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© The scientific sentence. 2010

Ladder Operators




The raising and lowering ladder operators J+ and J- 
respectively are defined by:

J+ = Jx + iJy, and
J- = Jx - iJy
They are self-adjoint: (J+)* = J- and (J-)* = J+.

Their commutation relations with Jz are: 
[Jz, J+] = JzJ+  - J+Jz = Jz(Jx + iJy) - (Jx + iJy)Jz = 
JzJx + iJzJy) - JxJz - iJyJz = JzJx - JxJz  + i(JzJy - JyJz)
= [Jz, Jx] + i[Jz,Jy]
We have seen that the following results:
[Lx, Ly] = iLz
[Ly, Lz] = iLx
[Lz, Lx] = iLy
That hold true for any angular momentum.

We denote by J an angular momentum  by J, that may representan 
orbital angular momentum (L)  or spin (S):

 [Jz, J+] = iJy + i x - iJx =  iJy + Jx = 
 (Jx + i Jy) = +  J+

Similarly,
[Jz, J-] = -  J-

[Jz, J+]  = +  J+
[Jz, J-] = -  J-


J2 = Jx2 + Jy2 + Jz2 
[J2, J+] = [Jx2 + Jy2 + Jz2 , J+] = 
[Jx2 + Jy2 + Jz2 , Jx + i Jy] = [Jx2 + Jy2 + Jz2 , Jx + i Jy] = 
Jx2Jx + i Jx2Jy + Jy2Jx + i Jy2Jy + Jz2Jx + i Jz2Jy 
- Jx2Jx + Jy2Jx - Jz2Jx  + iJx2Jy - iJy2Jy + iJz2Jy= 0

Similarly , [J2, J-] = 0


[J2, J+] = 0
[J2, J-] = 0


J+ J- = (Jx + i Jy) (Jx - iJy) = Jx2 - i Jx Jy + i Jy Jx + Jy2 = Jx2 - i [Jx, Jy] + Jy2 
= Jx2 + Jy2 - i iJz =  Jx2 + Jy2 + Jz = J2 - Jz2+ Jz 

Similarly,
J- J+ = (Jx - i Jy) (Jx + iJy) = Jx2 + i Jx Jy - i Jy Jx + Jy2 = Jx 2 + i [Jx, Jy]  + Jy2 
= Jx2 + Jy2 + i iJz =  Jx2 + Jy2 - Jz = J2 - Jz2 - Jz 


J+ J- =  J2 - Jz2 + Jz 
J- J+ =  J2 - Jz2 - Jz 


If is ζ an eigenfunction of both of J2 and Jz, so:
J2 ζ = α ζ and 
Jz ζ = β ζ

We have then:
J2 (J+ ζ) = J+(J2 ζ) = J+ (α ζ) = 
α J+ (ζ) 
And

Jz (J+ ζ) = +  J+ ζ + J+Jz ζ =
 J+ ζ + β J+ ζ = ( + β) J+ ζ

Similarly,

Jz (J- ζ) = -  J-ζ + J-Jz ζ =
-  J- ζ + β J-ζ = (- + β) J- ζ


J2 (J+ζ) = α J+ (ζ)
Jz (J+ζ) = ( + β) J+ζ
Jz (J-ζ) = (- + β) J-ζ


The ladder operators compined with Jz raise and 
lower the eigenvalues of Jz by . 
But the eigenvalues of J2  remain unchanged.

There must be limits for rising and lowering these eigenvalues. Let's set 
the top step of the ladder as ζt, and from the bottom as ζb.
Therefore:
J+ ζt = 0, and J- ζb = 0 

Let the eigenvalue of Jz at the top step be j which is the 
maximum eigenvalue for Jz, and g the eigenstate at the bottom,
 which is the minimum eigenvalue for Jz.

J2 = J- J+  + Jz 2 + Jz 
We have J2 = J- J+  + Jz 2 + Jz.
Then:
J2 ζt = J- J+ζt   + Jz 2 ζt + Jz ζt  = 0   + j22 ζt +  j ζt 
= 2 (j2 + j) ζt = 
2 j(j + 1) ζt
Therefore: α = 2 (j2 + j) = 2 j(j + 1)

Now from the bottom:
  
We have J2  = J+ J- +  Jz 2 - Jz, then:

J2 ζb = J+ J- ζb   + Jz 2 ζb - Jz ζb  = 0   + g22 ζb -  g ζb 
= 2 (g2 - g) ζb = 2 g(g - 1) ζb

Since:  α = 2 j(j + 1) = 2 g(g + 1), therefore g = - j.

J2 ζt = 2 j(j + 1) ζt
J2 ζb = 2 j(j + 1) ζb

The eigenvalues of Jz are β = m . The ivalue "m" 
is equal to one step within the ladder of the "j"s, of the 
range 2j + 1 (from - j to + j). The integer m varies from j to j. 
If the number of steps is N, we have j = N/2, that is either 
integer or half integer.

The eigenfunction ζ is characterized by the quantum 
numbers j and m. We write ζjm

J2jm) = j(j + 1)2 ζjm
Jzjm) = m m ζjm
m: from - j to + j


We have already :

J2 (J+ ζ) = α J+ (ζ)
Jz (J+ ζ) = ( + β) J+ ζ
Jz (J- ζ) = (- + β) J- ζ
And
α = 2 (j2 + j)
β = m

That is:

J2 (J+ ζjm) = α J+jm)
J2 (J- ζjm) = α J+jm)
Jz (J+ ζjm) = (m + 1) J- ζjm)
Jz (J- ζjm) = (m -1) J- ζjm

Let's represent the eigenfunction ζjm by the eigenket 
|j,m> and write:

J2 J+ |j,m> = 2 (j2 + j)J+ |j,m> 
J2 J- |j,m>  = 2 (j2 + j) J- |j,m> 
Jz J+ |j,m>  = (m + 1) J- |j,m> 
Jz J- |j,m>  = (m -1) J- |j,m> 


Now we are forcing J+ and J- to satisty the 
following equations:
J+ |j,m>  = ajm+|j,m+1> 
J- |j,m>  = ajm-|j,m-1>
And determine ajm+ and ajm-.

We have:
(J+ |j,m>)* = <(J+)T| =  a*jm+<j,m+1|; so:
|J+ |j,m>|2 = <j,m|J-J+ |j,m>

We have already:
J+ J- =  J2 - Jz 2 + Jz 
J- J+ =  J2 - Jz 2 - Jz 

Therefore:
|J+ |j,m>|2 = <j,m|J2 - Jz 2 - Jz |j,m>
= 2 j(j + 1) - m22 - m2<j,m|j,m> = 2 [j(j + 1) - m(m + 1)]

We have also:
|J+ |j,m>|2 = |ajm+|2<j,m+1|j,m+1> = |ajm+|2
(|j,m+1> are orthogonal)

Then:
 2 [j(j + 1) - m(m + 1)] = |ajm+|2. 
Therefore:
 
ajm+ = [j(j + 1) - m(m + 1)]1/2

Similarly:
ajm- = [j(j + 1) - m(m - 1)]1/2

Thus:


J+ |j,m>  = [j(j + 1) - m(m + 1)]1/2|j,m+1> 
J- |j,m>  = [j(j + 1) - m(m - 1)]1/2|j,m-1>


  


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