[Trilinos-Users] Anasazi: Matrix exponential for time evolution ?

Bochev, Pavel B pbboche at sandia.gov
Wed Mar 24 20:48:54 MDT 2010


David,

You may have already seen that paper, but just in case, for a rather comprehensive summary of algorithms and techniques, 

# Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later
# Cleve Moler and Charles Van Loan
# SIAM Review, Vol. 45, No. 1 (Mar., 2003), pp. 3-49 

is a good starting point, as well as the original from 1978.

Pavel Bochev


________________________________________
From: trilinos-users-bounces at software.sandia.gov [trilinos-users-bounces at software.sandia.gov] On Behalf Of David Hochstuhl [Davidhochstuhl at web.de]
Sent: Wednesday, March 24, 2010 3:46 PM
To: trilinos-users at software.sandia.gov
Subject: [Trilinos-Users] Anasazi: Matrix exponential for time evolution ?

Hello,



I have not only a single question but rather need a whole algorithm...lets start:



I am trying to propagate the Schrödinger equation in time,

i d/dt C = H(t) C

where H is a sparse matrix.



The initial state to this propagation was found by solving the time-independent Schrödinger equation

H C = E C

with Anasazi.





Ok, for the time evolution, I basically need to perform the following steps (or a combination of the two):



(i) Add a time-dependent pertubation D(t) to H,

     i.e. H(t) = H + D(t)

    (D is for instance the action of an electromagnetic field



(ii) For a given state C(t), apply the matrix exponential of -I*H(t) to obtain C(t+dt)

      C(t+dt) = exp(-I*H(t)*dt) C(t)





This can be done for instance with a simple implementation of the Lanczos algorithm, that people

usually call "short iterative Lanczos". However, the Anasazi solvers are of course much better than my

hand-written Lanczos agorithm. So, is there any way to perform the above procedure in an efficient way.



I thought about it yet, but I didn't find an efficient solution, and

before I start coding I wanted to hear what the trilinos cracks mean.

so, thanks in advance,



David






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