Ground-State Particles
 Tampan, Paminggir people. Lampung region of Sumatra, near Semangka Bay, 19th century, 64 x 64 cm. Photograph by D Dunlop. From the library of Darwin Sjamsudin, Jakarta.

Let particle P be characterized by some repetitive chain of events

$\Psi ^{\sf{P}} = \left( \sf{\Omega}_{\sf{1}}, \sf{\Omega}_{\sf{2}}, \sf{\Omega}_{\sf{3}} \ \ldots \ \right)$

where each orbital cycle is a bundle of $N$ quarks

$\sf{\Omega}^{\sf{P}} = \left\{ \sf{q}^{\sf{1}}, \sf{q}^{\sf{2}} \ldots \sf{q}^{\it{i}} \ldots \sf{q}^{\it{N}} \right\}$

and each quark is described by its phase $\delta _{\theta} = \pm 1$ which depends on the reference frame. Use the phase to sort quarks into a pair of compound quarks $\mathcal{S}_{\LARGE{\circ}}$ and $\mathcal{S}_{\LARGE{\bullet}}$ called phase-components so that

$\sf{\Omega}^{\sf{P}} = \left\{ \mathcal{S}_{\LARGE{\circ}}, \mathcal{S}_{\LARGE{\bullet}} \right\}$

These two phase-components are out of phase with each other, so

$\delta_{\theta} \left( \mathcal{S}_{\LARGE{\circ}} \right) = - \, \delta_{\theta} \left( \mathcal{S}_{\LARGE{\bullet}} \right)$

Definition: if P has perfect phase symmetry then both sets are composed from the same selection of quarks

$\mathcal{S}_{\LARGE{\circ}} = \mathcal{S}_{\LARGE{\bullet}}$

and we say that P is in its ground state. By this definition, we cannot make a ground-state model of P unless all of $\sf{\Omega}$'s quark coefficients are integer multiples of two. Now consider assessing the radius $\overline{\rho}$ of a particle in its ground-state. Since $\mathcal{S}_{\LARGE{\circ}} = \mathcal{S}_{\LARGE{\bullet}}$ we can drop the subscript to write

$\overline{\rho} \left( \mathcal{S}_{\LARGE{\circ}} \right) = \overline{\rho} \left( \mathcal{S}_{\LARGE{\bullet}} \right) = \overline{\rho}^{\mathcal{S}}$

Then recall that $\overline{\rho}$ is defined from sums and differences of quark coefficients, and quarks are indestructible. So by the of addition and subtraction $\overline{\rho}^{\sf{P}} = \overline{\rho} \left( \mathcal{S}_{\LARGE{\circ}} \right) + \overline{\rho} \left( \mathcal{S}_{\LARGE{\bullet}} \right) =2 \overline{\rho}^{\mathcal{S}}$ and the radius of P is twice the radius of $\mathcal{S}$. Let us also evaluate the wavevector of a free particle P which is given by the sum

\begin{align} \overline{ \kappa }^{ \sf{P}} = \frac{2 \pi k_{\sf{F}}}{hc} \sum_{i=1}^{N} \delta _{\theta}^{\, i} \, \bar{\rho}^{i} = \frac{2 \pi k_{\sf{F}}}{hc} \left[ \delta_{\theta} \left(\mathcal{S}_{\LARGE{\circ}} \right) \overline{\rho} \left( \mathcal{S}_{\LARGE{\circ}} \right) + \delta_{\theta} \left( \mathcal{S}_{\LARGE{\bullet}} \right) \overline{\rho} \left( \mathcal{S}_{\LARGE{\bullet}} \right) \vphantom{\Sigma^{2}} \right] = \frac{2 \pi k_{\sf{F}}}{hc} \left[ \delta_{\theta} \left(\mathcal{S}_{\LARGE{\circ}} \right) + \delta_{\theta} \left( \mathcal{S}_{\LARGE{\bullet}} \right) \vphantom{\Sigma^{2}} \right] \overline{\rho}^{\mathcal{S}} \end{align}

But the two components are out-of-phase, so $\delta_{\theta} \left( \mathcal{S}_{\LARGE{\circ}} \right) = - \, \delta_{\theta} \left( \mathcal{S}_{\LARGE{\bullet}} \right)$ and $\overline{ \kappa }^{ \sf{P}} = (0, 0, 0)$. Thus the wavenumber of a particle in its ground-state is always zero.

 Next step: photons.
page revision: 185, last edited: 06 Jan 2018 00:47
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