[Wien] LMMAX again

[Peter Blaha]

I'm not sure if it is really that difficult, but anyway:

For general symmetry, for l=1 there are 3 (complex) spherical harmonics 
with m=0,+ and -1, right ? 

Of course, also "real spherical hamonics" must have EXACTLY 3 combinations.

They are called   (1 0); (1 1) and (-1 1).!! A combination (1 -1) would be
a fourth combination and thus some redundend linear combination.

So without any symmetry (pointgroup 1) the list contains all possible
combinations (but not redundand ones!), namely 3 for l=1; 5 for l=2;...
lm: 0 0  1 0  1 1 -1 1  2 0  2 1 -2 1  2 2 -2 2

You never see any negative m!

For pointgroup  m (that's yours!?) the LM list starts with:
0 0  1 1 -1 1  2 0  2 2 -2 2  3 1 -3 1  3 3 -3 3  4 0  4 2 -4 2  4 4 -4 4

and already from this list it is obvious that for "odd" l you should remove
all "even" m values (1 0; 3 0; 3 2; -3 2; guess what comes for l=7 ?)

On the other hand for even l you must remove the "odd" m; (2 1; -2 1; ...)


  
> Thanks a lot to your patient explanations. However, I still have problems with it. According to Table7.39 in the UG, for example, there are four possible choices for l=1 with point group "M", (+-l,l-2m), they are (1,1),(-1,1) for m=0 and (1,-1),(-1,-1) for m=1. It is obvious that only (1,1) and (-1,1) are listed in the case.in2 file. I don't understand why (1,-1)(-1,-1) are not included? I think I have to understand this first before I try to  translate the lm definitions up to higher combinations given in Table7.39.


                                      P.Blaha
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Peter BLAHA, Inst.f. Materials Chemistry, TU Vienna, A-1060 Vienna
Phone: +43-1-58801-15671             FAX: +43-1-58801-15698
Email: blaha at theochem.tuwien.ac.at    WWW: http://info.tuwien.ac.at/theochem/
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