Recently, I've been learning about Hund's rules from Steve Simon's The Oxford Solid State Basics.
Hund's rules tell us about how electrons are arranged in a partially-filled shell in an atom. According to the textbook, one consequence of Hund's rules is that the multi-electron system is an eigenstate of the $L^2$, $S^2$ and $J^2$ angular momentum operators. This claim is used extensively in the textbook to study the magnetic properties of atoms.
Unfortunately, I'm having trouble making sense of this statement. It seems like this statement – at least the part about the $J^2$ operator – fails for some of the simplest examples that I can come up with.
I would be grateful if you could check my understanding and tell me where I have made a mistake.
First, I'll recap my understanding of Hund's rules.
Hund's rules (paraphrased). When filling a electron shell:
- Start by adding electrons into spin-down states, in descending order of $l_z$ quantum number.
- Add the remaining electrons into spin-up states, in ascending order of $l_z$ quantum number.
Example: Filling the $2p$ states in an atom (i.e. filling the electron states with $n = 2$ and $l = 1$).
\begin{align}
& \text{Zero electrons:} & & \left| \cdot \right> \\
& \text{One electron:} & & \left| 1, \downarrow \right> \\
& \text{Two electrons:} & & \mathcal A \left( \left| 1, \downarrow \right> \otimes \left| 0, \downarrow \right> \right)\\
& \text{Three electrons:} & & \mathcal A \left( \left| 1, \downarrow \right> \otimes \left| 0, \downarrow \right> \otimes \left| -1, \downarrow \right> \right) \\
& \text{Four electrons:} & & \mathcal A \left( \left| 1, \downarrow \right> \otimes \left| 0, \downarrow \right> \otimes \left| -1, \downarrow \right> \otimes \left| -1, \uparrow \right> \right) \\
& \text{Five electrons:} & & \mathcal A \left( \left| 1, \downarrow \right> \otimes \left| 0, \downarrow \right> \otimes \left| -1, \downarrow \right> \otimes \left| -1, \uparrow \right> \otimes \left| 0, \uparrow \right> \right) \\
& \text{Six electrons:} & & \mathcal A \left( \left| 1, \downarrow \right> \otimes \left| 0, \downarrow \right> \otimes \left| -1, \downarrow \right> \otimes \left| -1, \uparrow \right> \otimes \left| 0, \uparrow \right> \otimes \left| 1, \uparrow \right> \right) \\
\end{align}
Here, $\mathcal A$ denotes the antisymmetrisation operator, enforcing Fermi-Dirac statistics. For example,
$$ \mathcal A \left( \left| 1, \downarrow \right> \otimes \left| 0, \downarrow \right> \right) = \tfrac{1}{\sqrt{2}} \left( \left| 1, \downarrow \right> \otimes \left| 0, \downarrow \right> – \left| 0, \downarrow \right> \otimes \left| 1, \downarrow \right> \right).$$
It is clear that any multi-electron state that satisfies Hund's rules is an eigenstate of $L_z$, $S_z$ and $J_z$. (Here, I'm using $L_z$, $S_z$ and $J_z$ to denote the $z$-components of the orbital, spin and total angular momenta of all of the electrons combined.)
For example, if $| \psi \rangle = \mathcal A \left( \left| 1, \downarrow \right> \otimes \left| 0, \downarrow \right> \right)$, then it is clear that $L_z | \psi \rangle = \hbar |\psi \rangle $, $S_z |\psi\rangle = -\hbar |\psi\rangle$ and $J_z |\psi\rangle = 0$.
However, Steve Simon's book appears to make the more surprising claim that any multi-electron state satisfying Hund's rules is also an eigenstate of $L^2$, $S^2$ and $J^2$!
Claim: Suppose that $|\psi \rangle$ is a multi-electron state satisfying Hund's rules, with $$L_z |\psi \rangle = \hbar l_z |\psi\rangle, \qquad \qquad S_z |\psi \rangle = \hbar s_z |\psi\rangle.$$
Then:
- $L^2 |\psi\rangle = \hbar^2 |l_z| (|l_z| + 1) |\psi\rangle$.
- $S^2 |\psi\rangle = \hbar^2 |s_z| (|s_z| + 1) |\psi\rangle$.
- $J^2 |\psi\rangle = \hbar^2 |l_z \pm s_z| (|l_z \pm s_z| + 1) |\psi\rangle$, where the sign is $-$ if the shell is less than half-filled or $+$ if the shell is more than half-filled.
