5.2 The Hy­dro­gen Mol­e­cule

This sec­tion uses sim­i­lar ap­prox­i­ma­tions as for the hy­dro­gen mol­e­c­u­lar ion of chap­ter 4.6 to ex­am­ine the neu­tral H$_2$ hy­dro­gen mol­e­cule. This mol­e­cule has two elec­trons cir­cling two pro­tons. It will turn out that in the ground state, the pro­tons share the two elec­trons, rather than each be­ing as­signed one. This is typ­i­cal of co­va­lent bonds.

Of course, share is a vague term, but the dis­cus­sion will show what it re­ally means in terms of the six-di­men­sion­al elec­tron wave func­tion.

5.2.1 The Hamil­ton­ian

Just like for the hy­dro­gen mol­e­c­u­lar ion of chap­ter 4.6, for the neu­tral mol­e­cule the Born-​Op­pen­heimer ap­prox­i­ma­tion will be made that the pro­tons are at given fixed points. So the prob­lem sim­pli­fies to just find­ing the wave func­tion of the two elec­trons, $\Psi({\skew0\vec r}_1,{\skew0\vec r}_2)$, where ${\skew0\vec r}_1$ and ${\skew0\vec r}_2$ are the po­si­tions of the two elec­trons 1 and 2. In terms of scalar ar­gu­ments, the wave func­tion can be writ­ten out fur­ther as $\Psi(x_1,y_1,z_1,x_2,y_2,z_2)$.

In the Hamil­ton­ian, fol­low­ing the New­ton­ian anal­ogy the ki­netic and po­ten­tial en­ergy op­er­a­tors sim­ply add:

$$H = - \frac{\hbar^2}{2m_{\rm e}} \left( \nabla_1^2 + \na... ...rac1{\vert{\skew0\vec r}_1 - {\skew0\vec r}_2\vert} \right) %$$ (5.3)

In this ex­pres­sion, the Lapla­cians of the first two, ki­netic en­ergy, terms are with re­spect to the po­si­tion co­or­di­nates of the two elec­trons:

$$\nabla^2_1 = \frac{\partial^2}{\partial x_1^2} + \frac{\p... ...rtial^2}{\partial y_2^2} + \frac{\partial^2}{\partial z_z^2}.$$

The next four terms in the Hamil­ton­ian (5.3) are the at­trac­tive po­ten­tials be­tween the elec­trons and the pro­tons, with $r_{1{\rm {l}}}$, $r_{2{\rm {l}}}$, $r_{1{\rm {r}}}$, and $r_{2{\rm {r}}}$ be­ing the dis­tances be­tween elec­trons 1 and 2 and the left, re­spec­tively right pro­ton. The fi­nal term rep­re­sents the re­pul­sive po­ten­tial be­tween the two elec­trons.

Key Points

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The Hamil­ton­ian for the 6-di­men­sion­al elec­tron wave func­tion has been writ­ten down.

5.2.1 Re­view Ques­tions

1.

Ver­ify that the re­pul­sive po­ten­tial be­tween the elec­trons is in­fi­nitely large when the elec­trons are at the same po­si­tion.

Note: You might there­fore think that the wave func­tion needs to be zero at the lo­ca­tions in six-di­men­sion­al space where ${\skew0\vec r}_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew0\vec r}_2$. Some au­thors re­fer to that as a Coulomb hole.” But the truth is that in quan­tum me­chan­ics, elec­trons are smeared out due to un­cer­tainty. That causes elec­tron 1 to “see elec­tron 2 at all sides, and vice-versa, and they do there­fore not en­counter any un­usu­ally large po­ten­tial when the wave func­tion is nonzero at ${\skew0\vec r}_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew0\vec r}_2$. In gen­eral, it is just not worth the trou­ble for the elec­trons to stay away from the same po­si­tion: that would re­duce their un­cer­tainty in po­si­tion, in­creas­ing their un­cer­tainty-de­manded ki­netic en­ergy.

So­lu­tion hmola-a

2.

Note that the to­tal ki­netic en­ergy term is sim­ply a mul­ti­ple of the six-di­men­sion­al Lapla­cian op­er­a­tor. It treats all Carte­sian po­si­tion co­or­di­nates ex­actly the same, re­gard­less of which di­rec­tion or which elec­tron it is. Is this still the case if other par­ti­cles are in­volved?

