Quantum Mechanics
Schrodinger equation
Quantum Mechanics
Propagators : Pg
Quantum Simple Harmonic Oscillator QSHO
Quantum Mechanics
Simulation With GNU Octave
© 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 selfadjoint: (J_{+})^{*} = J_{} and (J)^{*} = J_{+}.
Their commutation relations with J_{z} are:
[J_{z}, J_{+}] = JzJ_{+}  J_{+}J_{z} = J_{z}(Jx + iJy)  (Jx + iJy)J_{z} =
JzJx + iJzJy)  JxJz  iJyJz = JzJx  JxJz + i(JzJy  JyJz)
= [J_{z}, Jx] + i[J_{z},Jy]
We have seen that the following results:
[L_{x}, L_{y}] = iℏL_{z}
[L_{y}, L_{z}] = iℏL_{x}
[L_{z}, L_{x}] = iℏL_{y}
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):
[J_{z}, J_{+}] = iℏJy + i x  iℏJx = iℏJy + ℏJx =
ℏ(Jx + i Jy) = + ℏ J_{+}
Similarly,
[J_{z}, J] =  ℏ J_{}
[J_{z}, J_{+}] = + ℏ J_{+}
[J_{z}, J] =  ℏ J_{}
J^{2} = Jx^{2} + Jy^{2} + J_{z}^{2}
[J^{2}, J_{+}] = [Jx^{2} + Jy^{2} + J_{z}^{2} , J_{+}] =
[Jx^{2} + Jy^{2} + J_{z}^{2} , Jx + i Jy] = [Jx^{2} + Jy^{2} + J_{z}^{2} , Jx + i Jy] =
Jx^{2}Jx + i Jx^{2}Jy + Jy^{2}Jx + i Jy^{2}Jy + J_{z}^{2}Jx + i J_{z}^{2}Jy
 Jx^{2}Jx + Jy^{2}Jx  J_{z}^{2}Jx + iJx^{2}Jy  iJy^{2}Jy + iJz^{2}Jy= 0
Similarly , [J^{2}, J_{}] = 0
[J^{2}, J_{+}] = 0
[J^{2}, J] = 0
J_{+} J_{} = (Jx + i Jy) (Jx  iJy) = Jx^{2}  i J_{x} J_{y} + i J_{y} J_{x} + J_{y}^{2} = J_{x}^{2}  i [J_{x}, J_{y}] + J_{y}^{2}
= J_{x}^{2} + J_{y}^{2}  i iℏJ_{z} = J_{x}^{2} + J_{y}^{2} + ℏJ_{z} = J^{2}  J_{z}^{2}+ ℏJ_{z}
Similarly,
J_{} J_{+} = (J_{x}  i J_{y}) (J_{x} + iJ_{y}) = J_{x}^{2} + i J_{x} J_{y}  i J_{y} J_{x} + J_{y}^{2} = J_{x} 2 + i [J_{x}, J_{y}] + J_{y}^{2}
= J_{x}^{2} + J_{y}^{2} + i iℏJ_{z} = J_{x}^{2} + J_{y}^{2}  ℏJ_{z} = J^{2}  J_{z}^{2}  ℏJ_{z}
J_{+} J_{} = J^{2}  J_{z}^{2} + ℏJ_{z}
J_{} J_{+} = J^{2}  J_{z}^{2}  ℏJ_{z}
If is ζ an eigenfunction of both of J^{2} and J_{z}, so:
J^{2} ζ = α ζ and
J_{z} ζ = β ζ
We have then:
J^{2} (J_{+} ζ) = J_{+}(J^{2} ζ) = J_{+} (α ζ) =
α J_{+} (ζ)
And
J_{z} (J_{+} ζ) = + ℏ J_{+} ζ + J_{+}J_{z} ζ =
ℏ J_{+} ζ + β J_{+} ζ = (ℏ + β) J_{+} ζ
Similarly,
J_{z} (J_{} ζ) =  ℏ J_{}ζ + J_{}J_{z} ζ =
 ℏ J_{} ζ + β J_{}ζ = (ℏ + β) J_{} ζ
J^{2} (J_{+}ζ) = α J_{+} (ζ)
J_{z} (J_{+}ζ) = (ℏ + β) J_{+}ζ
J_{z} (J_{}ζ) = (ℏ + β) J_{}ζ
The ladder operators compined with J_{z} raise and
lower the eigenvalues of J_{z} by ℏ.
But the eigenvalues of J^{2} remain unchanged.
There must be limits for rising and lowering these eigenvalues. Let's set
the top step of the ladder as ζ_{t}, and from the bottom as ζ_{b}.
