5.8.2 So­lu­tion mat­for-b


Find the eigen­states for the same prob­lem, but now in­clud­ing spin.

As sec­tion 5.7 showed, the an­ti­sym­met­ric wave func­tion with spin con­sists of a sum of six Slater de­ter­mi­nants. Ig­nor­ing the highly ex­cited first and sixth de­ter­mi­nants that have the elec­trons around the same nu­cleus, the re­main­ing $C$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4 Slater de­ter­mi­nants can be writ­ten out ex­plic­itly to give the two-par­ti­cle states

\begin{eqnarray*}\psi^{\rm S}_1 = \frac{\psi_{\rm {l}}\psi_{\rm {r}}{\uparrow}{\...
... \psi_{\rm {r}}\psi_{\rm {l}}{\downarrow}{\downarrow}}{\sqrt{2}}

Note that the Hamil­ton­ian does not in­volve spin, to the ap­prox­i­ma­tion used in most of this book, so that, fol­low­ing the tech­niques of sec­tion 5.5, an in­ner prod­uct like $H_{23}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\langle\psi^{\rm S}_2\vert H\psi^{\rm S}_3\rangle$ can be writ­ten out like

\begin{eqnarray*}H_{23} & = & \frac 12\langle\psi_{\rm {l}}\psi_{\rm {r}}{\uparr...
...w}- (H\psi_{\rm {r}}\psi_{\rm {l}}){\uparrow}{\downarrow}\rangle

and then mul­ti­plied out into in­ner prod­ucts of match­ing spin com­po­nents to give

H_{23} = -\frac 12\langle\psi_{\rm {l}}\psi_{\rm {r}}\vert H...
...}\psi_{\rm {l}}\vert H\psi_{\rm {l}}\psi_{\rm {r}}\rangle = L.

The other 15 ma­trix co­ef­fi­cients can be found sim­i­larly, and most will be zero.

If you do not have ex­pe­ri­ence with lin­ear al­ge­bra, you may want to skip this ques­tion, or bet­ter, just read the so­lu­tion. How­ever, the four eigen­vec­tors are not that hard to guess; maybe eas­ier to guess than cor­rectly de­rive.


Eval­u­at­ing all 16 in­ner prod­ucts $H_{{{\underline k}}k}$ as above, the ma­trix eigen­value prob­lem is found to be

\begin{array}{ccccccccl} (J+L) a_1 & + & 0 a_2 & + & 0 a_3 &...
...& + & 0 a_2 & + & 0 a_3 & + & (J+L) a_4 & = & E a_4

Af­ter bring­ing every­thing to the left hand side, the de­ter­mi­nant of the re­sult­ing ma­trix can again be set to zero, and the pos­si­ble en­ergy val­ues found. From those the eigen­vec­tors can be de­duced. It turns out that the larger eigen­value is triply de­gen­er­ate, so the three cor­re­spond­ing eigen­vec­tors are not unique; more than one ac­cept­able choice ex­ists for them. How­ever, you do have to nor­mal­ize them to unit length and en­sure that they are mu­tu­ally or­thog­o­nal.

Rather than go­ing through that math, it is quicker just to guess the eigen­vec­tors. One guess that works is

(a_1,a_2,a_3,a_4) = (1,0,0,0).

Just sub­sti­tute $a_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and $a_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a_4$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 into the equa­tions above and see that they are sat­is­fied pro­vided that the en­ergy $E$ has the ex­cited value $J+L$. The cor­re­spond­ing eigen­func­tion $\psi_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi^{\rm S}_1$ has, ac­cord­ing to its de­f­i­n­i­tion above, both elec­trons spin-up and in the ex­cited an­ti­sym­met­ric spa­tial state $(\psi_{\rm {l}}\psi_{\rm {r}}-\psi_{\rm {r}}\psi_{\rm {l}})$$\raisebox{.5pt}{$/$}$$\sqrt{2}$.

A sim­i­lar guess that works is

(a_1,a_2,a_3,a_4) = (0,0,0,1).

This cor­re­sponds to the eigen­func­tion $\psi_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi^{\rm S}_4$ in which both elec­trons are spin-down, and again in the an­ti­sym­met­ric spa­tial state. The third one that works is

(a_1,a_2,a_3,a_4) = (0,1,1,0)/\sqrt{2}

where the scale fac­tor $\sqrt{2}$ is only needed to en­sure that the vec­tor is of unit length. The eigen­state $\psi_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(\psi^{\rm S}_2+\psi^{\rm S}_3)$$\raisebox{.5pt}{$/$}$$\sqrt{2}$ cor­re­sponds, ac­cord­ing to the de­f­i­n­i­tions of $\psi^{\rm S}_2$ and $\psi^{\rm S}_3$ above, to the elec­trons be­ing in the an­ti­sym­met­ric spa­tial state times the ${\left\vert 1\:0\right\rangle}$ triplet state of sec­tion 5.5.6. The en­ergy is again the el­e­vated value $J+L$.

The fi­nal eigen­vec­tor

(a_1,a_2,a_3,a_4) = (0,1,-1,0)/\sqrt{2},

gives the eigen­state $\psi_4$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(\psi^{\rm S}_2-\psi^{\rm S}_3)$$\raisebox{.5pt}{$/$}$$\sqrt{2}$. This cor­re­sponds to the elec­trons be­ing in the sym­met­ric spa­tial state times the sin­glet spin state. The en­ergy is the ground state value $J-L$.