So­lu­tion hmola-b

5.2.2 Ini­tial ap­prox­i­ma­tion to the low­est en­ergy state

The next step is to iden­tify an ap­prox­i­mate ground state for the hy­dro­gen mol­e­cule. Fol­low­ing the same ap­proach as in chap­ter 4.6, it will first be as­sumed that the pro­tons are rel­a­tively far apart. One ob­vi­ous ap­prox­i­mate so­lu­tion is then that of two neu­tral atoms, say the one in which elec­tron 1 is around the left pro­ton in its ground state and elec­tron 2 is around the right one.

To for­mu­late the wave func­tion for that, the short­hand no­ta­tion $\psi_{\rm {l}}$ will again be used for the wave func­tion of a sin­gle elec­tron that in the ground state around the left pro­ton and $\psi_{\rm {r}}$ for one that is in the ground state around the right hand one:

$$\psi_{\rm {l}}({\skew0\vec r}) \equiv \psi_{100}(\vert{\ske... ...psi_{100}(\vert{\skew0\vec r}- {\skew0\vec r}_{\rm {rp}}\vert)$$

where $\psi_{100}$ is the hy­dro­gen atom ground state (4.40), and ${\skew0\vec r}_{\rm {lp}}$ and ${\skew0\vec r}_{\rm {rp}}$ are the po­si­tions of the left and right pro­tons.

The wave func­tion that de­scribes that elec­tron 1 is in the ground state around the left pro­ton and elec­tron 2 around the right one will be ap­prox­i­mated to be the prod­uct of the sin­gle elec­tron states:

$$\psi({\skew0\vec r}_1,{\skew0\vec r}_2) = \psi_{\rm {l}}({\skew0\vec r}_1) \psi_{\rm {r}}({\skew0\vec r}_2)$$

Tak­ing the com­bined wave func­tion as a prod­uct of sin­gle elec­tron states is re­ally equiv­a­lent to an as­sump­tion that the two elec­trons are in­de­pen­dent. In­deed, for the prod­uct state, the prob­a­bil­ity of find­ing elec­tron 1 at po­si­tion ${\skew0\vec r}_1$ and elec­tron 2 at ${\skew0\vec r}_2$ is:

$$\vert\psi_{\rm {l}}({\skew0\vec r}_1)\vert^2 { \rm d}^3 {\... ...\rm {r}}({\skew0\vec r}_2)\vert^2 { \rm d}^3 {\skew0\vec r}_2$$

or in words:

$$\begin{array}{l} \mbox{[probability of finding 1 at ${\ske... ... at ${\skew0\vec r}_2$ unaffected by where 1 is]} \end{array}$$

Such prod­uct prob­a­bil­i­ties are char­ac­ter­is­tic of sta­tis­ti­cally in­de­pen­dent quan­ti­ties. As a sim­ple ex­am­ple, the chances of get­ting a three in the first throw of a die and a five in the sec­ond throw are $\frac16\times\frac16$ or 1 in 36. Throw­ing the three does not af­fect the chances of get­ting a five in the sec­ond throw.

Key Points

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When the pro­tons are well apart, an ap­prox­i­mate ground state is that of two neu­tral atoms.

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Sin­gle elec­tron wave func­tions for that case are $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

The com­plete wave func­tion for that case is $\psi_{\rm {l}}({\skew0\vec r}_1)\psi_{\rm {r}}({\skew0\vec r}_2)$, as­sum­ing that elec­tron 1 is around the left pro­ton and elec­tron 2 around the right one.

5.2.2 Re­view Ques­tions

1.

If elec­tron 2 does not af­fect where elec­tron 1 is likely to be, how would a grey-scale pic­ture of the prob­a­bil­ity of find­ing elec­tron 1 look?

So­lu­tion hmolb-a

2.

When the pro­tons are close to each other, the elec­trons do af­fect each other, and the wave func­tion above is no longer valid. But sup­pose you were given the true wave func­tion, and you were once again asked to draw the blob show­ing the prob­a­bil­ity of find­ing elec­tron 1 (us­ing a plot­ting pack­age, say). What would the big prob­lem be?

