I'm currently studying atomic term symbols. I wanted to try it on a simple atomic carbon with the electron configuration $1s^22s^22p^2$.

I know, that only open-shell electrons are involved in the term symbol classification, so that leaves us working with solely 2 electrons in $2p$ sub-shell.

First of all, we have ${6}\choose{2}$$= \frac{6!}{2!(6-2)!}=15$ possible microstates for the $1s^22s^22p^2$ configuration.

The microstates are listed in the following table: enter image description here

And, if I understand it correctly, we can categorize them into several "subsets" labeled with the corresponding term symbols.

The atomic term symbol is defined as ${}^{2S+1}L_J$.

In our case, the possible values of S, L and J are following: $$\max(M_L) = \max(L) = +2 \Rightarrow L = 0,1,2$$ $$\max(M_S) = \max(S) = +1 \Rightarrow S = 0, 1$$ $$J = L+S, L+S-1, \ldots, |L-S| = 3,2,1,0$$

Considering possible values of $S$, multiplicity can be $1$ or $3$, i.e. singlet or triplet. Values of $L$ enable symbols $S,P,D$.

So, our presumed set of states is ${}^3D, {}^1D, {}^3P, {}^1P, {}^3S, {}^1S$.

${}^3D$ "contains" $S=1$ and $L=2$. In that situation, both electrons would have to be spin-up and positioned together in the orbital with $m_l = +1$. That is impossible, as it contradicts Pauli exlusion principle. I.e. the state ${}^3D$ does not exist.

${}^1D$ is possible ($L=2,1,0, S=0$), so we can compute the number of corresponding microstates $N$ with the formula $N = (2L+1)(2S+1) = 5\cdot 1 = 5$.

Using the same formula, ${}^3P$ is going to contain 9 microstates.

With the "substraction method" described in this video, we can arrive to the conclusion, that both ${}^1P$ and ${}^3S$ won't exist.

And finally, ${}^1S$ will contain the last one state.


Now I know the possible term symbols, but I'm not sure, which microstates belong to them specifically.

It's clear, that microstates 13 and 15 will belong to ${}^1D$, but how can I determine it for the other states? ${}^1D$ should contain 3 more microstates with $S=0$ and $m_L$ being equal to -1, 0 and +1, but there are multiple candidates for every configuration.

So how could I distinguish between, e.g. microstates 7 and 9 or 8, 11 and 14?

I don't understand this point, as even $J=2$ for ${}^1D$, so it can't be used to distinguish among the "similar" microstates and to assign them properly.

  • 2
    $\begingroup$ There is no direct assignment. Please take a look at the very last part of my answer here chemistry.stackexchange.com/questions/100413/… $\endgroup$
    – Feodoran
    Nov 30 '18 at 16:35
  • $\begingroup$ @Feodoran So, the sets of states belonging to different term symbols are basically overlapping? $\endgroup$
    – Eenoku
    Nov 30 '18 at 16:48
  • 1
    $\begingroup$ Yes, because it is a basis set expansion. $\endgroup$
    – Feodoran
    Nov 30 '18 at 16:52
  • $\begingroup$ @Feodoran Great! Would you write a short asnwer, so I could accept it? $\endgroup$
    – Eenoku
    Nov 30 '18 at 17:12

To add on to what Feodoran has mentioned, you might be interested in looking at the Clebsch–Gordan coefficients, which are essentially the coefficients of the microstates in the term symbols. You can look these up in e.g. the appendices of Atkins Molecular Quantum Mechanics, 5th ed. Note that each term symbol is a collection of states, so a $D$ term (for example) which has $L = 2$ would comprise five states: one each with $M_L = -2, -1, 0, +1, +2$. These five states can in general be denoted as $|L, M_L\rangle$.

$L$, which is a total angular momentum, is obtained by a coupling of two individual angular momenta. For two electrons in p-orbitals, these individual angular momenta are $l_1 = 1$ (with $m_{l1} = -1, 0, +1$) and $l_2 = 1$ (with $m_{l2} = -1, 0, +1$). The microstates in the table specify both values of $m_{l1}$ and $m_{l2}$, so they are simultaneous eigenstates of $m_{l1}$ and $m_{l2}$, i.e. they can be denoted by $|m_{l1},m_{l2}\rangle$.

Now, in general each coupled representation $|L, M_L\rangle$ will be a linear combination of the microstates, or uncoupled representations $|m_{l1},m_{l2}\rangle$. It is in general not possible to construct a state which is simultaneously an eigenstate of $L, M_L, m_{l1}$, and $m_{l2}$, because the relevant operators do not commute (refer to Atkins or any other QM text on coupling of angular momentum). There are examples of one-to-one correspondence between the coupled and uncoupled representations – for example, microstate 13 perfectly corresponds to $|L = 2, M_L = +2\rangle$, but it is not general, so:

$$ |L, M_L\rangle = \sum_{m_{l1},m_{l2}} c^{L, M_L}_{m_{l1},m_{l2}} |m_{l1},m_{l2}\rangle$$

The Clebsch–Gordan coefficients are simply the $c^{L, M_L}_{m_{l1},m_{l2}}$ in the expansion above. Usually the notation is slightly different – so instead of $(L, M_L, m_{l1}, m_{l2})$, it is usually $(j, m_j, m_{j1}, m_{j2})$ – but the concept is entirely the same.

