I Addition of angular momenta

[alebruna]

TL;DR Summary

Theoretical exercise on the addition of an angular momentum operator and a spin operator, derivation of eigenvalues and basis. Normalization of the eigenstates.

While studying for my exam I came across the addition of two angular momenta, in the simple case of $J=L+S$ where $L$ is the angular momentum operator and $S$ is the spin (in this case a fermion with $s=1/2$). I have some doubts on the derivation of the basis and the eigenvalues.

From the start we know that $J=L+S$ and by this $J^2=(L+S)^2$. We should start by the commutator $[J_m,J_n]$

$[J_m,J_n]=[L_m+S_m,L_n+S_n]=[L_m,L_n]+[S_m,S_n]=i\hbar\epsilon_{jmn}L_j+i\hbar\epsilon_{jmn}S_j=i\hbar\epsilon_{jmn}(L_j+S_j)=$
$=i\hbar\epsilon_{jmn}J_j$

that shows that $J$ is still an angular momentum. By this we know that all the properties holding for both $L$ and $S$ apply also for $J$, for example: $[J,J^2]=0$.
Now we can study the commutators $[J_m,L_n],\ [J_m,L^2],\ [J_m,S_n],\ [J_m,S^2],\ [J^2, L_n],\ [J^2,S_n],\ [J^2,L^2],$$\ [J^2,S^2]$:

$[J_m,L_n]=[L_m+S_m,L_n]=[L_m,L_n]\neq 0$
$[J_m,L^2]=\sum_n [J_m, L_n^2]=\sum_n\left( L_n[J_m,L_n]+[J_m,L_n]L_n\right)=\sum_n (i\hbar\epsilon_{jmn}L_n L_j+i\hbar\epsilon_{jmn}L_j L_n)=...=0$
$[J_m,S_n]=...\neq 0$
$[J_m,S^2]=...=0$
$[J^2,L_n]=[L^2+S^2+2L\cdot S,L_n]=[L^2,L_n]+\sum_m[L_m,L_n]=0+i\hbar\epsilon_{jmn}L_j\neq 0$
$[J^2,S_n]=...\neq 0$
$[J^2,L^2]=[L^2+S^2+2L\cdot S,L^2]=[L,L^2]\cdot S=0$
$[J^2,L^2]=...=0$

Since $J^2,\ J_m,\ L^2,\ S^2$ all commute, for the third theorem of hermitian operators they feature a common base of eigenstates. Since $J$ has the form of an angular momentum we can assume that $J_3$ is associated to an eigenvalue $\hbar m_j$ and $J^2$ is associated to an eigenvalue $\hbar j(j+1)$.
Hence the basis associated to operators $J_m,\ J^2,\ L^2,\ S^2$ is in Dirac notation: $|j, m_j, l, s\rangle$.
About the properties of the eigenvalues:

$J_3\left|j, m_j, l, s\right\rangle=(L_3+S_3)\left|j, m_j, l, s\right\rangle=(\hbar m+\hbar m_s)\left|j, m_j, l, s\right\rangle=\hbar m_j \left|j, m_j, l, s\right\rangle \rightarrow m_j=m+m_s$
$J^2\left|j, m_j, l, s\right\rangle=\hbar^2j(j+1)\left|j, m_j, l, s\right\rangle \ \text{with}\ j>0 \\$
$L^2\left|j, m_j, l, s\right\rangle=\hbar^2l(l+1)\left|j, m_j, l, s\right\rangle \\$
$S^2\left|j, m_j, l, s\right\rangle=\hbar^2s(s+1)\left|j, m_j, l, s\right\rangle \\ \rightarrow l-s< j<l+s$
also $-j\leq m_j\leq j$

We can see that the new basis of eigenvalues features a degeneracy for $m+1,\ s=1/2$ and $m,\ s=-1/2$ this means that the new basis can be written as
$$
\left|j, m_j, l, s\right\rangle=\sum_{m,m_s}^\ast C(m,m_s)\left|l,m\right\rangle \left|s,m_s\right\rangle
$$
The superscript * means that $m+m_s$ is constant.
Using spectrum generating algebra we can define $J_+=L_++S_+$ and $J_-=L_--S_-$.

$J_+\left|l,l\right\rangle\left|s,\frac{1}{2}\right\rangle=0 \\$
$J_-\left|l,-l\right\rangle\left|s,-\frac{1}{2}\right\rangle=0 \\$

Taking the two states $j=l+1/2$ and $j=l-1/2$ can give us the coefficients for the double series.
$$
\begin{cases}
\left|l-1/2, m_j, l, s\right\rangle=\alpha_-\left|l,m_j-1/2\right\rangle \left|s,1/2\right\rangle+\beta_-\left|l,m_j+1/2\right\rangle \left|s,-1/2\right\rangle \\
\left|l+1/2, m_j, l, s\right\rangle=\alpha_+\left|l,m_j-1/2\right\rangle \left|s,1/2\right\rangle+\beta_+\left|l,m_j+1/2\right\rangle \left|s,-1/2\right\rangle
\end{cases}
$$
Normalizing gets us that
$$
\alpha_\pm=\pm\sqrt{\frac{l\pm m_j+1/2}{2l+1}}=\pm\beta_\mp
$$
This should conclude the derivation. Is this exercise correct? If not can anyone point out how this is wrong and why?
Some source material I found online are linked below.

