Adding any \bgroup\color{black}$\ell$\egroup plus spin \bgroup\color{black}${1\over 2}$\egroup.

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

\begin{displaymath}\bgroup\color{black} J^2\psi_{jm_j}=j(j+1)\hbar^2\psi_{jm_j} \egroup\end{displaymath}

We choose to write the the quantum number \bgroup\color{black}$m_j$\egroup as \bgroup\color{black}$m+{1\over 2}$\egroup. This is really just the defintion of the dummy variable \bgroup\color{black}$m$\egroup. (Other choices would have been possible.)

The z component of the total angular momentum is just the sum of the z components from the orbital and the spin.

\begin{displaymath}\bgroup\color{black} m_j=m_l+m_s \egroup\end{displaymath}

There are only two product states which have the right \bgroup\color{black}$m_j=m+{1\over 2}$\egroup. If the spin is up we need \bgroup\color{black}$Y_{\ell m}$\egroup and if the spin is down, \bgroup\color{black}$Y_{\ell (m+1)}$\egroup.

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


We will find the coefficients \bgroup\color{black}$\alpha$\egroup and \bgroup\color{black}$\beta$\egroup so that \bgroup\color{black}$\psi$\egroup will be an eigenstate of

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

So operate on the right hand side with \bgroup\color{black}$J^2$\egroup.

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

And operate on the left hand side.

\begin{displaymath}\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\end{displaymath}

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

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

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

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

Collecting \bgroup\color{black}$\alpha$\egroup terms on the LHS and \bgroup\color{black}$\beta$\egroup terms on the RHS, we get two equations.

\left(j(j+1)-\ell(\ell + 1)-{3\over 4}-m\right)\alpha
&=& \sq...
&=& \left[j(j+1)-\ell(\ell + 1)-{3\over 4}+(m+1)\right]\beta

Now we just cross multiply so we have one equation with a common factor of \bgroup\color{black}$\alpha\beta$\egroup.

\left[j(j+1)-\ell(\ell+1)-{3\over 4}+(m+1)\right] \egroup\end{displaymath}

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

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

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 \bgroup\color{black}$\ell$\egroup by \bgroup\color{black}$-\ell-1$\egroup.

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

These are now simple to solve

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

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

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

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


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

\begin{displaymath}\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\end{displaymath}

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

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

\begin{displaymath}\bgroup\color{black}\beta=\sqrt{\ell-m\over \ell+m+1}
\sqrt{\ell+m+1\over{2\ell+1}}=\sqrt{\ell-m\over{2\ell+1}} \egroup\end{displaymath}

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 or Clebsch-Gordan coefficients of the form,

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

we get.


\left<\left(\ell+{1\over 2}\right)\left(m+{1\over 2}\right)\el...
...left\vert \ell (m+1) {1\over 2} {1\over 2}\right.\right>
&=& 0


\left<\left(\ell - {1\over 2}\right)\left(m+{1\over 2}\right)\...
...} {-1\over 2}\right.\right>
&=& -\sqrt{\ell+m+1\over{2\ell+1}}

Jim Branson 2013-04-22