D.37 Forces by par­ti­cle ex­change de­riva­tions

D.37.1 Clas­si­cal en­ergy min­i­miza­tion

The en­ergy min­i­miza­tion in­clud­ing a se­lec­ton is es­sen­tially the same as the one for only spo­ton and fo­ton field. That one has been dis­cussed in chap­ter A.22.1 and in de­tail in {A.2}. So only the key dif­fer­ences will be listed here.

The en­ergy to min­i­mize is now

$$\frac{\epsilon_1}{2}\int \left(\nabla\varphi\right)^2 { \r... ...vec r}-{\skew0\vec r}_{\rm {e}})\Big){ \rm d}^3{\skew0\vec r}$$

So the only real dif­fer­ence in the vari­a­tional analy­sis is

$$s_{\rm {p}}\delta^3_\varepsilon({\skew0\vec r}-{\skew0\vec ... ...}\delta^3_\varepsilon({\skew0\vec r}-{\skew0\vec r}_{\rm {p}})$$

That means that the Pois­son equa­tion now be­comes

$$- \nabla^2\varphi({\skew0\vec r}) = \frac{s_{\rm {p}}}{\ep... ... \delta^3_\varepsilon({\skew0\vec r}-{\skew0\vec r}_{\rm {e}})$$

Since the Pois­son equa­tion is lin­ear, the so­lu­tion is $\varphi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\varphi^{\rm {p}}+\varphi^{\rm {e}}$. Here $\varphi^{\rm {p}}$ is the fo­ton field (A.107) pro­duced by the spo­ton as be­fore, and $\varphi^{\rm {e}}$ is a sim­i­lar ex­pres­sion, but us­ing the se­lec­ton sarge and dis­tance from the se­lec­ton:

$$\varphi^{\rm {p}} = \frac{s_{\rm {p}}}{4 \pi \epsilon_1 \ve... ...\epsilon_1 \vert{\skew0\vec r}- {\skew0\vec r}_{\rm {e}}\vert}$$

The en­ergy low­er­ing is now

$$-{\textstyle\frac{1}{2}} \int \Big(\varphi^{\rm {p}}({\skew... ...ec r}-{\skew0\vec r}_{\rm {e}})\Big) { \rm d}^3{\skew0\vec r}$$

Mul­ti­ply­ing out, you get, of course, the en­ergy low­er­ings for the spo­ton and se­lec­ton in iso­la­tion. But you also get two ad­di­tional in­ter­ac­tion terms be­tween these sarges. These two terms are equal; the se­lec­ton field $\varphi^{\rm {e}}$ eval­u­ated at the po­si­tion of the spo­ton times spo­ton sarge is the same as the spo­ton field $\varphi^{\rm {p}}$ at the se­lec­ton times se­lec­ton sarge. So it is seen that each term con­tributes half to the Koulomb en­ergy as claimed in the text.

The fo­ton field en­ergy is still half of the par­ti­cle-field in­ter­ac­tion en­er­gies and of op­po­site sign. That is why the en­ergy change is half of what you would ex­pect from the in­ter­ac­tion of the par­ti­cles with each other’s field: the other half is off­set by changes in field en­ergy.

D.37.2 Quan­tum en­ergy min­i­miza­tion

This de­riva­tion in­cludes the se­lec­ton in the spo­ton-fo­tons sys­tem an­a­lyzed in {A.22.3}. Since the analy­sis is es­sen­tially un­changed, only the key dif­fer­ences will be high­lighted.

If an se­lec­ton is added to the sys­tem, the sys­tem wave func­tion be­comes

$$\psi_{\varphi\rm {pe}} = C_0 \psi_{\rm {p}} \psi_{\rm {e}} ... ...s \quad\quad \vert C_0\vert^2 + \vert C_1\vert^2 + \ldots = 1$$

The de­mon can hold the se­lec­ton in its other hand. The Hamil­ton­ian will now of course in­clude a term for the se­lec­ton in iso­la­tion, as well as an in­ter­ac­tion with the fo­ton field. These are com­pletely anal­o­gous to the cor­re­spond­ing spo­ton terms.

