Prerequisite Orbital angular momentum
, tensor product space
Consider the angular momentum spaces of the two systems, the basis is $ \left\lvert l_1, m_1 \right\rangle $ and $ \left\lvert l_2, m_2 \right\rangle $, where the dimensions of the fixed space of $l_1, l_2$ are $2l_1+1$ and $2l_2+1$ respectively. The angular momentum operators of the two spaces are
\begin{equation}
L_1^2 \qquad L_{1x} \qquad L_{1y} \qquad L_{1z}
\end{equation}
\begin{equation}
L_2^2 \qquad L_{2x} \qquad L_{2y} \qquad L_{2z}
\end{equation}
We use these two spaces to generate a $(2l_1+1)(2l_2+1)$-dimensional tensor product space, with a base of $ \left\lvert l_1, m_1 \right\rangle \otimes \left\lvert l_2, m_2 \right\rangle $ (or denoted as $ \left\lvert l_1, m_1, l_2, m2 \right\rangle $). A set of obvious Complete Set of Commutable Operators (CSCO) in the tensor product space is
\begin{equation}
\{L_{1z}, L_{2z}\}
\end{equation}
In other words, only $m_1, m_2$ two quantum numbers are needed to uniquely determine a basis.
Define the total angular momentum operator of the two systems on the tensor product space as1
\begin{equation} \begin{aligned}
&L_x = L_{1x} \otimes I + I \otimes L_{2x} \\
&L_y = L_{1y} \otimes I + I \otimes L_{2y} \\
&L_z = L_{1z} \otimes I + I \otimes L_{2z}
\end{aligned} \end{equation}
\begin{equation}
L^2 = L_x^2 + L_y^2 + L_z^2
\end{equation}
It can be proved that the newly added transaction relationship is
\begin{equation}
[{L^2},{L_z}] = [{L^2},{L_1^2}] = [{L^2},{L_2^2}] =
[{L_z},{L_{1z}}] = [{L_z},{L_{2z}}] = 0
\end{equation}
From this, we can get another set of CSCO as
\begin{equation}
\{L^2, L_z\}
\end{equation}
That is, we can think of another set of basis for the tensor product space. Let the eigenvalues (quantum numbers) of the two operators be $L$ and $M$ respectively, this set of basis can be recorded as $ \left\lvert L, M \right\rangle $. First of all, from the exchange relationship, $ \left\lvert l_1, m_1, l_2, m_2 \right\rangle $ is already the eigenvector of $L_z$, and each eigenvalue $M$ corresponds to an eigensubspace (because of degeneration), which is called “$M$ subspace” in the following. The dimension of the subspace $N_M$ is the number of different $m_1,m_2$ combinations satisfying $m_1 + m_2 = M$ (
fig. 1 left). For example, when $M = -1/2$ is in the figure, $N_M = 4$.
Fig. 1：Add angular momentum to two particles with angular quantum numbers $l_1 = 3/2$ and $l_2 = 2$ respectively. Each square in the figure represents a basis in the tensor product space. In the figure on the left, each base is in the form of $ \left\lvert l_1, m_1, l_2, m_2 \right\rangle $, and each base crossed by a diagonal line is opened into a $M$ subspace. On the right, each base is in the form of $ \left\lvert L, M \right\rangle $. The basis transformation is performed independently in each $M$ subspace, that is, a diagonal line in the left image becomes a row in the right image.
\begin{equation}
N_M =
\begin{cases}
l_1 + l_2 - \left\lvert M \right\rvert + 1 &( \left\lvert M \right\rvert > \left\lvert l_1 - l_2 \right\rvert ) \\
2\min\{l_1, l_2\} + 1 &( \left\lvert M \right\rvert \leqslant \left\lvert l_1 - l_2 \right\rvert )
\end{cases}
\end{equation}
The maximum value of $m_1$ in each set of bases in the $M$ subspace is
\begin{equation}
\max \left\{m_1 \right\} =
\begin{cases}
l_1 \quad &(M \geqslant l_1 - l_2) \\
l_2 + M &(M < l_1 - l_2)
\end{cases}
\end{equation}
For example, when $M = 1/2$ is in the figure, the maximum value of $m_1$ is $3/2$.