 

[member 666967]

\textbf{1. Angular Momentum and Spin Operators:}

You correctly started from the fact that you have two angular momentum operators:

\[
\mathbf{L} \quad \text{(orbital angular momentum)}, \quad \mathbf{S} \quad \text{(spin angular momentum)},
\]

and you define the total angular momentum operator as

\[
\mathbf{J} = \mathbf{L} + \mathbf{S}.
\]

Because both \(\mathbf{L}\) and \(\mathbf{S}\) satisfy the angular momentum commutation relations,

\[
[J_i, J_j] = i \hbar \epsilon_{ijk} J_k,
\]

it follows that \(\mathbf{J}\) also behaves like an angular momentum operator.

---

\textbf{2. Commutation and Common Eigenbasis:}

You mentioned studying the commutators involving \(J^2\), \(L^2\), and \(S^2\):

\[
[J^2, L^2] = 0, \quad [J^2, S^2] = 0, \quad [L^2, S^2] = 0.
\]

Since these operators commute pairwise, they can share a common eigenbasis. This is essential in constructing the joint eigenstates of these observables.

---

\textbf{3. Quantum Numbers and Dirac Notation:}

The standard notation for the joint eigenstates of \(J^2\) and \(J_z\) is

\[
|j, m_j\rangle,
\]

where \(j\) is the total angular momentum quantum number and \(m_j\) is its projection along the \(z\)-axis.

Similarly, the eigenstates of \(\mathbf{L}^2\) and \(L_z\) are

\[
|l, m_l\rangle,
\]

and for \(\mathbf{S}^2\) and \(S_z\),

\[
|s, m_s\rangle.
\]

For a spin-\(\frac{1}{2}\) particle (like a fermion), \(s = \frac{1}{2}\) and \(m_s = \pm \frac{1}{2}\).

---

\textbf{4. Addition of Angular Momenta:}

When combining \(\mathbf{L}\) and \(\mathbf{S}\), the possible total angular momentum quantum numbers \(j\) satisfy

\[
j = |l - s|, |l - s| + 1, \ldots, l + s.
\]

For example, if \(l = 1\) and \(s = \frac{1}{2}\), then \(j\) can be \(\frac{1}{2}\) or \(\frac{3}{2}\).

This gives the \emph{degeneracy} you mentioned: multiple \(m_j\) values correspond to each \(j\).

---

\textbf{5. Basis Transformation and Clebsch-Gordan Coefficients:}

The original basis is the \emph{uncoupled} basis:

\[
|l, m_l\rangle |s, m_s\rangle,
\]

where angular momenta are considered separately. The new basis after addition is the \emph{coupled} basis:

\[
|j, m_j; l, s\rangle,
\]

which are eigenstates of \(\mathbf{J}^2\) and \(J_z\).

The transformation between these two bases is given by the Clebsch-Gordan coefficients \(C^{j,m_j}_{l,m_l;s,m_s}\):

\[
|j, m_j; l, s\rangle = \sum_{m_l, m_s} C^{j,m_j}_{l,m_l;s,m_s} |l, m_l\rangle |s, m_s\rangle.
\]

These coefficients can be found in tables or calculated using standard recursion relations.

---

\textbf{6. Spectrum Generating Algebra and Ladder Operators:}

You also correctly introduced the ladder operators:

\[
J_\pm = J_x \pm i J_y,
\]

which raise or lower the \(m_j\) value by one:

\[
J_\pm |j, m_j\rangle = \hbar \sqrt{j(j+1) - m_j(m_j \pm 1)} |j, m_j \pm 1\rangle.
\]

The use of ladder operators is fundamental in constructing the entire multiplet of states starting from the highest weight state \( |j, j\rangle \).

---

\textbf{7. Normalization:}

The normalization condition

\[
\langle j, m_j; l, s | j, m_j; l, s \rangle = 1
\]

ensures that these states form an orthonormal basis. When constructing states via Clebsch-Gordan coefficients, the coefficients themselves satisfy orthogonality and normalization relations ensuring this.

---

\textbf{8. Is the Exercise Correct?}

Your exercise and derivation are essentially correct and follow the standard textbook approach for adding angular momenta in quantum mechanics. The key points are:

\begin{itemize}
\item Verify the commutation relations hold for \(\mathbf{J}\).
\item Identify the allowed \(j\) values.
\item Understand the transformation between uncoupled and coupled bases via Clebsch-Gordan coefficients.
\item Use ladder operators to generate all states.
\item Normalize states properly.
\end{itemize}

Your summary touches on all of these points appropriately.

---

\textbf{Additional Suggestions:}

If you want to deepen your understanding, I recommend:

\begin{itemize}
\item Reviewing a detailed example, e.g., adding \(l=1\) and \(s=1/2\).
\item Studying explicit Clebsch-Gordan tables.
\item Working through problems that require expanding states in both bases.
\item Consulting standard textbooks such as Sakurai's \emph{Modern Quantum Mechanics} or Cohen-Tannoudji's \emph{Quantum Mechanics}.
\end{itemize}