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		<id>https://e-learning.pan-training.eu/wiki/index.php?title=Phonons,_quantum_mechanical_treatment&amp;diff=968&amp;oldid=prev</id>
		<title>ucph&gt;Tommy: /* *Quantum mechanical formalism in three dimensions */</title>
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		<updated>2019-08-28T23:10:50Z</updated>

		<summary type="html">&lt;p&gt;&lt;span dir=&quot;auto&quot;&gt;&lt;span class=&quot;autocomment&quot;&gt;*Quantum mechanical formalism in three dimensions&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;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 &amp;#039;&amp;#039;phonons&amp;#039;&amp;#039;, which has wavenumber \({\bf q}\) and energy \(\hbar \omega_q\). &lt;br /&gt;
&lt;br /&gt;
==*The harmonic oscillator==&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
We consider a particle of mass \(M\) in a one-dimensional periodic potential. The Hamiltonian is given as&lt;br /&gt;
&lt;br /&gt;
\begin{equation}\label{eq:HOhamil} &lt;br /&gt;
H = \dfrac{p^2}{2M} + \dfrac{K x^2}{2} ,  &lt;br /&gt;
\end{equation}&lt;br /&gt;
&lt;br /&gt;
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&lt;br /&gt;
&lt;br /&gt;
\begin{align}&lt;br /&gt;
a^\dagger &amp;amp;= \dfrac{1}{\sqrt{2\hbar\omega_0 M}} &lt;br /&gt;
                   \left( M \omega_0 x + i p \right) . \label{dummy915753937}\\&lt;br /&gt;
a &amp;amp;= \dfrac{1}{\sqrt{2\hbar\omega_0 M}} &lt;br /&gt;
                   \left( M \omega_0 x - i p \right) .\label{dummy1648971532}&lt;br /&gt;
\end{align}  &lt;br /&gt;
&lt;br /&gt;
These operators are normalized so that &lt;br /&gt;
&lt;br /&gt;
\begin{equation}  \label{dummy1883154453}&lt;br /&gt;
[a,a^\dagger] = 1 .  &lt;br /&gt;
\end{equation}&lt;br /&gt;
&lt;br /&gt;
This commutation relation show that \(a\) and \(a^\dagger\) are boson operators. We can obtain the position and momentum operators:&lt;br /&gt;
&lt;br /&gt;
\begin{align}&lt;br /&gt;
x &amp;amp;= \sqrt{\dfrac{\hbar}{2\omega_0 M}} \left(a^\dagger + a\right) . \label{dummy1083051975}\\&lt;br /&gt;
p &amp;amp;= -i \sqrt{\dfrac{\hbar\omega_0 M}{2}} \left(a^\dagger - a\right) .\label{dummy102554818}&lt;br /&gt;
\end{align}&lt;br /&gt;
&lt;br /&gt;
By substitution into equation \eqref{eq:HOhamil}, we reach&lt;br /&gt;
&lt;br /&gt;
\begin{equation}  \label{dummy2070323948}&lt;br /&gt;
H = \hbar \omega_0 \left( a^\dagger a + \frac{1}{2}\right) ,&lt;br /&gt;
\end{equation}&lt;br /&gt;
&lt;br /&gt;
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 &amp;#039;&amp;#039;zero point energy&amp;#039;&amp;#039;. The zero point energy is related to a &amp;quot;smearing&amp;quot; of the ground state, due to Heisenberg&amp;#039;s uncertainty principle, which in this case is denoted &amp;#039;&amp;#039;zero point motion&amp;#039;&amp;#039;.&lt;br /&gt;
&lt;br /&gt;
==*The one-dimensional quantum model==&lt;br /&gt;
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}:&lt;br /&gt;
&lt;br /&gt;
\begin{equation}\label{eq:H_1d_phonon} &lt;br /&gt;
H = \displaystyle\sum_j \left[ \dfrac{p_j^2}{2M} &lt;br /&gt;
                 + \dfrac{K}{2} \left(u_j-u_{j+1} \right)^2 \right] .  &lt;br /&gt;
\end{equation}&lt;br /&gt;
&lt;br /&gt;
We now perform a Fourier transform of the displacement operators:&lt;br /&gt;
&lt;br /&gt;
\begin{align}&lt;br /&gt;
u_q &amp;amp;= \dfrac{1}{\sqrt{N}} \displaystyle\sum_j \exp(i q j a) u_j , \label{dummy801882678}\\&lt;br /&gt;
u_j &amp;amp;= \dfrac{1}{\sqrt{N}} \displaystyle\sum_q \exp(- i q j a) u_q ,\label{dummy1978718886}&lt;br /&gt;
\end{align}&lt;br /&gt;
&lt;br /&gt;
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&lt;br /&gt;
&lt;br /&gt;
\begin{equation}\label{eq:H_1d_phonon2} &lt;br /&gt;
H = \displaystyle\sum_{jqq&amp;#039;} \exp\big(-i(q+q&amp;#039;)ja \big)  &lt;br /&gt;
    \biggr[ \dfrac{p_q p_{q&amp;#039;}}{2MN} &lt;br /&gt;
    + \dfrac{K}{2N}\left(u_q u_{q&amp;#039;}\{1-\exp(-iqa)\}\{1-\exp(-iq&amp;#039;a)\}\right)\biggr].  \end{equation}&lt;br /&gt;
