What's the relationship between Term Symbol and Wave Functions?

Question

I'm trying to represent the Term Symbols in terms of wavefunctions and/or arrows for the excited state of hydrogen atom $2p^1$: $^2P_{3/2}$ and $^2P_{1/2}$.

Ideas I have:

The sum of $2J+1=4$ and $2J+1=2$ says how many different configurations contribute to that Term Symbol number, in this case I have six different configurations in the p orbital:

$\underline{\uparrow}$ $\underline{}$ $\underline{}$, $\underline{}$ $\underline{\uparrow}$ $\underline{}$, $\underline{}$ $\underline{}$ $\underline{\uparrow}$, $\underline{\downarrow}$ $\underline{}$ $\underline{}$, $\underline{}$ $\underline{\downarrow}$ $\underline{}$ and $\underline{}$ $\underline{}$ $\underline{\downarrow}$.

Now, the term symbol $^2P_{3/2}$ has only $2J+1=4$ of those six configurations. How do I know which of those pick?. Once I have chosen those configurations, how does their wave function look? is a symmetric linear combination? For example, if the states were
$\underline{\uparrow}$ $\underline{}$ $\underline{}$, $\underline{}$ $\underline{\uparrow}$ $\underline{}$, $\underline{}$ $\underline{}$ $\underline{\uparrow}$ and $\underline{\downarrow}$ $\underline{}$ $\underline{}$, then, would one of the two ($2S+1=2$) wave functions have the form: $\Psi=\frac{1}{2}(\Phi_{21-1}|\alpha>+\Phi_{210}|\alpha>+\Phi_{211}|\alpha>+\Phi_{21-1}|\beta>)$? where $|\alpha>=\uparrow$ and $|\beta>=\downarrow$.

Answer 1

You can write the states $\Psi_{nljm_j}$ (which are eigenvectors of $L^2$, $S^2$, $J^2$ and $J_z$) as certain superpositions of the states $\Phi_{nlm_lm_s}$ (which are eigenvectors of $L^2$, $S^2$, $L_z$ and $S_z$). But unfortunately this angular momentum coupling is quite an elaborate task, and usually text books on quantum mechanics take at least one whole chapter to explain it. You have already figured out the number of configurations for the terms. The final result (with the coefficients taken from the table of Clebsch-Gordan coefficients, section 2.3) is this.

The term $^2P_{3/2}$ (i.e. with $s=\frac{1}{2},l=1,j=\frac{3}{2}$) has $4$ configurations ($m_j=+\frac{3}{2},+\frac{1}{2},-\frac{1}{2},-\frac{3}{2}$): $$\begin{align} \Psi_{21\frac{3}{2}+\frac{3}{2}} &= \Phi_{21+1\uparrow} \\ \Psi_{21\frac{3}{2}+\frac{1}{2}} &= \sqrt{\frac{1}{3}}\Phi_{21+1\downarrow} + \sqrt{\frac{2}{3}}\Phi_{210\uparrow} \\ \Psi_{21\frac{3}{2}-\frac{1}{2}} &= \sqrt{\frac{1}{3}}\Phi_{21-11\uparrow} + \sqrt{\frac{2}{3}}\Phi_{210\downarrow} \\ \Psi_{21\frac{3}{2}-\frac{3}{2}} &= \Phi_{21-1\downarrow} \end{align}$$

And the term $^2P_{1/2}$ (i.e. with $s=\frac{1}{2},l=1,j=\frac{1}{2}$) has $2$ configurations ($m_j=+\frac{1}{2},-\frac{1}{2}$): $$\begin{align} \Psi_{21\frac{1}{2}+\frac{1}{2}} &= \sqrt{\frac{2}{3}}\Phi_{21+1\downarrow} - \sqrt{\frac{1}{3}}\Phi_{210\uparrow} \\ \Psi_{21\frac{1}{2}-\frac{1}{2}} &= - \sqrt{\frac{2}{3}}\Phi_{21-11\uparrow} + \sqrt{\frac{1}{3}}\Phi_{210\downarrow} \end{align}$$

Answer 2

Tne Term Symbol tells you how to couple the two different angular mmomenta involved. $^2P_{3/2}$ means $L=1$, $S=1/2$ and $J=3/2$. Thus the states involved must have the required angular momentum properties. $J=3/2$ has 4 states with different $z$-components. One of these is $J_z=+3/2$. There is only one way to make this state which is $|1,1>|1/2,1/2>$, where the notation shows the wavefunctions of the orbital angular momentum $|l,m_s>$ and spin $|s,m_s>$. Similarly there is but one way to make a $J_z=-3/2$ state. However there are two ways of making $J_z=1/2$ or $J_z=-1/2$ states. The one belonging to $J=3/2$ is found by applying the lowering operator $J_-=L_-+S_-$ to the $J_z=3/2$ state, and the $J_z=-1/2$ state by applying the lowering operator a second time, or the raising operator to the $J_z=-3/2$ state. The orthogonal linear combinations of $J_z=\pm1/2$ states belong to $J=1/2$.