# Total number of states for a two electron system

Let's say, I have the following electronic configuration, $$1s2s$$, and I'm trying to find all the possible states.

By looking at the configuration, it is obvious, that in the $$|l_1,m_{l_1},m_{s_1},l_2,m_{l_2},m_{s_2}\rangle$$ basis, we have a total of $$4$$ possible states. Since total spin $$(S)$$ is either $$0$$ or $$1$$, w should have one singlet and one triplet giving us a total of 4 states. There are exactly 4 possible electronic configurations.

$$\sum_i (2S_i +1)(2L_i+1) = 4$$

However in the basis $$|L,S,m_L,m_S\rangle$$, we get four possible states, as $$m_S = 0(singlet);0,1,-1(triplet)\space\space\space m_L=0$$. Hence there are four possible configurations.

Similarly, if we represent this in the coupled basis $$|L,S,J,m_j\rangle$$, we have the two cases : $$L,S=0$$ and $$L=0,S=1$$. In the first case, the only value of $$J$$ is $$0$$, and hence the value of $$m_j$$ is also $$0$$. In the second case, $$L\lt S$$, so, the number of possible $$J$$ values is $$(2L+1) = 1$$. Hence we obtain two-term symbol states $$^1S_0^1$$. Thus we are obtaining only two states, in the coupled basis.

How can we have four possible electronic configurations, or four possible wavefunctions given by slater determinants, but only 2 possible states in the coupled basis ? Shouldn't the number of term symbols be equal to the no. of possible electronic configurations or the no. of states in the uncoupled basis?

Shouldn't $$\sum_i (2J_i +1) = (2L+1)(2S+1)$$ be true ? If this is not true, then where are those $$2$$ states, that are included in the configuration and in the uncoupled basis, but not in the term symbols or coupled basis ?

What am I missing here ? What is the actual no. of states in this system ?

Let's enumerate basis states. Since $$l=m_l=0$$, I am going to use $$n\in(1,2)$$. The two spatial wave functions are as $$\psi_{n_1,n_2}$$ have (anti)symmetric combinations:

$$\psi_S = \frac 1{\sqrt 2}(\psi_{1,2}+\psi_{2,1})$$ $$\psi_A = \frac 1{\sqrt 2}(\psi_{1,2}-\psi_{2,1})$$

The spins have 1 antisymmetric combination:

$$\chi_0^0 = \frac 1{\sqrt 2}(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle)$$

and 3 antisymmetric:

$$\chi_1^0 = \frac 1{\sqrt 2}(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle)$$ $$\chi_1^1 = |\uparrow\uparrow\rangle$$ $$\chi_1^{-1} = |\downarrow\downarrow\rangle$$

The allowed states are the totally antisymmetric ones:

$$\psi_S\chi_0^0$$ $$\psi_A\chi_1^m\,\,\,m\in(-1,0,1)$$

of which there are 4 (though the total number of states is 8).

Since I symmetrized the spin states, this is already in the coupled basis. Since $$L=0$$, $$J=S$$. I think your counting went wrong when you failed to enumerate the spatial wave functions.

• Thanks, I get this finally. I had made a really stupid mistake. According to my book, when L less than S, multiplicity of J is $2L+1$, so in this case only one value of J is possible. Then I made the error of considering J must be equal to L. I see it now. Since J is equal to 1, we get three values of $m_j$. Hence we get our four term symbols. Commented Aug 17, 2021 at 9:55
• The term symbols should be $^1S_0^0 , ^3S_1^{-1}, ^3S_1^0, ^3S_1^1$. this is correct I hope. Commented Aug 17, 2021 at 9:56