So­lu­tion hmolb-b

5.2.3 The prob­a­bil­ity den­sity

For mul­ti­ple-par­ti­cle sys­tems like the elec­trons of the hy­dro­gen mol­e­cule, show­ing the mag­ni­tude of the wave func­tion as grey tones no longer works since it is a func­tion in six-di­men­sion­al space. You can­not vi­su­al­ize six-di­men­sion­al space. How­ever, at every spa­tial po­si­tion ${\skew0\vec r}$ in nor­mal space, you can in­stead show the “prob­a­bil­ity den­sity” $n({\skew0\vec r})$, which is the prob­a­bil­ity per unit vol­ume of find­ing ei­ther elec­tron in a vicin­ity ${\rm d}^3{\skew0\vec r}$ of the point. This prob­a­bil­ity is found as

$$n({\skew0\vec r}) = \int \vert\Psi({\skew0\vec r}, {\skew0... ...w0\vec r}_1, {\skew0\vec r})\vert^2{ \rm d}^3{\skew0\vec r}_1$$ (5.4)

since the first in­te­gral gives the prob­a­bil­ity of find­ing elec­tron 1 at ${\skew0\vec r}$ re­gard­less of where elec­tron 2 is, (i.e. in­te­grated over all pos­si­ble po­si­tions for elec­tron 2), and the sec­ond gives the prob­a­bil­ity of find­ing 2 at ${\skew0\vec r}$ re­gard­less of where 1 is. Since ${\rm d}^3{\skew0\vec r}$ is van­ish­ingly small, the chances of find­ing both par­ti­cles in it at the same time are zero.

The prob­a­bil­ity den­sity $n({\skew0\vec r})$ for state $\psi_{\rm {l}}({\skew0\vec r}_1)\psi_{\rm {r}}({\skew0\vec r}_2)$ with elec­tron 1 around the left pro­ton and elec­tron 2 around the right one is shown in fig­ure 5.1. Of course the prob­a­bil­ity den­sity for the state $\psi_{\rm {r}}({\skew0\vec r}_1)\psi_{\rm {l}}({\skew0\vec r}_2)$ with the elec­trons ex­changed would look ex­actly the same.

Fig­ure 5.1: State with two neu­tral atoms.

Key Points

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The prob­a­bil­ity den­sity is the prob­a­bil­ity per unit vol­ume of find­ing an elec­tron, whichever one, near a given point.

5.2.3 Re­view Ques­tions

1.

Sup­pose, given the wave func­tion $\psi_{\rm {l}}({\skew0\vec r}_1)\psi_{\rm {r}}({\skew0\vec r}_2)$, that you found an elec­tron near the left pro­ton. What elec­tron would it prob­a­bly be? Sup­pose you found an elec­tron at the point halfway in be­tween the pro­tons. What elec­tron would that likely be?

So­lu­tion hmolc-a

5.2.4 States that share the elec­trons

This sec­tion will ex­am­ine the states where the pro­tons share the two elec­trons.

The first thing is to shorten the no­ta­tions a bit. So, the state $\psi_{\rm {l}}({\skew0\vec r}_1)\psi_{\rm {r}}({\skew0\vec r}_2)$ which de­scribes that elec­tron 1 is around the left pro­ton and elec­tron 2 around the right one will be in­di­cated by $\psi_{\rm {l}}\psi_{\rm {r}}$, us­ing the con­ven­tion that the first fac­tor refers to elec­tron 1 and the sec­ond to elec­tron 2. In this con­ven­tion, the state where elec­tron 1 is around the right pro­ton and elec­tron 2 around the left one is $\psi_{\rm {r}}\psi_{\rm {l}}$, short­hand for $\psi_{\rm {r}}({\skew0\vec r}_1)\psi_{\rm {l}}({\skew0\vec r}_2)$. It is of course phys­i­cally the same thing as $\psi_{\rm {l}}\psi_{\rm {r}}$; the two elec­trons are iden­ti­cal.

The every pos­si­ble com­bi­na­tion idea of com­bin­ing every pos­si­ble state for elec­tron 1 with every pos­si­ble state for elec­tron 2 would sug­gest that the states $\psi_{\rm {l}}\psi_{\rm {l}}$ and $\psi_{\rm {r}}\psi_{\rm {r}}$ should also be in­cluded. But these states have the elec­trons around the same pro­ton, and that is not go­ing to be en­er­get­i­cally fa­vor­able due to the mu­tual re­pul­sion of the elec­trons. So they are not use­ful for find­ing a sim­ple ap­prox­i­mate ground state of low­est en­ergy.

States where the elec­trons are no longer as­signed to a par­tic­u­lar pro­ton can be found as lin­ear com­bi­na­tions of $\psi_{\rm {l}}\psi_{\rm {r}}$ and $\psi_{\rm {r}}\psi_{\rm {l}}$:

$$\psi = a \psi_{\rm {l}}\psi_{\rm {r}} + b \psi_{\rm {r}}\psi_{\rm {l}}$$ (5.5)

In such a com­bi­na­tion each elec­tron has a prob­a­bil­ity of be­ing found about ei­ther pro­ton, but wher­ever it is found, the other elec­tron will be around the other pro­ton.

