Character Decomposition of Tensor Products

Welcome to the seventh lesson in our module on Representation Theory. In our previous session, you learned how to construct character tables from the ground up, using the powerful orthogonality relations and other structural rules. This skill allows us to map out the fundamental building blocks—the irreducible representations—for any finite symmetry group.

Today, we will address a scenario that is central to physics: describing a system composed of multiple parts. When we combine two quantum systems, for example, the new state space is the tensor product of the individual spaces. We will explore how the symmetry representations combine in this process and, crucially, how to decompose the resulting—and generally reducible—representation back into its irreducible components. This technique is the foundation for understanding everything from the addition of angular momentum in atoms to selection rules for interactions in condensed matter and particle physics.

1. The Tensor Product of Representations

Imagine a composite quantum system, such as two electrons in a molecule or an elementary particle with both spatial and spin degrees of freedom. If a symmetry group $G$ acts on each component part, how does it act on the system as a whole? The mathematical framework for this is the tensor product (or direct product) of representations.

Let's say one part of our system transforms according to the representation $D^{(μ)}$ in a vector space $V$ , and the other part transforms according to $D^{(ν)}$ in a space $W$ . The combined system lives in the tensor product space $V \otimes W$ . A group operation $g \in G$ acts on a basis vector $v_{i} \otimes w_{j}$ of this new space by acting on each component simultaneously:

g \cdot (v_{i} \otimes w_{j}) = (g \cdot v_{i}) \otimes (g \cdot w_{j})

This defines a new representation, $D^{(μ \otimes ν)}$ , whose matrices are the Kronecker product of the individual representation matrices: $D^{(μ \otimes ν)} (g) = D^{(μ)} (g) \otimes D^{(ν)} (g)$ . The dimension of this new representation is the product of the individual dimensions, $d_{μ \otimes ν} = d_{μ} \times d_{ν}$ .

Fortunately, we don't need to work with these large matrices. The character of the tensor product representation follows a remarkably simple rule: it is the product of the individual characters.

χ^{(μ \otimes ν)} (g) = χ^{(μ)} (g) \times χ^{(ν)} (g)

This is because the trace of a Kronecker product of matrices is the product of their traces. This simple formula is our primary tool for this lesson.

For a formal definition and a discussion of the physical meaning of direct sums versus direct products, the following notes are an excellent resource.

[PDF] 5 Irreducible representations - Xie Chen

These lecture notes by Prof. Xie Chen from Caltech clearly distinguish between direct sum and direct product representations, which is a common point of confusion.

First, read the section on physical meaning to solidify your intuition. Then, read the section on the direct product. Focus on the definition of the product representation and its character, as given in equation (23). The example for D_3 anticipates the calculation we will perform shortly.

2. Decomposing the Product: The Clebsch-Gordan Series

The tensor product of two irreducible representations is, in general, reducible. Our main task is to decompose this product representation into a direct sum of irreps. This decomposition is often called a Clebsch-Gordan series for finite groups.

D^{(μ)} \otimes D^{(ν)} = k ⨁ c_{k} D^{(k)} or, in terms of characters, χ^{(μ)} \times χ^{(ν)} = k \sum c_{k} χ^{(k)}

The coefficients $c_{k}$ tell us how many times the irrep $D^{(k)}$ appears in the product.

The method for finding these coefficients is exactly the one we mastered in the fifth lesson for decomposing any reducible representation. The only new step is first calculating the character of our product representation.

The procedure is as follows:

Using the group's character table, identify the character vectors $χ^{(μ)}$ and $χ^{(ν)}$ for the two irreps you are multiplying.
Compute the character of the product representation, $χ^{(prod)}$ , by taking the element-wise product of the two characters: $χ^{(prod)} (C) = χ^{(μ)} (C) \times χ^{(ν)} (C)$ for each conjugacy class $C$ .
Apply the standard reduction formula to find the multiplicity $c_{k}$ of each irrep $χ^{(k)}$ in $χ^{(prod)}$ : $c_{k} = \frac{1}{N _{G}} C \sum N_{C} χ^{(k)} (C)^{*} χ^{(prod)} (C) = \frac{1}{N _{G}} C \sum N_{C} χ^{(k)} (C)^{*} χ^{(μ)} (C) χ^{(ν)} (C)$ where $N_{G}$ is the order of the group and $N_{C}$ is the number of elements in class $C$ .

