Consider the possible values of $S$ and $L$ for carbon configuration $1s^2 2s^2 2p^2$ and the corresponding rapresentations with arrows indicating the spins (consider only $S=0,L=0$ and $S=0,L=2$, as for $S=1,L=1$ some are missing).

enter image description here

I can't understand which one of the two representations indicated in red belong to $S=0,L=0$ state and which one to the $S=0,L=2$ state (I took a guess making the picture).

The spin wavefunction is in all cases antysymmetric ($S=0$) and equal to $$\chi =\left(\, |\!\uparrow_1 \downarrow_2\rangle-|\!\downarrow_1 \uparrow_2\rangle\,\right)\div\sqrt 2$$

Therefore the spatial wavefunction $(L)$ must be symmetric for Pauli principle and in the second red configuration is

$$\Psi =p_0 (1) p_0 (2)$$ in the first one is

$$\Psi= \left[p_- (1) p_+ (2)+p_+ (1) p_- (2)\right]\div\sqrt2$$

The wavefunctions are different so there must be a difference between the two red configurations, but which of them should belong to $S=0,L=2$ and the other to $S=0,L=0$ and why?

  • 2
    $\begingroup$ You need to make a table with the columns as $m_l$ and $m_s$ and also the totals (sum of $m_l$ which gives values -2,-1,0,1,2, and $m_s$, values -1,0,1) then remove those not possible by Pauli principle (those no good and also duplicates). There are 15 possible configurations in all. You can then identify those you need. The are 5 with L=2, 9 with L=1 and 1 with L=0. $\endgroup$
    – porphyrin
    Aug 10, 2018 at 8:56

1 Answer 1


You are separating the spatial and spin part of your two wavefunctions. I don't think that is a good idea here, because it does only work for one of the two configurations.

The wavefunction of a configuration is given by a Slater determinant build up from spin orbitals. In the following I will neglected the normalization constant of $\frac{1}{\sqrt2}$ for simplicity. The wavefunction for the first configuration ($\uparrow - \downarrow$) is

\begin{align} \Psi_1 &= \begin{vmatrix} p_-(1)\alpha(1) & p_+(1)\beta(1)\\ p_-(2)\alpha(2) & p_+(2)\beta(2)\\ \end{vmatrix}\\ &= p_-(1)\alpha(1)p_+(2)\beta(2) - p_+(1)\beta(1)p_-(2)\alpha(2) \end{align}

Note that this differs from your suggested wavefunction by the spin part of the orbitals, and there is no way to separate them out.

And for the other configuration ($-\uparrow\downarrow -$) we have \begin{align} \Psi_2 &= \begin{vmatrix} p_0(1)\alpha(1) & p_0(1)\beta(1)\\ p_0(2)\alpha(2) & p_0(2)\beta(2)\\ \end{vmatrix}\\ &= p_0(1)\alpha(1)p_0(2)\beta(2) - p_0(1)\beta(1)p_0(2)\alpha(2) \end{align}

Both, $\Psi_1$ and $\Psi_2$ look quite similar now. However, since the spatial part of both orbitals in $\Psi_2$ is the same here, we can factor them out

\begin{equation} \Psi_2 = p_0(1)p_0(2)\underbrace{[\alpha(1)\beta(2)-\beta(1)\alpha(2)]}_{\chi} \end{equation}

And we get what you have suggested for this configuration. But this does only work for $\Psi_1$!

Now, why is $\Psi_1$ assigned to $L=2$, while $\Psi_2$ is $L=0$?

I assume your confusion is because if you add up the orbital quantum numbers $m_l$ you get $0$ in both cases ($1-1=0$ and $0+0=0$). However, those sums give you $M_L$, not $L$. The many-electron quantum number $L$ has $2L+1$ components with quantum numbers $M_L=-L,-L+1,\dots 0, \dots L-1,L$. So for $L=2$ we have $M_L=-2,-1,0,1,2$, much like there are three orbitals $p_-$, $p0$ and $p+$ for $l=1$

So $\Psi_1$ would be the $M_L=0$ component of $L=2$, while $\Psi_2$ is the $M_L=0$ component of $L=0$.

But there is one issue with such assignments: You cannot really do it, as the configurations get mixed together. The point is, that the atom as a whole needs to be spherical symmetric (unless there is some external electric or magnetic field). But the individual $p$ orbitals are not. For example $p_0$ is (in the usual convention) aligned along the $z$-axis and zero in the $xy$ plane.

Mathematically speaking the configurations form a basis (Hilbert space) in which the actual electronic states are expanded. This is known as the configuration interaction method.

So for example the electronic state with $L=0$ would be \begin{equation} \Psi_2 = \frac{1}{\sqrt3}(|\uparrow\downarrow - -\rangle + |-\uparrow\downarrow - \rangle + |--\uparrow\downarrow \rangle) \end{equation}

where each $|\dots\rangle$ denotes a Slater determinant.


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