Eigenvalues of Kinetic Moments Calculation

Now we would like to calculate the kinetic moment eigenvalues. And 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 the both measured quantities and the eigenfunction Y. One needs to know about the Commutativity relations for the following derivation. Firstly, we know that 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 h Y = m² h² Y

(Lx² + Ly²)Y = (L² −Lz²)Y = h² (l - m²)Y

Since the total kinetic moment should be greater than its component it should be 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. Demonstrating it one can write down the following table:

L± L² Y = L± h²λY | because of the commutativity relation →
L² (L± Y) = h² l (L± Y) | (L± Y) can be the eigenfunction of L², having the eigenvalue l unchanged
L± Lz Y = L± m h Y | because of the commutativity relation →
Lz (L± Y) = (± h L± + L±Lz)Y = (± h L± + m h L±) |
Lz (L± Y) = h (m ± 1) (L± Y) | (L± Y) can be the eigenfunction of Lz having the eigenvalue m increased or decreased on 1

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+Y mmax = 0 L-Y mmin = 0

Multiplication of L+Y mmax by L−to the left:

0 = L−L+Y mmax = (Lx− i Ly)(Lx + i Ly)Y mmax = (Lx² + Ly² +i [LxLy − LyLx])Y mmax = (L² −Lz² - h Lz)Y mmax,

where we have used the Commutativity relation LxLy- LyLx = i h Lz. We can write down the following using the corresponding eigenvalues:

0 = L−L+Y mmax = h² (l- mmax² − mmax)Y mmax

and multiplying L+ by L-Y mmin = 0 one can obtain:
0 = L+ L−Y mmin = h² (l- mmin² + mmin)Y mmin

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 came to this solution 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. But we'll 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 are the eigenfunction Y. Or we would like to find the eigenfunctions of the following:

L² Yl,m = l (l + 1)h² Yl,m Lz Yl,m = m h Yl,m