Clebsch-Gordan Coefficients: Simple Cases

Welcome to the next lesson in our study of SU(2) representations. Last time, we established the Clebsch-Gordan series, which tells us which irreducible representations, labeled by total angular momentum $J$ , arise when we combine two systems with angular momenta $j_{1}$ and $j_{2}$ . We saw that the tensor product space decomposes into a direct sum of these irreps:

j_{1} \otimes j_{2} = J = ∣ j_{1} - j_{2} ∣ ⨁ j_{1} + j_{2} J

This answered the "what" of angular momentum addition. Today, we address the "how". Our goal is to explicitly construct the basis states $∣ J, M ⟩$ of the coupled representation as linear combinations of the states $∣ j_{1}, m_{1}; j_{2}, m_{2} ⟩$ from the uncoupled (tensor product) basis. The coefficients of this transformation are the celebrated Clebsch-Gordan coefficients, and this lesson will equip you with the methods to compute them for simple, yet fundamental, physical systems.

The Change of Basis and Clebsch-Gordan Coefficients

The relationship between the two bases is a change of basis, which can be written as:

∣ j_{1}, j_{2}; J, M ⟩ = m_{1}, m_{2} \sum ⟨ j_{1}, m_{1}; j_{2}, m_{2} ∣ j_{1}, j_{2}; J, M ⟩ ∣ j_{1}, m_{1}; j_{2}, m_{2} ⟩

The coefficients $⟨ j_{1}, m_{1}; j_{2}, m_{2} ∣ j_{1}, j_{2}; J, M ⟩$ are the Clebsch-Gordan (CG) coefficients. Often, the fixed $j_{1}$ and $j_{2}$ values are suppressed for brevity, leading to the more common notation $⟨ m_{1}, m_{2} ∣ J, M ⟩$ .

Before we dive into calculations, it's useful to understand the fundamental properties of these coefficients, which stem from the properties of angular momentum operators and the orthonormality of the bases. The following document provides a concise summary of these rules.

[PDF] II : Addition of Angular Momentum - Clebsch-Gordan Coefficients

This document by Dipan Kumar Ghosh provides a formal introduction to the addition of angular momentum. We will use it to establish the key properties of the CG coefficients.

Please read Section 2, "Properties of C-G Coeﬃcients". This section establishes four key properties: The M selection rule: The coefficient is non-zero only if M = m_1 + m_2. This greatly reduces the number of terms in the summation. Reality: By a standard phase convention (the Condon-Shortley convention), the CG coefficients are chosen to be real numbers. Orthonormality: Since the transformation between the two orthonormal bases is unitary (and real, thus orthogonal), the coefficients satisfy two important sum rules, given in equations (11) and (12). The triangle inequality: The total spin J must satisfy |j_1 - j_2| \le J \le j_1 + j_2, which we already discovered from the Clebsch-Gordan series.

With these properties in mind, let's explore the primary method for calculating the coefficients: the ladder operator technique.

The Ladder Operator Method

The most systematic way to compute CG coefficients is by using ladder operators. The total angular momentum raising and lowering operators can be written in two ways: acting on the coupled states or on the uncoupled states.

J^_{\pm} = J^_{1 \pm} + J^_{2 \pm}

The core of the method is to start with a state that is easy to identify—the "highest weight state" of the largest $J$ multiplet—and then repeatedly apply the lowering operator $J^_{-}$ to generate all other states in that multiplet.

Example 1: Two Spin-1/2 Particles ( $j_{1} = 1/2, j_{2} = 1/2$ )

Let's revisit the decomposition we found in the previous lesson: $1/2 \otimes 1/2 = 1 \oplus 0$ . We need to find the explicit forms for the $J = 1$ (triplet) and $J = 0$ (singlet) states.

The video below provides an excellent pedagogical derivation using a combination of ladder operators and orthogonality arguments.

Deriving Clebsch–Gordan Coefficients | Angular Momentum in Quantum Mechanics

The video "Deriving Clebsch–Gordan Coefficients" from the channel Pretty Much Physics provides a clear, self-contained derivation for the spin-1/2 case.

