Solution of the exercises

section 5 The Hamiltonian
  1. Fast and simple: take some book on crystal fields [22] and look up which parameters are nonzero for the orthorhombic crystal field ($B_2^0$,$B_2^2$,$B_4^0$,$B_4^2$,$B_4^4$,$B_6^0$,$B_6^2$,$B_6^4$,$B_6^6$).

    Detailed Calculation: take the point group $D_2^h$, the Stevens Operators $O_l^m$ taken as a vector form a representation of this point group, which is reducible. Follow the procedure outlined in the book [62] to split this representation into irreducible parts. The basis vectors of the unit representation may then be linear combined with some arbitrary crystal field parameters to give the most general crystal field. Note that the basis vectors of the unit representation can be obtained efficiently by constructing the projection operator (eq. 4.51 in [62]) into the subspace transforming according to the irreducible unit representation.

  2. Unfortunately no table is available for all relevant problems. Some cases are given in [63]. 2 ways are possible to solve the problem. The accurate group theoretical approach which follows the calculation outlined above for the crystal field and needs a lot of knowledge in group theory. However a more intuitive calculation can be made by writing down the most general two ion interaction and subsequent introduction of symmetry elements which will then give relation between parameters. As an example in the primitive orthorhombic lattice there are 2 nearest neighbours, which are equivalent.

    In general the bilinear two ion interaction has the form $-\frac{1}{2} \sum_{ij} {\mathbf J}_i^{alpha}{\mathcal J}_{\alpha\beta}(ij){\mathbf J}_j^{\beta}$. $i$ and $j$ number the different positions of the magnetic ions in the lattice. Without loss of generality the interaction constants ${\mathcal J}_{\alpha\beta}(ij)$ can be chosen such that ${\mathcal J}_{\alpha\beta}(ij)={\mathcal J}_{\beta\alpha}(ji)$ (because the expression is symmetric in angular momentum components any anisotropic part of the interaction tensor does not contribute to the interaction energy).

    If $i$ and $j$ are nearest neighbours on a orthorhombic lattice, they are situated on one of the crystallographic axes, for example at [$\pm$100]. The off-diagonal components of the corresponding interaction tensor ${\mathcal J}_{\alpha\beta}(\pm100)$ vanish, because spin-configurations such as a moment (m00) on [000] and (0m0) on [100] must have the same magnetic energy as (m00) on [000] and (0-m0) on [100]. Furthermore the spin-configuration with (m00) on [000] and (m'00) on [100] must have the same magnetic energy as (m00) on [000] and (m'00) on [-100]. This and similar considerations lead to the conclusion that the interaction tensor must be the same for [-100] and [+100]. Therefore the most general form of the interaction ${\mathcal J}_{\alpha\beta}(\pm100)$ is

    \begin{displaymath}
{\mathcal J}_{\alpha\beta}(\pm100)=\left(
\begin{array}{...
... & 0 & {\mathcal J}_{cc}(\pm100) \\
\end{array}
\right)
\end{displaymath} (224)

section 6.1 Using so1ion separately
section 7.2 starting a simulation

  1. In the crystallographic basis there are two ions.
  2. ... the simulation should run after typing mcphas
section 7.4 output files
  1. at 2.8 Tesla
  2. (0.6 0 0), (2/3 0 0) and (0.625 0 0)
section 9 McDisp - the calculation program for magnetic excitations

the inspection of the file mcdisp.mf shows that the calculation is performed at 3.2 Tesla field parallel to $b$, i.e. in the ferromagnetically aligned state. Due to the two ions in the crystallographic unit cell in general 2 modes are expected for each crystal field transition. After running mcdisp the results may be inspected in file ./results/mcdisp.qom, graphically using the program disp. It turns out the the soft mode (gap at (2/310) shows the largest intensity.

Martin Rotter 2017-01-10