Orbital Angular Momentum
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Quantum Numbers for Atoms
principal $\begin{align} {\mathrm{n}} \; \equiv \frac{ n^{\mathsf{d}} }{4} \end{align}$
azimuthal $\begin{align} \ell \; \equiv \frac{ N^{\sf{U}} + N^{\sf{D}} + \left| \, N^{\sf{U}} - N^{\sf{D}} \right| }{8} - \frac{ n^{\sf{d}}}{2} \end{align}$
spin angular
$\begin{align} s \; \equiv \frac{ n^{\mathsf{u}} + n^{\mathsf{\overline{u}}} -3 n^{\mathsf{d}}+ n^{\mathsf{\overline{d}}} }{ 8 } \end{align}$
total atomic
angular momentum
$\begin{align} j \; \equiv \frac{ \, \left| \, N^{\mathsf{U}} - N^{\mathsf{D}} \, \right| \, }{8} \end{align}$

The direct discussion of visual sensation is to be curtailed as we objectify descriptions and apply the hypothesis of spatial isotropy. To accomplish this, here are four quantum numbers to replace the four coefficients of rotating quarks that represent achromatic sensations. So now, instead of describing some shade of grey, the new numbers are said to specify the rotational state of an atom.

The quantum numbers $\mathrm{n}$, $\ell$, $s$ and $j$ also bring traditional mechanical features into a description. They are defined from an atom's quark coefficients as noted by $n$ and $N$. Recall that the italic letter $n$ is employed for quark coefficients whereas the upright font $\rm{n}$ is reserved for the principal quantum number. Note that $\mathrm{n}$ and $s$ are defined from simple sums and differences of quark coefficients. So they are always conserved because quarks are indestructible.

Some Useful Identities

$j = \ell - s$

$j = 𝘑$

$\begin{align} \left| \, N^{\mathsf{U}} - N^{\mathsf{D}} \, \right| = \delta _{z} \left( N^{\mathsf{U}} - N^{\mathsf{D}} \right) = \delta _{z} N^{\mathsf{U}} - \delta _{z} N^{\mathsf{D}} = 8 \hspace{2px} j \end{align}$

Definition: We use $Ł$ to note a quantum of orbital angular momentum. Absorbing or emitting $Ł$ changes the azimuthal quantum number by $\Delta \ell = \pm 1$ without altering $\rm{n}$ or $j$. This particle is used to explain the Lamb shiftXlink.png that is observed in the spectrum of hydrogen. So $Ł$ is called the Lamb particle and is given by

$Ł \leftrightarrow 2 {\sf{u}} + 2 {\sf{\bar{u}}} + 4 {\sf{\bar{d}}} + 2 {\mathbf{l}} + 2 {\mathbf{\bar{l}}}$

These quarks are arranged in out-of-phase pairs such that

$Ł = \left\{ {\sf{P}}_{\LARGE{\circ}}, {\sf{P}}_{\LARGE{\bullet}} \right\}$


${\sf{P}}_{\LARGE{\circ}} = \left\{ \{ {\sf{u}}, {\sf{\bar{d}}}, {\mathbf{l}} \}, \{ {\sf{\bar{u}}}, {\sf{\bar{d}}}, {\mathbf{\bar{l}}} \} \right\}$

and $\delta_{ \theta}$ notes the phase so that

$\delta_{\theta} \left( {\sf{P}}_{\LARGE{\circ}} \right) =- \delta_{\theta} \left( {\sf{P}}_{\LARGE{\bullet}} \right) = \pm 1$

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