A rigorous derivation of Hund's rules

Question

I was really looking forward to the atomic physics portion of my quantum mechanics class and was pretty disappointed with Griffith's hand-wavey qualitative description that was scarcely above the chem101 level. I've seen several other great posts about Hund's rules on the forums but I want to ask about a rigorous demonstration.

My first instinct would be to go through every atom one by one and calculate the expectation value of the Hamiltonian for each orbital. I'm hoping there is a more clever way to leave me undoubtably convinced.

Answer 1

As already pointed out out in the comments, a rigorous proof of Hund's rules cannot be found because Hund's rules are not exact laws to begin with. However, it is worth discussing under which circumstances they are expected to work. I will only discuss Hund's first rule, which states that

in a set of many-electron states arising from the same configuration, the lowest energy state is the one with the highest $S$ value.

Note that originally, Hund formulated this rule only for ground states of atoms; a lot of confusion surrounding its usefulness has originated from people applying it inconsiderately for molecules or for excited states (which is fine in many cases but one must be aware of the exceptions).

We shall first discuss those systems that can be well described by a single configuration, and then those which can not.

Systems well represented by a single symmetry adapted configuration

In this case, one might expect acceptable results from (non-degenerate) first-order perturbation theory (PT), or from the Hartree-Fock (HF) approximation. Let us start with two-electron systems: the Hamiltonian reads $$ H=h_1+h_2+W \ , $$ where $$ h_i=-\frac{1}{2}\nabla_i^2+U(\boldsymbol{r}_i) \ \ \ , \ \ \ W=\frac{1}{|\boldsymbol{r}_1-\boldsymbol{r}_2|} \ . $$ When neglecting $W$, the singlet and triplet solutions are degenerate. For the time being, let us assume that their spatial symmetry can be represented by the simple linear combinations $$ \begin{aligned} |^1\Phi_{pq}\rangle&=\frac{1}{\sqrt{2}}\left(|\phi_p\phi_q\rangle+|\phi_q\phi_p\rangle\right)\otimes|0,0\rangle \ , \\ |^3\Phi_{pq}\rangle&=\frac{1}{\sqrt{2}}\left(|\phi_p\phi_q\rangle-|\phi_q\phi_p\rangle\right)\otimes|1,M_S\rangle \ , \end{aligned} $$ and that these are all the states that can arise from the configuration $(p)(q)$. One could think of e.g. the singly excited $(1s)(nl)$ configurations of a He-like system (for $n>1$). Treating $W$ as a perturbation, we have $$ \begin{aligned} ^{1,3}E_{pq}^{(1)} &= \langle^{1,3}\Phi_{pq}|W|^{1,3}\Phi_{pq}\rangle \\ &= \langle\phi_p\phi_q|W|\phi_p\phi_q\rangle\pm \langle\phi_p\phi_q|W|\phi_q\phi_p\rangle \\ &= J_{pq}\pm K_{pq} \ . \end{aligned} $$ Up to first order, we have thus found $$ ^1E_{pq}^{[1]}-{^3E}_{pq}^{[1]}=2K_{pq}>0 \ , $$ in accordance with Hund's rule (see Addendum 1). However, trusting a first-order PT result is not always a good idea (see Addendum 2). A more reassuring result is that the same inequality holds for the HF energy of certain excited states, too[1]. To see this, consider the singlet/triplet energy functionals $$ {^{1,3}E}[\phi]= \langle\phi_1\phi_2|H|\phi_1\phi_2\rangle\pm \langle\phi_1\phi_2|H|\phi_2\phi_1\rangle \ , $$ and two sets of orthonormal HF orbitals $\{\phi^{(+)}_1,\phi^{(+)}_2\}$, $\{\phi^{(-)}_1,\phi^{(-)}_2\}$ minimizing ${^{1}E}[\phi]$ and ${^{3}E}[\phi]$, respectively. Then we simply find $$ \begin{aligned} {^1E}_{\text{HF}}={^{1}E}[\phi^{(+)}]>{^{3}E}[\phi^{(+)}]>{^{3}E}[\phi^{(-)}]={^3E}_{\text{HF}} \ . \end{aligned} $$ The first inequality follows from our previous considerations, while the second one is due to the usual variational property ($\{\phi^{(+)}_1,\phi^{(+)}_2\}$ are not the optimal orbitals of ${^3E}[\phi]$). While this is a more trustworthy result than our previous one based on first-order PT, its applicability is more limited, since HF works only for the lowest energy state of each symmetry species, or for the lowest energy configuration of a given type (for example, it can be used to demonstrate Hund's rule for the configuration $(1s)(2p)$, but not for $(1s)(3p)$). Nevertheless, these results make it rather believable that Hund's rule is valid for singly excited states of two-electron systems (indeed, experiments found no violation of the rule for such states).