The eigen­func­tion must be nor­mal­ized, which not­ing that $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ are real and nor­mal­ized pro­duces

$$\langle \psi \vert \psi \rangle_6 = \langle a\psi_{\rm {l}... ...2ab \langle \psi_{\rm {l}} \vert \psi_{\rm {r}} \rangle^2 = 1$$ (5.6)

as­sum­ing that $a$ and $b$ are real. As a re­sult, only the ra­tio $a$$\raisebox{.5pt}{$/$}$​$b$ can be cho­sen freely. The prob­a­bil­ity den­sity of the com­bi­na­tion can be found to be:

$$n = \psi_{\rm {l}}^2+\psi_{\rm {r}}^2 + 2ab \langle \psi_{\... ...\rm {r}} \rangle (\psi_{\rm {l}}^2+\psi_{\rm {r}}^2) \right\}$$ (5.7)

The most im­por­tant com­bi­na­tion state is the one with $b$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a$:

$$\psi({\skew0\vec r}_1,{\skew0\vec r}_2) = a \left[ \psi_... ...r}}({\skew0\vec r}_1)\psi_{\rm {l}}({\skew0\vec r}_2) \right]$$ (5.8)

This state is called “sym­met­ric with re­spect to ex­chang­ing elec­tron 1 with elec­tron 2,” or more pre­cisely, with re­spect to re­plac­ing ${\skew0\vec r}_1$ by ${\skew0\vec r}_2$ and vice-versa. Such an ex­change does not change this wave func­tion at all. If you change ${\skew0\vec r}_1$ into ${\skew0\vec r}_2$ and vice-versa, you still end up with the same wave func­tion. In terms of the hy­dro­gen ground state wave func­tion, it may be writ­ten out fully as

$$\fbox{$\displaystyle \Psi \approx a \left[ \psi_{100}(\v... ...{\skew0\vec r}_2-{\skew0\vec r}_{\rm{lp}}\vert) \right] $} %$$ (5.9)

with $\psi_{100}(r)$ $\vphantom0\raisebox{1.5pt}{$\equiv$}$ ${e}^{-r/a_0}$$\raisebox{.5pt}{$/$}$​$\sqrt{{\pi}a_0^3}$, where $a_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.53 Å is the Bohr ra­dius, and ${\skew0\vec r}_1$, ${\skew0\vec r}_2$, ${\skew0\vec r}_{\rm {lp}}$, and ${\skew0\vec r}_{\rm {rp}}$ are again the po­si­tion vec­tors of the elec­trons and pro­tons.

The prob­a­bil­ity den­sity of this wave func­tion looks like fig­ure 5.2. It has in­creased like­li­hood for elec­trons to be found in be­tween the pro­tons, com­pared to fig­ure 5.1 in which each pro­ton had its own elec­tron.

Fig­ure 5.2: Sym­met­ric shar­ing of the elec­trons.

The state with $b$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$$a$,

$$\psi({\skew0\vec r}_1,{\skew0\vec r}_2) = a \left[ \psi_... ...r}}({\skew0\vec r}_1)\psi_{\rm {l}}({\skew0\vec r}_2) \right]$$ (5.10)

is called an­ti­sym­met­ric with re­spect to ex­chang­ing elec­tron 1 with elec­tron 2: swap­ping ${\skew0\vec r}_1$ and ${\skew0\vec r}_2$ changes the sign of wave func­tion, but leaves it fur­ther un­changed. As seen in fig­ure 5.3, the an­ti­sym­met­ric state has de­creased like­li­hood for elec­trons to be found in be­tween the pro­tons.

Fig­ure 5.3: An­ti­sym­met­ric shar­ing of the elec­trons.

Key Points

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In state $\psi_{\rm {l}}\psi_{\rm {r}}$, the elec­tron num­bered 1 is around the left pro­ton and 2 around the right one.

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In state $\psi_{\rm {r}}\psi_{\rm {l}}$, the elec­tron num­bered 1 is around the right pro­ton and 2 around the left one.

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In the sym­met­ric state $a(\psi_{\rm {l}}\psi_{\rm {r}}+\psi_{\rm {r}}\psi_{\rm {l}})$ the pro­tons share the elec­trons equally; each elec­tron has an equal chance of be­ing found around ei­ther pro­ton. In this state there is in­creased prob­a­bil­ity of find­ing an elec­tron some­where in be­tween the pro­tons.

