Re: formation of lowering operator and raising operator
- To: mathgroup at smc.vnet.net
- Subject: [mg119026] Re: formation of lowering operator and raising operator
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Fri, 20 May 2011 06:36:07 -0400 (EDT)
Clear[a] a[0][x_List] := 0 x a[k_][x_List] /; Length@x >= k := (x - UnitVector[Length@x, k]) Sqrt@Max[0, k - 1] a[3]@{0, 1, 2, 3} {0, Sqrt[2], Sqrt[2], 3 Sqrt[2]} a[2]@{0, 1, 2, 3} {0, 0, 2, 3} a[0]@{0, 1, 2, 3} {0, 0, 0, 0} Bobby On Thu, 19 May 2011 06:40:24 -0500, tarun dutta <tarunduttaz 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 > 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 > -- DrMajorBob at yahoo.com