These results can be immediately generalized for many-electron systems with two valence electrons and an arbitrary number of core electrons doubly filling orbitals of lower energy. As long as we are only interested in the singlet-triplet splitting, the two valence electrons can be treated independently of the core electrons to a good approximation (the effect of the core electrons is essentially a partial shielding of the nuclei anyway).

Previously we assumed that for a given two-electron configuration, all states could be written as the simple two-term linear combinations above, which is in general true for singly excited states of the form $(1s)(nl)$. Things are getting more difficult when dealing with doubly excited atomic configurations. Let us take the $(2p)^2$ configuration of a He-like atom as an example, which gives rise to the following states: $$ \begin{aligned} |{^3P}_{g}\rangle &= \frac{1}{\sqrt{2}}\Big(|2p_{0}2p_{+1}\rangle-|2p_{+1}2p_0\rangle\Big)\otimes|1,M_S\rangle \ , \\ |{^1D}_{g}\rangle&= \frac{1}{\sqrt{2}}\Big(|2p_{0}2p_{+1}\rangle+|2p_{+1}2p_{0}\rangle\Big)\otimes|0,0\rangle \ , \\ |{^1S}_g\rangle&=\frac{1}{\sqrt{3}} \Big(|2p_{+1}2p_{-1}\rangle+|2p_{-1}2p_{+1}\rangle -|2p_{0}2p_{0}\rangle\Big)\otimes|0,0\rangle \ . \end{aligned} $$ In the case of $^3P_g$ and $^1D$, we chose the $M=1$ component for convenience. The first-order PT energies read $$ \begin{aligned} E^{(1)}({^3P}_g) =&\langle2p_{0}2p_{+1}|W|2p_{0}2p_{+1}\rangle-\langle2p_{0}2p_{+1}|W|2p_{+1}2p_{0}\rangle \ , \\ E^{(1)}({^1D}) =&\langle2p_{0}2p_{+1}|W|2p_{0}2p_{+1}\rangle+\langle2p_{0}2p_{+1}|W|2p_{+1}2p_{0}\rangle \ , \\ E^{(1)}({^1S}) =&\frac{1}{3}\big(2\langle2p_{+1}2p_{-1}|W|2p_{+1}2p_{-1}\rangle+2\langle2p_{+1}2p_{-1}|W|2p_{-1}2p_{+1}\rangle \\ &+\langle2p_{0}2p_{0}|W|2p_{0}2p_{0}\rangle -4\langle2p_{+1}2p_{-1}|W|2p_{0}2p_{0}\rangle \big) \ . \end{aligned} $$ Interpreting $\langle2p_{0}2p_{+1}|W|2p_{+1}2p_{0}\rangle$ as an exchange term, we can immediately see that $E^{[1]}({^3P}_g)<E^{[1]}({^1D})$. But the various terms of the $^1S$ energy do not fit into this interpretation, and the complete energy order must be determined by direct computation. Looking up a table of Clebsch-Gordan coefficients and carrying out the integrals over angles lead to $$ \begin{aligned} E^{(1)}({^3P}_g) &=F_0-5F_2 \ , \\ E^{(1)}({^1D}) &=F_0+F_2 \ , \\ E^{(1)}({^1S}) &=F_0+10F_2 \ , \end{aligned} $$ where $$ F_L=\frac{1}{(2L+1)^2} \int_0^{\infty}\mathrm{d}r_1r_1^2R_{21}^2(r_1) \int_0^{\infty}\mathrm{d}r_2r_2^2R_{21}^2(r_2) \frac{r_<^L}{r_>^{L+1}} \ , $$ and $R_{21}(r)$ is the radial part of the $2p$ orbital. Due to the positivity of the integrands (hence $F_L>0$), $E^{[1]}({^1D})<E^{[1]}({^1S})$ now follows, in agreement with Hund's second rule. Since $(2p)^2$ is the lowest configuration of $(np^2)$ type, our previous considerations about HF energies can be directly applied, leading to $$ E_{\text{HF}}({^3P}_g)<E_{\text{HF}}({^1D})<E_{\text{HF}}({^1S}) \ . $$ See Addendum 3 for an important subtlety.

