We now would like to calculate the kinetic moment eigenvalues. We know
that one can measure only the squared kinetic moment L² and one of its
components (we're talking about Lz) simultaneously. We will try to
find the eigenvalues of both measured quantities and the eigenfunction Y. One needs to understand the Commutativity
relations for the following derivation. First, we know that the squared
kinetic moment eigenvalues h²λ have been given. This is
not a restriction since λ can be any possible number taking
h² in a way that λ would be dimensionless. The
corresponding derivation is true for the z-component of of kinetic moment, the
eigenvalue of which we will assign as mh.
L² Y
= (Lx² +
Ly² + Lz²)Y = h² λ Y
Lz² Y = Lz (LzY) = Lz m
hY = m²
h² Y
(Lx² +
Ly²)Y = (L² −Lz²)Y = h²
(l - m²)Y
Since the total kinetic moment should be greater than its component, we have the following:
l - m² ≥ 0 m² ≤ λ
Similar to the harmonic oscillator, when factorizing Lx² + Ly² we can introduce two operators:
L+ = Lx + i Ly L− = Lx − i Ly
It follows from the well-known Commutativity relations for L², Lx, Ly, Lz:
[L², L±] = 0
; [Lx, Ly] =
i h Lz
[Lz, L±] = −
[L±, Lz] = ±
h L±
As for the harmonic oscillator operators L+ and L− are
producing and annihilating operators, correspondingly. By demonstrating this, we can
produce the following table:
L±
L² Y = L±
|
because of the commutativity relation → |
L² (L± Y) =
|
(L±
Y) can be the eigenfunction
of L², |
L±
Lz Y = L± m
|
because of the commutativity relation → |
Lz (L±
Y) = (±
|
|
Lz (L±
Y) = |
(L±
Y) can be the eigenfunction
of Lz having the
eigenvalue |
L+ increases and L− decreases the eigenvalue of Lz on 1.
From the relation l-m² ≥ 0 and correspondingly m² ≤ λ it follows that it should be maximum mmax and minimum mmin for which it should be the following:
L+Ymmax = 0 L-Ymmin = 0
Multiplication of L+Ymmax by L−to the left:
0 = L−L+Ymmax = (Lx− i Ly)(Lx + i Ly)Ymmax =
(Lx² + Ly² +i [LxLy
− LyLx])Ymmax = (L² −Lz²
- hLz)Ymmax,
where we have used the Commutativity
relation LxLy- LyLx
= i h Lz. We can write the following using
the corresponding eigenvalues:
0 = L−L+Ymmax = h² (l- mmax² −
mmax)Ymmax
l - mmax² − mmax = 0 | } | mmax² + mmax =
mmin² − mmin oder mmax² − mmin² + mmax + mmin = 0 |
l - mmin² + mmin = 0 |
The transformation gives finally (mmax + mmin)(mmax− mmin) + (mmax + mmin) = 0
(mmax + mmin)(mmax− mmin + 1) = 0
The second term (mmax− mmin
+ 1) ¹ 0 can never be 0 in this
product because mmax ≥ mmin ! And finally:
mmin = − mmax |
since L+ increases the eigenvalue
on 1 the difference mmax − mmin should be integer:
mmin= − mmax and mmax−
mmin = integer number:
mmax = integer number/2 ³ 0 |
We have found out from this quantum mechanical solution that the lowest kinetic moment can have zero meaning (along z-axis) m = ½. This kinetic moment is the electron spin or intrinsic kinetic moment of electron. We have come to this solution by applying only operators and the commutativity relation.
If we assign the maximum value of m by l, mmax = l
then we will obtain since λ = mmax(mmax+1) and l ≥ 0 the
following relations:
− l ≤ m ≤ l
λ = l (l + 1) |
We characterize the electron spin by s, i.e. for spin s = 1/2 mmax = +1/2 and mmin = -1/2 assign two possible alignments along z-axis (it's often said: ms = +1/2, -1/2) (or: Spin up and Spin down ¯). If one wants to find the coordinate image of the eigenfunction then he (or she) will have troubles. The eigenfunctions is likely to be represented by matrixes. We will talk about it a bit later. Now we would like to leave half-integer kinetic moment and talk a bit about integer kinetic moments for which there is the eigenfunction Y. We would also like to find the eigenfunctions of the following:
L² Yl,m = l
(l + 1) Lz Yl,m = m
|
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