Carbon has a $\mathrm{p^2}$ configuration, and within the Russell-Saunders coupling scheme, we have

$$\begin{align} s_1 = s_2 &= \frac{1}{2} & S &= 1,0 \\ l_1 = l_2 &= 1 & L &= 2,1,0 \\ \end{align}$$

which gives rise to $^3D$, $^1D$, $^3P$, $^1P$, $^3S$, and $^1S$ terms. Out of these six, only $^1D$, $^3P$, and $^1S$ are allowed by the Pauli principle.

It's not hard to show that the $^3D$ term is forbidden, since this term has to contain one state with $m_{l1} = m_{l2} = 1$, which would correspond to a spatial wavefunction that is symmetric with respect to interchange of the two electrons. Pairing this with a triplet spin wavefunction, which is also symmetric, would lead to an overall symmetric wavefunction, which is not allowed.

But that's all the textbooks say - they don't describe why the $^1P$ term is forbidden, for example. I know how to construct a table of microstates, and I know I can use the direct product table of the full rotation group to find that $P \times P = S + [P] + D$, and I know that both of these methods show that the allowed term symbols are only $^1D$, $^3P$, and $^1S$.

Essentially, I am looking for a more intuitive explanation of why the $^1P$ and $^3S$ states are forbidden that preferably links back to the requirement that the wavefunction be antisymmetric - if that is possible.

Alternatively, an explanation of why $P \times P = S + [P] + D$ would also work. I know that from the Clebsch-Gordan series we get $S$, $P$, and $D$ terms, but not why the $P$ state should be antisymmetric. I do know the general idea of how to obtain the vector coupling coefficients (by repeated application of raising/lowering operators), and these do show that the $P$ state is antisymmetric, but I'd like a more interpretative answer than just "that's how the maths works out".


1 Answer 1


Okay, actually I managed to find somewhat of an explanation myself.

We could construct a state with $M_L = 1$ and $M_S = 0$, for example: $$\begin{align} \psi = \frac{1}{2}\left[\mathrm{2p_1}(1)\mathrm{2p_0}(2) + \mathrm{2p_0}(1)\mathrm{2p_1}(2)\right]\left[\alpha(1)\beta(2) - \beta(1)\alpha(2)\right] \end{align}$$

where $\mathrm{2p}_m$ represents the 2p orbital with $m_l = m$.

However, such a combination of $(M_L,M_S)$ can only occur once, since the two electrons are indistinguishable. In other words, the following two other wavefunctions are not permissible because permuting the two electrons changes the wavefunction: $$\begin{align} \psi &= \frac{1}{\sqrt{2}}\left[\mathrm{2p_1}(1)\mathrm{2p_0}(2)\right]\left[\alpha(1)\beta(2) - \beta(1)\alpha(2)\right] \\ \psi &= \frac{1}{\sqrt{2}}\left[\mathrm{2p_0}(1)\mathrm{2p_1}(2)\right]\left[\alpha(1)\beta(2) - \beta(1)\alpha(2)\right] \end{align}$$

and the $^1D$ state, which has $S = 0 \Rightarrow M_S = 0$ and $L = 2 \Rightarrow M_L = 2,1,0,-1,-2$, already has to encompass one state with such a combination of $(M_L,M_S)$. Therefore, a state with $(M_L,M_S) = (1,0)$ cannot contribute to a $^1P$ term.

It is logically not the strongest argument (sufficiency vs necessity), but it is kind of what I was looking for. If anybody can provide another viewpoint though I will be glad to accept the answer.


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