[Pw_forum] IFC's and eigendisplacements in example06
Ömer Koç
efeomer76 at gmail.com
Thu Dec 13 08:54:11 CET 2007
Dear PWSCF users,
My questions are about example06
1) Could you give me some analysis and explanations on the range of the
interatomic force constants (IFC's) for AlAs in example06.
On the Forum, there are some informations and explanations but i don't
understand them. I need more explanations.
alas444.fc
2 2 2 10.5000000 0.0000000 0.0000000 0.0000000 0.0000000
0.0000000
1 'Al ' 24590.7655930491
2 'As ' 68285.4024548272
1 1 0.0000000 0.0000000 0.0000000
2 2 0.2500000 0.2500000 0.2500000
T
13.7439581 0.0000000 0.0000000
0.0000000 13.7439581 0.0000000
0.0000000 0.0000000 13.7439581
1
2.5582103 0.0000000 0.0000000
0.0000000 2.5582103 0.0000000
0.0000000 0.0000000 2.5582103
2
-2.5582103 0.0000000 0.0000000
0.0000000 -2.5582103 0.0000000
0.0000000 0.0000000 -2.5582103
4 4 4
1 1 1 1
1 1 1 2.17004984418E-01
2 1 1 -5.06952740903E-03
3 1 1 9.99492789182E-05
4 1 1 -5.06952740903E-03
1 2 1 1.07967422105E-02
2 2 1 -6.07584921925E-04
3 2 1 -1.61474518909E-04
4 2 1 -5.06952740903E-03
1 3 1 4.18136269798E-03
2 3 1 -6.07584921925E-04
3 3 1 9.99492789182E-05
4 3 1 -6.07584921925E-04
1 4 1 1.07967422105E-02
2 4 1 -5.06952740903E-03
3 4 1 -1.61474518909E-04
4 4 1 -6.07584921925E-04
1 1 2 -5.06952740903E-03
2 1 2 -1.61474518909E-04
3 1 2 -6.07584921925E-04
…………………………………………………………………………
2) Is there a meaning of ratio of displacement of the two ions (Al and As)
in the eigendisplacement pattern?
The calculated vibrational eigenvectors at X point are below. How can we
evaluate this data?
q = 0.0000 0.0000 1.0000
**************************************************************************
omega( 1) = 2.841119 [THz] = 94.770143 [cm-1]
( 0.611885 0.039292 -0.353070 -0.018348 0.000000 0.000000
) Al
( -0.352406 -0.018313 0.610733 0.039218 0.000000 0.000000 )
As
omega( 2) = 2.841119 [THz] = 94.770143 [cm-1]
( 0.353530 0.003400 0.613143 -0.001588 0.000000 0.000000
)
( 0.611989 -0.001585 0.352865 0.003394 0.000000 0.000000
)
omega( 3) = 6.563802 [THz] = 218.946309 [cm-1]
( 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
)
( 0.000000 0.000000 0.000000 0.000000 1.000000 0.000000
)
omega( 4) = 10.442067 [THz] = 348.312171 [cm-1]
( 0.753056 0.016289 -0.563430 0.002731 0.000000 0.000000
)
( 0.203284 -0.000985 -0.271700 -0.005877 0.000000 0.000000
)
omega( 5) = 10.442067 [THz] = 348.312171 [cm-1]
( -0.563178 -0.017050 -0.753227 -0.002855 0.000000 0.000000
)
( 0.271762 0.001030 0.203193 0.006152 0.000000 0.000000
)
omega( 6) = 12.209736 [THz] = 407.275646 [cm-1]
( 0.000000 0.000000 0.000000 0.000000 -0.138259 0.990396
)
( 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
)
**************************************************************************
diagonalizing the dynamical matrix ...
Many thanks in advance
Ömer.
--
Ömer Koç
100. Yıl University
Dept of Physics Van, Turkey
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.quantum-espresso.org/pipermail/users/attachments/20071213/5978166a/attachment.html>
More information about the users
mailing list