# Distinguish between two possible configurations for angular momentum in carbon atom

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).

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?

• 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. Aug 10, 2018 at 8:56

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

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

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 $$$$\Psi_2 = \frac{1}{\sqrt3}(|\uparrow\downarrow - -\rangle + |-\uparrow\downarrow - \rangle + |--\uparrow\downarrow \rangle)$$$$

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