So the en­ergy to be min­i­mized for the ground state be­comes

$$E = E_{\rm {p}} + E_{\rm {e}} + \vert C_1\vert^2 \hbar \om... ... {e}}\right\rangle} \Big\vert\vert C_1\vert\cos(\alpha+\beta)$$

If this is min­i­mized as in {A.22.3}, the en­ergy is

$$E = E_{\rm {p}} + E_{\rm {e}} - \frac{1}{2\epsilon_1{\cal V... ...c r}_{\rm {e}}}{\left\vert\psi_{\rm {e}}\right\rangle} \vert^2$$

The square ab­solute value of a quan­tity can be found as the prod­uct of that quan­tity times its com­plex con­ju­gate. That gives the same en­ergy low­er­ing as for the lone spo­ton, and a sim­i­lar term for a lone se­lec­ton. How­ever, there is an ad­di­tional term

$$- \frac{s_{\rm {p}}s_{\rm {e}}}{2\epsilon_1{\cal V}k^2} \l... ... r}_{\rm {p}}}{\left\vert\psi_{\rm {p}}\right\rangle} \right)$$

If you write out the in­ner prod­uct in­te­grals over the se­lec­ton co­or­di­nates ex­plic­itly, this be­comes

$$- \int s_{\rm {e}} \psi_{\rm {e}}^*({\skew0\vec r}_{\rm {e}... ...({\skew0\vec r}_{\rm {e}}) { \rm d}^3{\skew0\vec r}_{\rm {e}}$$

Summed over all ${\vec k}$, the sec­ond term in­side the square brack­ets gives the same an­swer as the first; that is be­cause op­po­site ${\vec k}$ val­ues ap­pear equally in the sum­ma­tion. Look­ing at the first term, the sum­ma­tion over ${\vec k}$ pro­duces again the spo­ton po­ten­tial $\varphi^{\rm {p}}_{\rm {cl}}$, but now eval­u­ated at the po­si­tion of the se­lec­ton. That then shows the ad­di­tional en­ergy low­er­ing to be

$$- \int s_{\rm {e}} \psi_{\rm {e}}^*({\skew0\vec r}_{\rm {e}... ..._{\rm {e}}({\skew0\vec r}_{\rm {e}}) { \rm d}^3{\skew0\vec r}$$

Ex­cept for the dif­fer­ences in no­ta­tion, that is the same se­lec­ton-spo­ton in­ter­ac­tion en­ergy as found in {A.22.1}.

D.37.3 Rewrit­ing the La­grangian

The rules of en­gage­ment are as fol­lows:

• The Carte­sian axes are num­bered us­ing an in­dex $i$, with $i$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, 2, and 3 for $x$, $y$, and $z$ re­spec­tively.
• Also, $r_i$ in­di­cates the co­or­di­nate in the $i$ di­rec­tion, $x$, $y$, or $z$.
• De­riv­a­tives with re­spect to a co­or­di­nate $r_i$ are in­di­cated by a sim­ple sub­script $i$.
• If the quan­tity be­ing dif­fer­en­ti­ated is a vec­tor, a comma is used to sep­a­rate the vec­tor in­dex from dif­fer­en­ti­a­tion ones.
• In­dex ${\overline{\imath}}$ is the num­ber im­me­di­ately fol­low­ing $i$ in the cyclic se­quence ...123123...and ${\overline{\overline{\imath}}}$ is the num­ber im­me­di­ately pre­ced­ing $i$.
• If $i$ is al­ready been used for some­thing else, $j$ can be used the same way.
• Time de­riv­a­tives are in­di­cated by a sub­script t.

Con­sider first the square mag­netic field:

$${\cal B}^2 = \sum_i (A_{{\overline{\overline{\imath}}},{\ov... ...}} - A_{{\overline{\imath}},{\overline{\overline{\imath}}}})^2$$

Ex­pand­ing out the square, that is equiv­a­lent to

$${\cal B}^2 = \sum_i ( A_{{\overline{\overline{\imath}}},{\o... ...ath}}} A_{{\overline{\imath}},{\overline{\overline{\imath}}}})$$

The sum­ma­tion in­dices can now be cycli­cally re­de­fined to give an equiv­a­lent sum over $i$ equal to

$${\cal B}^2 = \sum_i (A_{i,{\overline{\overline{\imath}}}}^2... ...ine{\overline{\imath}}}} A_{{\overline{\overline{\imath}}},i})$$

The terms can be com­bined in sets of three as

$${\cal B}^2 = A_{i,j}^2 - A_{i,j}A_{j,i}$$

Here sum­ma­tion over $i$ and $j$ is now un­der­stood.

