How to specify atomic carbon terms in the coupled and uncoupled representation?

So, we know, that the atomic carbon in the electronic configuration $$1s^22s^22p^2$$ has the following terms

$${}^1S, {}^1D, {}^3P$$

My question is - how can I correctly specify these terms in the terms of coupled and uncoupled representations?

My attempt

So, in the case of terms, we're only considering the orbital angular momentum, not the spin. Because of that, we can describe the single terms in the coupled representation $$\left|L, M_L\right>$$ which correspond with the linear combination of microstates, i.e. the uncoupled representations $$\left|m_{l1}, m_{l2}\right>$$ using Clebsch-Gordan coefficients.

For $${}^1S$$ term it's pretty easy, as $$L=0$$ and $$M_L=0$$ (as described in this answer):

\begin{align}{}^1S: |L = 0, M_L = 0\rangle &= \frac{1}{\sqrt 3} |m_{l1}= 1, m_{l2} = -1\rangle + \frac{1}{\sqrt 3} |-1, 1\rangle - \frac{1}{\sqrt 3} |0, 0\rangle\\ &= \frac{1}{\sqrt{3}} \left| 8 \right> + \frac{1}{\sqrt{3}} \left| 11 \right> - \frac{1}{\sqrt{3}}\left| 14 \right >\end{align}

In the last expression there are wavefunctions specified with indices from the microstate table below.

But further it gets somewhat more tricky - both $${}^3P$$ and $${}^1D$$ will contain multiple states. $$P$$ corresponds with $$L=1$$ and so $$M_L \in \left\{ -1, 0, 1 \right\}$$. I suppose, that its coupled representations are $$\left| L=1, M_L=-1\right>, \left| L=1, M_L=0\right>, \left| L=1, M_L=1\right>$$.

\begin{align} {}^3P: \left| L=1, M_L=-1\right> &= \frac{1}{\sqrt{2}}\left| -1, 0 \right> + \frac{1}{\sqrt{2}}\left| 0, -1 \right>\\ &= \frac{1}{\sqrt{2}}\left| 2 \right> + \frac{1}{\sqrt{2}}\left|5 \right>\\ \left| L=1, M_L=0\right> &= \frac{1}{\sqrt{2}}\left| 1, -1 \right> - \frac{1}{\sqrt{2}}\left| -1, 1 \right>\\ &= \frac{1}{\sqrt{2}}\left| 3 \right> - \frac{1}{\sqrt{2}}\left|6 \right>\\ \left| L=1, M_L=1\right> &= \frac{1}{\sqrt{2}}\left| 1, 0 \right> - \frac{1}{\sqrt{2}}\left| 0, 1 \right>\\ &= \frac{1}{\sqrt{2}}\left| 1 \right> - \frac{1}{\sqrt{2}}\left|4 \right> \end{align}

$${}^1D$$ corresponds with $$L=2$$ and $$M_L \in \left\{ -2,-1,0,1,2 \right\}$$.

\begin{align} {}^1D:\left| L = 2, M_L = -2 \right> &= \left| -1, -1 \right> = \left| 15\right>\\ \left| L = 2, M_L = -1 \right> &= \frac{1}{\sqrt{2}}\left| 0, -1 \right> + \frac{1}{\sqrt{2}}\left|-1, 0 \right> \\ &= \frac{1}{\sqrt{2}}\left| 10 \right> + \frac{1}{\sqrt{2}}\left| 12 \right>\\ \left| L = 2, M_L = 0 \right> &= \frac{1}{\sqrt{6}}\left|1, -1 \right> + \sqrt{\frac{2}{3}}\left| 0, 0 \right> + \frac{1}{\sqrt{6}}\left| -1, 1 \right> \\ &= \frac{1}{\sqrt{6}}\left| 8 \right> + \sqrt{\frac{2}{3}}\left|14 \right> + \frac{1}{\sqrt{6}}\left| 11 \right> \\ \left| L = 2, M_L = 1 \right> &= \frac{1}{\sqrt{2}}\left| 1, 0 \right> + \frac{1}{\sqrt{2}}\left| 0, 1 \right>\\ &= \frac{1}{\sqrt{2}}\left|7 \right> + \frac{1}{\sqrt{2}}\left|9 \right> \\ \left| L = 2, M_L = 2 \right> &= \left| 1, 1 \right> = \left| 13 \right> \end{align}

Is this the right approach or do I understand it incorrectly?

