Re: formation of lowering operator and raising operator
- To: mathgroup at smc.vnet.net
- Subject: [mg119035] Re: formation of lowering operator and raising operator
- From: Sseziwa Mukasa <mukasa at gmail.com>
- Date: Fri, 20 May 2011 06:37:45 -0400 (EDT)
On May 19, 2011, at 7:40 AM, 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..
a[i_,ket_]:=Sqrt[i-1] (ket-IntegerDigits[2^(i+1),2,Length[ket]])
a[0,ket_]:=ConstantArray[0,Length[ket]]
b[i_,ket_]:=Sqrt[i+1] (ket+IntegerDigits[2^(i+1),2,Length[ket])