A nice application of the above result is the prediction of the ground state of the carbon atom. Using our previous "separated core" approximation to add 4 core electrons yields the configuration $(1s)^2(2s)^2(2p)^2$ with the same order of energy levels, so we can now see why the ground state of C is ${^3P}_g$.

Things are getting considerably worse when dealing with higher and higher doubly excited atomic configurations containing two open shells of $l_1,l_2\geq1$; most of the atomic anti-Hund cases are related to such systems. It turns out that Hund's rule is more or less replaced by a so-called alternating rule, which states

for a given two-electron configuration yielding two states with the same $L$, ${^1E}>{^3E}$ in case of natural parity states, and ${^3E}>{^1E}$ in case of unnatural parity states,

natural and unnatural parity referring to the parity quantum number of the state being equal to $(-1)^{L}$ and $(-1)^{L+1}$, respectively. The rule essentially states that natural parity states follow Hund's rule, while unnatural parity states do not, leading to e.g. $E(^3P_g)>E(^1P_g)$ in the $(2p)(3p)$ configuration.

The justification of this rule is quite tedious and relies on a bunch of $3j$ and $6j$ identities, so I will not attempt to do it here; a good and detailed discussion of the problem can be found in Refs. [2,3], and references cited therein. Also, see Chapter VII. of Ref. [4] for examples.

Systems with strong configuration mixing

Since Hund's rule always refers to states formed from the same single configuration, it is not surprising that its predictions fail for strongly correlated atoms or molecules whose wave functions are a mixture of multiple configurations of similar weight.

The failure of Hund's rule is well demonstrated by the lowest-lying excited $D$ states of Mg, as $E({^3D})-E({^1D})\approx1554.0 \, \text{cm}^{-1}$ (value taken from the NIST database[5]). This is a consequence of the strong mixing of $(3s)(3d)$ and $(3p)^2$ configurations, as can be understood from the following crude model. The Ne-like 10-electron core is neglected, and the two valence electrons are described either by the two singlet configurations $$ |^1D\rangle=\frac{1}{\sqrt{2}}\Big(|3s,3d_{+2}\rangle+|3d_{+2},3s\rangle\Big)\otimes|0,0\rangle \ , $$ $$ |^1D'\rangle=|3p_{+1},3p_{+1}\rangle\otimes|0,0\rangle \ , $$ or by the triplet configuration $$ |^3D\rangle=\frac{1}{\sqrt{2}}\Big(|3s,3d_{+2}\rangle-|3d_{+2},3s\rangle\Big)\otimes|1,M_S\rangle \ . $$ Treating $W$ as a perturbation, we must use degenerate PT to find the singlet-triplet splitting, since the zeroth-order energy of $^1D$ and $^1D'$ are equal. Strictly speaking, we should also include $(3p)(3d)$ and $(3d)^2$ configurations in the degenerate blocks of $W$, as these can also form $^1D$ and $^3D$ states; however, their coupling is expected to be small, so they are neglected.

