Re: formation of lowering operator and raising operator
- To: mathgroup at smc.vnet.net
- Subject: [mg119077] Re: formation of lowering operator and raising operator
- From: Ray Koopman <koopman at sfu.ca>
- Date: Sat, 21 May 2011 06:47:45 -0400 (EDT)
- References: <ir2vj9$gsd$1@smc.vnet.net>
On May 19, 4:40 am, tarun dutta <tarundut... at gmail.com> wrote:
> i have a basis like" ket={0,1,2,3}"
> define lowering operator as "Ai "where i can vary from 0 to 4
In general, 0 <= i <= Length@ket ?
> 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;
That decremented the value in position i.
>
> similarly, A2{0,1,2,3} will give result as Sqrt[1]{0,0,2,3} here i=2
That, too, decremented the value in position i.
>
> In general Ai{0,1,2,....i,,,,3,,,}===sqrt[i]{0,1,2,...i-1,...3...}
But that decremented the value in position i+1.
Which position should be decremented, i or i+1 ?
If it's i+1 then what should happen when i = Length@ket ?
>
> 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}
That incremented the value in position i+1.
Does i in Bi refer to the same position that i in Ai does?
What should B0 give?
>
> how will i construct it in mathematica?
> any help will be much appreciated..
> regards,
> tarun dutta