D.24 Hy­dro­gen mol­e­cule ground state and spin

The pur­pose of this note is to ver­ify that the in­clu­sion of spin does not change the spa­tial form of the ground state of the hy­dro­gen mol­e­cule. The low­est ex­pec­ta­tion en­ergy $\langle{E}\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\langle\psi_{\rm {gs}}\vert H\psi_{\rm {gs}}\rangle$, char­ac­ter­iz­ing the cor­rect ground state, only oc­curs if all spa­tial com­po­nents $\psi_{\pm\pm}$ of the ground state with spin,

$$\psi_{\rm gs} = \psi_{++} {\uparrow}{\uparrow}+ \psi_{+-} ... ...+} {\downarrow}{\uparrow}+ \psi_{-} {\downarrow}{\downarrow},$$

are pro­por­tional to the no-spin spa­tial ground state $\psi_{{\rm {gs}},0}$.

The rea­son is that the as­sumed Hamil­ton­ian (5.3) does not in­volve spin at all, only spa­tial co­or­di­nates, so, for ex­am­ple,

$$\left(H\psi_{++}{\uparrow}{\uparrow}\right) \equiv H\left... ...S_{z2})\right) = \left(H\psi_{++}\right){\uparrow}{\uparrow}$$

and the same for the other three terms in $H\psi_{\rm {gs}}$. So the ex­pec­ta­tion value of en­ergy be­comes

$$\begin{array}{l} \left\langle{E}\right\rangle = \langle \... ...(H\psi_{-}\right){\downarrow}{\downarrow} \rangle \end{array}$$

Be­cause of the or­tho­nor­mal­ity of the spin states, this mul­ti­plies out into in­ner prod­ucts of match­ing spin states as

$$\left\langle{E}\right\rangle = \langle\psi_{++}\vert H\psi... ... H\psi_{-+}\rangle + \langle\psi_{-}\vert H\psi_{-}\rangle.$$

In ad­di­tion, the wave func­tion must be nor­mal­ized, $\langle\psi_{\rm {gs}}\vert\psi_{\rm {gs}}\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, or

$$\langle\psi_{++}\vert\psi_{++}\rangle + \langle\psi_{+-}\v... ...\psi_{-+}\rangle + \langle\psi_{-}\vert\psi_{-}\rangle = 1.$$

Now when $\psi_{++}$, $\psi_{+-}$, $\psi_{-+}$, and $\psi_{-}$ are each pro­por­tional to the no-spin spa­tial ground state $\psi_{{\rm {gs}},0}$ with the low­est en­ergy $E_{\rm {gs}}$, their in­di­vid­ual con­tri­bu­tions to the en­ergy will be given by $\left\langle\vphantom{H\psi_{\pm\pm}}\psi_{\pm\pm}\hspace{-\nulldelimiterspace}... ...e{.03em}\right.\!\left\vert\vphantom{\psi_{\pm\pm}}H\psi_{\pm\pm}\right\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_{\rm {gs}}\left\langle\vphantom{\psi_{\pm\pm}}\psi_{\pm\pm}\hspace{-\nulldeli... ...ce{.03em}\right.\!\left\vert\vphantom{\psi_{\pm\pm}}\psi_{\pm\pm}\right\rangle$, the low­est pos­si­ble. Then the to­tal en­ergy $\left\langle{E}\right\rangle$ will be $E_{\rm {gs}}$. Any­thing else will have more en­ergy and can there­fore not be the ground state.

It should be pointed out that to a more ac­cu­rate ap­prox­i­ma­tion, spin causes the elec­trons to be some­what mag­netic, and that pro­duces a slight de­pen­dence of the en­ergy on spin; com­pare ad­den­dum {A.39}. This note ig­nored that, as do most other de­riva­tions in this book.