Including Phonons and Crystal-Field Phonon interactions

In this section we discuss, how lattice dynamics may be considered in the framework of the Hamiltonian (223). We will see that this corresponds to a system of coupled Einstein oscillators. One such oscillator can be modelled by setting up a sipf file with the module phonon. Coupling has to be done in mcdisp.j. Rephrasing lattice dynamics in this way allows to couple phonons to the crystal field.

A three dimensional Einstein oscillator (for atom $i$) in a solid can be described by the following Hamiltonian

  $$\hat H_E(i)=\frac{a_0^2{\hat \mathbf w_i}^2}{2m_i} - \frac{1}{2} {\hat \mathbf u}^T_i \overline{K}(ii) {\hat \mathbf u}_i$$ (66)

Here $\hat \mathbf u$ is the dimensionless displacement vector ( $\hat \mathbf u_i={\hat \mathbf P}_i/a_0=\Delta {\hat \mathbf r}_i/a_0$, with the Bohr radius $a_0=0.5219$ Å), $m_i$ the mass of the atom $i$, $\hat \mathbf w_i=d\hat \mathbf u_i/dt= \mathbf p_i/a_0$ the conjugate momentum to $\hat \mathbf u_i$ and $\overline{K}(ii)$ the Matrix describing the restoring force.

Coupling such oscillators leads to the Hamiltonian

  $$\hat H_{phon}=\sum_i \hat H_E(i) -\frac{1}{2} \sum_{i\neq i'} {\hat \mathbf u}_i^T \overline{K}(ii') {\hat \mathbf u}_{i'}$$ (67)

Note that our coupling constants $K_{\alpha\beta}(ii')=-A_{\alpha\beta}(ii')$, where $A_{\alpha\beta}$ are the second-order derivatives of the potential energy as defined e.g. in [30, page 99].

In a mean field type of theory the phonon single ion module has thus to solve the Hamiltonian

  $$\hat H_E=\frac{a_0^2{\hat \mathbf w}^2}{2m} - \frac{1}{2} {\hat \mathbf u}^T \overline{K} {\hat \mathbf u} - {\mathbf F}^T {\hat \mathbf u}$$ (68)

Here the force $\mathbf F$ corresponds to the exchange field $\mathbf H_{xc}$ and $\hat \mathbf u$ to the general operator $\hat \mathbf I$ and $\overline{K}(nn')$ to $\mathcal J(nn')$ of equation (223), respectively. The single ion Hamiltonian (71) can be solved by transforming it to normal coordinates (main axis of the Einstein oscillator) using the transformation matrix $\overline{S}$, which diagonalises $\overline{K}=\overline{S}^T\overline{\Omega}\overline{S}$:

  $$\hat \mathbf u'=\overline{S} \hat \mathbf u +\overline{\Omega}^{-1} \overline{S} \mathbf F$$ (69)

  $$\hat H_E=\frac{a_0^2{\hat \mathbf w}^{'2}}{2m}-\frac{1}{2} {\ha... ...'T} \overline{\Omega} {\hat \mathbf u'} -\frac{1}{2}\mathbf F^T \mathbf u_0$$ (70)

Due to the action of the force $\mathbf F$ the equilibrium position of the oscillator is $\mathbf u_0=-\overline{S}^T\overline{\Omega}^{-1}\overline{S}\mathbf F$ (it is the task of the function Icalc to return this equilibrium position), the energies correspond to the three elements of the diagonal matrix $\overline{\Omega}$, i.e. $\Omega_{11}=-m a_0^2 (\Delta_1 /\hbar)^2$, $\Omega_{22}=-m a_0^2 (\Delta_2 /\hbar)^2$, $\Omega_{33}=-m a_0^2 (\Delta_3 /\hbar)^2$. In order to run mcdisp we have to calculate the transition matrix elements:

The single ion susceptibility for such a transition, e.g. $\Delta_1$ - corresponds to

	$$\overline{\chi}^0$$	$=$	$$\sum_{\nu\mu}\frac{\langle \nu\vert\hat \mathbf u\vert\mu\rangle\... ...ert\hat \mathbf u^T\vert\nu \rangle}{\Delta_1 -\hbar \omega} (p_{\nu} -p_{\mu})$$	(71)
		$=$	$$\sum_{\nu\mu}\frac{\langle \nu\vert\overline{S}^T \hat \mathbf u'... ...f u'^T \overline{S}\vert\nu \rangle}{\Delta_1 -\hbar \omega} (p_{\nu} -p_{\mu})$$	(72)

Because the different components of $\mathbf u'$ commute and the Hamiltonian (73) is separable, for the transition $\Delta_1$ only the terms with $u_1'$ in the nominator contribute:

	$$\chi^0_{\alpha\beta}$$	$=$	$$S^T_{\alpha1}\sum_{\nu\mu}\frac{\langle \nu\vert\hat u_1'\vert\mu... ...hat u_1'\vert\nu \rangle}{\Delta_1 -\hbar \omega} (p_{\nu} -p_{\mu}) S_{1\beta}$$	(73)
		$=$	$$S^T_{\alpha1}S_{1\beta}\frac{\hbar^2}{2ma_0^2\Delta_1}\left(\frac{1}{\Delta_1-\hbar\omega}+\frac{1}{\Delta_1+\hbar\omega}\right )$$	(74)

In order to derive the last result we had to express $\hat u_1'$ in terms of ladder operators $\hat u_1'=a_0^{-1} \hbar/\sqrt{2m\Delta_1}(\hat a+\hat a^{\dagger})$ and apply $\hat a^{\dagger}\vert\nu\rangle=\sqrt{\nu+1}\vert\nu+1\rangle$, $\hat a\vert\nu\rangle=\sqrt{\nu}\vert\nu-1\rangle$ and $\sum_{\nu=0}^{\infty}(p_{\nu}-p_{\nu+1})(\nu+1)=1$, $p_{\nu}=exp(-\nu\Delta_1/kT)(1-exp(-\Delta_1/kT))$. This shows that the single ion susceptibility of our atom can be written as a sum of three effective transitions (with temperature independent susceptibility)

$$\chi^0_{\alpha\beta}$$

$=$

$$\sum_{n=1,2,3} S^T_{\alpha n}S_{n\beta}\frac{\hbar^2}{2ma_0^2\Delta_n} \left(\frac{1}{\Delta_n-\hbar\omega}+\frac{1}{\Delta_n+\hbar\omega}\right )$$

(75)

Thus the module phonon has to provide in it's function du1calc these three transitions (=number of transitions).