D.21 So­lu­tion of the hy­dro­gen mol­e­c­u­lar ion

The key to the vari­a­tional ap­prox­i­ma­tion to the hy­dro­gen mol­e­c­u­lar ion is to be able to ac­cu­rately eval­u­ate the ex­pec­ta­tion en­ergy

= \langle a \psi_{\rm {l}} +...
... {r}}\vert
H\vert a \psi_{\rm {l}} + b \psi_{\rm {r}} \rangle

This can be mul­ti­plied out and sim­pli­fied by not­ing that $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ are eigen­func­tions of the par­tial Hamil­to­ni­ans. For ex­am­ple,

H\psi_{\rm {l}} = E_1 \psi_{\rm {l}}
- \frac{e^2}{4\pi\epsilon_0}\frac{1}{r_{\rm {r}}} \psi_{\rm {l}}

where $E_1$ is the -13.6 eV hy­dro­gen atom ground state en­ergy. The ex­pres­sion can be fur­ther sim­pli­fied by not­ing that by sym­me­try

\langle \psi_{\rm {r}}\vert r_{\rm {l}}^{-1}\psi_{\rm {r}}\...
...ngle \psi_{\rm {r}}\vert r_{\rm {r}}^{-1}\psi_{\rm {l}}\rangle

and that $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ are real, so that the left and right sides of the var­i­ous in­ner prod­ucts can be re­versed. Also, $a$ and $b$ are re­lated by the nor­mal­iza­tion re­quire­ment

a^2 + b^2 + 2 ab \langle\psi_{\rm {l}}\vert\psi_{\rm {r}}\rangle = 1

Clean­ing up the ex­pec­ta­tion en­ergy in this way, the re­sult is

\left\langle{E}\right\rangle &= & E_1 - \\
&& \frac{e^2}{4 ...
...r_{\rm {r}}^{-1} \psi_{\rm {l}} \right\rangle

which in­cludes the pro­ton to pro­ton re­pul­sion en­ergy (the 1$\raisebox{.5pt}{$/$}$$d$). The en­ergy $E_1$ is the $\vphantom{0}\raisebox{1.5pt}{$-$}$13.6 eV amount of en­ergy when the pro­tons are far apart.

Nu­mer­i­cal in­te­gra­tion is not needed; the in­ner prod­uct in­te­grals in this ex­pres­sion can be done an­a­lyt­i­cally. To do so, take the ori­gin of a spher­i­cal co­or­di­nate sys­tem $(r,\theta,\phi)$ at the left pro­ton, and the axis to­wards the right one, so that

r_{\rm {l}} = \vert{\skew0\vec r}-{\skew0\vec r}_{\rm {lp}}...
...ew0\vec r}_{\rm {rp}}\vert = \sqrt{d^2+r^2-2 d r\cos(\theta)}.

In those terms,

\psi_{\rm {l}} = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0}
...{\sqrt{\pi a_0^3}}
e^{-\sqrt{d^2+r^2-2 d r\cos(\theta)}/a_0}.

Then in­te­grate an­gles first us­ing ${ \rm d}^3{\skew0\vec r}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r^2\sin(\theta){\rm d}{r}{ \rm d}\theta{ \rm d}\phi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-r^2{\rm d}{r}{ \rm d}\cos(\theta){ \rm d}\phi$. Do not for­get that $\sqrt{x^2}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert x\vert$, not $x$, e.g. $\sqrt{(-3)^2}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3, not $\vphantom{0}\raisebox{1.5pt}{$-$}$3. More de­tails are in [25, pp. 305-307].

The over­lap in­te­gral turns out to be

\left\langle \psi_{\rm {l}} \vert\psi_{\rm {r}} \right\rang...
... + \frac{d}{a_0} + \frac13 \left(\frac{d}{a_0}\right)^2\right]

and pro­vides a mea­sure of how much the re­gions of the two wave func­tions over­lap. The di­rect in­te­gral is

\left\langle \psi_{\rm {l}}\vert r_{\rm {r}}^{-1} \psi_{\rm...
...rac{1}{d} - \left[\frac{1}{a_0}+\frac{1}{d}\right] e^{-2d/a_0}

and gives the clas­si­cal po­ten­tial of an elec­tron den­sity of strength $\vert\psi_{\rm {l}}\vert^2$ in the field of the right pro­ton, ex­cept for the fac­tor $\vphantom{0}\raisebox{1.5pt}{$-$}$$e^2$$\raisebox{.5pt}{$/$}$$4\pi\epsilon_0$. The “ex­change in­te­gral” is

\left\langle \psi_{\rm {l}} \vert r_{\rm {l}}^{-1} \psi_{\r...
\left[\frac{1}{a_0}+\frac{d}{a_0^2}\right] e^{-d/a_0}.

and is some­what of a twi­light term, since $\psi_{\rm {l}}$ sug­gests that the elec­tron is around the left pro­ton, but $\psi_{\rm {r}}$ sug­gests it is around the right one.