Adding any $\ell$ plus spin ${1\over 2}$.

We wish to write the states of total angular momentum $j$ in terms of the product states $Y_{\ell m}\chi_\pm$. We will do this by operating with the $J^2$ operator and setting the coefficients so that we have eigenstates.

We choose to write the the quantum number $m_j$ as $m+{1\over 2}$. There are only two terms which have the right $m_j$. If the spin is up we need $Y_{\ell m}$ and if the spin is down, $Y_{\ell (m+1)}$.

$$\bgroup\color{black}\psi_{j(m+{1\over 2})}=\alpha Y_{\ell m}\chi_++\beta Y_{\ell (m+1)}\chi_-\egroup$$

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We will find the coefficients $\alpha$ and $\beta$ so that $\psi$ will be an eigenstate of

$$\bgroup\color{black} J^2 = (\vec{L}+\vec{S})^2 = L^2 + S^2 + 2L_zS_z + L_+S_- + L_-S_+. \egroup$$

So operate on the right hand side with $J^2$.

$$\begin{eqnarray*} & J^2\psi_{j,m+{1\over 2}} = \alpha\hbar^2 \left[ \ell(\ell +... ...qrt{\ell(\ell + 1) - (m+1)m} \sqrt{(1)} Y_{lm}\chi_+ \right] \\ \end{eqnarray*}$$

And operate on the left hand side.

$$\bgroup\color{black} J^2\psi_{j,m+{1\over 2}} = j(j+1)\hbar^2... ...\alpha Y_{lm}\chi_+ + \beta Y_{\ell,(m+1)}\chi_- \right)\egroup$$

Since the two terms are orthogonal, we can equate the coefficients for each term, giving us two equations. The $Y_{\ell m}\chi_+$ term gives

$$\bgroup\color{black}\alpha j(j+1) = \alpha \left( \ell(\ell +... ...4} + m \right) + \beta \sqrt{ \ell(\ell + 1) -m(m+1)}. \egroup$$

The $Y_{\ell (m+1)}\chi_-$ term gives

$$\bgroup\color{black}\beta j(j+1) = \beta \left( \ell(\ell+1) ... ...(m+1) \right) + \alpha \sqrt{ \ell(\ell + 1) -m(m+1)}. \egroup$$

Collecting $\alpha$ terms on the LHS and $\beta$ terms on the RHS, we get

$$\bgroup\color{black} \left(j(j+1)-\ell(\ell + 1)-{3\over 4}-m\right)\alpha = \sqrt{(\ell-m)(\ell+m+1)}\beta\egroup$$

$$\bgroup\color{black}\sqrt{(\ell-m)(\ell+m+1)}\alpha = \left[j(j+1)-\ell(\ell + 1)-{3\over 4}+(m+1)\right]\beta .\egroup$$

Now we just cross multiply so we have one equation with a common factor of $\alpha\beta$.

$$\bgroup\color{black}(\ell-m)(\ell+m+1)=\left[j(j+1)-\ell(\ell... ...ght] \left[j(j+1)-\ell(\ell+1)-{3\over 4}+(m+1)\right] \egroup$$

While this equation looks like a mess to solve, if we notice the similarity between the LHS and RHS, we can solve it if

$$\bgroup\color{black}\ell=j(j+1)-\ell(\ell+1)-{3\over 4}.\egroup$$

If we look a little more carefully at the LHS, we can see that another solution (which just interchanges the two terms in parentheses) is to replace $\ell$ by $-\ell-1$.

$$\bgroup\color{black}-\ell-1=j(j+1)-\ell(\ell+1)-{3\over 4}.\egroup$$

These are now simple to solve

$$\bgroup\color{black}j(j+1)=\ell(\ell+1) + \ell +{3\over 4} \Rightarrow\qquad j=\ell+{1\over 2} \egroup$$

$$\bgroup\color{black}j(j+1)=\ell(\ell+1) - \ell-1 +{3\over 4} \Rightarrow\qquad j=\ell-{1\over 2} \egroup$$

So these are (again) the two possible values for $j$. We now need to go ahead and find $\alpha$ and $\beta$.

Plugging $j=\ell+{1\over 2}$ into our first equation,

$$\bgroup\color{black}(\ell-m)\alpha=\sqrt{(\ell-m)(\ell+m+1)}\beta\egroup$$

we get the ratio between $\beta$ and $\alpha$. We will normalize the wave function by setting $\alpha^2+\beta^2=1$. So lets get the squares.

$$\bgroup\color{black}\beta^2={(\ell-m)^2\over{(\ell-m)(\ell+m+1)}}\alpha^2 ={(\ell-m)\over{(\ell+m+1)}}\alpha^2\egroup$$

$$\bgroup\color{black}\alpha^2+\beta^2=1 \Rightarrow\qquad {\ell+m+1+\ell-m \over {\ell+m+1}}\alpha^2=1\egroup$$

$$\bgroup\color{black}\alpha=\sqrt{\ell+m+1\over{2\ell +1}} \egroup$$

$$\bgroup\color{black}\beta={\ell-m\over{\sqrt{(\ell-m)(\ell+m+... ...qrt{\ell+m+1\over{2\ell+1}}=\sqrt{\ell-m\over{2\ell+1}} \egroup$$

So we have completed the calculation of the coefficients. We will make use of these in the hydrogen atom, particularly for the anomalous Zeeman effect.

Writing this in the notation of matrix elements, we get

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$$\bgroup\color{black}\left< jm_j\ell s\vert\ell m_\ell s m_s\right> \egroup$$

$$\bgroup\color{black}\left<\left(\ell+{1\over 2}\right)\left(m... ...right.\right> = \alpha = \sqrt{\ell+m+1\over{2\ell+1}} \egroup$$

$$\bgroup\color{black}\left<\left(\ell+{1\over 2}\right)\left(m... ...2}\right.\right> = \beta = \sqrt{\ell-m\over{2\ell+1}} \egroup$$

$$\bgroup\color{black}\left<\left(\ell+{1\over 2}\right)\left(m... ...t\vert \ell m {1\over 2} {-1\over 2}\right.\right> = 0 \egroup$$

$$\bgroup\color{black}\left<\left(\ell+{1\over 2}\right)\left(m... ...ert \ell (m+1) {1\over 2} {1\over 2}\right.\right> = 0 \egroup$$

Similarly

$$\bgroup\color{black}\left<\left(\ell - {1\over 2}\right)\left... ...{1\over 2}\right.\right> = \sqrt{\ell-m\over{2\ell+1}} \egroup$$

$$\bgroup\color{black}\left<\left(\ell - {1\over 2}\right)\left... ...over 2}\right.\right> = -\sqrt{\ell+m+1\over{2\ell+1}} \egroup$$