For ThCr2Si2 structure (space group I4/mmm) sample, e.g. CeMn2Ge2, Mn atoms occupy the 4d site - site position (0, 0.5, 0.25). It was said that this sample has a ferromagnetic spiral structure (see attached figure) on the Mn site.
http://www.sciencedirect.com/science/article/pii/0925838894013780
How to create a spiral magnetic structure based on irreducible representation analysis, e.g. in CeMn2Ge2 assuming that the propagation vector along c-axis? What are the important points that should be noted?
You have a propagation wave vecktor (0,0,k) in a host crystal with point group D_4h = D_4 x i (magnetic moments are invariant under space inversion, so we factor out space inversion = i).
Now, under which point symmetry elements of D_4 is (0,0,k) invariant?
All the rotations about z, none of the others. This leaves us with point group C_4.
C_4 has 3 IRs, labelled A, B, and E. If you take a closer look you will find that E is actually composed of 2 1D-IRs that are complex conjugates of each other. These will be your spiral structures.
http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=204&option=4
Hi Detlefs, thanks very much. Can you provide some more detailed explanation how to reduce D_4h to D_4 and then to C_4?
D_4h is simply the product group of D_4 and i (the proper name is probably C_i). In other words, when you remove inversion symmetry (and all operations containing inversion) from D_4h you end up with D_4, which is a subgroup of D_4h.
D_4 contains rotations about (001), (100) and (110). Rotations about (100) and (110) do not leave (00k) invariant, therefore have to be thrown out. What remains is the subgroup C_4. When you add space inversion back in you get C_4h.
The next step is then to put a "test moment" (mx, my, mz) onto one of the Mn sites, apply all the symmetries of C_4h multiplied by the IR of choice, and see what moments result on the other Mn sites - you have to apply the symmetry operation to the moments and the Mn positions. You need inversion symmetry to link the (xy,0.25) to the (xy,0.75) sites.
Hi Detlefs, there are still two questions:
1. In the first step there are a lot of different subgroups. Why it should remove inversion symmetry to get D_4. And late this inversion symmetry was added back in C_4?
2 In the last step to test moment, does the operation of inversion symmetry on the magnetic moment reverse the magnetic moment direction?
1) You don't have to, but I find it easier to separate the even and odd operations.
2) When putting the inversion back in, you double the number of IRs - for each IR in C_4 you get an even and an odd IR in C_4h where, essentially, the inversion does not or does invert the magnetic moment. You want the E_g (even), otherwise the (xy 0.25) ferromagnetic moment would be opposite the (xy 0.75) ferromagnetic moment and the net moment would vanish.
In relation to one of your other questions:
I4/mmm is not a chiral space group. Therefore you expect symmetry-equivalent left- and right-handed domains to exist. One domain type has to be the mirror image of the other. These are the two complex-conjugate partners in the E (or E_g) IR.
There are chiral space groups where you find just one helicity in the magnetic structure.
- Badges
- Science topic
- Similar topics
- Physical Sciences
- Thermodynamics