[Pw_forum] rotational ASR in nanotube
baroni at sissa.it
Fri Aug 27 18:07:05 CEST 2004
Dear Gang Wu:
I am not sure that what I am going to suggest is the simplest thing you
can do, but it will work (I think).
You have a matrix (the dynamical matrix) which you know should have a
zero eigenvalue and you also
know the eigenvector (because of you previous analysis). The matrix
that is actually produced by the code
satisfies these criteria only approximately.
What is the most general expression of a (simmetric) matrix, say A,
having a_0 as an eigenvalue and |u0> as an eigenvector?
A = a_0 |u0><u0| + (1-|uo><u0|) ANYTHING (1-|u0><u0|)
where ANYTHING is any symmetric matrix.
Can you see now how to choose , a_0, |u_0>, and ANYTHING in such a way
as to make A as close as possible to your calculated dynamical matrix?
This, I think would be the answer to your problem.
Take a thought at this and revert to us with the solution!
On Aug 27, 2004, at 1:36 PM, WU Gang wrote:
> Dear Mr. Baroni,
> Many thanks for your reply. With your kindly help, I have gotten the
> criterion of rotational ASR and a formula which can check whether my
> force constant matrix obeys this criterion. But how can I fix the
> original force constant matrix to make it obey the rotational ASR? Do
> I need to add a Lagrange constraint? But I have no idea how to add
> this. Would you please give me some advice? Thank you very much!
> Gang Wu
> On Wed, 25 Aug 2004 12:44:55 +0200, Stefano Baroni <baroni at sissa.it>
>> On Aug 25, 2004, at 12:22 PM, WU Gang wrote:
>>> Hello All.
>>> Recently I calculated the phonon-dispersion curves for nanotubes, but
>>> I found that at Gamma point, there is always some imaginary phonon
>>> frequencies, even when I applied translational acoustic summing rule
>>> (ASR) on the final force constant matrix. After look up some
>>> reference, a so-called rotational ASR seems to be important in reduce
>>> these unstable frequencies.
>> Nice problem of physics!
>>> But What does this rule mean in nanotube?
>> It simply means that you cannot pay any energy to twisting a tube
>> around its axis.
>> In free space, this is trivially rotational invariance. With periodic
>> boundary conditions,
>> of course, such a twist would correspond to a non-trivial periodic
>> which has got
>> zero frequencies. The same, by the way would hold or any molecule.
>>> And how can I apply this rule to the result force constant matrix?
>> Once you understand what physically this "rotational ASR" means,
>> it should not be difficult to figure out how to impose it onto the
>> dynamical matrix that
>> you calculate. We would leave this to you as an exercice (Hint: figure
>> out which atomic
>> displavement pattern would correspond to a rigid twist of the tube and
>> see how the ASR is
>> imposed in the translational case).
>>> Thank you very much!
>> You are most welcome!
>> Stefano Baroni
>> Stefano Baroni --- SISSA & DEMOCRITOS National Simulation Center
>> via Beirut 2-4 34014 Trieste Grignano / [+39] 040 3787 406 (tel) -528
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Stefano Baroni --- SISSA & DEMOCRITOS National Simulation Center
via Beirut 2-4 34014 Trieste Grignano / [+39] 040 3787 406 (tel) -528
Please, if possible, don't send me MS Word or PowerPoint attachments
Why? See: http://www.gnu.org/philosophy/no-word-attachments.html
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