}^\dagger-b_{-k}\right)</math> Directly substituting these definitions for <math>Q_k</math> and <math>\Pi_k</math> into the wavevector space Hamiltonian, as it is defined above, and simplifying then results in the Hamiltonian taking the form:<ref name=girvinYang /> :<math>\mathcal{H} =\sum_k \hbar\omega_k \left({b_k}^\dagger b_k+\tfrac12\right)</math> This is known as the second quantization technique, also known as the occupation number formulation, where ''n<sub>k</sub>'' = ''b<sub>k</sub>''<sup>†</sup>''b<sub>k</sub>'' is the occupation number. This can be seen to be a sum of N independent oscillator Hamiltonians, each with a unique wave vector, and compatible with the methods used for the quantum harmonic oscillator (note that ''n<sub>k</sub>'' is [[Hermitian matrix|hermitian]]).<ref name=ashcrroftMermin /> When a Hamiltonian can be written as a sum of commuting sub-Hamiltonians, the energy eigenstates will be given by the products of eigenstates of each of the separate sub-Hamiltonians. The corresponding [[energy]] [[Spectrum of an operator|spectrum]] is then given by the sum of the individual eigenvalues of the sub-Hamiltonians.<ref name=ashcrroftMermin /> As with the quantum harmonic oscillator, one can show that ''b<sub>k</sub>''<sup>†</sup> and ''b<sub>k</sub>'' respectively create and destroy a single field excitation, a phonon, with an energy of ''ħω<sub>k</sub>''.<ref name=ashcrroftMermin /><ref name=girvinYang /> Three important properties of phonons may be deduced from this technique. First, phonons are [[boson]]s, since any number of identical excitations can be created by repeated application of the creation operator ''b<sub>k</sub>''<sup>†</sup>. Second, each phonon is a "collective mode" caused by the motion of every atom in the lattice. This may be seen from the fact that the creation and annihilation operators, defined here in momentum space, contain sums over the position and momentum operators of every atom when written in position space. (See [[position and momentum space]].)<ref name=ashcrroftMermin /> Finally, using the ''position–position [[correlation function]]'', it can be shown that phonons act as waves of lattice displacement.{{citation needed|date=August 2020}} This technique is readily generalized to three dimensions, where the Hamiltonian takes the form:<ref name=ashcrroftMermin /><ref name=girvinYang /> :<math>\mathcal{H} = \sum_k \sum_{s = 1}^3 \hbar \, \omega_{k,s} \left( {b_{k,s}}^\dagger b_{k,s} + \tfrac12 \right).</math> This can be interpreted as the sum of 3N independent oscillator Hamiltonians, one for each wave vector and polarization.<ref name=ashcrroftMermin />
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