The phonon Hamiltonian is accustomed by
\mathbf{H} = \frac{1}{2}\sum_{\alpha}(p_{\alpha}^{2} + \omega^{2}_{\alpha}q_{\alpha}^{2} -\frac{1}{2}\hbar\omega_{\alpha})
In agreement of the operators, these are accustomed by
\mathbf{H} = \sum_{\alpha}\hbar\omega_{\alpha}a_{\alpha}^{\dagger}a_{\alpha}
Here, in cogent the Hamiltonian (quantum mechanics) in abettor formalism, we accept not taken into annual the \frac{1}{2}\hbar \omega_{q} term, back if we yield an complete filigree or, for that amount a continuum, the \frac{1}{2}\hbar\omega_{q} agreement will add up giving an infinity. Hence, it is "renormalized" by putting the agency of \frac{1}{2}\hbar\omega_{q} to 0 arguing that the aberration in activity is what we admeasurement and not the complete amount of it. Hence, the \frac{1}{2}\hbar\omega_{q} agency is absent in the abettor formalised announcement for the Hamiltonian.
The arena accompaniment aswell alleged the "vacuum state" is the accompaniment composed of no phonons. Hence, the activity of the arena accompaniment is 0. When, a arrangement is in accompaniment |n_{1}n_{2}n_{3}...\rangle, we say there are nα phonons of blazon α. The nα are alleged the activity amount of the phonons. Activity of a individual phonon of blazon α getting \hbar \omega_{q}, the absolute activity of a accepted phonon arrangement is accustomed by n_{1}\hbar\omega_{1} + n_{2}\hbar\omega_{2}+ .... In added words, the phonons are non-interacting. The activity of conception and abolishment operators are accustomed 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,
a_{\alpha}|n_{1}...n_{\alpha -1}n_{\alpha}n_{\alpha +1}...\rangle = \sqrt{n_{\alpha}}|n_{1}...,n_{\alpha -1},(n_{\alpha}-1),n_{\alpha+1},...\rangle
i.e. a^{\dagger}_{\alpha} creates a phonon of blazon α while aα annihilates. Hence, they are appropriately the conception and abolishment abettor for phonons. Analogous to the Breakthrough harmonic oscillator case, we can ascertain atom amount abettor as N = \sum_{\alpha}a_{\alpha}^{\dagger}a_{\alpha}. The amount abettor commutes with a cord of articles of the conception and abolishment operators if, the amount of a's are according to amount of a^{\dagger}'s.
Phonons are bosons since, |\alpha,\beta\rangle = |\beta, \alpha\rangle i.e. they are symmetric beneath exchange.7
\mathbf{H} = \frac{1}{2}\sum_{\alpha}(p_{\alpha}^{2} + \omega^{2}_{\alpha}q_{\alpha}^{2} -\frac{1}{2}\hbar\omega_{\alpha})
In agreement of the operators, these are accustomed by
\mathbf{H} = \sum_{\alpha}\hbar\omega_{\alpha}a_{\alpha}^{\dagger}a_{\alpha}
Here, in cogent the Hamiltonian (quantum mechanics) in abettor formalism, we accept not taken into annual the \frac{1}{2}\hbar \omega_{q} term, back if we yield an complete filigree or, for that amount a continuum, the \frac{1}{2}\hbar\omega_{q} agreement will add up giving an infinity. Hence, it is "renormalized" by putting the agency of \frac{1}{2}\hbar\omega_{q} to 0 arguing that the aberration in activity is what we admeasurement and not the complete amount of it. Hence, the \frac{1}{2}\hbar\omega_{q} agency is absent in the abettor formalised announcement for the Hamiltonian.
The arena accompaniment aswell alleged the "vacuum state" is the accompaniment composed of no phonons. Hence, the activity of the arena accompaniment is 0. When, a arrangement is in accompaniment |n_{1}n_{2}n_{3}...\rangle, we say there are nα phonons of blazon α. The nα are alleged the activity amount of the phonons. Activity of a individual phonon of blazon α getting \hbar \omega_{q}, the absolute activity of a accepted phonon arrangement is accustomed by n_{1}\hbar\omega_{1} + n_{2}\hbar\omega_{2}+ .... In added words, the phonons are non-interacting. The activity of conception and abolishment operators are accustomed 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,
a_{\alpha}|n_{1}...n_{\alpha -1}n_{\alpha}n_{\alpha +1}...\rangle = \sqrt{n_{\alpha}}|n_{1}...,n_{\alpha -1},(n_{\alpha}-1),n_{\alpha+1},...\rangle
i.e. a^{\dagger}_{\alpha} creates a phonon of blazon α while aα annihilates. Hence, they are appropriately the conception and abolishment abettor for phonons. Analogous to the Breakthrough harmonic oscillator case, we can ascertain atom amount abettor as N = \sum_{\alpha}a_{\alpha}^{\dagger}a_{\alpha}. The amount abettor commutes with a cord of articles of the conception and abolishment operators if, the amount of a's are according to amount of a^{\dagger}'s.
Phonons are bosons since, |\alpha,\beta\rangle = |\beta, \alpha\rangle i.e. they are symmetric beneath exchange.7
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