Lorentz Representation Restriction to \(SO(3)\)
Question. Let us review the following:
\[
(j^+, j^-) \longrightarrow (j^+ + j^-) \oplus (j^+ + j^- - 1) \oplus (j^+ + j^- - 2) \oplus \cdots \oplus |j^+ - j^-|
\]
on restricting \(SO(3,1)\) to \(SO(3)\)?
This is a really important and elegant idea, so let’s unpack it carefully and make the logic completely transparent.
1. The key structural idea
The Lorentz algebra satisfies
\[
\mathfrak{so}(3,1) \cong \mathfrak{su}(2)_+ \oplus \mathfrak{su}(2)_-
\]
This means every finite-dimensional irreducible representation is labelled by two spins:
\[
(j^+, j^-)
\]
You can think of this as:
- one “left” angular momentum \(\mathbf{J}_+\),
- one “right” angular momentum \(\mathbf{J}_-\).
2. What happens when we restrict to rotations \(SO(3)\)?
Rotations are generated by
\[
J_i = J_{+i} + J_{-i}
\]
This is the crucial step. It tells you that the physical rotation generator is the sum of two angular momenta.
3. Interpretation as a two-particle system
Now comes the key insight:
- Treat \(\mathbf{J}_+\) as the angular momentum of particle “+”,
- Treat \(\mathbf{J}_-\) as the angular momentum of particle “−”.
Then
\[
\mathbf{J} = \mathbf{J}_+ + \mathbf{J}_-
\]
So the problem becomes: what are the possible total angular momenta when you combine spins \(j^+\) and \(j^-\)?
4. Use standard angular momentum addition
From ordinary quantum mechanics,
\[
j^+ \otimes j^- = (j^+ + j^-) \oplus (j^+ + j^- - 1) \oplus \cdots \oplus |j^+ - j^-|
\]
This is just the Clebsch–Gordan decomposition.
5. What the formula means
Therefore, when you restrict a Lorentz irreducible representation \((j^+, j^-)\) to the rotation subgroup \(SO(3)\), it becomes
\[
(j^+, j^-) \longrightarrow (j^+ + j^-) \oplus (j^+ + j^- - 1) \oplus \cdots \oplus |j^+ - j^-|
\]
Each term on the right is an ordinary spin-\(j\) representation of \(SO(3)\).
6. Why this works
The deeper reason is that the Lorentz group splits into two independent \(SU(2)\) factors, while rotations sit diagonally inside them. In effect, the rotation subgroup is embedded through the combination
\[
J_i = J_{+i} + J_{-i}.
\]
So restricting the Lorentz representation to \(SO(3)\) is the same as taking the tensor product \(j^+ \otimes j^-\) and decomposing it under the diagonal \(SU(2)\).
7. Concrete examples
Example 1: \(\left(\tfrac12, 0\right)\)
\[
\frac12 \otimes 0 = \frac12
\]
Hence
\[
\left(\tfrac12, 0\right) \rightarrow \tfrac12.
\]
A Weyl spinor remains spin-\(\tfrac12\) under spatial rotations.
Example 2: \(\left(\tfrac12, \tfrac12\right)\)
\[
\frac12 \otimes \frac12 = 1 \oplus 0
\]
Hence
\[
\left(\tfrac12, \tfrac12\right) \rightarrow 1 \oplus 0.
\]
This is the vector representation, which splits into a spin-1 part (the spatial components) and a spin-0 part (the time component).
Example 3: \((1,0)\)
\[
1 \otimes 0 = 1
\]
Hence
\[
(1,0) \rightarrow 1.
\]
So this representation is pure spin-1 under rotations.
8. Intuition summary
- Lorentz representations are labelled by two spins \((j^+, j^-)\).
- Rotations see only the sum \(\mathbf{J}_+ + \mathbf{J}_-\).
- So the restriction problem becomes ordinary angular momentum addition.
- The result is a ladder of spins running from \(j^+ + j^-\) down to \(|j^+ - j^-|\).
9. One-line takeaway
Restricting \((j^+, j^-)\) to \(SO(3)\) is exactly the same as adding two angular momenta \(j^+\) and \(j^-\).
This is additional material to complement page 455 of Group Theory in a Nutshell by Anthony Zee.
Created with assistance from ChatGPT.