Using the coefficients for $j_1 = j_2 = 1$ in Atkins (which is relevant to the $\mathrm{p^2}$ configuration of atomic carbon), we find that the $^1S$ term is given by

$$|L = 0, M_L = 0\rangle = \underbrace{\frac{1}{\sqrt 3} |m_{l1}= 1, m_{l2} = -1\rangle + \frac{1}{\sqrt 3} |m_{l1}= -1, m_{l2} = 1\rangle}_{\text{a combination of microstates 8 and 11}} - \underbrace{\frac{1}{\sqrt 3} |m_{l1}= 0, m_{l2} = 0\rangle}_{\text{microstate 14}}$$

It's slightly more confusing because there is a spin component as well, which the coefficients do not cover, and this is why I've given a handwavy "combination of microstates 8 and 11". This "combination" is needed to generate the singlet spin wavefunction $2^{-1/2}(|\alpha\beta\rangle - |\beta\alpha\rangle)$.

As an aside, the familiar singlet and triplet spin wavefunctions

$$\underbrace{|\alpha\alpha\rangle, |\beta\beta\rangle, \frac{1}{\sqrt 2}(|\alpha\beta\rangle + |\beta\alpha\rangle)}_{\text{triplet spin wavefunctions}}, \underbrace{\frac{1}{\sqrt 2}(|\alpha\beta\rangle + |\beta\alpha\rangle)}_{\text{singlet spin wavefunction}}$$

are also coupled representations. These four wavefunctions are, respectively,

$$|S=1, M_S=+1\rangle, |S=1, M_S=-1\rangle, |S=1, M_S=0\rangle, |S=0, M_S=0\rangle$$

(Different notation again, but it is the same concept.) The uncoupled representations are the states in which each electron has been assigned a spin, i.e. $|\alpha\alpha\rangle, |\alpha\beta\rangle, |\beta\alpha\rangle, |\beta\beta\rangle$. For example:

$$|m_{s1} = +1/2, m_{s2} = +1/2\rangle = |\alpha\alpha\rangle$$

and so we can see a simpler example of how the coefficients link the two representations:

$$\begin{align} |S=1, M_S=+1\rangle &= |m_{s1} = +1/2, m_{s2} = -1/2 \rangle \\ |S=1, M_S=-1\rangle &= |m_{s1} = -1/2, m_{s2} = -1/2 \rangle \\ |S=1, M_S=0\rangle &= \frac{1}{\sqrt 2}|m_{s1} = +1/2, m_{s2} = -1/2 \rangle + \frac{1}{\sqrt 2}|m_{s1} = -1/2, m_{s2} = +1/2 \rangle \\ |S=0, M_S=0\rangle &= \frac{1}{\sqrt 2}|m_{s1} = +1/2, m_{s2} = -1/2 \rangle - \frac{1}{\sqrt 2}|m_{s1} = -1/2, m_{s2} = +1/2 \rangle \\ \end{align}$$


Part of your confusion is that the table you show is not the best way to show the information. If you make a table as below and fill in the values and remove those due to the Pauli principle then it is easier to partition up values. ( I'll leave you to complete it) $$\begin{array}{cccc}\\ m_{l1} & m_{l2} & m_{s1} & m_{s2} & M_L & M_s\\ \hline 1 & 1 & + & + &\text{no good Pauli}&\text{no good Pauli} \\ 1 & 1 & + & - & 2 & 0\\ 1 & 1 & - & + & \text{same as previous one} \\ 1 & 0 & + & + & 1 & 1\\ 0 & 1 & + & + & \text{same as previous one}\\ 1 & 0 & + & - & 1 & 0\\ 1 & 0 & - & + & 1 & 0\\ etc& &\text{and no more Pauli conflicts}&\\ \end{array}$$

When you are done there will be 15 valid combinations. Arrange these on a 3 by 5 grid. Three for the $M_s$ and five for the $M_L$ and at each intersection put the number of times these values occur. For example there is 1 value with $M_L=1, M_s=-1$ and 3 with both values of zero.

There is only one value with $M_L=2 , M_s=0$ so this must belong to a $^1D$ state so remove one column with $M_S=0$ (as there a 5 values in the series i.e. $M_L=-2,-1,0,1,2,\;M_s=0$) , what remains is a 3 by 3 array with one element in each except at the centre where there are two. One of these center ones must be the $^1S$ as both $M_L=M_s=0$, those remaining are the $^3P$.


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