Please watch the entire video (0:00-6:24). It walks through the following key steps: Identifying Unique States: The states with the highest and lowest total M values, M=+1 and M=-1, are unique. This immediately tells us |1, 1\rangle = |\frac{1}{2},\frac{1}{2}; \frac{1}{2},\frac{1}{2}\rangle and |1, -1\rangle = |\frac{1}{2},-\frac{1}{2}; \frac{1}{2},-\frac{1}{2}\rangle. The CG coefficients for these are simply 1. Handling Degeneracy: The M=0 subspace is two-dimensional, spanned by |\uparrow\downarrow\rangle and |\downarrow\uparrow\rangle. The coupled states |1, 0\rangle and |0, 0\rangle must be two orthogonal linear combinations of these. Using J^2: The video applies the total angular momentum squared operator, \hat{J}^2 = (\vec{J}_1 + \vec{J}_2)^2, to find the specific combinations that correspond to the eigenvalues J(J+1)\hbar^2 for J=1 and J=0. Orthogonality and Phase: Once the |1, 0\rangle state is found, the |0, 0\rangle state is determined by orthogonality. The final signs are fixed by the Condon-Shortley phase convention. After completing the derivation, we find the famous triplet and singlet states: Triplet States (J=1): \begin{aligned} |1, 1\rangle &= |\uparrow\uparrow\rangle \\ |1, 0\rangle &= \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle) \\ |1, -1\rangle &= |\downarrow\downarrow\rangle \end{aligned} Singlet State (J=0): |0, 0\rangle = \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle) Notice that the triplet states are symmetric under particle exchange, while the singlet state is antisymmetric. This symmetry property has profound consequences in quantum statistics. Example 2: Spin-1 and Spin-1/2 (j_1 = 1, j_2 = 1/2) This case is slightly more complex and serves as a better illustration of the full power of the ladder operator method. From the Clebsch-Gordan series, we know 1 \otimes 1/2 = 3/2 \oplus 1/2. The composite 6-dimensional space splits into a 4-dimensional J=3/2 subspace and a 2-dimensional J=1/2 subspace. The following video provides an exemplary step-by-step calculation.

After completing the derivation, we find the famous triplet and singlet states:

Triplet States (J=1): $∣1, 1 ⟩ ∣1, 0 ⟩ ∣1, - 1 ⟩ = ∣ ↑↑ ⟩ = \frac{1}{2} (∣ ↑↓ ⟩ + ∣ ↓↑ ⟩) = ∣ ↓↓ ⟩$
Singlet State (J=0): $∣0, 0 ⟩ = \frac{1}{2} (∣ ↑↓ ⟩ - ∣ ↓↑ ⟩)$

Notice that the triplet states are symmetric under particle exchange, while the singlet state is antisymmetric. This symmetry property has profound consequences in quantum statistics.

Example 2: Spin-1 and Spin-1/2 ( $j_{1} = 1, j_{2} = 1/2$ )

This case is slightly more complex and serves as a better illustration of the full power of the ladder operator method. From the Clebsch-Gordan series, we know $1 \otimes 1/2 = 3/2 \oplus 1/2$ . The composite 6-dimensional space splits into a 4-dimensional $J = 3/2$ subspace and a 2-dimensional $J = 1/2$ subspace.

The following video provides an exemplary step-by-step calculation.

Clebsch-Gordan Coefficients Step-by-Step (Spin 1/2 + Spin 1)

The video "Clebsch-Gordan Coefficients Step-by-Step (Spin 1/2 + Spin 1)" from Let's Get Complex is a perfect guide for this calculation.

Please watch from 2:52 to the end. The video follows the canonical procedure: Start at the Top: Identifies the highest weight state |3/2, 3/2\rangle, which must be |1, 1\rangle \otimes |1/2, 1/2\rangle. Descend the Ladder: Repeatedly applies the lowering operator J_- = J_{1-} + J_{2-} to generate the entire J=3/2 multiplet: |3/2, 1/2\rangle, |3/2, -1/2\rangle, and |3/2, -3/2\rangle. Pay close attention to how the operator acts on both sides of the equation and how the coefficients are determined. Find the Next Multiplet: The highest state of the next multiplet, |1/2, 1/2\rangle, must lie in the M=1/2 subspace. It is found by requiring it to be orthogonal to the |3/2, 1/2\rangle state already found. Phase Convention: The Condon-Shortley convention is used to fix the relative sign in the expression for |1/2, 1/2\rangle. The convention states that the coefficient of the uncoupled state with the largest m_1 (in this case, |1,1; 1/2, -1/2\rangle) is taken to be positive. Complete the Second Ladder: The final state, |1/2, -1/2\rangle, is found by applying J_- one last time.

