# Degeneracy of states when Spin-Orbit coupling is taken into account

Before I ask this question I must include some necessary background information as follows so that the question makes sense:

In a simple quantum mechanical treatment of Hydrogen, all the states with the same value of principle quantum number $$n$$ are degenerate as they have the same energy. The total degeneracy of all the states with a given value of is $$2n^2$$ and this can be shown as follows:

First, lets consider (for reference mostly) the relevant quantum numbers for Hydrogen in this problem:

$$n$$: Principle quantum number ($$n\ge 1$$; integer values)

$$\ell$$: Orbital angular momentum quantum number ($$0\le\ell\le n-1$$; integer values)

$$m_{\ell}$$: Magnetic quantum number ($$-\ell\le m_{\ell}\le \ell$$; integer values)

$$m_s$$: Spin projection quantum number ($$-s\le m_s\le s$$; for our case $$m_s=\pm \frac12$$)

The table for the allowed quantum numbers are: From the table above you can easily see that for a given $$n$$ the sum of the associated $$m_{\ell}$$ values for each $$\ell$$ will give $$n^2$$. Or written more compactly since each $$\ell$$ has $$2\ell + 1$$ values for $$m_{\ell}$$:

$$\sum_{\ell=0}^{\ell=n-1}(2\ell + 1)= n^2$$

Now we note that there are two possible values for $$m_s$$ (which I didn't bother to put in the table as it is tedious) so we multiply the above expression by $$2$$ to get the answer of $$2n^2$$.

The above was done in the without considering Spin-Orbit coupling. When spin-orbit coupling is taken into account, each $$n$$, $$\ell$$ state gives rise to two levels with total angular momentum $$j=\ell \pm \frac12$$

Show that the total number of states with the same values of $$n$$ and $$\ell$$ is unchanged. This illustrates the general point that recoupling the angular momenta never changes the total number of states available.

First I will quote the solution to this and then explain which part I don't understand:

The state $$n,\ell$$ has a degeneracy of $$2(2\ell+1)$$. When it is split into two states with $$j=\ell-\frac12$$ and $$j=\ell +\frac12$$, the degeneracies of these states are $$\color{red}{2\ell}$$ $$\,\color{red}{\text{&}}$$ $$\,\color{red}{2\ell + 2}$$, so the total degeneracy is still $$4\ell +2$$. Since this holds for each value of $$\ell$$, it holds for the complete set of states having the same value of $$n$$.

Firstly, I understand why the state $$n,\ell$$ has a degeneracy of $$2(2\ell+1)$$, this is because there are two possible values of $$m_{s}$$ for a given $$\ell$$ and there are $$2\ell +1$$ values of $$m_{\ell}$$ which multiply to give $$2(2\ell+1)$$.

The part of the solution I don't understand is marked in red.

After the $$n,\ell$$ state splits into $$2$$ states why must these states have degeneracies $$\color{red}{2\ell}$$ and $$\,\color{red}{2\ell + 2}$$?

Put in another way; Why not split as $$\ell$$ and $$3\ell +2$$ for example which sum to give the required $$4\ell +2$$ or $$\ell + 1$$ and $$3\ell +1$$?

I fear I am missing something very simple here.

Let's start from the beginning then. $\overrightarrow{L}$ and $\overrightarrow{S}$ are vectors like any other, but they can only take integer length values that we call $\ell$ and $s$.
$m_l$ and $m_s$ are the projection of the vector in the z direction usually. $m_\ell$ can only take integer values from $-\ell$ when the vector $\overrightarrow{L}$ is parallel to z but into the opposite direction and as it gets rotated it gains +1 because it's only integers (quantum mechanics) when it's orthogonal to the z axis it's $0$ then when it's parallel it is $\ell$. so it can take $(2\ell +1)$ values.
So then $j$ is defined as $\overrightarrow{J}=\overrightarrow{L}+\overrightarrow{S}$ and $m_j$ is the projection. $j$ can take specific values $j=|\ell-s|,(|\ell-s|+1),..,|\ell+s|$ depending on the angle between these two vectors.
$m_j$ itself has $(2j+1)$ values. Where you can replace $j$ by its value.
• " they can only take integer length values that we call ℓ and s. " No, $\vec{S}$ has a half-integer quantum number, $s=\frac{1}{2}$. May 27, 2017 at 19:52
• It's what I commented, every $j$ state has degeneracy $2j+1$ and when $j=\ell + 1/2$ you replace $j$ by it's value in $2j+1=2(\ell +1/2)+1=2\ell +1+1$ how more clearly do you want? May 28, 2017 at 11:45