Kinetic Moment Eigenvalues Calculation

Now let's calculate kinetic moment eigenvalues. Moreover to the best of our knowledge we can precisely measure squared kinetic moment L² and one of its components (let it be Lz) simultaneously. We would like to find eigenvalues of the two measured quantities and also its eigenfunction Y. To make the following calculation one needs knowledge of commutativity relations. The initial point of our calculation is that we are given the squared kinetic moment h²λ eigenvalues. There are no special restrictions since quantity λ can be either number. Moreover h² is chosen in a way that λ would be non-dimensional. The above-mentioned consideration is also true for z-component of kinetic moment eigenvalue of which we know to be 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 then one can write down:

l - m² ≥ 0 m² ≤ λ

Similar to harmonic oscillator one can factorize Lx² + Ly² and introduce two new operators:

L+ = Lx + i Ly L− = Lx − i Ly

And also one can obtain the following expressions for L², Lx, Ly, Lz from commutativity relations:

[L², L±] = 0 ; [Lx, Ly] = i h Lz

[Lz, L±] = − [L±, Lz] = ± h L±

As for harmonic oscillator L+, L− are formation and annihilation operators, correspondingly. Proving the last statement we can write down:

L± L² Y = L± h²λY | according to commutativity law →
L² (L± Y) = h² l (L± Y) | (L± Y) can be commutatited to L², moreover its eigenvalue l is unchanged
L± Lz Y = L± m h Y | according to commutativity law →
Lz (L± Y) = (± h L± + L±Lz)Y = (± h L± + m h L±) |
Lz (L± Y) = h (m ± 1) (L± Y) |

L+ increases, L− decreases the eigenvalue of Lz on 1.

According to expression l-m² ≥ 0 and m² ≤ λ it follows there should be maximum mmax and minimum mmin that are given by:

L+Y mmax = 0 L-Y mmin = 0

Multiplying L− on L+Y mmax

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. And with the corresponding eigenvalues one can obtain:

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

And finally we obtain by multiplication L+ on L-Y mmin = 0:
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

This transformation gives finally (mmax + mmin)(mmax− mmin) + (mmax + mmin) = 0 and

(mmax + mmin)(mmax− mmin + 1) = 0

The second factor in this product can never be (mmax− mmin + 1) ¹ 0, since mmax ≥ mmin ! Thus one will obtain:

mmin = − mmax

Since L+ eigenvalue is more greater by 1 the difference mmax − mmin must be integer figure: because of mmin= − mmax and mmax− mmin = integer figure, and finally:

mmax = integer number/2 ³ 0

And now we can obtain quantum mechanical solution that states the smallest kinetic moment isn't equal to 0 since value (along z-axis) m = ½. This kinetic moment is spin or intrinsic kinetic moment of electron. We have come to this solution applying only operators and commutativity relations.

If we assign the maximum value of l as mmax = l then we'll obtain (since λ = mmax(mmax+1) and l ≥ 0):

− l ≤ m ≤ l λ = l (l + 1)

We assign the electron spin as s, i.e. for spin s = 1/2 and mmax = +1/2 and mmin = -1/2 mean all possible spin orientation relative to z-axis (it is common said that: ms = +1/2, -1/2) or spin up and Spin down ¯. If we yield to temptation and try to find out location image of eigenfunction then we will fail. The eigenfunctions are like to be represented by matrixes. Now let's consider integer values of kinetic moment for which we also try to find out location image of eigenfunction Y. In other words we are trying to find out the eigenfunctions of:

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