D.56 Explanation of Hund’s first rule

Hund’s first rule of spin-alignment applies because electrons in atoms prefer to go into spatial states that are antisymmetric with respect to electron exchange. Spin alignment is then an unavoidable consequence of the weird antisymmetrization requirement.

To understand why electrons want to go into antisymmetric spatial states, the interactions between the electrons need to be considered. Sweeping them below the carpet as the discussion of atoms in chapter 5.9 did is not going to cut it.

To keep it as simple as possible, the case of the carbon atom will be considered. As the crude model of chapter 5.9 did correctly deduce, the carbon atom has two 1s electrons locked into a zero-spin singlet state, and similarly two 2s electrons also in a singlet state. Hund’s rule is about the final two electrons that are in 2p states. As far as the simple model of chapter 5.9 was concerned, these electrons can do whatever they want within the 2p subshell.

To go one better than that, the correct interactions between the two 2p electrons will need to be considered. To keep the arguments manageable, it will still be assumed that the effects of the 1s and 2s electrons are independent of where the 2p electrons are.

Call the 2p electrons $\alpha$ and $\beta$. Under the stated conditions, their Hamiltonian takes the form

\begin{displaymath}
H_\alpha + H_\beta + V_{\alpha \beta}
\end{displaymath}

where $H_\alpha$ and $H_\beta$ are the single-electron Hamiltonians for the electrons $\alpha$ and $\beta$, consisting of their kinetic energy, their attraction to the nucleus, and the repulsion by the 1s and 2s electrons. Note that in the current analysis, it is not required that the 1s and 2s electrons are treated as located in the nucleus. Lack of shielding can be allowed now, but it must still be assumed that the 1s and 2s electrons are unaffected by where the 2p electrons are. In particular, $H_\alpha$ is assumed to be be independent of the position of electron $\beta$, and $H_\beta$ independent of the position of electron $\alpha$. The mutual repulsion of the two 2p electrons is given by $V_{\alpha\beta}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^2$$\raisebox{.5pt}{$/$}$$4\pi\epsilon_0\vert{\skew0\vec r}_\alpha-{\skew0\vec r}_\beta\vert$.

Now assume that electrons $\alpha$ and $\beta$ appropriate two single-electron spatial 2p states for themselves, call them $\psi_1$ and $\psi_2$. For carbon, $\psi_1$ can be thought of as the 2p$_z$ state and $\psi_2$ as the 2p$_x$ state, The general spatial wave function describing the two electrons takes the generic form

\begin{displaymath}
a\psi_1({\skew0\vec r}_1)\psi_2({\skew0\vec r}_2) + b\psi_2({\skew0\vec r}_1)\psi_1({\skew0\vec r}_2).
\end{displaymath}

The two states $\psi_1$ and $\psi_2$ will be taken to be orthonormal, like p$_z$ and p$_x$ are, and then the normalization requirement is that $\vert a\vert^2+\vert b\vert^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.

The expectation value of energy is

\begin{displaymath}
\langle a\psi_1\psi_2 + b\psi_2\psi_1
\vert H_\alpha + H...
...V_{\alpha\beta} \vert
a\psi_1\psi_2 + b\psi_2\psi_1\rangle.
\end{displaymath}

That can be multiplied out and then simplified by noting that in the various inner product integrals involving the single-electron Hamiltonians, the integral over the coordinate unaffected by the Hamiltonian is either zero or one because of orthonormality. Also, the inner product integrals involving $V_{\alpha\beta}$ are pairwise the same, the difference being just a change of names of integration variables.

The simplified expectation energy is then:

\begin{displaymath}
E_{\psi_1} + E_{\psi_2}
+ \langle\psi_1\psi_2\vert V_{\a...
...gle\psi_1\psi_2\vert V_{\alpha\beta}\vert\psi_2\psi_1\rangle.
\end{displaymath}

The first two terms are the single-electron energies of states $\psi_1$ and $\psi_2$. The third term is the classical repulsion between between two electron charge distributions of strengths $\vert\psi_1\vert^2$ and $\vert\psi_2\vert^2$. The electrons minimize this third term by going into spatially separated states like the 2p$_x$ and 2p$_z$ ones, rather than into the same spatial state or into greatly overlapping ones.

The final one of the four terms is the interesting one for Hund’s rule; it determines how the two electrons occupy the two states $\psi_1$ and $\psi_2$, symmetrically or antisymmetrically. Consider the detailed expression for the inner product integral appearing in the term:

\begin{displaymath}
\langle\psi_1\psi_2\vert V_{\alpha\beta}\vert\psi_2\psi_1\...
... r}_1)
{\,\rm d}^3{\skew0\vec r}_1{\rm d}^3{\skew0\vec r}_2
\end{displaymath}

where $f({\skew0\vec r}_1,{\skew0\vec r}_2)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_2({\skew0\vec r}_1)\psi_1({\skew0\vec r}_2)$.

The sign of this inner product can be guesstimated. If $V_{\alpha\beta}$ would be the same for all electron separation distances, the integral would be zero because of orthonormality of $\psi_1$ and $\psi_2$. However, $V_{\alpha\beta}$ favors positions where ${\skew0\vec r}_1$ and ${\skew0\vec r}_2$ are close to each other; in fact $V_{\alpha\beta}$ is infinitely large if ${\skew0\vec r}_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew0\vec r}_2$. At such a location $f({\skew0\vec r}_1,{\skew0\vec r}_2)f^*({\skew0\vec r}_2,{\skew0\vec r}_1)$ is a positive real number, so it tends to have a positive real part in regions it really counts. That means the inner product integral should have the same sign as $V_{\alpha\beta}$; it should be repulsive.

And since this integral is multiplied by $a^*b+b^*a$, the energy is smallest when that is most negative, which is for the antisymmetric spatial state $a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$b$. Since this state takes care of the sign change in the antisymmetrization requirement, the spin state must be unchanged under particle exchange; the spins must be aligned. More precisely, the spin state must be some linear combination of the three triplet states with net spin one. There you have Hund’s rule, as an accidental byproduct of the Coulomb repulsion.

This leaves the philosophical question why for the two electrons of the hydrogen molecule in chapter 5.2 the symmetric state is energetically most favorable, while the antisymmetric state is the one for the 2p electrons. The real difference is in the kinetic energy. In both cases, the antisymmetric combination reduces the Coulomb repulsion energy between the electrons, and in the hydrogen molecule model, it also increases the nuclear attraction energy. But in the hydrogen molecule model, the symmetric state achieves a reduction in kinetic energy that is more than enough to make up for it all. For the 2p electrons, the reduction in kinetic energy is nil. When the positive component wave functions of the hydrogen molecule model are combined into the symmetric state, they allow greater access to fringe areas farther away from the nuclei. Because of the uncertainty principle, less confined electrons tend to have less indeterminacy in momentum, hence less kinetic energy. On the other hand, the 2p states are half positive and half negative, and even their symmetric combination reduces spatial access for the electrons in half the locations.