&lt;br /&gt;
The sum over \(j\) depends only on the first exponential function. This is summed out to give \(N \delta_{q,-q&amp;#039;}\). Next, the sum over \(q&amp;#039;\) is easily performed yielding  &lt;br /&gt;
&lt;br /&gt;
\begin{align} \label{eq:H_1d_phonon3}&lt;br /&gt;
H &amp;amp;= \displaystyle\sum_{q} \left[ \dfrac{p_q p_{-q}}{2M} &lt;br /&gt;
    + K \left(u_q u_{-q}\{1-\cos(qa)\}\right)\right]  \\&lt;br /&gt;
 &amp;amp;= \displaystyle\sum_{q} \left[ \dfrac{p_q p_{-q}}{2M} &lt;br /&gt;
    + \dfrac{M}{2} \left(u_q u_{-q}\omega_q^2 \right)\right], \nonumber&lt;br /&gt;
\end{align}  &lt;br /&gt;
&lt;br /&gt;
where we in the last step have used the expression for the classical vibration frequency, \(\omega_q\), from equation \ref{eq:dispersion_1atom}.&lt;br /&gt;
&lt;br /&gt;
We now define the Fourier transformed creation and annihilation operators as&lt;br /&gt;
&lt;br /&gt;
\begin{align}&lt;br /&gt;
a_q^\dagger &amp;amp;= \dfrac{1}{\sqrt{2\hbar\omega_q M}} &lt;br /&gt;
                   \left( M \omega_q u_q + i p_q \right) . \label{dummy1154685165}\\&lt;br /&gt;
a_q &amp;amp;= \dfrac{1}{\sqrt{2\hbar\omega_q M}} &lt;br /&gt;
                   \left( M \omega_q u_q - i p_q \right) .\label{dummy32026784}&lt;br /&gt;
\end{align}  &lt;br /&gt;
&lt;br /&gt;
It can be shown that these operators describe bosons, through &lt;br /&gt;
&lt;br /&gt;
\begin{equation} \label{dummy1259932973}&lt;br /&gt;
[a_q,a_q^\dagger] = 1 .   &lt;br /&gt;
\end{equation}&lt;br /&gt;
&lt;br /&gt;
The operators \(u_q\) and \(p_q\) are given by:&lt;br /&gt;
&lt;br /&gt;
\begin{align}&lt;br /&gt;
u_q &amp;amp;= \sqrt{\dfrac{\hbar}{2\omega_q M}} \left(a_q^\dagger + a_q\right) . \label{dummy964522178}\\&lt;br /&gt;
p_q &amp;amp;= -i \sqrt{\dfrac{\hbar\omega_q M}{2}} \left(a_q^\dagger - a_q\right) .\label{dummy187797063}&lt;br /&gt;
\end{align} &lt;br /&gt;
&lt;br /&gt;
Inserting into the Hamiltonian in equation \ref{eq:H_1d_phonon3}, we reach the final solution&lt;br /&gt;
&lt;br /&gt;
\begin{equation}  \label{dummy1773738828}&lt;br /&gt;
H = \displaystyle\sum_q \left(a_q^\dagger a_q + \dfrac{1}{2}\right) \hbar \omega_q . &lt;br /&gt;
\end{equation}&lt;br /&gt;
&lt;br /&gt;
This shows that the system is described by \(N\) independent oscillator modes (phonons), distinguished by their wave number \(q\).&lt;br /&gt;
&lt;br /&gt;
==*Quantum mechanical formalism in three dimensions==&lt;br /&gt;
&lt;br /&gt;
The general features of three-dimensional phonons is described in [[Lattice vibrations, classical treatment]].&lt;br /&gt;
&lt;br /&gt;
To make a quantum mechanical descriptiopn, one needs the displacement operators to&lt;br /&gt;
include the polarization vector, \({\bf e}_{q,p}\):  &lt;br /&gt;
&lt;br /&gt;
\begin{equation}&lt;br /&gt;
{\bf u}_j(t) = \sqrt{\frac{\hbar}{2 M N}} &lt;br /&gt;
   \sum_{q,p} \frac{{\bf e}_{q,p}}{\sqrt{\omega_{q,p}}} &lt;br /&gt;
  \times \left[ a_{q,p} \exp(i({\bf q}\cdot {\bf r}_j - \omega_{q,p} t))&lt;br /&gt;
       + a^\dagger_{q,p} \exp(-i({\bf q}\cdot {\bf r}_j - \omega_{q,p} t))&lt;br /&gt;
\right] ,&lt;br /&gt;
\end{equation}&lt;br /&gt;
&lt;br /&gt;
where \(p\) is the polarization index, which runs over 3 times &lt;br /&gt;
the number of atoms in the unit cell.&lt;br /&gt;
&lt;br /&gt;
In this version of the notes, we make no attempt of solving this problem.&lt;br /&gt;
In stead, we just show the final result for the phonon Hamiltonian:&lt;br /&gt;
\begin{equation}&lt;br /&gt;
H = \sum_{q,p} \hbar \omega_{q,p} &lt;br /&gt;
   \left( a^\dagger_{q,p} a_{q,p} + \frac{1}{2} \right) .&lt;br /&gt;
\end{equation}&lt;br /&gt;
This shows that also in three dimensions the lattice vibrations are quantized &lt;br /&gt;
in terms of Bosonic phonons, with energies depending upon both \({\bf q}\) and \(p\).&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references /&amp;gt;&lt;/div&gt;</summary>
		<author><name>ucph&gt;Tommy</name></author>
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