Table of microstates

I'll try to illustrate the procedure for two of the nine states that form the term symbol $$^3P$$. This term symbol has $$S = 1$$ and $$L = 1$$, which makes for three possible values of $$M_S$$ and three possible values of $$M_L$$, for a total of nine states. These states are "coupled" in both the spin and the spatial dimensions:

$$|S,M_S,L,M_L\rangle = |1,+1,1,+1\rangle, |1,0,1,+1\rangle, \cdots$$

The microstates in the table are states which are "uncoupled" in both dimensions:

$$|m_{s1},m_{s2},m_{l1},m_{l2}\rangle = |{+1/2}, {+1/2}, {+1}, {+1}\rangle, \cdots$$

To go from the term symbols ("doubly coupled" - don't use this terminology, I made it up) to the microstates ("doubly uncoupled"), we have to convert both spin and spatial components from coupled to uncoupled representations.

Thankfully, the spin and spatial components are separable, so we can write

$$|S,M_S,L,M_L\rangle = |S,M_S\rangle \cdot |L,M_L\rangle$$

Uncoupling the spin part is simple enough:

\begin{align} |S=1,M_S=+1\rangle &= |\alpha(1)\alpha(2)\rangle \\ |S=1,M_S=0\rangle &= \frac{1}{\sqrt 2}\left\{|\alpha(1)\beta(2)\rangle + |\beta(1)\alpha(2)\rangle\right\} \\ |S=1,M_S=-1\rangle &= |\beta(1)\beta(2)\rangle \\ \end{align}

where $$|\alpha\rangle$$ denotes $$m_s = +1/2$$ and $$|\beta\rangle$$ denotes $$m_s = -1/2$$, as usual.

For the spatial parts, you will need to use the Clebsch–Gordan coefficients to expand $$|L,M_L\rangle$$ in terms of the "uncoupled representations" $$|m_{l1},m_{l2}\rangle$$. You already seem to understand this. For illustrative purposes we'll use

$$|L = 1, M_L = +1\rangle = \frac{1}{\sqrt 2}(|m_{l1} = +1, m_{l2} = 0\rangle - |m_{l1} = 0, m_{l2} = +1\rangle)$$

and we'll simplify the notation by denoting $$m_l = +1, 0, -1$$ as $$p^+, p^0, p^-$$ respectively, so:

$$|L = 1, M_L = +1\rangle = \frac{1}{\sqrt 2}\left\{|p^+(1)p^0(2)\rangle - |p^0(1)p^+(2)\rangle\right\}$$

Now, say we wish to find the microstate corresponding to $$|S,M_S,L,M_L\rangle = |1,+1,1,+1\rangle$$. What we need to do is to expand both spin and spatial components in terms of their respective uncoupled representations, then multiply them together again.

\begin{align} &|S=1,M_S=+1\rangle \cdot |L = 1, M_L = +1\rangle \\ &\qquad = |\alpha(1)\alpha(2)\rangle \cdot \frac{1}{\sqrt 2}\left\{|p^+(1)p^0(2)\rangle - |p^0(1)p^+(2)\rangle\right\} \\ &\qquad = \frac{1}{\sqrt 2} \left\{|p^+(1)\alpha(1)p^0(2)\alpha(2)\rangle - |p^0(1)\alpha(1)\,p^+(2)\alpha(2)\rangle\right\} \end{align}

Notice how this is a (suitably antisymmetrised) state where one electron is in $$p^+$$ with spin up, and the other electron is in $$p^0$$ with spin up. This corresponds exactly to microstate 1.

Although the microstates are uncoupled states, it doesn't mean they aren't antisymmetrised. By virtue of quantum indistinguishability, they have to be antisymmetrised: that's why most of the microstates are linear combinations of "composite states" such as $$|p^+(1)\alpha(1)p^0(2)\alpha(2)\rangle$$ (such a state on its own is not physically permissible, whereas the microstates are physically permissible states).

