Quantum Mechanics
Schrodinger equation
Quantum Mechanics
Propagators : Pg
Quantum Simple Harmonic Oscillator QSHO
Quantum Mechanics
Simulation With GNU Octave
© The scientific sentence. 2010
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| 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] = iℏLz
[Ly, Lz] = iℏLx
[Lz, Lx] = iℏLy
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+] = iℏJy + i x - iℏJx = iℏJy + ℏJx =
ℏ(Jx + i Jy) = + ℏ J+
Similarly,
[Jz, J-] = - ℏ J-
[Jz, J+] = + ℏ J+
[Jz, J-] = - ℏ 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 iℏJz = Jx2 + Jy2 + ℏJz = J2 - Jz2+ ℏJz
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 iℏJz = Jx2 + Jy2 - ℏJz = J2 - Jz2 - ℏJz
J+ J- = J2 - Jz2 + ℏJz
J- J+ = J2 - Jz2 - ℏJz
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) = + ℏ J+ ζ + J+Jz ζ =
ℏ J+ ζ + β J+ ζ = (ℏ + β) J+ ζ
Similarly,
Jz (J- ζ) = - ℏ J-ζ + J-Jz ζ =
- ℏ J- ζ + β J-ζ = (-ℏ + β) J- ζ
J2 (J+ζ) = α J+ (ζ)
Jz (J+ζ) = (ℏ + β) J+ζ
Jz (J-ζ) = (-ℏ + β) J-ζ
The ladder operators compined with Jz raise and
lower the eigenvalues of Jz by ℏ.
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 ℏj which is the
maximum eigenvalue for Jz, and ℏg the eigenstate at the bottom,
which is the minimum eigenvalue for Jz.
J2 = J- J+ + Jz 2 + ℏJz
We have J2 = J- J+ + Jz 2 + ℏJz.
Then:
J2 ζt = J- J+ζt + Jz 2 ζt + ℏJz ζt = 0 + j2ℏ2 ζt + ℏℏ j ζt
= ℏ2 (j2 + j) ζt =
ℏ2 j(j + 1) ζt
Therefore: α = ℏ2 (j2 + j) = ℏ2 j(j + 1)
Now from the bottom:
We have J2 = J+ J- + Jz 2 - ℏJz, then:
J2 ζb = J+ J- ζb + Jz 2 ζb - ℏJz ζb = 0 + g2ℏ2 ζb - ℏℏ g ζb
= ℏ2 (g2 - g) ζb = ℏ2 g(g - 1) ζb
Since: α = ℏ2 j(j + 1) = ℏ2 g(g + 1), therefore g = - j.
J2 ζt = ℏ2 j(j + 1) ζt
J2 ζb = ℏ2 j(j + 1) ζb
The eigenvalues of Jz are β = m ℏ. 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
J2 (ζjm) = j(j + 1)ℏ2 ζjm
Jz (ζjm) = m ℏm ζjm
m: from - j to + j
We have already :
J2 (J+ ζ) = α J+ (ζ)
Jz (J+ ζ) = (ℏ + β) J+ ζ
Jz (J- ζ) = (-ℏ + β) J- ζ
And
α = ℏ2 (j2 + j)
β = mℏ
That is:
J2 (J+ ζjm) = α J+ (ζjm)
J2 (J- ζjm) = α J+ (ζjm)
Jz (J+ ζjm) = ℏ(m + 1) J- ζjm)
Jz (J- ζjm) = ℏ(m -1) J- ζjm
Let's represent the eigenfunction ζjm by the eigenket
|j,m> and write:
J2 J+ |j,m> = ℏ2 (j2 + j)J+ |j,m>
J2 J- |j,m> = ℏ2 (j2 + j) J- |j,m>
Jz J+ |j,m> = ℏ(m + 1) J- |j,m>
Jz J- |j,m> = ℏ(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 + ℏJz
J- J+ = J2 - Jz 2 - ℏJz
Therefore:
|J+ |j,m>|2 = <j,m|J2 - Jz 2 - ℏJz |j,m>
= ℏ2 j(j + 1) - m2ℏ2 - mℏ2<j,m|j,m> = ℏ2 [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:
ℏ2 [j(j + 1) - m(m + 1)] = |ajm+|2.
Therefore:
ajm+ = ℏ[j(j + 1) - m(m + 1)]1/2
Similarly:
ajm- = ℏ[j(j + 1) - m(m - 1)]1/2
Thus:
J+ |j,m> = ℏ[j(j + 1) - m(m + 1)]1/2|j,m+1>
J- |j,m> = ℏ[j(j + 1) - m(m - 1)]1/2|j,m-1>
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