Clebsch-Gordan Series for SU(2) Tensor Products

The Problem of Combining Angular Momenta

In classical mechanics, combining two angular momenta, ${\vec{J}}_1$ and ${\vec{J}}_2$, is straightforward: the total angular momentum is simply their vector sum, ${\vec{J}}_{total}={\vec{J}}_1+{\vec{J}}_2$. The magnitude and direction of ${\vec{J}}_{total}$ can be found using standard vector addition.

In quantum mechanics, this simple picture breaks down. The components of an angular momentum operator do not commute with each other (e.g., $[J_x,J_y]=i\hslash J_z$). This is a manifestation of the uncertainty principle: we cannot simultaneously know the precise values of all three components of an angular momentum vector. Consequently, we cannot simply add the components of ${\vec{J}}_1$ and ${\vec{J}}_2$ as if they were classical vectors. We need a new formalism that respects the rules of quantum mechanics.

The video below provides an excellent introduction, contrasting the classical approach with the operator-based formalism required in quantum mechanics.

The Tensor Product Space

Let's consider two quantum systems, 1 and 2, described by Hilbert spaces $H_1$ and $H_2$. Suppose system 1 is in a state $|j_1,m_1⟩$ from an irrep of spin $j_1$, and system 2 is in a state $|j_2,m_2⟩$ from an irrep of spin $j_2$. The composite system is described by a state in the tensor product space $H=H_1\otimes H_2$.

A basis for this new space can be formed by taking all possible tensor products of the basis vectors from $H_1$ and $H_2$. These basis states are written as $|j_1,m_1⟩\otimes |j_2,m_2⟩$, or more compactly as $|j_1,m_1;j_2,m_2⟩$. The dimension of this composite space is the product of the individual dimensions: $\dim (H)=(2j_1+1)(2j_2+1)$. This basis is often called the uncoupled basis because it's built from states where the individual angular momenta of the subsystems are well-defined.

Now, how do operators corresponding to physical observables act on this new space?

1. An operator ${\hat{O}}_1$ that acts only on system 1 becomes ${\hat{O}}_1\otimes I_2$, where $I_2$ is the identity on $H_2$.
2. An operator ${\hat{O}}_2$ that acts only on system 2 becomes $I_1\otimes {\hat{O}}_2$.
3. The total angular momentum operator $\vec{J}$ is the sum of the individual operators: $$\vec{J}={\vec{J}}_1+{\vec{J}}_2\equiv {\vec{J}}_1\otimes I_2+I_1\otimes {\vec{J}}_2$$

This rule for adding operators in a tensor product space might seem non-intuitive. It arises directly from the way Lie algebra generators combine when we take the tensor product of the Lie group representations they generate. An infinitesimal rotation on the composite system is the product of the infinitesimal rotations on the subsystems. The rule for the generators follows from taking the derivative of this product, as the following video explains masterfully.

Decomposing the Product Space: A Canonical Example

The uncoupled basis states $|j_1,m_1;j_2,m_2⟩$ are eigenstates of ${\hat{J}}_1^2,{\hat{J}}_{1z},{\hat{J}}_2^2,{\hat{J}}_{2z}$. However, they are generally not eigenstates of the total angular momentum squared, ${\hat{J}}^2=({\vec{J}}_1+{\vec{J}}_2{)}^2$.

Since the composite system is also a physical system with rotational symmetry, its states must also organize themselves into irreducible representations of SU(2). This means there must exist a different basis, the coupled basis, whose states are simultaneous eigenstates of ${\hat{J}}^2$ and ${\hat{J}}_z$, labeled as $|J,M⟩$.

Finding this new basis is equivalent to decomposing the reducible tensor product representation into a direct sum of irreducible representations. Let's work through the foundational example: combining two spin-1/2 particles.

The uncoupled space $H=H_{1/2}\otimes H_{1/2}$ has dimension $2\times 2=4$. The basis states are:

Using up/down arrow notation: $|\uparrow \uparrow ⟩,|\uparrow \downarrow ⟩,|\downarrow \uparrow ⟩,|\downarrow \downarrow ⟩$.

The total $J_z$ operator is $J_z=J_{1z}+J_{2z}$. Its eigenvalues $M$ are just the sum of the individual $m_1$ and $m_2$:

• For $|\uparrow \uparrow ⟩$: $M=\frac{1}{2}+\frac{1}{2}=1$
• For $|\uparrow \downarrow ⟩$: $M=\frac{1}{2}-\frac{1}{2}=0$
• For $|\downarrow \uparrow ⟩$: $M=-\frac{1}{2}+\frac{1}{2}=0$
• For $|\downarrow \downarrow ⟩$: $M=-\frac{1}{2}-\frac{1}{2}=-1$

We see that we have one state with $M=1$, one with $M=-1$, and two with $M=0$. The state with the highest $M$ value, $|\uparrow \uparrow ⟩$, must be a highest-weight state. Since $M_{max}=1$, this state must belong to an irrep with total spin $J=1$. This is the state $|J=1,M=1⟩$.

