A.33 Ex­pla­na­tion of the Lon­don forces

To fully un­der­stand the de­tails of the Lon­don forces, it helps to first un­der­stand the pop­u­lar ex­pla­na­tion of them, and why it is all wrong. To keep things sim­ple, the ex­am­ple will be the Lon­don at­trac­tion be­tween two neu­tral hy­dro­gen atoms that are well apart. (This will also cor­rect a small er­ror that the ear­lier dis­cus­sion of the hy­dro­gen mol­e­cule made; that dis­cus­sion im­plied in­cor­rectly that there is no at­trac­tion be­tween two neu­tral hy­dro­gen atoms that are far apart. The truth is that there re­ally is some Van der Waals at­trac­tion. It was ig­nored be­cause it is small com­pared to the chem­i­cal bond that forms when the atoms are closer to­gether and would dis­tract from the real story.)

Fig­ure A.23: Pos­si­ble po­lar­iza­tions of a pair of hy­dro­gen atoms.

The pop­u­lar ex­pla­na­tion for the Lon­don force goes some­thing like this: “Sure, there would not be any at­trac­tion be­tween two dis­tant hy­dro­gen atoms if they were per­fectly spher­i­cally sym­met­ric. But ac­cord­ing to quan­tum me­chan­ics, na­ture is un­cer­tain. So some­times the elec­tron clouds of the two atoms are some­what to the left of the nu­clei, like in fig­ure A.23 (b). This po­lar­iza­tion [di­pole cre­ation] of the atoms turns out to pro­duce some elec­tro­sta­tic at­trac­tion be­tween the atoms. At other times, the elec­tron clouds are some­what to the right of the nu­clei like in fig­ure A.23 (c); it is re­ally the same thing seen in the mir­ror. In cases like fig­ure A.23 (a), where the elec­tron clouds move to­wards each other, and (b), where they move away from each other, there is some re­pul­sion be­tween the atoms; how­ever, the wave func­tions be­come cor­re­lated so that (b) and (c) are more likely than (a) and (d). Hence a net at­trac­tion re­sults.”

Be­fore ex­am­in­ing what is wrong with this ex­pla­na­tion, first con­sider what is right. It is per­fectly right that fig­ure A.23 (b) and (c) pro­duce some net at­trac­tion be­tween the atoms, and that (a) and (d) pro­duce some re­pul­sion. This fol­lows from the net Coulomb po­ten­tial en­ergy be­tween the atoms for given po­si­tions of the elec­trons:

$$V_{\rm {lr}} = \frac{e^2}{4\pi\epsilon_0} \left( \frac{1}... ...}}} - \frac{1}{r_{\rm {r}}} + \frac{1}{r_{\rm {lr}}} \right)$$

where $e$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.6 10$\POW9,{-19}$ C is the mag­ni­tude of the charges of the pro­tons and elec­trons, $\epsilon_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 8.85 10$\POW9,{-12}$ C$\POW9,{2}$/J m is the per­mit­tiv­ity of space, $d$ is the dis­tance be­tween the nu­clei, $r_{\rm {l}}$ is the dis­tance be­tween the left elec­tron and the right nu­cleus, $r_{\rm {r}}$ the one be­tween the right elec­tron and the left nu­cleus, and $r_{\rm {lr}}$ is the dis­tance be­tween the two elec­trons. If the elec­trons charges are dis­trib­uted over space ac­cord­ing to den­si­ties $n_{\rm {l}}({\skew0\vec r}_{\rm {l}})$ and $n_{\rm {r}}({\skew0\vec r}_{\rm {r}})$, the clas­si­cal po­ten­tial en­ergy is

$$V_{\rm {lr}} = \frac{e^2}{4\pi\epsilon_0} \int_{{\rm all}... ... d}^3{\skew0\vec r}_{\rm {l}}{\rm d}^3{\skew0\vec r}_{\rm {r}}$$

(Since the first, 1/$d$, term rep­re­sents the re­pul­sion be­tween the nu­clei, it may seem strange to in­te­grate it against the elec­tron charge dis­tri­b­u­tions, but the charge dis­tri­b­u­tions in­te­grate to one, so they dis­ap­pear. Sim­i­larly in the sec­ond and third term, the charge dis­tri­b­u­tion of the un­in­volved elec­tron in­te­grates away.)