Therefore:
J_{+} ζ_{t} = 0, and J_{} ζ_{b} = 0
Let the eigenvalue of J_{z} at the top step be ℏj which is the
maximum eigenvalue for J_{z}, and ℏg the eigenstate at the bottom,
which is the minimum eigenvalue for J_{z}.
J^{2} = J_{} J_{+} + J_{z} 2 + ℏJ_{z}
We have J^{2} = J_{} J_{+} + J_{z} 2 + ℏJ_{z}.
Then:
J^{2} ζ_{t} = J_{} J_{+}ζ_{t} + J_{z} 2 ζ_{t} + ℏJ_{z} ζ_{t} = 0 + j^{2}ℏ^{2} ζ_{t} + ℏℏ j ζ_{t}
= ℏ^{2} (j^{2} + j) ζ_{t} =
ℏ^{2} j(j + 1) ζ_{t}
Therefore: α = ℏ^{2} (j^{2} + j) = ℏ^{2} j(j + 1)
Now from the bottom:
We have J^{2} = J_{+} J_{} + J_{z} 2  ℏJ_{z}, then:
J^{2} ζ_{b} = J_{+} J_{} ζ_{b} + J_{z} 2 ζ_{b}  ℏJ_{z} ζ_{b} = 0 + g^{2}ℏ^{2} ζ_{b}  ℏℏ g ζ_{b}
= ℏ^{2} (g^{2}  g) ζ_{b} = ℏ^{2} g(g  1) ζ_{b}
Since: α = ℏ^{2} j(j + 1) = ℏ^{2} g(g + 1), therefore g =  j.
J^{2} ζ_{t} = ℏ^{2} j(j + 1) ζ_{t}
J^{2} ζ_{b} = ℏ^{2} j(j + 1) ζ_{b}
The eigenvalues of J_{z} 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 ζ_{j}^{m}
J^{2} (ζ_{j}^{m}) = j(j + 1)ℏ^{2} ζ_{j}^{m}
J_{z} (ζ_{j}^{m}) = m ℏm ζ_{j}^{m}
m: from  j to + j
We have already :
J^{2} (J_{+} ζ) = α J_{+} (ζ)
J_{z} (J_{+} ζ) = (ℏ + β) J_{+} ζ
J_{z} (J_{} ζ) = (ℏ + β) J_{} ζ
And
α = ℏ^{2} (j^{2} + j)
β = mℏ
That is:
J^{2} (J_{+} ζ_{j}^{m}) = α J_{+} (ζ_{j}^{m})
J^{2} (J_{} ζ_{j}^{m}) = α J_{+} (ζ_{j}^{m})
J_{z} (J_{+} ζ_{j}^{m}) = ℏ(m + 1) J_{} ζ_{j}^{m})
J_{z} (J_{} ζ_{j}^{m}) = ℏ(m 1) J_{} ζ_{j}^{m}
Let's represent the eigenfunction ζ_{j}^{m} by the eigenket
j,m> and write:
J^{2} J_{+} j,m> = ℏ^{2} (j^{2} + j)J_{+} j,m>
J^{2} J_{} j,m> = ℏ^{2} (j^{2} + j) J_{} j,m>
J_{z} J_{+} j,m> = ℏ(m + 1) J_{} j,m>
J_{z} J_{} j,m> = ℏ(m 1) J_{} j,m>
Now we are forcing J_{+} and J_{} to satisty the
following equations:
J_{+} j,m> = a_{jm+}j,m+1>
J_{} j,m> = a_{jm}j,m1>
And determine a_{jm+} and a_{jm}.
We have:
(J_{+} j,m>)^{*} = <(J_{+})^{T} = a^{*}_{jm+}<j,m+1; so:
J_{+} j,m>^{2} = <j,mJ_{}J_{+} j,m>
We have already:
J_{+} J_{} = J^{2}  J_{z} 2 + ℏJ_{z}
J_{} J_{+} = J^{2}  J_{z} 2  ℏJ_{z}
Therefore:
J_{+} j,m>^{2} = <j,mJ^{2}  J_{z} 2  ℏJ_{z} j,m>
= ℏ^{2} j(j + 1)  m^{2}ℏ^{2}  mℏ^{2}<j,mj,m> = ℏ^{2} [j(j + 1)  m(m + 1)]
We have also:
J_{+} j,m>^{2} = a_{jm+}^{2}<j,m+1j,m+1> = a_{jm+}^{2}
(j,m+1> are orthogonal)
Then:
ℏ^{2} [j(j + 1)  m(m + 1)] = a_{jm+}^{2}.
Therefore:
a_{jm+} = ℏ[j(j + 1)  m(m + 1)]^{1/2}
Similarly:
a_{jm} = 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,m1>

