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