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formation of lowering operator and raising operator

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  formation of lowering operator and raising operator tarun dutta <tarunduttaz@gmail.com> - 2011-05-19 11:40 +0000
    Re: formation of lowering operator and raising operator Peter Breitfeld <phbrf@t-online.de> - 2011-05-20 10:34 +0000
    Re: formation of lowering operator and raising operator Ray Koopman <koopman@sfu.ca> - 2011-05-21 10:47 +0000

#2494 — formation of lowering operator and raising operator

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

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#2499

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

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#2561

On May 19, 4:40 am, tarun dutta <tarundut...@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

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