5.2.4.2 So­lu­tion hmold-b

Ques­tion:

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?

An­swer:

For any state, the prob­a­bil­ity of find­ing elec­tron 1 near a given $z$, re­gard­less of where elec­tron 2 is, is found by set­ting $z_1$ equal to $z$ and in­te­grat­ing over all pos­si­ble po­si­tions $z_2$ for elec­tron 2. For the two-di­men­sion­al state $\psi_{\rm {l}}\psi_{\rm {r}}$ shown in the left col­umn of fig­ure 5.2, you are then in­te­grat­ing over ver­ti­cal lines in the top pic­ture; imag­ine mov­ing all blank ink ver­ti­cally to­wards the $z_1$ axis and then set­ting $z_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $z$. The re­sult­ing curve is shown im­me­di­ately be­low. As ex­pected, elec­tron 1 is in this state most likely to be found some­where around $z_{\rm {lp}}$, the neg­a­tive po­si­tion of the left pro­ton. Re­gard­less of where elec­tron 2 is.

Fig­ure 5.2: Prob­a­bil­ity den­sity 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 $a(\psi_{\rm{l}}\psi_{\rm{r}}-\psi_{\rm{r}}\psi_{\rm{l}})$. From top to bot­tom, the top row of curves show the prob­a­bil­ity of find­ing elec­tron 1 near $z$ re­gard­less where elec­tron 2 is. The sec­ond row shows the prob­a­bil­ity of find­ing elec­tron 2 near $z$ re­gard­less where elec­tron 1 is. The third row shows the to­tal prob­a­bil­ity of find­ing ei­ther elec­tron near $z$, the sum of the pre­vi­ous two rows. The fourth row shows the same as the third, but as­sum­ing the true three-di­men­sion­al world rather than just the line through the nu­clei.

(Note that all pos­si­ble po­si­tions of elec­tron 2 should re­ally be found by in­te­grat­ing over all pos­si­ble po­si­tions in three di­men­sions, not just ax­ial ones. The fi­nal row in the fig­ure gives the to­tal prob­a­bil­i­ties when cor­rected for that. But the idea is the same, just harder to vi­su­al­ize.)

The prob­a­bil­ity den­sity at a given value of $z$ also needs to in­clude the pos­si­bil­ity of find­ing elec­tron 2 there. That prob­a­bil­ity is found by set­ting $z_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $z$ and then in­te­grat­ing over all pos­si­ble val­ues of $z_1$. You are now mov­ing the blank ink hor­i­zon­tally to­wards the $z_2$ axis, and then set­ting $z_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $z$. The re­sult­ing curve is shown in the sec­ond graph in the left col­umn of fig­ure 5.2. As ex­pected, elec­tron 2 is most likely to be found some­where around $z_{\rm {rp}}$, the pos­i­tive po­si­tion of the right pro­ton.

To get the prob­a­bil­ity den­sity, the chance of find­ing ei­ther pro­ton near $z$, you need to add the two curves to­gether. That is done in the third graph in the left col­umn of fig­ure 5.2. An elec­tron is likely to be some­where around each pro­ton. This graph looks ex­actly like the cor­rect three-di­men­sion­al curve shown in the bot­tom graph, but that is re­ally just a co­in­ci­dence.

The states $\psi_{\rm {r}}\psi_{\rm {l}}$ and $a(\psi_{\rm {l}}\psi_{\rm {r}}\pm\psi_{\rm {r}}\psi_{\rm {l}})$ can be in­te­grated sim­i­larly; they are shown in the sub­se­quent columns in fig­ure 5.2. Note how the line of zero wave func­tion in the an­ti­sym­met­ric case dis­ap­pears dur­ing the in­te­gra­tions. Also note that re­ally, the prob­a­bil­ity den­sity func­tions of the sym­met­ric and an­ti­sym­met­ric states are quite dif­fer­ent, though they look qual­i­ta­tively the same.