The order of the bases in linear algebra is very important, and the order will affect the values of coordinates and matrices. We usually order $ \left\lvert l_1, m_1, l_2, m_2 \right\rangle $ ($m_2 = M - m_1$) in descending order of $m_1$, and $ \left\lvert L, M \right\rangle $ in descending order of $L$.
It can be proved that in each $M$ subspace, $ \left\lvert L, M \right\rangle $ takes $L = l_1 + l_2, l_1 + l_2 - 1,\dots$ (total $N_M$). The minimum value of $L$ in all $M$ subspaces is $ \left\lvert l_1 - l_2 \right\rvert $, which is obtained when $ \left\lvert M \right\rvert \leqslant \left\lvert l_1 - l_2 \right\rvert $ (the last column on the right side of fig. 1 ), so the range of $L$ in the entire tensor product space is
\begin{equation}
\left\lvert l_1 - l_2 \right\rvert \leqslant L \leqslant l_1 + l_2
\end{equation}
If the $L$ of two particles are known, how much might their respective $l_1, l_2$ be? According to eq. 10 , fig. 2 can be drawn. For example, when $L = 0$ is $0053$, only $l_1 = l_2$ can be allowed.
Fig. 2：For a given $L$, the shadow part is allowed $l_1, l_2$, and the shadow extends to the upper right indefinitely.
CG coefficient
Since each $M$ subspace can be expanded by $N_M$ orthogonal normalized $ \left\lvert l_1, m_1, l_2, m_2 \right\rangle $ basis or $ \left\lvert L,M \right\rangle $ basis, it is necessary to discuss the unitary transformation matrix between them. The matrix element $ \left\langle l_1, m_1, l_2, m_2 \middle| L, M \right\rangle $ is called Clebsch-Gordan coefficient or simply CG coefficient. See "Clebsch-Gordan coefficient " for details.
\begin{equation}
\left\lvert L, M \right\rangle = \sum_{m_1, m_2} \left\langle l_1, m_1, l_2, m_2 \middle| L, M \right\rangle \cdot \left\lvert l_1, m_1, l_2, m_2 \right\rangle
\end{equation}
\begin{equation}
\left\lvert l_1, m_1, l_2, m_2 \right\rangle = \sum_{L} \left\langle L, M \middle| l_1, m_1, l_2, m_2 \right\rangle \cdot \left\lvert L, M \right\rangle
\end{equation}
Note that
eq. 11 needs to meet $M = m_1 + m_2$ to sum $m_1,m_2$, so the above two formulas are sums of $N_M$ terms. The CG coefficient can also be denoted by the other two symbols as
\begin{equation}
\begin{bmatrix}l_1 & l_2 & L\\ m_1 & m_2 & M\end{bmatrix} = C_{l_1, m_1, l_2, m_2}^{L, M} = \left\langle l_1, m_1, l_2, m_2 \middle| L, M \right\rangle
\end{equation}
To obtain the CG coefficient, we only need to diagonalize the $L^2$ matrix under the $ \left\lvert l_1, m_1, l_2, m_2 \right\rangle $ base in each $M$ subspace
\begin{equation}
L^2 = L_1^2 + L_2^2 + 2(L_{1x} L_{2x} + L_{1y} L_{2y} + L_{1z} L_{2z})
\end{equation}
Only $L_{1x} L_{2x} + L_{1y} L_{2y}$ is not a diagonal matrix. It can be expressed as
\begin{equation}
2 (L_{1x} L_{2x} + L_{1y} L_{2y} ) = L_{1+} L_{2-} + L_{1-} L_{2+}
\end{equation}
So based on $ \left\lvert l_1, m_1, l_2, m_2 \right\rangle $, the matrix elements of $L^2$ matrix are
\begin{equation} \begin{aligned}
&\quad \left\langle l_1, m'_1, l_2, m'_2 \right\rvert L^2 \left\lvert l_1, m_1, l_2, m_2 \right\rangle = \hbar ^2 \times\\
& \left[ \begin{aligned}
&\delta_{m'_1, m_1} \delta_{m'_2, m_2} [l_1(l_1 + 1) + l_2(l_2 + 1) + 2 m_1 m_2] \\
+ &\delta_{m'_1, m_1 + 1} \delta_{m'_2, m_2 - 1} \sqrt{l_1 (l_1 + 1) - m_1(m_1 + 1)} \sqrt{l_2 (l_2 + 1) - m_2(m_2 - 1)}\\
+ &\delta_{m'_1, m_1 - 1} \delta_{m'_2, m_2 + 1} \sqrt{l_1 (l_1 + 1) - m_1(m_1 - 1)} \sqrt{l_2 (l_2 + 1) - m_2(m_2 + 1)}\end{aligned} \right]
\end{aligned} \end{equation}
It can be seen that this is a three-diagonal matrix, and its $N_M$ orthogonal normalized eigen-column vectors are the coordinates of $ \left\lvert L, M \right\rangle $ ($L = l_1 + l_2, l_1 + l_2 - 1, \dots$) on the basis of $ \left\lvert l_1, m_1, l_2, m_2 \right\rangle $. Arranging these column vectors in the base order from left to right is $ \boldsymbol{\mathbf{U}} _M$.
