So­lu­tion hmold-a


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.


Here are the pic­tures, as­sum­ing the ori­gin is halfway in be­tween the pro­tons:

Fig­ure 5.1: Wave func­tions on the $z$-axis through the nu­clei. From left to right: $\psi_{\rm{l}}\psi_{\rm{r}}$, $\psi_{\rm{r}}\psi_{\rm{l}}$, the sym­met­ric com­bi­na­tion $a(\psi_{\rm{l}}\psi_{\rm{r}}+\psi_{\rm{r}}\psi_{\rm{l}})$, and the an­ti­sym­met­ric one $a(\psi_{\rm{l}}\psi_{\rm{r}}-\psi_{\rm{r}}\psi_{\rm{l}})$.
\begin{figure}\centering {}
... \multiput(-163,92)(108,0){4}{\makebox(0,0)[r]{$z_2$}}

These re­sults can be ex­plained as fol­lows: For the state $\psi_{\rm {l}}\psi_{\rm {r}}$, elec­tron 1 is around the left pro­ton, so its likely $z_1$ po­si­tions are clus­tered around the po­si­tion $z_{\rm {lp}}$ of that pro­ton, in­di­cated by a tick mark on the neg­a­tive $z_1$-​axis in fig­ure 5.1. Sim­i­larly elec­tron 2 is around the right pro­ton, so its $z_2$ po­si­tions are clus­tered around the pos­i­tive value $z_{\rm {rp}}$ in­di­cated by the tick mark on the pos­i­tive $z_2$ axis. This means the wave func­tion, $\Psi(0,0,z_1,0,0,z_2)$, will look as shown in the left pic­ture of fig­ure 5.1. It will be mostly in the quad­rant of neg­a­tive $z_1$ and pos­i­tive $z_2$.

Sim­i­larly $\psi_{\rm {r}}\psi_{\rm {l}}$ will look as the sec­ond pic­ture. Here the po­si­tions of elec­tron 1 clus­ter around the pos­i­tive po­si­tion of the right pro­ton and those of elec­tron 2 around the neg­a­tive po­si­tion of the left pro­ton.

When you av­er­age the two states sym­met­ri­cally, you get a two-blob pic­ture like the third pic­ture. Now it is elec­tron 1 around the left pro­ton and elec­tron 2 around the right one or vice-versa. But there is still al­most no prob­a­bil­ity of find­ing both pro­tons in the first quad­rant, both near the right pro­ton. Nor are you likely to find them in the third quad­rant, both near the left pro­ton.

If you av­er­age the first two states an­ti­sym­met­ri­cally, you get the fourth pic­ture. In the an­ti­sym­met­ric com­bi­na­tion, the wave func­tion is zero on the sym­me­try line be­tween the blobs.

You see that the states are re­ally dif­fer­ent when looked at in the full six-di­men­sion­al space.