4
$\begingroup$

Hund's first rule states:

The term with the largest possible value of $S$ for a given configuration has the lowest energy.

But, in practice, in order the find the ground state, we fill up the electrons in the subshells (following Pauli's exclusion principle) in such a way that $S_z$ is maximized, not $S$. While I agree that maximizing $S_z$ does indeed mean maximizing $S$. But Hund's prescription of maximizing $S$ does not mean that we have to maximize $S_z$, too. But that's what we do. We maximize $S_z$ even though Hund asks us to maximize $S$.

So my question is when we always maximize $S_z$, why then does the Hund's not rule tell us to maximize $S_z$ directly?

$\endgroup$

1 Answer 1

3
$\begingroup$

Hund's first rule is based on non-relativistic quantum mechanics without any spin-orbit interactions. Since the non-relativistic Hamiltonian (in the absence of magnetic fields) is spin-independent, all $2S+1$ states of a spin multiplet of given $S$ have the same energy. This means that you are always free to assume that you work with the highest weight member of the multiplet for which $M_S=\langle\hat{S}_z\rangle=S$ (all unpaired electrons being spin-up). So there is no contradiction: maximizing $M_S$ (basically maximizing the number of unpaired spin-up electrons in a configuration of given subshells) is equivalent to maximizing $S$. If you are interested, see this answer of mine for some further details concerning Hund's first rule.

Addressing the comments

About your $3d^5$ example (Mn$^{2+}$): this configuration indeed represents the maximal $M_S=5/2$ member of the $^6S$ ground state. But this ground state is sixfold degenerate! Its possible $M_S$ values range from $-5/2$ to $+5/2$, the lowest and highest $M_S$ members really being just five up/down "arrows in boxes": $$ |^6S_{M_S=+5/2}\rangle= {\cal{A}}(|3d_{-2} \, 3d_{-1} \, 3d_{0} \, 3d_{+1} \, 3d_{+2}\rangle)\otimes|\uparrow\uparrow\uparrow\uparrow\uparrow\rangle \ , $$ $$ |^6S_{M_S=-5/2}\rangle= {\cal{A}}(|3d_{-2} \, 3d_{-1} \, 3d_{0} \, 3d_{+1} \, 3d_{+2}\rangle)\otimes|\downarrow\downarrow\downarrow\downarrow\downarrow\rangle \ . $$ However, the other members cannot be represented like this, they are non-trivial linear combinations whose Clebsch-Gordan coefficients are determined by rotational+spin symmetry. Any of these states can be thought of as the ground state.

The allowed maximal value of $M_S$ is indeed maximized in the ground state. Again, looking at this example, the next highest $\hat{S}^2$ eigenstate would be $S=3/2$, meaning that its maximal $M_S$ value is $3/2<5/2$.

Maximizing $M_S$ is a generally easy way to arrive at the highest $S$ multiplet that can arise from populating given subshells, but apart from that, the highest $M_S$ member does not have any special role compared to any other of the $2S+1$ members of the given spin symmetry.

I stress that I only talked about atoms/molecules neglecting spin-orbit coupling and without external fields. In e.g. transition metal complexes, the $d$ orbitals of the metal ion are split by a "crystal field" of its surroundings, and become non-degenerate. In such cases, one can experience deviations from Hund's rule in cases where some of the orbital energies become so high, that instead lower lying orbitals become doubly populated (see e.g. [Fe(NO$_2$)$_6$]$^{3-}$ in wikipedia).

$\endgroup$
8
  • $\begingroup$ "So there is no contradiction: maximizing $M_S$ (basically maximizing the number of unpaired spin-up electrons in a configuration of given subshells) is equivalent to maximizing $S$." I agree that maximizing $M_S$ is equivalent to maximizing $S$. No doubt about that. But the converse need not be true. Prescription of maximizing $S$, does not unambiguously tell me what to do with $M_S$. And if all possible $M_S$ have same energy, then it is even more unambiguous why I should maximize $M_S$. $\endgroup$ Dec 8, 2023 at 15:02
  • $\begingroup$ @Solidification But once you determined the lowest state by maximizing $S$, you can choose whichever $M_S$ member of the multiplet you like, they all have the same energy in the non-relativistic framework. Working with the highest number of unpaired spin-up electrons (that is, the highest $M_S$) is just a useful guideline, that helps you to determine the spin symmetry by simply "putting arrows in boxes". In this way, you can do this without having to deal with the spin coupling via Clebsch-Gordan coefficients (inevitable when constructing lower $M_S$ symmetry... $\endgroup$ Dec 8, 2023 at 15:12
  • $\begingroup$ ... adapted configurations). To put it another way: if your maximum spin is $S$, then any any other, lower $\hat{S}^2$ eigenstate can only have a lower $M_S$ value at best (if your highest spin possibility is e.g. a quartet ($S=3/2$), then the maximal $M_S$ also equals 3/2, which of course cannot be the case for any doublet possibilities). $\endgroup$ Dec 8, 2023 at 15:21
  • $\begingroup$ So the pictorial representation in which the ground state of $nd^5$ configuration is drawn as one spin-up electron in each of the five subshells of $d$ (i.e., five upward arrows in five boxes) is not the only possible configuration according to Hund's rule? Do you agree? But I have a feeling that the maximum value of $M_S$ is the actual ground state (as is often used in magnetism texts). $\endgroup$ Dec 8, 2023 at 15:31
  • 1
    $\begingroup$ @Solidification I meant that the largest possible value that $M_S$ can take is maximized, which is not surprising because that maximal value equals $S$. I'm just trying to emphasize that what you see as maximizing $M_S$ in the "arrows in boxes" representation is really just a practical way to find the maximal $S$. You use the highest (or lowest, doesn't matter) $M_S$ because this is that member from which you can immediately read off the value of $S$ whose multiplet it belongs to. If you used four ups and one down ($M_S=3/2$), that could be the part of either a $S=5/2$ or a $S=3/2$ multiplet. $\endgroup$ Dec 8, 2023 at 17:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.