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simplifying an expression using non-commutative algebra

  • To: mathgroup at smc.vnet.net
  • Subject: [mg113203] simplifying an expression using non-commutative algebra
  • From: Kirill Igumenshchev <bribeme at gmail.com>
  • Date: Mon, 18 Oct 2010 05:48:42 -0400 (EDT)

Hi everyone,

I am new to this and the main reason is that i got really stuck on
smtn and i'd love to learn more. i posted this question also at
http://stackoverflow.com/questions/3955913/simplifying-and-rearranging-non-commutative-variables-in-mathematica

here we go:

this is a little complicated. i spent 2 weeks on this, so i'd love ur
input or suggestion how to figure it out or where to post.

in short, i have an expression that contains multiplications between
p1,p2,q1 and q2, and i'd like to use [qi,pi]=ii*hb, where i={1,2} to
get the expression to a symmetric form (pi^a*qi^b+qi^b*pi^a)/2.

so for example, for p2*q2*p2^2 i get (p2*q2^3+q2^3*p2)/2 +
1/2*ii*p2^2*hb using simplification and some replacements. but i
cannot simplify q2*q1^2*p2 although i inputed a rule q2*p2->
(p2*q2+q2*p2)/2 +ii/2*hb and that variables with 1s and 2s commute.

in more detail,

here is the mathematica code (i use a package
http://homepage.cem.itesm.mx/lgomez/quantum/):
http://dl.dropbox.com/u/8916126/post.nb

the code works when the index is either 1 or 2 but doesn't work when
both indexes are used:
p2*q2*q1*q2 gives p2*q1*q2^2, p2*q2*q2 can further be simplified  but
since there is q1, mathematica doesn't do it.

in even more detail:
in the end, i'm trying to write a mathematica code that can get
equations in appendix (eq. A2) in this papaer:
http://dl.dropbox.com/u/8916126/Prezhdo02_8704.pdf .
 and http://dl.dropbox.com/u/8916126/henonHeiles7.nb is the code that
i'm using. it is a little different because i couldn't get the post.nb
to run as well but post.nb would be ideal.

in the end i'd like to use it for other kind of hamiltonians upto 4th
power or even higher.

I understand that i might not be clear about this so let me know if i
can articulate better.


thank you,

--Kirill


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