For an uglier but more instructive case, let's look at $$|S,M_S,L,M_L\rangle = |1,0,1,+1\rangle$$.

\begin{align} &|S=1,M_S=0\rangle \cdot |L = 1, M_L = +1\rangle \\ &\qquad = \frac{1}{\sqrt 2}\left\{|\alpha(1)\beta(2)\rangle + |\beta(1)\alpha(2)\rangle\right\} \cdot \frac{1}{\sqrt 2}\left\{|p^+(1)p^0(2)\rangle - |p^0(1)p^+(2)\rangle\right\} \\ &\qquad = \frac{1}{2} \left\{ |p^+(1)\alpha(1)p^0(2)\beta(2)\rangle - |p^0(1)\beta(1)p^+(2)\alpha(2)\rangle + |p^+(1)\beta(1)p^0(2)\alpha(2)\rangle - |p^0(1)\alpha(1)p^+(2)\beta(2)\rangle \right\} \end{align}

Notice the first two terms in this expansion describe an (antisymmetrised) state where one spin-up electron is in $$p^+$$ and one spin-down electron is in $$p^0$$. This corresponds to microstate 7. Likewise, the second two terms correspond to microstate 9. All in all, we can write:

$$|S = 1, M_S = 0, L = 1, M_L = +1\rangle = \frac{1}{\sqrt 2}(|7\rangle + |9\rangle)$$

Using this procedure you can work your way through the rest of the states in $$^3P$$ (and indeed, $$^1S$$ and $$^1D$$ too). As a final example, let's look at one of the five $$^1D$$ states:

\begin{align} &|S = 0, M_S = 0, L = 2, M_L = 0\rangle \\ &\qquad = |S = 0, M_S = 0\rangle \cdot |L = 2, M_L = 0\rangle \\ &\qquad = \frac{1}{\sqrt 2}\left\{|\alpha(1)\beta(2)\rangle - |\beta(1)\alpha(2)\rangle\right\} \cdot \\ &\qquad\qquad\qquad \left[\frac{1}{\sqrt 6}\left\{|p^+(1)p^-(2)\rangle + |p^-(1)p^+(2)\rangle \right\} + \sqrt{\frac{2}{3}}|p^0(1)p^0(2)\rangle \right] \\ &\qquad = \frac{1}{\sqrt{12}}\left\{ |p^+(1)\alpha(1)p^-(2)\beta(2)\rangle - |p^-(1)\beta(1)p^+(2)\alpha(2)\rangle \right\} \\ &\qquad \qquad + \frac{1}{\sqrt{12}}\left\{ |p^-(1)\alpha(1)p^+(2)\beta(2)\rangle - |p^+(1)\beta(1)p^-(2)\alpha(2)\rangle \right\} \\ &\qquad \qquad + \frac{1}{\sqrt{3}}\left\{ |p^0(1)\alpha(1)p^0(2)\beta(2)\rangle - |p^0(1)\beta(1)p^0(2)\alpha(2)\rangle \right\} \\ \end{align}

The first line of this expansion corresponds to $$(\sqrt{1/6}|8\rangle)$$; the second to $$(\sqrt{1/6}|11\rangle)$$; and the third to $$\sqrt{2/3}|14\rangle$$. So, your expansion was correct; however, be careful, because the uncoupled representations of the spatial wavefunctions

$$|m_{l_1} = +1, m_{l2} = -1\rangle$$

do not themselves correspond to the microstates. Instead, it is only by multiplying this with the spin wavefunction can you obtain a wavefunction that corresponds to the microstates.

• Just one more question - I don't understand, why is there minus in $\left| S=0, M_S=0 \right>$ in the very last example? – Eenoku Jan 15 at 0:35
• @Eenoku that's the singlet spin wavefunction. Atkins MQM 5th ed section 4.12 describes this in detail, and the bottom bit of my previous answer sort of summarises this. But really, any textbook should cover it (sometimes it appears in the context of the helium atom). – orthocresol Jan 15 at 0:44
• No problem. Feel free to ask a new question if you want more detail (I think it’s too long for comments). – orthocresol Jan 15 at 1:45