Up to first order, the splitting formula reads $$ E(^3D)-E(^1D)\approx\langle^3D|\hat{W}|^3D\rangle-\lambda \ , $$ where $\lambda$ is the lower eigenvalue of $$ {\cal{W}}= \begin{bmatrix} \langle^1D|\hat{W}|^1D\rangle & \langle^1D|\hat{W}|^1D'\rangle \\ \langle^1D'|\hat{W}|^1D\rangle & \langle^1D'|\hat{W}|^1D'\rangle \end{bmatrix} \ . $$ The matrix elements of hydrogenic orbitals can be calculated analytically, leading to $$ \begin{aligned} \langle^{1,3}D|\hat{W}|^{1,3}D\rangle &= \frac{337\pm21}{4608}Z_{\text{eff}} \ , \\ \langle^{1}D'|\hat{W}|^{1}D'\rangle &= \frac{563}{7680}Z_{\text{eff}} \ , \\ \langle^{1}D|\hat{W}|^{1}D'\rangle &= \sqrt{6} \frac{35}{4608}Z_{\text{eff}} \ , \end{aligned} $$ where $Z_{\text{eff}}$ is the effective nuclear charge seen by the valence electrons. As a further approximation, the screening effect is assumed to be perfect, reducing the nuclear charge to $Z_{\text{eff}}\approx12-10=2$. After diagonalizing ${\cal{W}}$, we find $$ E(^3D)-E(^1D)\approx0.02362 \, E_h \approx5184.4 \, \text{cm}^{-1} , $$ which $-$ while grossly overestimating the size of the gap $-$ correctly predicts the energy of $^1D$ being lower. The coefficients of $|^1D\rangle$ and $|^1D'\rangle$ are $c\approx0.664$ and $c'\approx-0.747$, signalling a strong configuration mixing; without the coupling, we would have found $E(^1D)>E(^3D)$.

Similar violations of Hund's rule can be found in strongly correlated molecules as well.

Addendum 1: the proof of $K_{pq}>0$

Although this can be found in many places, I write down the proof of $K_{pq}>0$ for the sake of completeness: $$ \begin{aligned} K_{pq} &= \int\mathrm{d}^3r_1\int\mathrm{d}^3r_2 \frac{1}{|\boldsymbol{r}_1-\boldsymbol{r}_2|} \phi^*_p(\boldsymbol{r}_1)\phi^*_q(\boldsymbol{r}_2) \phi_q(\boldsymbol{r}_1)\phi_p(\boldsymbol{r}_2) \\ &= \int\mathrm{d}^3r_1\int\mathrm{d}^3r_2 \int\frac{\mathrm{d}^3k}{(2\pi)^3} \frac{4\pi}{k^2}e^{i\boldsymbol{k}(\boldsymbol{r}_1-\boldsymbol{r}_2)} \phi^*_p(\boldsymbol{r}_1)\phi^*_q(\boldsymbol{r}_2) \phi_q(\boldsymbol{r}_1)\phi_p(\boldsymbol{r}_2) \\ &= \int\frac{\mathrm{d}^3k}{(2\pi)^3} \frac{4\pi}{k^2} \left| \int\mathrm{d}^3r \phi_p(\boldsymbol{r})\phi^*_q(\boldsymbol{r})e^{-i\boldsymbol{k}\boldsymbol{r}}\right|^2 \\ &>0 \ . \end{aligned} $$ Note that nothing was assumed about the orbitals apart from normalizability.