The following resource provides the formal definition and formula for these coefficients.

[PDF] Representations of crystallographic groups

These slides from a summer school on crystallography offer a concise summary of the relevant theorem.

Please read the section titled "Reduction of direct products into irreps". It formally defines the Clebsch-Gordan coefficients c_{ij}^k and gives the explicit formula for their computation using characters, which is exactly the formula we have stated above.

3. Worked Example: $E \otimes E$ in the $C_{3 v}$ Group

Let's apply this method to the group $C_{3 v}$ . In the previous lesson, we constructed its character table:

$C_{3 v}$	$E$	$2 C_{3}$	$3 σ_{v}$
$A_{1}$	1	1	1
$A_{2}$	1	1	-1
$E$	2	-1	0

We will now compute the decomposition of the tensor product of the two-dimensional representation $E$ with itself: $E \otimes E$ . This would correspond, for instance, to a system of two identical particles whose states transform according to the $E$ irrep.

Step 1: Calculate the character of the product representation $Γ_{E \otimes E}$ .

We simply multiply the character row for $E$ with itself:

$χ^{(prod)} (E) = χ^{(E)} (E) \times χ^{(E)} (E) = 2 \times 2 = 4$
$χ^{(prod)} (C_{3}) = χ^{(E)} (C_{3}) \times χ^{(E)} (C_{3}) = (- 1) \times (- 1) = 1$
$χ^{(prod)} (σ_{v}) = χ^{(E)} (σ_{v}) \times χ^{(E)} (σ_{v}) = 0 \times 0 = 0$

So, the character vector for our $4 \times 4 = 16$ -dimensional reducible representation is $χ^{(E \otimes E)} = (4, 1, 0)$ .

Step 2: Decompose $χ^{(E \otimes E)}$ using the reduction formula.

Now we find how many times each irrep is contained in this product.

For $A_{1}$ :
$c_{A_{1}} = \frac{1}{6} [(1) (1)^{*} (4) + (2) (1)^{*} (1) + (3) (1)^{*} (0)] = \frac{1}{6} [4 + 2 + 0] = 1$
For $A_{2}$ :
$c_{A_{2}} = \frac{1}{6} [(1) (1)^{*} (4) + (2) (1)^{*} (1) + (3) (- 1)^{*} (0)] = \frac{1}{6} [4 + 2 + 0] = 1$
For $E$ :
$c_{E} = \frac{1}{6} [(1) (2)^{*} (4) + (2) (- 1)^{*} (1) + (3) (0)^{*} (0)] = \frac{1}{6} [8 - 2 + 0] = 1$

Result:
The decomposition is $E \otimes E = A_{1} \oplus A_{2} \oplus E$ .

This result shows that the 16-dimensional representation space of $E \otimes E$ breaks down into three invariant subspaces, transforming as $A_{1}$ , $A_{2}$ , and $E$ . As a check on the dimensions: $dim (A_{1}) + dim (A_{2}) + dim (E) = 1 + 1 + 2 = 4$ , which equals the dimension of the product representation $dim (E) \times dim (E) = 2 \times 2 = 4$ . Note that the dimension of the representation space is $4 \times 4 = 16$ , while the dimension of the representation itself is 4. This is a subtle point; the dimension of the representation refers to the size of the vectors it acts on, not the matrices.

Symmetrized and Antisymmetrized Products

A physically important subtlety arises when taking the tensor product of an irrep with itself, as in our $E \otimes E$ example. If the product describes two identical particles, the total state must obey quantum statistics: it must be either symmetric (for bosons) or antisymmetric (for fermions) under particle exchange. This means we are interested not in the full product $E \otimes E$ , but in its symmetrized square $[E]^{2}$ and antisymmetrized square ${E}^{2}$ .

There are specific formulae for the characters of these symmetrized products:

Symmetrized character: $χ_{[D]^{2}} (g) = \frac{1}{2} [(χ (g))^{2} + χ (g^{2})]$
Antisymmetrized character: $χ_{{D}^{2}} (g) = \frac{1}{2} [(χ (g))^{2} - χ (g^{2})]$

For $C_{3 v}$ , $(C_{3})^{2} = C_{3}^{2}$ (same class) and $(σ_{v})^{2} = E$ .