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In the an­ti­sym­met­ric state $a(\psi_{\rm {l}}\psi_{\rm {r}}-\psi_{\rm {r}}\psi_{\rm {l}})$ the pro­tons also share the elec­trons equally; each elec­tron has again an equal chance of be­ing found around ei­ther pro­ton. But in this state there is de­creased prob­a­bil­ity of find­ing an elec­tron some­where in be­tween the pro­tons.

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So, like for the mol­e­c­u­lar ion, at large pro­ton sep­a­ra­tions the weird trick of shuf­fling un­ob­serv­able wave func­tions around does again pro­duce dif­fer­ent phys­i­cal states with pretty much the same en­ergy.

5.2.4 Re­view Ques­tions

1.

Ob­vi­ously, the vi­sual dif­fer­ence be­tween the var­i­ous states is mi­nor. It may even seem counter-in­tu­itive that there is any dif­fer­ence at all: the states $\psi_{\rm {l}}\psi_{\rm {r}}$ and $\psi_{\rm {r}}\psi_{\rm {l}}$ are ex­actly the same phys­i­cally, with one elec­tron around each pro­ton. So why would their com­bi­na­tions be any dif­fer­ent?

The quan­tum dif­fer­ence would be much more clear if you could see the full six-di­men­sion­al wave func­tion, but vi­su­al­iz­ing six-di­men­sion­al space just does not work. How­ever, if you re­strict your­self to only look­ing on the $z$-​axis through the nu­clei, you get a draw­able $z_1,z_2$-​plane de­scrib­ing near what ax­ial com­bi­na­tions of po­si­tions you are most likely to find the two elec­trons. In other words: what would be the chances of find­ing elec­tron 1 near some ax­ial po­si­tion $z_1$ and elec­tron 2 at the same time near some other ax­ial po­si­tion $z_2$?

Try to guess these prob­a­bil­i­ties in the $z_1,z_2$-​plane as grey tones, (darker if more likely), and then com­pare with the an­swer.

So­lu­tion hmold-a

2.

Based on the pre­vi­ous ques­tion, how would you think the prob­a­bil­ity den­sity $n(z)$ would look on the axis through the nu­clei, again ig­nor­ing the ex­is­tence of po­si­tions be­yond the axis?

So­lu­tion hmold-b

5.2.5 Vari­a­tional ap­prox­i­ma­tion of the ground state

The pur­pose of this sec­tion is to find an ap­prox­i­ma­tion to the ground state of the hy­dro­gen mol­e­cule us­ing the rough ap­prox­i­ma­tion of the wave func­tion de­scribed in the pre­vi­ous sub­sec­tions.

Like for the hy­dro­gen mol­e­c­u­lar ion of chap­ter 4.6.6, the idea is that since the true ground state is the state of low­est en­ergy among all wave func­tions, the best among ap­prox­i­mate wave func­tions is the one with the low­est en­ergy. The ap­prox­i­mate wave func­tions are here of the form $a\psi_{\rm {l}}\psi_{\rm {r}}+b\psi_{\rm {r}}\psi_{\rm {l}}$; in these the pro­tons share the elec­trons, but in such a way that when one elec­tron is around the left pro­ton, the other is around the right one, and vice-versa.

A com­puter pro­gram is again needed to print out the ex­pec­ta­tion value of the en­ergy for var­i­ous val­ues of the ra­tio of co­ef­fi­cients $a$$\raisebox{.5pt}{$/$}$​$b$ and pro­ton-pro­ton dis­tance $d$. And worse, the ex­pec­ta­tion value of en­ergy for given $a$$\raisebox{.5pt}{$/$}$​$b$ and $d$ is a six-di­men­sion­al in­te­gral, and parts of it can­not be done an­a­lyt­i­cally; nu­mer­i­cal in­te­gra­tion must be used. That makes it a much more messy prob­lem, {D.23}.

You might just want to take it on faith that the bind­ing en­ergy, at the state of low­est en­ergy found, turns out to be 3.2 eV, at a pro­ton to pro­ton spac­ing of 0.87 Å, and that it oc­curs for the sym­met­ric state $a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $b$.

Key Points

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An ap­prox­i­mate ground state can be found for the hy­dro­gen mol­e­cule us­ing a vari­a­tional method much like that for the mol­e­c­u­lar ion.

