The energy shifts due to spin-orbit interaction are given by $$\Delta E_j = \frac{C}{2} (j(j+1)-l(l+1)-s(s+1))$$

If $l$, $s$ and $j$ are for the outer electron in sodium for example, then for $l>0$ there are only 2 sub levels. That means there are only two values of $j$. Why is this the case? If $j=l+s$, $l+s-1$,...$|l-s|$, why aren't there more possible values for higher values of $l$?


j varies from $l+s$ to $|l-s|$ as you said and since it varies in only integer steps, the next integer lower than $l+\frac{1}{2}$ is $l+\frac{1}{2}-1$ = $l-\frac{1}{2}$ = $l-\frac{1}{2}$...since electron spin is $\frac{1}{2}$...no more possible values of $j$!

  • $\begingroup$ $s=\frac{1}{2}$? I thought it was $\hbar^2 l(l+1)$? $\endgroup$ – user13948 Oct 18 '16 at 12:48
  • $\begingroup$ no. $s=\frac{1}{2}$, meaning electron spin is always $\frac{1}{2}$ and the azimuthal spin quantum number '$m_s$' can have a value of $+\frac{1}{2}$ and $-\frac{1}{2}$......$\hbar ^2$$l(l+1)$ is the value of $\vec {L^2}$ for a particular measurement, it has nothing to do with $s$ $\endgroup$ – Prasad Mani Oct 18 '16 at 12:51
  • $\begingroup$ So $s=\frac{1}{2}$ for electrons. Clearly very important not to confuse that with $L$. Thank you. $\endgroup$ – user13948 Oct 18 '16 at 12:56
  • $\begingroup$ exactly. so no matter what value of $l$ you have, there are only two possible $j$ for a single valence electron system (sodium, hydrogen etc)......$l+\frac{1}{2}$ and $l-\frac{1}{2}$; if there are two or more electrons, then you have to write down all possible couplings called LS coupling where you will see more than 2 values of $j$ $\endgroup$ – Prasad Mani Oct 18 '16 at 12:59

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