Raising and lowering operator method
How to get the eigenvalue and eigenvector from the general form of the $L^2$ matrix eq. 16 ? A clever way is to start with the only two non-degenerate $M$ spaces (refer to [14] ), which are obtained by eq. 14 and eq. 15
\begin{equation}
\begin{aligned}
L^2 \left\lvert l_1, \pm l_1, l_2, \pm l_2 \right\rangle &= \hbar^2l_1(l_1 + 1) + \hbar^2l_2(l_2 + 1) + 2\hbar^2l_1 l_2\\
&= \hbar^2(l_1 + l_2)(l_1 + l_2 + 1)
\end{aligned}
\end{equation}
So let the quantum number of $L^2$ be $L$ and the eigenvalue is $\hbar^2L(L+1)$, then $ \left\lvert l_1, m_1 = \pm l_1, l_2, m_2 = \pm l_2 \right\rangle $ corresponds to $L = l_1 + l_2$. and so
\begin{equation}
\left\lvert l_1, m_1 = \pm l_1, l_2, m_2 = \pm l_2 \right\rangle = \left\lvert L = l_1 + l_2, M = \pm (l_1 + l_2) \right\rangle
\end{equation}
Next, use the descending operator $L_- = L_-^{(1)} + L_-^{(2)}$ of $L_z$ for $ \left\lvert l_1, l_1, l_2, l_2 \right\rangle $ to get the first basis from the left of the second line of fig. 1 (no normalization is considered for now)
\begin{equation}
\left\lvert L = l_1 + l_2, M = l_1 + l_2 - 1 \right\rangle = L_- \left\lvert l_1, l_1, l_2, l_2 \right\rangle = \left\lvert l_1, l_1 - 1, l_2, l_2 \right\rangle + \left\lvert l_1, l_1, l_2, l_2 - 1 \right\rangle
\end{equation}
But the $M = l_1 + l_2 - 1$ subspace is a $2$-dimensional subspace, so we can easily find another orthogonal normalization basis $ \left\lvert l_1, l_1 - 1, l_2, l_2 \right\rangle + \left\lvert l_1, l_1, l_2, l_2 - 1 \right\rangle $, and this is $ \left\lvert L = l_1 + l_2 - 1, M = l_1 + l_2 - 1 \right\rangle $.
We can use the subspace operator again on the two bases in the $M = l_1 + l_2 - 1$ subspace to get the first two bases $ \left\lvert l_1 + l_2, M \right\rangle $ and $ \left\lvert l_1 + l_2 - 1, M \right\rangle $ in $M = l_1 + l_2 - 2$. But this is a three-dimensional subspace, so the third orthogonal basis $ \left\lvert l_1 + l_2 - 2, M \right\rangle $ can be found. By analogy, all $(2l_1+1)(2l_2+1)$ bases of $ \left\lvert L,M \right\rangle $ can be found.
1. ^ Note that sometimes for convenience we will directly write it in the form of $L_x = L_{1x} + L_{2x}$. Strictly speaking, this is incorrect, because the operator on the left side of the equal sign The space is the tensor product of the space where the two operators on the right side of the equal sign are located, not the addition of the two operators in the same space.