Addendum 2: a famous non-explanation of Hund's rule

A seemingly intuitive but wrong explanation of Hund's first rule motivates ${^1E}>{^3E}$ with the Fermi hole. Let $^1\Psi$ and $^3\Psi$ denote the exact wave functions associated with a given configuration. Since the spatial part of a triplet wave function is antisymmetric under exchange of electron coordinates, then $$ \lim_{\boldsymbol{r}_2\rightarrow\boldsymbol{r}_1}{^3\Psi}(\boldsymbol{r}_1,\boldsymbol{r}_2)=0 \ , $$ with the obvious assumption of $^3\Psi$ being continuous. Then it seems reasonable that the value of $^3\Psi$ is small wherever $|\boldsymbol{r}_1-\boldsymbol{r}_2|$ is small, which would mean $\langle{^3\Psi}|W|{^3\Psi}\rangle<\langle{^1\Psi}|W|{^1\Psi}\rangle$, explaining Hund's rule. However, this "reasonable assumption" turned out to be wrong, when people in the 1960s started doing highly accurate variational calculations for two-electron atoms and realized that in most cases $\langle{^3\Psi}|W|{^3\Psi}\rangle>\langle{^1\Psi}|W|{^1\Psi}\rangle$! This wrong explanation can sometimes still be found in textbooks. For further details, see Refs. [2,3].

This is why I am not particularly fond of the first-order PT proof of Hund's rule: it is mathematically valid, but by demonstrating $\langle{^3\Phi}|W|{^3\Phi}\rangle<\langle{^1\Phi}|W|{^1\Phi}\rangle$, it basically shows Hund's rule to be correct for the wrong reason. The HF-based proof does not have this problem as it references the total singlet/triplet HF energies; unfortunately, it is applicable to much less states than the PT.

Addendum 3: on the boundedness of excited HF energies

A subtle but important issue is that the variational theorem does not hold for all of our excited HF energies, meaning that they are not necessarily upper bounds to their respective exact energies. The variational theorem holds for excited states only if we can ensure the orthogonality of our trial function to all lower energy exact states. On the HF level, this can be guaranteed only for the lowest energy state of a given symmetry species. For this reason, $E_{\text{HF}}({^3P}_g)>E_{\text{exact}}({^3P}_g)$ must hold for the $^3P_g$ state of the $(2p)^2$ configuration. On the other hand, the HF states $^1S$ and $^1D$ originating from the same configuration may overlap with lower energy exact states associated with $(1s)(ns)$ and $(1s)(nd)$, so they can possibly fall below the exact energies. Similarly, the singly excited $^1S$ HF state arising from $(1s)(2s)$ can overlap with the exact ground state.

The results of the actual computations for He are shown in the table.

\begin{array}{lrr} & E_{\text{exact}} \, / \, E_h & E_{\text{HF}} \, / \, E_h \\ \hline {^3P}_g & -0.711 & -0.701 \\ {^1D} & -0.699 & -0.669 \\ {^1S} & -0.619 & -0.623 \\ \end{array}

Only the $^1S$ HF energy falls slightly below the exact one. The exact ${^3P}_g$ energy was taken from Ref. [6], the $^1S$ and $^1D$ ones from Ref. [7].

It is important to see that these issues do not alter our conclusions, as the established inequalities do not depend on $E_{\text{HF}}>E_{\text{exact}}$. Also, due to the restricted form of the trial functions, even if there is a variational collapse, it is expected to be small. Therefore, the HF energies are still expected to be good approximations, and they can be used to guess at the order of the exact energy levels.

Further details on variational excited state calculations and on the HF calculation for the ${^3P}_g$ state can be found here.

References

[1]: Colpa: Mol. Phys. 28 2 581 (1974)

[2]: Morgan, Kutzelnigg: J. Phys. Chem. 97 2425 (1993)

[3]: Kutzelnigg, Morgan: Z. Phys. D 36 197 (1996)

[4]: Condon, Shortley: The Theory of Atomic Spectra

[5]: NIST: Atomic Spectra Database (Mg I)

[6]: Duan, Gu, Ma: Phys. Lett. A 283 229 (2001)

[7]: Callaway: Phys. Lett. A 66 3 201 (1978)

Answer 2

Hund's rule is the interplay between the Pauli exclusion principle and the Coulomb interaction, which was first pointed out by Heitler and London in 1926 and later extended by Heisenberg to put forward his celebrated theory of quantum magnetism. It may be possible to give the derivation of Hund's rule using group representation theory.

