Bead panel from a baby carrier, Kayan or Kenyah people. Borneo 20th century, 27 x 27 cm. From the Teo Family collection, Kuching. Photograph by D Dunlop. |

Let particle P be formed from a ground-state particle combined with a photon. Definition: P is called an **excited** particle, or a particle in an excited state. Characterize P 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 quarks

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

Sort the quarks in $\sf{\Omega}$ by their phase into a pair of compound quarks $\left\{ \mathcal{S}_{\LARGE{\circ}}, \mathcal{A}_{\LARGE{\circ}} \right\}$ and $\left\{ \mathcal{S}_{\LARGE{\bullet}}, \mathcal{A}_{\LARGE{\bullet}} \right\}$ called *phase components* so that

$\sf{\Omega}^{\sf{P}} = \left\{ \left\{ \mathcal{S}_{\LARGE{\circ}}, \mathcal{A}_{\LARGE{\circ}} \right\}, \ \left\{ \mathcal{S}_{\LARGE{\bullet}}, \mathcal{A}_{\LARGE{\bullet}} \right\} \vphantom{{\Sigma^{2}}^{2}} \right\}$

Since an excited particle is the union of a ground-state with a photon, there will be a symmetric part $\mathcal{S}_{\LARGE{\circ}} = \mathcal{S}_{\LARGE{\bullet}}$ and an anti-symmetric part $\mathcal{A}_{\LARGE{\circ}} = \overline{\mathcal{A} _{\LARGE{\bullet}} }$ in this partition. And, we can assess the wavelength $\lambda$ of P by simply extending the analysis used for photons to obtain

$\begin{align} \lambda^{\sf{P}} = \frac { hc }{ 2 k_{\sf{F}} \rho^{\mathcal{A}} } \end{align}$

where $\rho^{\mathcal{A}}$ is the radius of an anti-symmetric phase-component. We can also establish the work $W$ that is required to build P by continuing the same analysis used for ground-state particles to get

$W ^{\sf{P}} = 2 k_{\sf{F}} \, \rho^{\mathcal{S}}$

where $\rho^{\mathcal{S}}$ is the radius of a symmetric phase-component. Note that these quantities may vary independently of each other because the quarks in $\mathcal{S}$ and $\mathcal{A}$ do not have any special relationship to each other. But overall, excited particles are characterized by a **mean radius** defined by the scalar sum

$\begin{align} \tilde{\rho} ^{\, \sf{P}} \equiv \frac{ \; \rho ^{\mathcal{S}} + \rho ^{\mathcal{A}} }{2} \end{align}$