Including Phonons, Electrical Polarisation and Crystal-Field Phonon interactions

EWALD: I think you are now ready for the really interesting subjects - I mean phonons and their interactions.

SIMPLICIUS: Ewald , you know, in the past 40 years there always was money for electronic studies and practically no money for phonons and lattice dynamics. So why should we suddenly go back and study phonons. You can predict phonon dynamics with DFT methods nowadays - even in exotic matter. Are you serious about studying phonons ?

EWALD: I know, the mainstream of solid state research is based on the Born-Oppenheimer adiabatic approximation and thus phonons are considered of minor importance. However, BCS theory on superconductivity is a good example on how phonon may enter suddenly into the theory with unexpected effects. Similar, in magnetism we usually neglect phonons. However, there are some examples which demonstrate, that for example for the interpretation of crystal field spectra it may be necessary to take into account compound modes having both phononic and magnetic character and correspondingly can be observed in magnetic and nuclear scattering of neutrons.

ORLANDO: According to my simple view of this subject, aren't we talking about the origin of magnetostriction ?

EWALD: Yes indeed, I did not think this way. You are right, magnetostriction would be zero if there was not some couling between the electronic and lattice degrees of freedom.

SIMPLICIUS: So how can phonons be handled ?

EWALD: Let me first show you, how lattice dynamics may be considered in the framework of the Hamiltonian (253). 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 then 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

  $$\hat H_E(i)=\frac{a_0^2{\hat \mathbf w_i}^2}{2m_i} - \frac{1}{2} {\hat \mathbf u}^T_i \overline{K}(ii) {\hat \mathbf u}_i$$ (86)

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

  $$\hat H_{phon}=\sum_i \hat H_E(i) -\frac{1}{2} \sum_{i\neq i'} {\hat \mathbf u}_i^T \overline{K}(ii') {\hat \mathbf u}_{i'}$$ (87)

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 [32, page 99].

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

  $$\hat H_E=\frac{a_0^2{\hat \mathbf w}^2}{2m} - \frac{1}{2} {\hat \mathbf u}^T \overline{K} {\hat \mathbf u} - {\mathbf F}^T {\hat \mathbf u}$$ (88)

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 (253), respectively. The single ion Hamiltonian (91) can be solved by transforming it to normal coordinates (main axis of the Einstein oscillator) using the transformation matrix $\overline{S}$, which diagonalises $\overline{K}=\overline{S}^T\overline{\Omega}\overline{S}$:

  $$\hat \mathbf u'=\overline{S} \hat \mathbf u +\overline{\Omega}^{-1} \overline{S} \mathbf F$$ (89)

  $$\hat H_E=\frac{a_0^2{\hat \mathbf w}^{'2}}{2m}-\frac{1}{2} {\ha... ...'T} \overline{\Omega} {\hat \mathbf u'} -\frac{1}{2}\mathbf F^T \mathbf u_0$$ (90)

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

	$$\overline{\chi}^0$$	$=$	$$\sum_{\nu\mu}\frac{\langle \nu\vert\hat \mathbf u-\hat \mathbf u_... ...-\hat \mathbf u_0^T\vert\nu \rangle}{\Delta_1 -\hbar \omega} (p_{\nu} -p_{\mu})$$	(91)
		$=$	$$\sum_{\nu\mu}\frac{\langle \nu\vert\overline{S}^T \hat \mathbf u'... ...f u'^T \overline{S}\vert\nu \rangle}{\Delta_1 -\hbar \omega} (p_{\nu} -p_{\mu})$$	(92)

Because the different components of $\mathbf u'$ commute and the Hamiltonian (93) is separable, for the transition $\Delta_1$ only the terms with $u_1'$ in the nominator contribute:

	$$\chi^0_{\alpha\beta}$$	$=$	$$S^T_{\alpha1}\sum_{\nu\mu}\frac{\langle \nu\vert\hat u_1'\vert\mu... ...hat u_1'\vert\nu \rangle}{\Delta_1 -\hbar \omega} (p_{\nu} -p_{\mu}) S_{1\beta}$$	(93)
		$=$	$$S^T_{\alpha1}S_{1\beta}\frac{\hbar^2}{2ma_0^2\Delta_1}\left(\frac{1}{\Delta_1-\hbar\omega}+\frac{1}{\Delta_1+\hbar\omega}\right )$$	(94)

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)

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

$=$

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

(95)

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