5.8.1 So­lu­tion mat­for-a

Ques­tion:

As a rel­a­tively sim­ple ex­am­ple, work out the above ideas for the $Q$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 hy­dro­gen mol­e­cule spa­tial states $\psi^{\rm S}_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_{\rm {l}}\psi_{\rm {r}}$ and $\psi^{\rm S}_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_{\rm {l}}\psi_{\rm {r}}$. Write the ma­trix eigen­value prob­lem and iden­tify the two eigen­val­ues and eigen­vec­tors. Com­pare with the re­sults of sec­tion 5.3.

As­sume that $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ have been slightly ad­justed to be or­tho­nor­mal. Then so are $\psi^{\rm S}_1$ and $\psi^{\rm S}_2$ or­tho­nor­mal, since the var­i­ous six-di­men­sion­al in­ner prod­uct in­te­grals, like

$$\begin{eqnarray*}\lefteqn{ \langle\psi^{\rm S}_1\vert\psi^{\rm S}_2\rangle\equiv... ... r}_2) { \rm d}^3 {\skew0\vec r}_1 { \rm d}^3 {\skew0\vec r}_2 \end{eqnarray*}$$

can ac­cord­ing to the rules of cal­cu­lus be fac­tored into three-di­men­sion­al in­te­grals as

$$\begin{eqnarray*}\lefteqn{\langle\psi^{\rm S}_1\vert\psi^{\rm S}_2\rangle} \ &&... ..._{\rm {r}}\rangle\langle\psi_{\rm {r}}\vert\psi_{\rm {l}}\rangle \end{eqnarray*}$$

which is zero if $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ are or­tho­nor­mal.

Also, do not try to find ac­tual val­ues for $H_{11}$, $H_{12}$, $H_{21}$, and $H_{22}$. As sec­tion 5.2 noted, that can only be done nu­mer­i­cally. In­stead just re­fer to $H_{11}$ as $J$ and to $H_{12}$ as $\vphantom{0}\raisebox{1.5pt}{$-$}$$L$:

$$\begin{eqnarray*}& H_{11} \equiv\langle\psi^{\rm S}_1\vert H\psi^{\rm S}_1\rangl... ...i_{\rm {r}}\vert H\psi_{\rm {r}}\psi_{\rm {l}}\rangle\equiv - L. \end{eqnarray*}$$

Next note that you also have

$$\begin{eqnarray*}& H_{22} \equiv\langle\psi^{\rm S}_2\vert H\psi^{\rm S}_2\rangl... ...}}\psi_{\rm {l}}\vert H\psi_{\rm {l}}\psi_{\rm {r}}\rangle = - L \end{eqnarray*}$$

be­cause they are the ex­act same in­ner prod­uct in­te­grals; the dif­fer­ence is just which elec­tron you num­ber 1 and which one you num­ber 2 that de­ter­mines whether the wave func­tions are listed as $\psi_{\rm {l}}\psi_{\rm {r}}$ or $\psi_{\rm {r}}\psi_{\rm {l}}$.

An­swer:

Us­ing the ab­bre­vi­a­tions $J$ and $L$, the ma­trix eigen­value prob­lem be­comes

$$\begin{array}{rcl} J a_1 - L a_2 & = & E a_1 \ - L a_1 + J a_2 & = & E a_2 \end{array}$$

or tak­ing every­thing to the left hand side,

$$\begin{array}{rcl} \left(J-E\right) a_1 - L a_2 & = & 0 \ - L a_1 + \left(J-E\right) a_2 & = & 0 \end{array}.$$

For this ho­mo­ge­neous sys­tem of equa­tions to have a so­lu­tion other than the triv­ial one $a_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, the de­ter­mi­nant of the ma­trix must be zero:

$$\left\vert \begin{array}{cc} \left(J-E\right) & - L \ - L &... ...rt = 0 \qquad\Longrightarrow\qquad\left(J-E\right)^2 - L^2 = 0$$

which al­lows for the two pos­si­bil­i­ties

$$J-E = L \qquad\mbox{or}\qquad J-E = -L.$$

So there are two en­ergy eigen­val­ues:

$$E_1 = J - L \qquad\mbox{and}\qquad E_2 = J + L.$$

In the first case, since ac­cord­ing to the equa­tions above

$$\left(J-E_1\right) a_1 - L a_2 = 0 \qquad\Longrightarrow\qquad L a_1 - L a_2 = 0$$

it fol­lows that $a_1$ and $a_2$ must be equal, pro­duc­ing the eigen­func­tion

$$a_1\left(\psi^{\rm S}_1+\psi^{\rm S}_2\right) = a_1 \left(\psi_{\rm {l}}\psi_{\rm {r}} + \psi_{\rm {r}}\psi_{\rm {l}}\right)$$

and nor­mal­iza­tion shows that $a_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1/$\sqrt{2}$, as­sum­ing you take it real and pos­i­tive, (and as­sum­ing that $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ are re­ally ad­justed to be or­tho­nor­mal as as­sumed in this par­tic­u­lar ques­tion).

So the first en­ergy eigen­state is

$$\mbox{eigenfunction: } \frac{1}{\sqrt{2}}\left(\psi_{\rm {l}... ...r}}\psi_{\rm {l}}\right) \qquad\mbox{eigenvalue: } E_1 = J - L$$

Sim­i­larly the sec­ond en­ergy eigen­state is

$$\mbox{eigenfunction: } \frac{1}{\sqrt{2}}\left(\psi_{\rm {l}... ...}}\psi_{\rm {l}}\right) \qquad\mbox{eigenvalue: } E_2 = J + L.$$

Com­par­ing with sec­tion 5.3, the first eigen­state can be rec­og­nized as the ground state in which the nu­clei share the elec­trons sym­met­ri­cally. The sec­ond eigen­state is the ex­cited-en­ergy an­ti­sym­met­ric state in which the elec­trons share the elec­trons an­ti­sym­met­ri­cally. Other states of this type, like sim­ply $\psi_{\rm {l}}\psi_{\rm {r}}$, say, have ex­pec­ta­tion en­ergy some­where in be­tween $E_1$ and $E_2$, but they are not eigen­states and do not have def­i­nite en­ergy.

Of course, the analy­sis here is ap­prox­i­mate. But as dis­cussed more in the ear­lier sec­tion 5.2, the true ground state is re­ally sym­met­ric, and the ex­cited en­ergy eigen­state re­ally is an­ti­sym­met­ric. Af­ter all, the Hamil­ton­ian com­mutes with the op­er­a­tion of swap­ping the elec­trons, (swap­ping the elec­trons does not do any­thing phys­i­cally,) so the en­ergy eigen­states must also be eigen­states of the swap­ping op­er­a­tor. The sym­met­ric state is an eigen­state of the swap­ping op­er­a­tor with eigen­value 1, it stays the same, and the an­ti­sym­met­ric state is an eigen­state with eigen­value $\vphantom{0}\raisebox{1.5pt}{$-$}$1; it changes sign un­der the swap.