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Re: Question about Hold

• To: mathgroup at smc.vnet.net
• Subject: [mg48826] Re: Question about Hold
• From: "Steve Luttrell" <steve_usenet at _removemefirst_luttrell.org.uk>
• Date: Fri, 18 Jun 2004 02:12:48 -0400 (EDT)
• References: <carjrn$r76$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

You could try the following ( I asssume C is a c-number and not an
operator):

OpProd[x___, A, B, y___] := OpProd[x, B, A, y] + C OpProd[x, y];

OpProd[A, B] then evaluates to C OpProd[] + OpProd[B, A]

OpProd[A, A, B] then evaluates to 2 C OpProd[A] + OpProd[B, A, A]

There are lots of variations on this sort of approach. The basic trick this
depends on is using OpProd[<list of operators in the product>] to hide from
Mathematica the fact that you have a "product" of operators, so you can then
get on with defining your own private set of rules for simplifying OpProd
(as above, for example).

Steve Luttrell

"Daohua Song" <ds2081 at columbia.edu> wrote in message
news:carjrn$r76$1 at smc.vnet.net...
> Hi,evryone,
>     I define a commute relation for quantum operators,i fact i learn this
> from the forum.Here is the problem.
>     Commute[A,B]=C.
>     Commute[A^2,B], i try to use Hold[A*A] instead of A^2, then
mathematica
> even won't evaluate it.!
>     Any suggestion to work on the power commutation?
>     Thanks
> Daohua
>

• Follow-Ups:
  • Re: Re: Question about Hold
    • From: Daohua Song <ds2081@columbia.edu>