Since it is as­sumed that the atoms are well apart, the in­te­grand above can be sim­pli­fied us­ing Tay­lor se­ries ex­pan­sions to give:

$$V_{\rm {lr}} = \frac{e^2}{4\pi\epsilon_0} \int_{{\rm all}... ...}^3{\skew0\vec r}_{\rm {l}} {\rm d}^3{\skew0\vec r}_{\rm {r}}$$

where the po­si­tions of the elec­trons are mea­sured from their re­spec­tive nu­clei. Also, the two $z$ axes are both taken hor­i­zon­tal and pos­i­tive to­wards the left. For charge dis­tri­b­u­tions as shown in fig­ure A.23, the $x_{\rm {l}}x_{\rm {r}}$ and $y_{\rm {l}}y_{\rm {r}}$ terms in­te­grate to zero be­cause of odd sym­me­try. How­ever, for a dis­tri­b­u­tion like in fig­ure A.23 (c), $n_{\rm {l}}$ and $n_{\rm {r}}$ are larger at pos­i­tive $z_{\rm {l}}$, re­spec­tively $z_{\rm {r}}$, than at neg­a­tive one, so the in­te­gral will in­te­grate to a neg­a­tive num­ber. That means that the po­ten­tial is low­ered, there is at­trac­tion be­tween the atoms. In a sim­i­lar way, dis­tri­b­u­tion (b) pro­duces at­trac­tion, while (a) and (d) pro­duce re­pul­sion.

So there is noth­ing wrong with the claim that (b) and (c) pro­duce at­trac­tion, while (a) and (d) pro­duce re­pul­sion. It is also per­fectly right that the com­bined quan­tum wave func­tion gives a higher prob­a­bil­ity to (b) and (c) than to (a) and (d).

So what is wrong? There are two ma­jor prob­lems with the story.

1.

En­ergy eigen­states are sta­tion­ary. If the wave func­tion os­cil­lated in time like the story sug­gests, it would re­quire un­cer­tainty in en­ergy, which would act to kill off the low­er­ing of en­ergy. True, states with the elec­trons at the same side of their nu­clei are more likely to show up when you mea­sure them, but to reap the ben­e­fits of this in­creased prob­a­bil­ity, you must not do such a mea­sure­ment and just let the elec­tron wave func­tion sit there un­chang­ing in time.

2.

The num­bers are all wrong. Sup­pose the wave func­tions in fig­ures (b) and (c) shift (po­lar­ize) by a typ­i­cal small amount $\varepsilon$. Then the at­trac­tive po­ten­tial is of or­der $\varepsilon^2$$\raisebox{.5pt}{$/$}$​$d^3$. Since the dis­tance $d$ be­tween the atoms is as­sumed large, the en­ergy gained is a small amount times $\varepsilon^2$. But to shift atom en­ergy eigen­func­tions by an amount $\varepsilon$ away from their ground state takes an amount of en­ergy $C\varepsilon^2$ where $C$ is some con­stant that is not small. So it would take more en­ergy to shift the elec­tron clouds than the di­pole at­trac­tion could re­cover. In the ground state, the elec­tron clouds should there­fore stick to their orig­i­nal cen­tered po­si­tions.

On to the cor­rect quan­tum ex­pla­na­tion. First the wave func­tion is needed. If there were no Coulomb po­ten­tials link­ing the atoms, the com­bined ground-state elec­tron wave func­tion would sim­ply take the form

$$\psi({\skew0\vec r}_{\rm {l}},{\skew0\vec r}_{\rm {r}}) = ... ...({\skew0\vec r}_{\rm {l}})\psi_{100}({\skew0\vec r}_{\rm {r}})$$

where $\psi_{100}$ is the ground state wave func­tion of a sin­gle hy­dro­gen atom. To get a suit­able cor­re­lated po­lar­iza­tion of the atoms, throw in a bit of the $\psi_{210}$ 2p$_z$ states, as fol­lows:

$$\psi({\skew0\vec r}_{\rm {l}},{\skew0\vec r}_{\rm {r}})= \... ...{\skew0\vec r}_{\rm {l}})\psi_{210}({\skew0\vec r}_{\rm {r}}).$$