The resource [PDF] II : Addition of Angular Momentum (ID LINK) also presents this calculation in "Example 4" on page 9, which you may find helpful as a written reference.

Generalizations and Tabulated Results

The ladder operator method is completely general. It can be used to derive recursion relations that relate any CG coefficient to its neighbors. For your reference, the general recursion relation derived from $J_{\pm}$ is:

(J \mp M) (J \pm M + 1) ⟨ m_{1}, m_{2} ∣ J, M \pm 1 ⟩ = (j_{1} \mp m_{1} + 1) (j_{1} \pm m_{1}) ⟨ m_{1} \mp 1, m_{2} ∣ J, M ⟩ + (j_{2} \mp m_{2} + 1) (j_{2} \pm m_{2}) ⟨ m_{1}, m_{2} \mp 1∣ J, M ⟩

Starting from $⟨ j_{1}, j_{1}; j_{2}, J - j_{1} ∣ J, J ⟩ = + 1$ , these relations can be used to generate all other coefficients.

Given your background, you may appreciate the most rigorous approach. The resource [PDF] Total Angular Momentum: Solved Problems (ID LINK) contains a complete algebraic derivation (Problem 5) for the general case of adding a spin-1/2 particle to an arbitrary spin $s_{2}$ . This involves setting up the eigenvalue equation for $S^^{2}$ on a general state and solving the resulting system of linear equations, a beautiful demonstration of the underlying linear algebra. I highly recommend working through it to solidify your understanding.

In practice, physicists rarely derive these coefficients from scratch. They are extensively tabulated. Below is a standard table of Clebsch-Gordan coefficients.

A table of Clebsch-Gordan coefficients, published by the Particle Data Group. Note the convention that a square-root is understood over every entry. Each sub-table corresponds to a specific $j_{1} \otimes j_{2}$ coupling. The rows and columns are indexed by $m_{1}$ and $m_{2}$ . The coefficients for different total $J$ values are listed in the cells.

Let's use the table to verify one of our results for $j_{1} = 1, j_{2} = 1/2$ . We found that:

∣ J = 3/2, M = 1/2 ⟩ = \frac{1}{3} ∣1, 1; 1/2, - 1/2 ⟩ + \frac{2}{3} ∣1, 0; 1/2, 1/2 ⟩

In the table, find the block for $1 \times 1/2$ . Look for the column corresponding to $J = 3/2$ . Go down to the rows where $m_{1} + m_{2} = 1/2$ .

For $m_{1} = 1, m_{2} = - 1/2$ , the entry is $1/3$ . Remembering the square root, this is $1/3$ .
For $m_{1} = 0, m_{2} = 1/2$ , the entry is $2/3$ . This gives $2/3$ .

The table matches our calculation perfectly.

Conclusion

In this lesson, we have mastered the techniques for computing Clebsch-Gordan coefficients, the fundamental numbers that bridge the uncoupled and coupled pictures of composite quantum systems.

Key Takeaways:

Clebsch-Gordan coefficients are the real-valued elements of the orthogonal transformation matrix between the uncoupled basis $∣ j_{1}, m_{1}; j_{2}, m_{2} ⟩$ and the coupled basis $∣ J, M ⟩$ .
They are non-zero only if $M = m_{1} + m_{2}$ and $∣ j_{1} - j_{2} ∣ \leq J \leq j_{1} + j_{2}$ .
The most powerful method for their calculation is the ladder operator method. It involves starting with the highest-weight state of a multiplet and systematically applying the lowering operator $J_{-}$ .
Orthogonality between states with the same $M$ but different $J$ is used to find the highest-weight states of the lower-J multiplets.
The Condon-Shortley phase convention is used to uniquely fix all signs, ensuring consistency across calculations and tables.

You are now able to construct the explicit total angular momentum eigenstates for any simple composite system. This skill is a prerequisite for a vast range of applications. One of the most important is the simplification of matrix elements for physical operators, which is the subject of the Wigner-Eckart theorem. We will explore this powerful theorem in our next lesson.

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