[Wien] Parity determination by group theory analysis question

Paul Fons paul-fons at aist.go.jp
Fri Jun 6 02:56:19 CEST 2014


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> 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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Dr. Paul Fons
Senior Research Scientist
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