While I seem to be able to prove the claims made about $L^2$ and $S^2$, it appears that the claim about $J^2$ fails on the simplest of examples. So I must have misunderstood something.
Proof that $L^2 |\psi\rangle = \hbar^2 |l_z| (|l_z| + 1) |\psi\rangle$:
Case: Shell is less than half-filled. The result follows from the operator identity
$$ L^2 = L_z^2 + \hbar L_z + L_- L_+,$$
since if $|\psi \rangle$ is a state satisfying Hund's rules where the shell is less than half filled, then $|\psi\rangle$ is annihilated by $L_+$.
To understand why $|\psi\rangle$ is annihilated by $L_+$, it's best to working through some concrete examples. Let's look at the example of the $l = 1$ shell, with two electrons placed in the shell. So $|\psi\rangle = \mathcal A \left( \left| 1, \downarrow \right> \otimes \left| 0, \downarrow \right> \right).$ Computing the action of $L_+$ on $|\psi\rangle$, we get
\begin{align} L_+ |\psi \rangle = \mathcal A L_+ \left( \left| 1, \downarrow \right> \otimes \left| 0, \downarrow \right> \right) = \mathcal A \left( \sqrt 2 \hbar \left| 1, \downarrow \right> \otimes \left| 1, \downarrow \right> \right) = 0.\end{align}
I encourage you to repeat this calculation for other multi-electron states satisfying Hund's rules with the shell less than half-filled – you'll soon pick up on the general pattern.
Case: Shell is more than half-filled. This time, the relevant operator identity is
$$ L^2 = L_z^2 – \hbar L_z + L_+ L_-.$$
If $|\psi \rangle$ is a state satisfying Hund's rules where the shell is more than half-filled, then $|\psi\rangle$ is annihilated by $L_-$. Again, feel free to check this on some examples.
Proof that $S^2 |\psi\rangle = \hbar^2 |s_z| (|s_z| + 1) |\psi\rangle$:
We use the operator identity
$$ S^2 = S_z^2 – \hbar S_z + S_+ S_-,$$
and we observe that if $|\psi \rangle$ is a multi-electron state satisfying Hund's rules, then $|\psi\rangle$ is annihilated by $S_-$, regardless of whether the shell is less than half-filled or more than half-filled.
Why the claim for $J^2 |\psi\rangle$ appears to be incorrect.
Consider the example of the $l=1$ shell with one electron. So $|\psi \rangle = |1, \downarrow \rangle$. Let's see what happens if we apply the operator
$ J^2 = J_z^2 + \hbar J_z + J_- J_+$
to $|\psi \rangle$. Clearly, $|\psi \rangle$ is an eigenstate of $J_z^2 + \hbar J_z $, with eigenvalue $\tfrac 3 4 \hbar^2 $. However,
$$ J_- J_+ |\psi \rangle = J_- \left(\hbar | 1, \uparrow \rangle \right) = \sqrt{2} \hbar^2 | 0, \uparrow \rangle + \hbar^2 | 1, \downarrow \rangle.$$
So our $|\psi\rangle$ is not an eigenstate of $J_- J_+$. Therefore, $|\psi\rangle$ is not an eigenstate of $J^2$ either, contradicting the claim in the book.
Where is my mistake? Did I misunderstand Hund's rules? Did I misinterpret the claim about the $J^2$ operator? Or is there is an error in my reasoning?
Best Answer
A physical example of your problem would be to calculate of the term symbol of boron in the ground state.
Where did you get your paraphrased Hund's rule? I don't think it is accurate. Normally, you assume $LS$ coupling (Russell–Saunders coupling), so $L,S,J$ are conserved quantum numbers.
In your example, you have one electron in a $p$ orbital, so necessarily $S=1/2$ and $L=1$ (non need to apply the first two rules). For the third rule, your orbital is less than half filled, so the vectors anti align and $J=L-S$. In your case, this means that $J=1/2$, ie the term symbol is $^2P_{1/2}$. Therefore, using the Clebsch Gordon coefficients, your ground state (with $j_z=1/2$, there is another corresponding to $j_z=-1/2$) is rather: $$ |\psi\rangle =\sqrt{\frac{2}{3}}|1,\downarrow\rangle-\sqrt{\frac{1}{3}}|0,\uparrow\rangle $$
Hope this helps.