5.2.6 Com­par­i­son with the ex­act ground state

The so­lu­tion for the ground state of the hy­dro­gen mol­e­cule ob­tained in the pre­vi­ous sub­sec­tion is, like the one for the mol­e­c­u­lar ion, pretty good. The ap­prox­i­mate bind­ing en­ergy, 3.2 eV, is not too much dif­fer­ent from the ex­per­i­men­tal value of 4.52 eV. Sim­i­larly, the bond length of 0.87 Å is not too far from the ex­per­i­men­tal value of 0.74 Å.

Qual­i­ta­tively, the ex­act ground state wave func­tion is real, pos­i­tive and sym­met­ric with re­spect to re­flec­tion around the sym­me­try plane and to ro­ta­tions around the line con­nect­ing the pro­tons, and so is the ap­prox­i­mate one. The rea­sons for these prop­er­ties are sim­i­lar as for the mol­e­c­u­lar ion; {A.8,A.9}.

One very im­por­tant new sym­me­try for the neu­tral mol­e­cule is the ef­fect of ex­chang­ing the elec­trons, re­plac­ing ${\skew0\vec r}_1$ by ${\skew0\vec r}_2$ and vice-versa. The ap­prox­i­mate wave func­tion is sym­met­ric (un­changed) un­der such an ex­change, and so is the ex­act wave func­tion. To un­der­stand why, note that the op­er­a­tion of ex­chang­ing the elec­trons com­mutes with the Hamil­ton­ian, (ex­chang­ing iden­ti­cal elec­trons phys­i­cally does not do any­thing). So en­ergy eigen­func­tions can be taken to be also eigen­func­tions of the “ex­change op­er­a­tor.” Fur­ther­more, the ex­change op­er­a­tor is a Her­mit­ian one, (tak­ing it to the other side in in­ner prod­ucts is equiv­a­lent to a sim­ple name change of in­te­gra­tion vari­ables,) so it has real eigen­val­ues. And more specif­i­cally, the eigen­val­ues can only be plus or mi­nus one, since swap­ping elec­trons does not change the mag­ni­tude of the wave func­tion. So the en­ergy eigen­func­tions, in­clud­ing the ground state, must be sym­met­ric un­der elec­tron ex­change (eigen­value one), or an­ti­sym­met­ric (eigen­value mi­nus one.) Since the ground state must be every­where pos­i­tive, (or more pre­cisely, of a sin­gle sign), a sign change due to swap­ping elec­trons is not pos­si­ble. So only the sym­met­ric pos­si­bil­ity ex­ists for the ground state.

One is­sue that does not oc­cur for the mol­e­c­u­lar ion, but only for the neu­tral mol­e­cule is the mu­tual re­pul­sion be­tween the two elec­trons. This re­pul­sion is re­duced when the elec­tron clouds start to merge, com­pared to what it would be if the clouds were more com­pact. (A sim­i­lar ef­fect is that the grav­ity force of the earth de­creases when you go down be­low the sur­face. To be sure, the po­ten­tial en­ergy keeps go­ing down, or up for elec­tron clouds, but not as much as it would oth­er­wise. Com­pare fig­ure 13.7.) Since the nu­clei are com­pact, it gives an ad­van­tage to nu­cleus-elec­tron at­trac­tion over elec­tron-elec­tron re­pul­sion. This in­creases the bind­ing en­ergy sig­nif­i­cantly; in the ap­prox­i­mate model from about 1.8 eV to 3.2 eV. It also al­lows the pro­tons to ap­proach more closely; {D.23}.

The ques­tion has been asked whether there should not be an “ac­ti­va­tion en­ergy” in­volved in cre­at­ing the hy­dro­gen mol­e­cule from the hy­dro­gen atoms. The an­swer is no, hy­dro­gen atoms are rad­i­cals, not sta­ble mol­e­cules that need to be taken apart be­fore re­com­bin­ing. In fact, the hy­dro­gen atoms at­tract each other even at large dis­tances due to Van der Waals at­trac­tion, chap­ter 10.1, an ef­fect lost in the ap­prox­i­mate wave func­tions used in this sec­tion. But hy­dro­gen atoms that fly into each other also have enough en­ergy to fly apart again; some of the ex­cess en­ergy must be ab­sorbed else­where to form a sta­ble mol­e­cule. Ac­cord­ing to web sources, hy­dro­gen mol­e­cule for­ma­tion in the uni­verse is be­lieved to typ­i­cally oc­cur on dust specks.

Key Points

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The ap­prox­i­mate ground state is pretty good, con­sid­er­ing its sim­plic­ity.