Almost all standard textbooks on magnetism discussed the origin of Hund's rule. I try to give you the essential physics without diving deep into calculation. Look into "Interacting Electrons and Quantum Magnetism" by Assa Auerbach for details. One can look into the Group theory by Hamarmash for a group theoretical approach.

Here, we follow the argument of Heitler and London.

Let's assume that two electrons are interacting via Coulomb interaction. The Hamiltonian of the system can be written as

\begin{equation} H = H_{0} + H_{U}. \end{equation} Where $H_{0}$ is the free Hamiltonian for which we know the eigenstates, and $H_{U}$ is the Coulomb repulsion between electrons, given by

$$ H_{U} = \sum_{i < j} \frac{1}{r_{i} -r_{j}} $$. Where $r$ is the radial coordinate of the electron.

Important: Note that $H_{U}$ is manifestly spin-independent. However, because of the antisymmetry of the many-electron wave function, the eigenenergies of $H_{U}$ depend on the spin. This is the basis of the multiple structures in atoms and of Hund’s first two rules.

Let's go further and assume two orbitals $a$ and $b$. Then, the two-electron Slater determinants with spins $\sigma$ and $\sigma'$ is given by

$$ \Psi_{a\sigma,b\sigma'}(r_{1},s_{1};r_{2},s_{2}) = \frac{1}{\sqrt{2}} \Bigl(\psi_{a}(r_{1})\psi_{b}(r_{2}) \sigma(s_{1})\sigma'(s_{2}) - \psi_{b}(r_{1})\psi_{a}(r_{2})\sigma'(s_{1})\sigma(s_{2})\Bigr) $$ Where the meaning of the coordinates is self-explanatory.

One can check that $\Psi_{a\sigma,b\sigma'}(r_{1},s_{1};r_{2},s_{2})$ is the eigenstate of the free Hamiltonian $H_{0}$ with eigenvalue $\varepsilon_{a} + \varepsilon_{b}$. But when the Coulomb interaction is turned on, the degeneracy is lifted. Let's assume that both electrons have the same spin to see how. In that case, we can factor out the spin-dependent part of $\Psi$ giving.

$$ \Psi_{s} = \Psi_{a\sigma,b\sigma} = \frac{1}{\sqrt{2}} \Bigl(\psi_{a}(r_{1})\psi_{b}(r_{2}) - \psi_{b}(r_{1})\psi_{a}(r_{2})\Bigr)\sigma(s_{1})\sigma(s_{2})$$

Doing the same for the opposite site gives.

$$ \Psi_{t} = \Psi_{a\sigma,b,-\sigma} = \frac{1}{\sqrt{2}} \Bigl(\psi_{a}(r_{1})\psi_{b}(r_{2}) + \psi_{b}(r_{1})\psi_{a}(r_{2})\Bigr)\sigma(s_{1})\sigma(s_{2})$$ With this preparation, we can write down the matrix elements for the $H_{U}$ in the $\Psi_{\uparrow\uparrow},\Psi_{\uparrow\downarrow},\Psi_{\downarrow\uparrow},\Psi_{\downarrow\downarrow}$ basis, which is given by

\begin{equation} \left(H_{U}\right)_{ab} = \begin{pmatrix} U_{ab} - J_{ab} & 0 & 0 & 0 \\ 0 & U_{ab} & -J_{ab} & 0 \\ 0 & -J_{ab} & U_{ab} & 0 \\ 0 & 0 & 0 & U_{ab} - J_{ab} \end{pmatrix} \end{equation}

