[Wien] Parity determination by group theory analysis question
Fecher, Gerhard
fecher at uni-mainz.de
Fri Jun 6 09:02:05 CEST 2014
You say that you don't have inversion symmetry, however, D3d has inversion symmetry (see your first e-mail)
Just some remarks from the reference given by irrep: Table 55 on page 58 in Koster et al [7]
D3d = D3 x Ci
even (+) and odd (-) parity mean even = s, d, g and odd = p, f, ... angular momentum
Gamma5- = Gamma5+ x Gamma1-
Gamma6- = Gamma6+ x Gamma1-
Gamma5+ = d3/2,-3/2 - i d3/2,+3/2 (and higher total and orbital angular momenta, g, i, ..., d5/2,+-3/2, g7/2,+-3/2, g9/2,+-3/2,)
Gamma6+ = -d3/2,3/2 - i d3/2,-3/2 (- " -)
Gamma1- = pzSz
Gamma5+ & Gamma5- (same for 6) would mean that you mix d states with p and/or f states (or higher angular momenta), e.g. p3/2,+-3/2, f5/2,+-3/2, f7/2,+-3/2
If you apply a magnetization along z then you should have C3i symmetry
However, I am not sure whether the parity assignment in irrep is always correct in the sense of Koster,
I experienced some different results for unoccupied states in cubic symmetry where the DOS told d states but the parity was assigned by irrep to odd.
Ciao
Gerhard
DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy:
"I think the problem, to be quite honest with you,
is that you have never actually known what the question is."
====================================
Dr. Gerhard H. Fecher
Institut of Inorganic and Analytical Chemistry
Johannes Gutenberg - University
55099 Mainz
________________________________________
Von: wien-bounces at zeus.theochem.tuwien.ac.at [wien-bounces at zeus.theochem.tuwien.ac.at]" im Auftrag von "Paul Fons [paul-fons at aist.go.jp]
Gesendet: Freitag, 6. Juni 2014 02:56
An: A Mailing list for WIEN2k users
Betreff: Re: [Wien] Parity determination by group theory analysis question
Hi,
Thank you for your note. I am most grateful for some help on the group theory question as there seem to be very few people in general who answer such questions. I should have mentioned I realize in retrospect that the calculation was carried out with spin-orbit coupling. The degeneracy in the spin is presumably due to the presence of inversion symmetry in the crystal structure (the crystal structure does have inversion symmetry). Would it be fair to conclude that this is a case of "accidental degeneracy" and that there are two sets of states (GM5+ + GM6+) and (GM5- + GM6-) and that the two pairs of states have parity 1 and -1 respectively?
Best wishes,
Paul
On Jun 4, 2014, at 18:08, Juan Manuel Perez Mato <wmppemam at lg.ehu.es<mailto:wmppemam at lg.ehu.es>> wrote:
Hi,
(GM5+ + GM6+ ) and (GM5- + GM6-) form two different "physically" irreducible representations of even and odd parity of D3d (as double group). Therefore the degeneracy of the bands corresponding to these two physically irreps is not forced by symmetry. Consequently the four-fold degeneracy of the band most probably comes from not considering the spin. I would bet that it will disappear if spin orbit coupling is introduced, and they will then split into two separate two-fold degenerate bands associated with the two different physically irreducible representations of different parity.
regards,
J. Manuel Perez-Mato
Fac. Ciencia y Tecnologia,
Universidad del Pais Vasco, UPV
48080 BILBAO,
Spain
tel. +34 946012473
fax. +34 946013500
***************************************************
El 04/06/2014, a las 09:29, Paul Fons escribió:
Hi,
I have been using the irreducible representations calculated by irrep to determine the parity of the wavefunction for particular k-points for each band. A typical case is shown below for the gamma point. As I am solving the system without spin, each band is at least two-fold degenerate. For most bands, for example the first band, the situation is clear as the irrep is G4+ that the parity is +. This also follows from the character of the (I) class being the same dimension as the irrep. I am a little more confused by what to do a band such as band 9 where there are two irreps spanning the same band, but presumably as the character of the (I) class is the same as that of the dimension of the irrep (and that both irreps are of the same parity, +) that the parity is again +. The situation I am writing to ask about, however is that of band 63 which is four-fold degenerate and is composed of irreps of both parities. How does one interpret the parity from the information below for band 63? Thanks for any advice.
The point group is D3d
12 symmetry operations in 6 classes
Table 55 on page 58 in Koster et al [7]
Table 42.4 on page 371 in Altmann et al [8]
E 2C3 3C2 I 2IC3 3IC2
G1+ A1g 1 1 1 1 1 1
G2+ A2g 1 1 -1 1 1 -1
G3+ Eg 2 -1 0 2 -1 0
G1- A1u 1 1 1 -1 -1 -1
G2- A2u 1 1 -1 -1 -1 1
G3- Eu 2 -1 0 -2 1 0
--------------------------------------------
G4+ E1/2g 2 1 0 2 1 0
G5+ 1E3/2g 1 -1 i 1 -1 i
G6+ 2E3/2g 1 -1 -i 1 -1 -i
G4- E1/2u 2 1 0 -2 -1 0
G5- 1E3/2u 1 -1 i -1 1 -i
G6- 2E3/2u 1 -1 -i -1 1 i
class, symmetry ops, exp(-i*k*taui)
E 10 (+1.00+0.00i)
2C3 2 7 (+1.00+0.00i)
3C2 1 8 9 (+1.00+0.00i)
I 3 (+1.00+0.00i)
2IC3 6 11 (+1.00+0.00i)
3IC2 4 5 12 (+1.00+0.00i)
bnd ndg eigval E 2C3 3C2 I 2IC3 3IC2
1 2 -7.239676 2.00+0.00i 1.00+0.00i 0.00-0.00i 2.00+0.00i 1.00-0.00i 0.00-0.00i =G4+
3 2 -7.239533 2.00+0.00i 1.00+0.00i -0.00+0.00i -2.00+0.00i -1.00-0.00i 0.00-0.00i =G4-
5 2 -6.641446 2.00+0.00i 1.00+0.00i 0.00+0.00i 2.00-0.00i 1.00+0.00i 0.00+0.00i =G4+
7 2 -6.641393 2.00-0.00i 1.00-0.00i -0.00-0.00i -2.00-0.00i -1.00+0.00i 0.00-0.00i =G4-
9 2 -6.641277 2.00+0.00i -2.00-0.00i -0.00+0.00i 2.00+0.00i -2.00-0.00i -0.00-0.00i =G5+ + G6+
63 4 -1.880979 4.00-0.00i -4.00+0.00i 0.00-0.00i 0.00-0.00i -0.00+0.00i -0.00+0.00i =G5+ + G6+ + G5- + G6-
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