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

Question

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 $m_s=\pm \frac12$)

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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 $2n^2$.

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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 $n$.

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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.

Answer 1

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.