In Page 63, Section 2.5 of Weinberg's QFT Volume 1, on "One-particle states", he considers the representation of homogeneous Lorentz transformation, $U(\Lambda, 0) \equiv U(\Lambda)$ $$ U(\Lambda) \Psi_{p, \sigma}=\sum_{\sigma^{\prime}} C_{\sigma^{\prime} \sigma}(\Lambda, p) \Psi_{\Lambda p, \sigma^{\prime}} $$ then he claims that,

In general, it may be possible by using suitable linear combinations of the $\Psi_{p, \sigma}$to choose the $\sigma$ labels in such a way that the matrix $C_{\sigma^{\prime} \sigma}(\Lambda, p)$ is block-diagonal; in other words, so that the $\Psi_{p, \sigma}$ with $\sigma$ within any one block by themselves furnish a representation of the inhomogeneous Lorentz group.

Now my question: the effect of $U(\Lambda)$ is to bring the state $\Psi_{p, \sigma}$ to $\Psi_{\Lambda p, \sigma^{\prime}}$, so the space that the $C_{\sigma'\sigma}$ acts is different for different $\Lambda$. But this "conclusion" is weird to me since I think the representation space of a group should be the same for the group elements.


1 Answer 1


This is an example of induced representation. Consider two groups, $K < G$. Let a representation $D(K)$ act in a vector (usually, Hilbert) space ${\mathbb{V}}$. Based on this, we now wish to construct a representation of $G$. In mathematics, this (so-called "induced") representation is denoted with $\operatorname{Ind}_K^GD$ or simply $D(K)\uparrow G$.

Such a representation will be a fiber bundle whose base is the quotient space $G/K$, with copies of ${\mathbb{V}}$ playing the role of fibers. Simply speaking, we take multiple copies of ${\mathbb{V}}$ and agree that the action of $G$ is two-fold. It permutes the unit vectors within each copy of ${\mathbb{V}}$, and it also permutes the copies of ${\mathbb{V}}$.

Does this answer your question?

  • $\begingroup$ So in this case, $G$ is the full Lorentz group, and $K$ is the Little group of it. The representation space of $K$ is $\mathbb{V}$. The $G$ operation can take vectors from one $\mathbb{V}$ to another, is that correct understanding? $\endgroup$
    – Jiahao Fan
    Sep 6, 2020 at 10:08
  • $\begingroup$ Well I doubt it since Weinberg's claim is that the representation is induced from the representation of the little group. $\endgroup$
    – Jiahao Fan
    Sep 6, 2020 at 12:35
  • $\begingroup$ @JiahaoFan Yes, you are right and I am wrong. The rep is induced from the little group (which is the group of 3-dim rotations) to the Lorentz group. $\endgroup$ Sep 6, 2020 at 19:41

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