For $\varepsilon$ $\raisebox{.3pt}{$>$}$ 0, it pro­duces the de­sired cor­re­la­tion be­tween the wave func­tions: $\psi_{100}$ is al­ways pos­i­tive, and $\psi_{210}$ is pos­i­tive if the elec­tron is at the pos­i­tive-$z$ side of its nu­cleus and neg­a­tive oth­er­wise. So if both elec­trons are at the same side of their nu­cleus, the prod­uct $\psi_{210}({\skew0\vec r}_{\rm {l}})\psi_{210}({\skew0\vec r}_{\rm {r}})$ is pos­i­tive, and the wave func­tion is in­creased, giv­ing in­creased prob­a­bil­ity of such states. Con­versely, if the elec­trons are at op­po­site sides of their nu­cleus, $\psi_{210}({\skew0\vec r}_{\rm {l}})\psi_{210}({\skew0\vec r}_{\rm {r}})$ is neg­a­tive, and the wave func­tion is re­duced.

Now write the ex­pec­ta­tion value of the en­ergy:

$$\left\langle{E}\right\rangle = \langle \sqrt{1-\varepsilo... ...si_{100}\psi_{100} + \varepsilon\psi_{210}\psi_{210} \rangle$$

where $H_{\rm {l}}$ and $H_{\rm {r}}$ are the Hamil­to­ni­ans of the in­di­vid­ual elec­trons and

$$V_{\rm {lr}} = \frac{e^2}{4\pi\epsilon_0} \frac{x_{\rm {l}}x_{\rm {r}}+y_{\rm {l}}y_{\rm {r}}-2z_{\rm {l}}z_{\rm {r}}}{d^3}$$

is again the po­ten­tial be­tween atoms. Work­ing out the in­ner prod­uct, not­ing that the $\psi_{100}$ and $\psi_{210}$ are or­tho­nor­mal eigen­func­tions of the atom Hamil­to­ni­ans $H_{\rm {l}}$ and $H_{\rm {r}}$ with eigen­val­ues $E_1$ and $E_2$, and that most $V_{\rm {lr}}$ in­te­grals are zero on ac­count of odd sym­me­try, you get

$$\left\langle{E}\right\rangle = 2 E_1 + 2 \varepsilon^2 (E_... ...}\vert z_{\rm {l}}z_{\rm {r}}\vert\psi_{210}\psi_{210}\rangle.$$

The fi­nal term is the sav­ior for de­riv­ing the Lon­don force. For small val­ues of $\varepsilon$, for which the square root can be ap­prox­i­mated as one, this en­ergy-low­er­ing term dom­i­nates the en­ergy $2\varepsilon^2(E_2-E_1)$ needed to dis­tort the atom wave func­tions. The best ap­prox­i­ma­tion to the true ground state is then ob­tained when the qua­dratic in $\varepsilon$ is min­i­mal. That hap­pens when the en­ergy has been low­ered by an amount

$$\frac{2}{E_2-E_1} \left( \frac{e^2}{4\pi\epsilon_0}\langl... ...100}\vert z\vert\psi_{210}\rangle^2 \right)^2 \frac{1}{d^6}.$$

Since the as­sumed eigen­func­tion is not ex­act, this vari­a­tional ap­prox­i­ma­tion will un­der­es­ti­mate the ac­tual Lon­don force. For ex­am­ple, it can be seen that the en­ergy can also be low­ered sim­i­lar amounts by adding some of the 2p$_x$ and 2p$_y$ states; these cause the atom wave func­tions to move in op­po­site di­rec­tions nor­mal to the line be­tween the nu­clei.

So what is the phys­i­cal mean­ing of the sav­ior term? Con­sider the in­ner prod­uct that it rep­re­sents:

$$\langle\psi_{100}\psi_{100}\vert V_{\rm {lr}}\vert\psi_{210}\psi_{210}\rangle.$$

That is the en­ergy if both elec­trons are in the spher­i­cally sym­met­ric $\psi_{100}$ ground state if both elec­trons are in the an­ti­sym­met­ric 2p$_z$ state. The sav­ior term is a twi­light term, like the ones dis­cussed ear­lier in chap­ter 5.3 for chem­i­cal bonds. It re­flects na­ture’s habit of do­ing busi­ness in terms of an un­ob­serv­able wave func­tion in­stead of ob­serv­able prob­a­bil­i­ties.