We can generate the rest of the $J=1$ multiplet by applying the lowering operator $J_{-}=J_{1-}+J_{2-}$. This procedure, along with the identification of the remaining state, is shown in detail in the next segment of the eigenchris video.

As the video demonstrates, the four uncoupled states recombine into two separate, irreducible multiplets:

1. A spin-1 triplet ($J=1$) with three states corresponding to $M=1,0,-1$.
2. A spin-0 singlet ($J=0$) with one state corresponding to $M=0$.

The tensor product space decomposes into a direct sum of these irreducible subspaces:

In the language of representations, we write this decomposition as:

Or, using dimensions:

The General Rule: The Clebsch-Gordan Series

This decomposition is an example of a general rule. When we take the tensor product of two SU(2) irreps with spins $j_1$ and $j_2$, the resulting reducible representation decomposes into a direct sum of irreps with total spin $J$ ranging from $|j_1-j_2|$ to $j_1+j_2$ in integer steps.

This is the Clebsch-Gordan series:

The following excerpt from Peter Woit's notes provides a concise, formal statement of this rule.

Let's check this formula with some examples:

• Two spin-1/2 particles: $j_1=1/2,j_2=1/2$. Total spin $J$ ranges from $|\frac{1}{2}-\frac{1}{2}|=0$ to $\frac{1}{2}+\frac{1}{2}=1$. So, $J=0,1$. The decomposition is $1/2\otimes 1/2=1\oplus 0$. This matches our detailed calculation.
• A spin-1 particle and a spin-1/2 particle: $j_1=1,j_2=1/2$. Total spin $J$ ranges from $|1-\frac{1}{2}|=\frac{1}{2}$ to $1+\frac{1}{2}=\frac{3}{2}$. So, $J=1/2,3/2$. The decomposition is $1\otimes 1/2=3/2\oplus 1/2$. Let's check dimensions: $(2\cdot 1+1)(2\cdot \frac{1}{2}+1)=3\times 2=6$. On the right side: $(2\cdot \frac{3}{2}+1)+(2\cdot \frac{1}{2}+1)=4+2=6$. The dimensions match.

Visualizing the Decomposition

Since you have an interest in computational work and modeling, you'll appreciate that there is a wonderfully simple graphical algorithm for finding these decompositions, which connects to the weight diagrams we will study for more complex algebras like $\mathfrak{su}(3)$.

Imagine the weight diagram for an irrep $j$ is just a set of nodes on a line at positions $-j,-j+1,\dots ,j$. To find the decomposition of $j_1\otimes j_2$, you take the weight diagram of $j_1$ and "add" the weight diagram of $j_2$ to each of its nodes. You then count how many nodes land on each position. The result gives you the structure of the decomposed irreps.

The image below shows the decomposition for $1\otimes 1=2\oplus 1\oplus 0$ using a more abstract tensor notation, which is another way to visualize how the nine components of the product space regroup into multiplets of dimension 5 ($J=2$), 3 ($J=1$), and 1 ($J=0$).

Conclusion

In this lesson, we have developed the formal procedure for combining angular momenta in quantum mechanics. This is a cornerstone of quantum theory with vast applications, from atomic and molecular physics (understanding term symbols and spectral lines) to particle physics (combining quark spins to form hadrons) and condensed matter physics (describing magnetic interactions and collective excitations).

The key step is to form the tensor product of the individual state spaces and then decompose this product space into a direct sum of irreducible representations. This decomposition is governed by the Clebsch-Gordan series, a simple and elegant rule that dictates the possible values of the total angular momentum $J$.

Key Takeaways:

• Combining quantum systems requires forming the tensor product of their Hilbert spaces, $H=H_1\otimes H_2$.
• Lie algebra generators for the composite system are formed by the rule $J_{total}=J_1\otimes I_2+I_1\otimes J_2$.
• The tensor product of two SU(2) irreps is generally reducible.
• This reducible representation decomposes into a direct sum of irreps according to the Clebsch-Gordan series:
  $j_1\otimes j_2={⨁}_{J=|j_1-j_2|}^{j_1+j_2}J$
• This decomposition corresponds physically to finding the possible values for the total angular momentum $J$ of the combined system.

We have now answered which irreps appear in the decomposition. The next natural question is, how are the new basis states $|J,M⟩$ explicitly constructed from the old basis states $|j_1,m_1;j_2,m_2⟩$? The coefficients of this change-of-basis transformation are the famous Clebsch-Gordan coefficients. In our next lesson, we will learn how to calculate them.