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Re: formation of lowering operator and raising operator

  • To: mathgroup at smc.vnet.net
  • Subject: [mg119020] Re: formation of lowering operator and raising operator
  • From: Peter Breitfeld <phbrf at t-online.de>
  • Date: Fri, 20 May 2011 06:35:02 -0400 (EDT)
  • References: <ir2vj9$gsd$1@smc.vnet.net>

tarun dutta wrote:

> i have a basis like" ket={0,1,2,3}"
> define lowering operator as "Ai "where i can vary from 0 to 4
> I need to operate it on the ''ket' such as
> A3{0,1,2,3} will give the result as Sqrt[2]{0,1,1,3}  ,here i=3;
>
> similarly,   A2{0,1,2,3} will give result as Sqrt[1]{0,0,2,3} here i=2
>
> In general Ai{0,1,2,....i,,,,3,,,}===sqrt[i]{0,1,2,...i-1,...3...}
>
> one constraint if A0{0,1,2,3} will give==Sqrt[0]{0,1,2,3}
> since number can not be negative within the basis..
>
> In the same way  if raising operator Bi operate on {0,1,2,3}
> as Bi{0,1,...i....,2,3}  will give sqrt[ i+1]{0,1,2.....i+1,...2,3}
>
>
> how will i construct it in mathematica?
> any help will be much appreciated..
> regards,
> tarun dutta
>

You may use ReplacePart. It's a little tricky, because Mathematica
counts list from 1:

A[i_][lst_] := 
 Sqrt[i] {First[lst], 
    ReplacePart[Rest[lst], i -> Rest[lst][[i]] - 1]} // Flatten
A[0][lst_] := 0*lst

ket = {0, 1, 2, 3};
Table[A[i][ket], {i, 0, 3}] // Column


Out=
 {0, 0, 0, 0},
 {0, 0, 2, 3},
 {0, Sqrt[2], Sqrt[2], 3 Sqrt[2]},
 {0, Sqrt[3], 2 Sqrt[3], 2 Sqrt[3]}

Your examples and your general rule differ. I made it for the general rule. 
-- 
_________________________________________________________________
Peter Breitfeld, Bad Saulgau, Germany -- http://www.pBreitfeld.de


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