LS coupling for identical electrons

[mk_gm1]

Let's say I'm considering the $$3p^2$$ electrons. From the Pauli Exclusion Principle, we know that two electrons cannot have the same state, which in this case means m_l and m_s cannot both be the same for each electron.

What this means is that the following 6 terms must not be allowed:

$$m_{l1} \hspace{0.1 in} m_{l2} \hspace{0.1 in} m_{s1} \hspace{0.1 in} m_{s2}$$

$$-1 \hspace{0.1 in} -1 \hspace{0.1 in} \downarrow \hspace{0.3 in} \downarrow$$
$$-1 \hspace{0.1 in} -1 \hspace{0.1 in} \uparrow \hspace{0.3 in} \uparrow$$
$$0 \hspace{0.4 in} 0 \hspace{0.3 in} \downarrow \hspace{0.3 in} \downarrow$$
$$0 \hspace{0.4 in} 0 \hspace{0.3 in} \uparrow \hspace{0.3 in} \uparrow$$
$$+1 \hspace{0.1 in} +1 \hspace{0.1 in} \downarrow \hspace{0.3 in} \downarrow$$
$$+1 \hspace{0.1 in} +1 \hspace{0.1 in} \uparrow \hspace{0.3 in} \uparrow$$

These correspond to $$M_L=\sum m_{li}<br /> = -2, -2, 0, 0, 2, 2$$ and $$M_S = \sum m_{si} = -1, 1, -1, 1, -1, 1$$ respectively.

My question is this - how does this lead to the conclusion that the allowed terms are ¹S, ¹D and ³P ? For example, there's a M_L = 0, M_S= -1 term in both ³S and ³P - why do we disallow one and not the other?

Also, what leads us to disallow ¹P (for which M_L=-1, 0, 1 and M_S=0)? Surely the only way to have M_S = 0 is to have $$\downarrow_1 \hspace{0.1 in} \uparrow_2$$ or vice versa, and hence $$m_{s1} \neq m_{s2}$$ and we have no violation of the Pauli Exclusion Principle?

 

[ManyNames]

What do you mean by ''why do we disallow one and not the other?''

 

[mk_gm1]

Ok let me try and clarify what I'm confused by. Have a look at the following table which I have taken from my notes:

In this table we are looking at part of the identical (np)² configuration. There are 15 possible configurations, all of which I haven't posted.

Take a look at the 2nd row. I can see why this state corresponds to the ³P term. For it to be a P term, it must have L = 1 where $$L \geq |M_L|$$. It clearly has this. For it to be a ³P term, the multiplicity 2S+1=3 i.e. S=1 where $$S \geq | M_S |$$. This is obvious.

Now if you look at the 4th row, I am not sure why it also has to be a ³P state . I understand that it can be (i.e. that the ³P term includes this state), but this could equally well be a ¹P state, no? This is because here S=0. Therefore why do we choose one and not the other?