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,

\begin{displaymath}
\psi_{\rm gs} =
\psi_{++} {\uparrow}{\uparrow}+ \psi_{+-...
...} {\downarrow}{\uparrow}+ \psi_{--} {\downarrow}{\downarrow},
\end{displaymath}

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,

\begin{displaymath}
\left(H\psi_{++}{\uparrow}{\uparrow}\right)
\equiv
H\l...
...z2})\right)
=
\left(H\psi_{++}\right){\uparrow}{\uparrow}
\end{displaymath}

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

\begin{displaymath}
\begin{array}{l}
\big\langle E\big\rangle = \langle
\p...
...i_{--}\right){\downarrow}{\downarrow}
\rangle
\end{array}
\end{displaymath}

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

\begin{displaymath}
\big\langle E\big\rangle =
\langle\psi_{++}\vert H\psi_{...
...\psi_{-+}\rangle +
\langle\psi_{--}\vert H\psi_{--}\rangle.
\end{displaymath}

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

\begin{displaymath}
\langle\psi_{++}\vert\psi_{++}\rangle +
\langle\psi_{+-}...
...si_{-+}\rangle +
\langle\psi_{--}\vert\psi_{--}\rangle = 1.
\end{displaymath}

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 $\big\langle\psi_{\pm\pm}\big\vert H\psi_{\pm\pm}\big\rangle $ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_{\rm {gs}}\big\langle\psi_{\pm\pm}\big\vert\psi_{\pm\pm}\big\rangle $, the lowest possible. Then the total energy $\big\langle E\big\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.38}. This note ignored that, as do most other derivations in this book.