Why does the ground state of a Li atom have a total spin of S = 1/2? I get that the total z component of the spin has to be either Ms = 1/2 or Ms = -1/2 because of the Pauli exclusion principle, but that could also correspond to a total spin of S = 3/2, so why is it always S = 1/2?

  • $\begingroup$ If it was really $S = 3/2$, then there would be a $M_S = +3/2$ state, but there isn't. $\endgroup$ Mar 16, 2020 at 8:00
  • $\begingroup$ But it doesn't have to, right? I mean, you could have a state with S = 3/2, Ms = +1/2 and it would be a valid state $\endgroup$
    – Marcosko
    Mar 16, 2020 at 8:03
  • 1
    $\begingroup$ It does have to. Anything with $S = 3/2$ will necessarily comprise four different states with $M_S = +3/2, +1/2, -1/2, -3/2$. You can't have some of them allowed and some of them forbidden by the Pauli exclusion principle. $\endgroup$ Mar 16, 2020 at 8:04
  • $\begingroup$ I see. Is that because the spin wave function must be either symmetric or antisymmetric? $\endgroup$
    – Marcosko
    Mar 16, 2020 at 8:07
  • $\begingroup$ Not quite, that's just how angular momentum works, and it isn't peculiar to spin angular momentum. The symmetrisation thing (we might as well focus on "antisymmetric" only, since we're talking about electrons here, i.e. fermions) is basically the same thing as the Pauli principle. $\endgroup$ Mar 16, 2020 at 8:08

2 Answers 2


Probably the easier way to explain it is to rely on the fact that if Li indeed had a ground state with $S = 3/2$, this would necessarily comprise four levels with $M_S = +3/2, +1/2, -1/2, -3/2$. This sort of splitting can be observed under a magnetic field, for example (the Zeeman effect).

Note that it's not possible for the Pauli principle to forbid $M_S = +3/2$ and $-3/2$ while simultaneously allowing $M_S = +1/2$ and $-1/2$. The Pauli principle doesn't mean that "this state can exist, but the electrons can't reach that state". It means that "this state simply cannot exist": because any state that violates the Pauli principle is equal to zero, which isn't a valid state. So it's not as if the $M_S = \pm 3/2$ states are there, and that the electrons simply choose to go into the $M_S = \pm 1/2$ states. The $\pm 3/2$ states are nonexistent. And consequently, because a $S = 3/2$ state necessitates the existence of $M_S = \pm 3/2$ states, it cannot exist either.

However, since that line of argument doesn't seem to fully convince you, you may want to go through the "exercise" of proving it. Unless I'm terribly wrong, this should be a completely foolproof (albeit terribly tedious) method. You can construct the full wavefunction for the Li atom as a Slater determinant (without loss of generality we choose $\alpha$ for the third spin):

$$\Psi = \frac{1}{\sqrt{6}} \begin{vmatrix} 1s(1)\alpha(1) & 1s(1)\beta(1) & 2s(1)\alpha(1) \\ 1s(2)\alpha(2) & 1s(2)\beta(2) & 2s(2)\alpha(2) \\ 1s(3)\alpha(3) & 1s(2)\beta(3) & 2s(2)\alpha(3) \\ \end{vmatrix}$$

so "all" we need to do is to apply $\hat{S}^2$ to this wavefunction and see what eigenvalue it returns. We know that

$$\hat{S}^2|\Psi\rangle = S(S+1)|\Psi\rangle$$

and so if $S = 1/2$ then the eigenvalue will be $3/4$; and if $S = 3/2$ the eigenvalue will be $15/4$. The $\hat{S}^2$ operator can be expanded as

$$\begin{align} \hat{S}^2 &= (\hat{S}_1 + \hat{S}_2 + \hat{S}_3)^2 \\ &= \hat{S}_1^2 + \hat{S}_2^2 + \hat{S}_3^2 + 2(\hat{S}_1 \cdot \hat{S}_2) + 2(\hat{S}_1 \cdot \hat{S}_3) + 2(\hat{S}_2 \cdot \hat{S}_3) \\ &= \hat{S}_1^2 + \hat{S}_2^2 + \hat{S}_3^2 + 2(\hat{S}_{1x} \hat{S}_{2x} + \hat{S}_{1y}\hat{S}_{2y} + \hat{S}_{1z}\hat{S}_{2z}) \\ &\quad + 2(\hat{S}_{1x} \hat{S}_{3x} + \hat{S}_{1y}\hat{S}_{3y} + \hat{S}_{1z}\hat{S}_{3z}) + 2(\hat{S}_{2x} \hat{S}_{3x} + \hat{S}_{2y}\hat{S}_{3y} + \hat{S}_{2z}\hat{S}_{3z}) \end{align}$$

We already know what the individual operators $\hat{S}_i^2$, $\hat{S}_{ix}$, $\hat{S}_{iy}$, and $\hat{S}_{iz}$ do to the basis states $|\alpha(i)\rangle$ and $|\beta(i)\rangle$. Bearing in mind that operators don't act on different degrees of freedom, you can work out what this monster operator does to the Li wavefunction.

If you look online, there are examples of how to perform this calculation for a two-spin system (e.g. "prove that the singlet and triplet wavefunctions are eigenstates of $\hat{S}^2$"). The extension to a three-spin system is not difficult, but is extremely tedious, which is why I gave up on writing it out in full. But I am fairly certain that at the end of the day you will find the eigenvalue of $3/4$.


In terms of how the electrons are arranged*, the reason you get only spin $1/2$ is because the most stable configuration has two of the electrons in the same 1s orbital, which by the Pauli Exclusion forces them to have opposite spins. This leaves only the third electron, sitting in the 2s orbital, to contribute a net spin.

You could get a spin of $3/2$ in an excited state where one of the electrons is promoted from the 1s orbital to a higher, otherwise empty one such as a 2p orbital. Then with all three electrons you may have them all with the same direction of spin leading to a total spin of $3/2$. However, in the case of lithium (and other alkali metals) the energy required to do this is much greater than that required to ionize the outer electron from the ground state, so you are more likely to get ionization rather than excitation to a state with spin $3/2$.

*I assume you mean electronic spin. The nucleus also has it's own spin, but this is often ignored by chemists because nuclear spin does not couple into chemical processes.


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