Where $U_{ab}$ is known as the Coulomb integral $$ U_{ab} = \int d^{3}r_{1} \int d^{3}r_{2} \frac{|\psi_{a}(r_{1})|^{2}|\psi_{a}(r_{2})|^{2}}{|r_{1} -r_{2}|} $$

and $J_{ab}$ is known as the famous exchange integral

$$ J_{ab} = \int d^{3}r_{1} \int d^{3}r_{2} \frac{\psi^{*}_{a}(r_{1})\psi_{b}(r_{1})\psi^{*}_{b}(r_{2})\psi_{a}(r_{2})}{|r_{1} -r_{2}|} $$

After diagonalizing the $\left(H_{U}\right)_{ab}$ one can find that the triplet states $\uparrow\uparrow,\downarrow\downarrow,\frac{1}{\sqrt{2}}(\uparrow\downarrow + \downarrow\uparrow)$ are the eigen states of $H_{U}$ with the eigenvalue

$$ \varepsilon_{\mathrm{triplet}} = U_{ab} - J_{ab} $$

and the singlet state $\frac{1}{\sqrt{2}}(\uparrow\downarrow - \downarrow\uparrow)$ is the eigenstate of $\left(H_{U}\right)_{ab}$ with eigenvalue

$$ \varepsilon_{\mathrm{singlet}} = U_{ab} + J_{ab} $$

Sign of $U$ and $J$: One can see quite clearly that the Coulomb integral $U_{ab}$ is having a positive integrated. While the sign of exchange integral $J_{ab}$ is not obvious. Nevertheless, by rewriting the integral using the convolution theorem as well as the Fourier transform, one can show that $J_{ab}$ is positive.

Thus, the triplet states are below the singlet state by an energy $2J_{ab}$.

Therefore, the lowest energy configuration of the two-electron system is a triplet or $\uparrow\uparrow,\downarrow\downarrow,\frac{1}{\sqrt{2}}(\uparrow\downarrow + \downarrow\uparrow)$. This is nothing but Hund's first rule: For an atomic shell, the lowest state will have maximum spin.

Addendum: Since $H_{U}$ only contains interactions within the system of electrons, it commutes with the total orbital momentum, $[H_{U}, L_{\mathrm{Total}}] = 0$, and also it commutes with the total spin $[H_{U}, S_{\mathrm{Total}}] = 0$. The eigenstates of $H_{0} + H_{U}$ can thus be classified by their quantum numbers $L$ and $S$.

Answer 3

I was really looking forward to the atomic physics portion of my quantum mechanics class and was pretty disappointed with Griffith's hand-wavey qualitative description

By "Griffith[] [sic]" I assume you mean D. J. Griffiths, the author of the undergraduate textbook "Introduction to Quantum Mechanics."

Griffiths' textbook is introductory and does not go into detail about multi-electron atoms, multiplets, and all that.

I've seen several other great posts about Hund's rules on the forums but I want to ask about a rigorous demonstration.

You should read the textbook by the other professor with a name like Griffiths, namely J. S. Griffith (no "s").

J. S. Griffith's "The Theory of Transition-Metal Ions" goes into great detail about Hartree-Fock, multiplets, and about how Hund's rule works. (This book is in the public domain, so you should have no qualms about "stealing" a PDF of it on the internet--hard copies are also available pretty cheap.) Griffith mentions Hund's rule on page 81. I would suggest carefully reading the first 81 pages of this textbook and then you will know more about Hund's rule than most professional physicists.

The textbook by Condon and Shortly titled "The Theory of Atomic Spectra" is another oldie, but a goodie. Those boys are famous in Atomic Physics circles.

Note also that the ground state multiplet as determined by Hund's rules is not always the correct ground state, especially in real solids at very high pressure where the compression of the atoms can cause the ground state to transition from high spin to low spin. See, for example, this article regarding high-spin low-spin transition from 1972. (This is just one article that came up on google, there are many others, some much more recent.)