$χ_{[E]^{2}}$ : $(\frac{1}{2} [2^{2} + 2], \frac{1}{2} [(- 1)^{2} + (- 1)], \frac{1}{2} [0^{2} + 2]) = (3, 0, 1)$
$χ_{{E}^{2}}$ : $(\frac{1}{2} [2^{2} - 2], \frac{1}{2} [(- 1)^{2} - (- 1)], \frac{1}{2} [0^{2} - 2]) = (1, 1, - 1)$

Decomposing these new characters, you will find:

$[E]^{2} = A_{1} \oplus E$
${E}^{2} = A_{2}$

Notice that $[E]^{2} \oplus {E}^{2} = A_{1} \oplus E \oplus A_{2}$ , which recovers our original result for $E \otimes E$ . This decomposition has direct physical consequences, for instance in determining the allowed electronic states of molecules.

4. A Look Ahead: Connection to SU(2) and Angular Momentum

The methods we've developed for finite groups have a powerful and direct parallel in the study of continuous groups (Lie groups), which are foundational to modern physics. A classic example is the addition of angular momentum in quantum mechanics, governed by the group SU(2).

The combination of two spin-1/2 particles (each transforming under the 2-dimensional representation of SU(2)) is described by the tensor product $D^{(1/2)} \otimes D^{(1/2)}$ . This product space decomposes into a 3-dimensional spin-1 representation (the triplet state) and a 1-dimensional spin-0 representation (the singlet state). This is famously written as:

D^{(1/2)} \otimes D^{(1/2)} = D^{(1)} \oplus D^{(0)}

Or, in terms of dimensions: $2 \otimes 2 = 3 \oplus 1$ . This decomposition is precisely what underpins phenomena like the hyperfine splitting of atomic levels and is related to the formation of Cooper pairs in superconductivity, a topic central to your interest in condensed matter physics.

The following video provides an excellent deep dive into this specific, crucial example. While it uses Lie algebra tools like ladder operators that we will cover formally in a later module, the conceptual picture of decomposing a product space should now feel familiar.

Spinors for Beginners 19: Tensor Product Representations of su(2) [Clebsch-Gordan coefficients]

The video "Spinors for Beginners 19" by eigenchris gives a beautiful explanation of tensor product representations for SU(2).

To build your physical intuition, please watch these segments: Introduction (0:00-2:48): This part connects the abstract idea of tensor products to the concrete physics of multi-particle systems and the addition of angular momentum. Decomposition of Two Spin-1/2 States (15:23-22:29): This is the core of the video, where the presenter explicitly constructs the spin-1 (triplet) and spin-0 (singlet) states from the tensor product of two spin-1/2 states. Visual Trick (32:46-33:48): This brief segment introduces a wonderfully intuitive visual method using "weight diagrams" to quickly determine the decomposition of SU(2) tensor products.

This preview shows that the group-theoretical machinery you are learning is not just for finite point groups, but is a universally applicable language for describing symmetry in physics. The coefficients that relate the product basis (e.g., $∣ ↑ ⟩ ∣ ↓ ⟩$ ) to the decomposed basis (e.g., $∣ J, M ⟩$ ) are the famous Clebsch-Gordan coefficients, which we will study in detail in Module 6.

Conclusion

In this lesson, you've added a vital tool to your arsenal: the ability to decompose tensor product representations. This allows us to predict the symmetry properties of composite systems, a task that arises constantly in quantum and condensed matter physics.

Key Takeaways:

The representation of a composite system is the tensor product of the representations of its parts.
The character of a tensor product representation is the product of the characters of the individual representations.
This product representation is generally reducible and can be decomposed into a direct sum of irreps using the standard reduction formula.
This procedure, the Clebsch-Gordan series, provides the blueprint for understanding how symmetries combine, with direct applications in areas like the addition of angular momentum.

You now understand how to analyze a reducible representation, whether it's given to you directly or constructed from a tensor product. In our next and final lesson of this module, we will complete our toolkit by learning how to use the projection formula. This powerful technique will allow us to go beyond characters and construct the actual basis vectors that transform according to a specific irreducible representation.

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