Phonons, quantum mechanical treatment: Difference between revisions
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Latest revision as of 22:15, 18 February 2020
With the discussion on lattice vibrations (last section) in mind, we now treat the problem of lattice vibrations quantum mechanically. We will show how the vibrations with frequencies \(\omega_q\) of the classical solution quantize into bosonic phonons, which has wavenumber \({\bf q}\) and energy \(\hbar \omega_q\).
*The harmonic oscillator
The solution to the phonon problem lends much from the standard example of one particle in a harmonic potential. Therefore, it is of value to summarize this solution here. A more thorough derivation is given in most textbooks on quantum mechanics.
We consider a particle of mass \(M\) in a one-dimensional periodic potential. The Hamiltonian is given as
\begin{equation}\label{eq:HOhamil} H = \dfrac{p^2}{2M} + \dfrac{K x^2}{2} , \end{equation}
where \(p\) is the momentum operator, \(p = i \hbar \partial/\partial x\), and \([x,p]=i\hbar\). We recall that the classical oscillator frequency is given by equation \eqref{eq:omega0} and define the creation and annihilation operators as
\begin{align} a^\dagger &= \dfrac{1}{\sqrt{2\hbar\omega_0 M}} \left( M \omega_0 x + i p \right) . \label{dummy915753937}\\ a &= \dfrac{1}{\sqrt{2\hbar\omega_0 M}} \left( M \omega_0 x - i p \right) .\label{dummy1648971532} \end{align}
These operators are normalized so that
\begin{equation} \label{dummy1883154453} [a,a^\dagger] = 1 . \end{equation}
This commutation relation show that \(a\) and \(a^\dagger\) are boson operators. We can obtain the position and momentum operators:
\begin{align} x &= \sqrt{\dfrac{\hbar}{2\omega_0 M}} \left(a^\dagger + a\right) . \label{dummy1083051975}\\ p &= -i \sqrt{\dfrac{\hbar\omega_0 M}{2}} \left(a^\dagger - a\right) .\label{dummy102554818} \end{align}
By substitution into equation \eqref{eq:HOhamil}, we reach
\begin{equation} \label{dummy2070323948} H = \hbar \omega_0 \left( a^\dagger a + \frac{1}{2}\right) , \end{equation}
where \(a^\dagger a\) is the number operator, \(n\), with eigenvalues being integers from zero and upwards. The oscillation is thus quantized with energy quanta of \(\hbar\omega_0\). Notice that even the ground state has a finite energy, \(E_{\rm gs} = \hbar \omega_0 / 2\). This is called the zero point energy. The zero point energy is related to a "smearing" of the ground state, due to Heisenberg's uncertainty principle, which in this case is denoted zero point motion.
*The one-dimensional quantum model
This lattice model is solved in much the same way as for the one-particle harmonic oscillator. We can write the Hamiltonian as the sum of the kinetic energy and a harmonic potential of the form in equation \eqref{eq:harmonic_potential}:
\begin{equation}\label{eq:H_1d_phonon} H = \displaystyle\sum_j \left[ \dfrac{p_j^2}{2M} + \dfrac{K}{2} \left(u_j-u_{j+1} \right)^2 \right] . \end{equation}
We now perform a Fourier transform of the displacement operators:
\begin{align} u_q &= \dfrac{1}{\sqrt{N}} \displaystyle\sum_j \exp(i q j a) u_j , \label{dummy801882678}\\ u_j &= \dfrac{1}{\sqrt{N}} \displaystyle\sum_q \exp(- i q j a) u_q ,\label{dummy1978718886} \end{align}
and similar for the \(p_j\) operators. We have here applied periodic Born-von Karman boundary conditions as in the classical case: \(u_{N+1} \equiv u_1\), and used that the equilibrium positions are given by \(r_j = j a\). Inserting this into equation \eqref{eq:H_1d_phonon}, we reach
\begin{equation}\label{eq:H_1d_phonon2} H = \displaystyle\sum_{jqq'} \exp\big(-i(q+q')ja \big) \biggr[ \dfrac{p_q p_{q'}}{2MN} + \dfrac{K}{2N}\left(u_q u_{q'}\{1-\exp(-iqa)\}\{1-\exp(-iq'a)\}\right)\biggr]. \end{equation}
The sum over \(j\) depends only on the first exponential function. This is summed out to give \(N \delta_{q,-q'}\). Next, the sum over \(q'\) is easily performed yielding
\begin{align} \label{eq:H_1d_phonon3} H &= \displaystyle\sum_{q} \left[ \dfrac{p_q p_{-q}}{2M} + K \left(u_q u_{-q}\{1-\cos(qa)\}\right)\right] \\ &= \displaystyle\sum_{q} \left[ \dfrac{p_q p_{-q}}{2M} + \dfrac{M}{2} \left(u_q u_{-q}\omega_q^2 \right)\right], \nonumber \end{align}
where we in the last step have used the expression for the classical vibration frequency, \(\omega_q\), from equation \ref{eq:dispersion_1atom}.
We now define the Fourier transformed creation and annihilation operators as
\begin{align} a_q^\dagger &= \dfrac{1}{\sqrt{2\hbar\omega_q M}} \left( M \omega_q u_q + i p_q \right) . \label{dummy1154685165}\\ a_q &= \dfrac{1}{\sqrt{2\hbar\omega_q M}} \left( M \omega_q u_q - i p_q \right) .\label{dummy32026784} \end{align}
It can be shown that these operators describe bosons, through
\begin{equation} \label{dummy1259932973} [a_q,a_q^\dagger] = 1 . \end{equation}
The operators \(u_q\) and \(p_q\) are given by:
\begin{align} u_q &= \sqrt{\dfrac{\hbar}{2\omega_q M}} \left(a_q^\dagger + a_q\right) . \label{dummy964522178}\\ p_q &= -i \sqrt{\dfrac{\hbar\omega_q M}{2}} \left(a_q^\dagger - a_q\right) .\label{dummy187797063} \end{align}
Inserting into the Hamiltonian in equation \ref{eq:H_1d_phonon3}, we reach the final solution
\begin{equation} \label{dummy1773738828} H = \displaystyle\sum_q \left(a_q^\dagger a_q + \dfrac{1}{2}\right) \hbar \omega_q . \end{equation}
This shows that the system is described by \(N\) independent oscillator modes (phonons), distinguished by their wave number \(q\).
*Quantum mechanical formalism in three dimensions
The general features of three-dimensional phonons is described in Lattice vibrations, classical treatment.
To make a quantum mechanical descriptiopn, one needs the displacement operators to include the polarization vector, \({\bf e}_{q,p}\):
\begin{equation} {\bf u}_j(t) = \sqrt{\frac{\hbar}{2 M N}} \sum_{q,p} \frac{{\bf e}_{q,p}}{\sqrt{\omega_{q,p}}} \times \left[ a_{q,p} \exp(i({\bf q}\cdot {\bf r}_j - \omega_{q,p} t)) + a^\dagger_{q,p} \exp(-i({\bf q}\cdot {\bf r}_j - \omega_{q,p} t)) \right] , \end{equation}
where \(p\) is the polarization index, which runs over 3 times the number of atoms in the unit cell.
In this version of the notes, we make no attempt of solving this problem. In stead, we just show the final result for the phonon Hamiltonian: \begin{equation} H = \sum_{q,p} \hbar \omega_{q,p} \left( a^\dagger_{q,p} a_{q,p} + \frac{1}{2} \right) . \end{equation} This shows that also in three dimensions the lattice vibrations are quantized in terms of Bosonic phonons, with energies depending upon both \({\bf q}\) and \(p\).