Re: formation of lowering operator and raising operator
- To: mathgroup at smc.vnet.net
- Subject: [mg119059] Re: formation of lowering operator and raising operator
- From: tarun dutta <tarunduttaz at gmail.com>
- Date: Sat, 21 May 2011 06:44:27 -0400 (EDT)
On May 20, 3:36 pm, DrMajorBob <btre... at austin.rr.com> wrote:
> 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 <tarundut... at gmail.com>
> wrote:
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> > 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
>
> --
> DrMajor... at yahoo.com
thanks bob,peter and mukasa.but there is something wrong in operation
such that i do not make you understand.first of all
a[0][x_List] := 0 x....this is not right, all elements of vector will
not be zero.
here is the right operation...
a[i]@{a,b,c,d}...if i=2 then a[i] will be a[2] and now
a[2]@{a,b,c,d},,,the result will be Sqrt[b]{a,b-1,c,d} and sqrt[b]
will be outside of this vector as display."i" determines the position
which element of vector will be decreasing by 1.
now if i=0 then a[i] will operate on the element which is in the
zeroth position of vector..e.g.
a[0]@{a,b,c,d}===result ==Sqrt[a]{a-1,b,c,d}....now if a=0 then this
element will remain 0,will not be 0-1and also will be no change in the
rest of the element..e.g.
a[0]@{a=0,b,c,d}===sqrt[a=0]{a-1,b,c,d}==sqrt[0]{0-1,b,c,d}==sqrt[0]
{0,b,c,d}
Sqrt[] will not be multiplied with the vector.
thanks again for replying.
regards,
tarun