The square elec­tric field is

$${\cal E}^2=(-\varphi_i-A_{i,t})^2 = A_{i,t}^2 + 2 A_{i,t}\varphi_i + \varphi_i^2$$

All to­gether, that gives

$$\begin{eqnarray*} {\cal E}^2 - c^2 {\cal B}^2 & = & \displaystyle A_{i,t}^2 - ... ... - 2 A_{i,i}\varphi_t + c^2 A_{i,j}A_{j,i} - c^2 A_{i,i}A_{j,j} \end{eqnarray*}$$

as can be ver­i­fied by mul­ti­ply­ing out and sim­pli­fy­ing. The right hand side in the first line is the self-ev­i­dent elec­tro­mag­netic La­grangian den­sity, ex­cept for the fac­tor $\epsilon_0$$\raisebox{.5pt}{$/$}$​2. The sec­ond line is the square of the Lorentz con­di­tion quan­tity. The fi­nal line can be writ­ten as a sum of pure de­riv­a­tives:

$$2 (A_i \varphi_i)_t - 2 (A_i \varphi_t)_i + c^2 (A_i A_{j,i})_j - c^2 (A_i A_{j,j})_i$$

Pure de­riv­a­tives do not pro­duce changes in the ac­tion, as the changes in the po­ten­tials dis­ap­pear on the bound­aries of in­te­gra­tion.

D.37.4 Coulomb po­ten­tial en­ergy

The Coulomb po­ten­tial en­ergy be­tween charged par­ti­cles is typ­i­cally de­rived in ba­sic physics. But it can also eas­ily be ver­i­fied from the con­ven­tional elec­tro­mag­netic en­ergy (A.143). In the steady case, there is only the elec­tric field, due to the Coulomb po­ten­tial. The en­ergy may then be writ­ten as

$$\frac{\epsilon_0}{2} \int{\cal E}_{\rm {C}}^2{ \rm d}^3{\s... ...skew0\vec r};t)\rho({\skew0\vec r};t){ \rm d}^3{\skew0\vec r}$$

where the first equal­ity comes from the de­f­i­n­i­tion of the elec­tric field, the sec­ond from in­te­gra­tion by parts and the third one from the first Maxwell equa­tion. Sub­sti­tu­tion of the Coulomb po­ten­tial in terms of the charge dis­tri­b­u­tion as given in {A.22.8},

$$\varphi_{\rm {C}}({\skew0\vec r};t) = \int_{{\rm all }{\un... ...line{\skew0\vec r}}\vert}{ \rm d}^3{\underline{\skew0\vec r}}$$

now gives the Koulomb po­ten­tial en­ergy $V_{\rm {C}}$ for a con­tin­u­ous charge dis­tri­b­u­tion:

$$V_{\rm C} = {\textstyle\frac{1}{2}} \int_{{\rm all }{\skew... ...} { \rm d}^3{\skew0\vec r}{\rm d}^3{\underline{\skew0\vec r}}$$

For point charges, the charge dis­tri­b­u­tion is by de­f­i­n­i­tion

$$\rho({\skew0\vec r};t) = \sum_{i=1}^I q_i \delta^3({\skew0\vec r}-{\skew0\vec r}_i)$$

Here $\delta^3$ is the three-di­men­sion­al delta func­tion, ${\skew0\vec r}_i$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew0\vec r}_i(t)$ the po­si­tion of point charge $i$, and $q_i$ its charge.

Re­call that the delta func­tion picks out the value at ${\skew0\vec r}_i$ from what­ever it is in­te­grated against. Us­ing this twice on the Coulomb po­ten­tial en­ergy above,

$$V_{\rm C} = {\textstyle\frac{1}{2}} \sum_{i=1}^I \int_{{\rm... ...n_0\vert{\skew0\vec r}_i-{\skew0\vec r}_{{\underline i}}\vert}$$

That is the Coulomb po­ten­tial en­ergy $V_{\rm {C}}$ for point charges.

Note again that phys­i­cally all the en­ergy is in­side the elec­tro­mag­netic field. There is no en­ergy of in­ter­ac­tion of the charged par­ti­cles with the field. If equal charges move closer to­gether, they in­crease the en­ergy in the elec­tro­mag­netic field. That re­quires work.