Template: Atomic Transition Forces

Here is a repertoire of little kicks and torques used to get hydrogen to jump from state to state. These field quanta are generically noted by $\mathscr{F}$. There are no electrochemical quarks in these fields. Energies are typically stated in micro electron-volts. The most important characteristic for classifying these fields is their helicity, $\delta_{z} \,$. The atomic quantum numbers $\, \ell$ and $s$ are also relevant. The letter $Ł$ notes a frequently used collection of quarks that is called the Lamb quantum.


Recall that the internal energy of quark $\sf{z}$ is noted by $U ^{\sf{z}}$. Then a conjugate difference $\; \Delta \hspace{-2px} U^{\sf{Z}}$, and a conjugate mean $\tilde{U}^{\sf{Z}}$, describe the relationship between quarks and anti-quarks

$\begin{align} \Delta \hspace{-2px} U^{\sf{Z}} \equiv \frac{U^{\sf{\overline{z}}} - \, U^{\sf{z}}}{2} \end{align}$ $\begin{align} \tilde{U}^{\sf{Z}} \equiv \frac{U^{\sf{\overline{z}}} + \, U^{\sf{z}}}{2} \end{align}$

Usually we assume that the internal-energies of down-quarks are small enough to be completely negligible. Then we write $\Delta \hspace{-2px} U^{\sf{D}} \! =0$ and $\tilde{U}^{\sf{D}} \! =0$. But the conjugate mean for down quarks, as found from hydrogen observations, is $\; \tilde{U}^{\sf{D}}= -27$ (µeV). Also, we generally make an assumption of conjugate symmetry so that $\Delta \hspace{-2px} U^{\sf{Z}} \! =0$ and $\tilde{U}^{\sf{Z}} \! = U^{\sf{\overline{z}}} = U^{\sf{z}}$. But again, these assumptions are not good enough for hydrogen where the differences shown in the accompanying table provide a more accurate description of fine structure in the spectrum. The energy of $\mathscr{F}$ is also a function of its spin angular momentum quantum number, $s \,$. The constant of proportionality is $\; k_{\sf{spin}} = -91$ (µeV), so the dependence is slight. But for field quanta, quarks and anti-quarks are paired, mass and charge are always absent, and small effects are noticeable. Let $N$ note the quark coefficients of $\mathscr{F}$. Then the internal energy is obtained from a sum of these small terms

$\begin{align} U ( \mathscr{F} ) \; \equiv \; s k_{\sf{spin}} \; + \; N^{\sf{D}} \hspace{1px} \tilde{U}^{\sf{D}} \; + \; \sum_{\zeta =1}^{16} N^{\zeta} \Delta \hspace{-2px} U^{\zeta} \end{align}$

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