[Pw_forum] Imposed symmetry

[Stefano Baroni]

On Feb 6, 2009, at 10:12 AM, Gabriele Sclauzero wrote:

Gabriele:

> Stefano Baroni wrote:
>> parameters along energy derivatives (forces and stress). When the  
>> ground
>> state is nondegenerate, energy derivatives have the same symmetry  
>> as the
>> the geometry of the system.
>
> Please Stefano help me to understand this point, which is not clear  
> to me. If the ground
> state is degenerate (you meant degenerate at a fixed external  
> potential, i.e. fixed atomic
> geometry, am I right?)

YES

> you could reach a lower energy by lowering the symmetry: do you
> think this is possible in practical calculations.

This is possible in reality (in fact, the Jahn-Teller effect is  
exactly this). Whether or not this is possible "in practice" depends  
on how faithfully "practice" mimics "reality" (out of the metaphor: on  
how calculations are done).

> I thought that you could only end up in
> a configuration with higher symmetry than the starting one, not lower.

it depends ... when the nuclear geometry is such that the ground state  
is degenerate, it may happen that two or more Born-Oppenheimer  
surfaces (each one corresponding to a different degenerate state)  
intersect at that geometry in such a way that the gradients along each  
one of the surfaces does not vanish. In the nondegenerate case,  
symmetry trivially imposes that forces along symmetry-breaking  
directions vanish. In the degenerate case, the situation is more  
tricky: symmetry imposes that the SUM of the forces corresponding to  
different degenerate states vanish. Which implies that you can lower  
the energy by breaking the symmetry along one of the intersecting BO  
energy surfaces

> From the theoretical point of view, is it correct to apply the  
> standard formulation of
> HK theorem to a system with degenerate GS (is the HK mapping still  
> valid?)

this problem used to be a very popular and controversial one, some  
time ago. I do not know the "politically correct" answer. Let me give  
you mine.

1) strictly, the usual derivation of DFT does not hold for degenerate  
states
2) That derivation can be easily extended to a (arbitrarily small)  
finite temperature (Mermin's DFT does that)
3) at finite temperature, degeneracy does not matter. DFT holds, the  
Hellmann-Feynman theorem applies. only one BO surface exists, no  
symmetry breaking distortions would occur
4) at zero temperature complications arise and the BO surface would  
develop a cusp, corresponding to different forces along different  
directions. common sense is restored "in practice" by considering  
infinitesimal distortions of the high-simmetry geometry. the limit of  
the forces as the distortion goes to zero would depend on dierction,  
which is exactly a amnifestation of the Jahn-Teller effect

(or at least, so it seems to me)

Ciao
Stefano


---
Stefano Baroni - SISSA  &  DEMOCRITOS National Simulation Center -  
Trieste
http://www.sissa.it/~baroni / [+39] 040 3787 406 (tel) -528 (fax) /  
stefanobaroni (skype)

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de la pensée - Jean Piaget

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