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I was reading Zee's book "Quantum field theory in a nutshell" and came to the chapter regarding representations and algebra.

He writes that the representations of $SO(3,1)$ are $(0,0),(0, 0),( \frac{1}{2} , 0), (0, \frac{1}{ 2} ), (1, 0), (0, 1), ( \frac{1} {2} , \frac{1}{ 2} )$ ... He then continues that the 1D representation $(0,0)$ is the Lorentz scalar, etc.

So my questions are

  • I was wondering, what does it mean for $(j^+,j^-)$ to be a representation of $SO(3,1)$?

  • what does it mean (or how can you tell) when the representation $(0,0)$ is a Lorentz scalar, or when the representation $(\frac{1}{2},\frac{1}{2})$ is a Lorentz vector etc.?

  • How do you count the dimensions? e.g. for $(0,0)$ it is 1 dimensional representation.

My only knowledge in group theory is from an introductory course in quantum field theory as well as in particle physics. I may need a not so rigorous explanation for now.

Qmechanic
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Gradient137
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2 Answers2

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Upon complexification, $so(3,1)$ breaks up into two commuting $su(2)\oplus su(2)$ so finite dimensional, non-unitary irreps of $so(3,1)$ can be labelled by two “angular momentum-like” quantum numbers $(j^+,j^-)$ (both real so that $2j^\pm+1$ is integer).

Clearly if $j=0$ is an $su(2)$ scalar, then $(j^+,j^-)=(0,0)$ will also be a scalar as it is left invariant under both $su(2)$’s in $su(2)\oplus su(2)$. Moreover, since you are given it is one-dimensional, it must go to itself under group transformation (since the vector space is 1d), and is thus a scalar. The dimension of $(j^+,j^-)$ is $(2j^++1)(2j^-+1)$ and the adjoint is $(1,0)+(0,1)$.

There is a subtle point when dealing with complexification and real forms. When going back to the real form $so(3,1)$, some irreps which are reducible in the complex field are irreducible over the reals: the adjoint is an example of this.

ZeroTheHero
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  • Oh I see, the explanation for $SU(2)$ hits the spot for my second question. I do get the fact that $j^\pm$ is in $SU(2)$ that's where we get $SU(2)\oplus SU(2)$. – Gradient137 Oct 14 '18 at 05:17
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I’ll try to contribute a casual explanation, maybe someone else can write a more rigorous answer if needed.

(1) The Lorentz group is isomorphic to two copies of $SL(2,\mathbb C)$. The representations of $SL(2,\mathbb C)$ can each be labeled by a half-integer number. Therefore the pairs $(m,n)$ arise.

(2+3) The dimension of a rep of $SL(2,\mathbb C)$ is $(2m+1)$, and the dimensions for the Lorentz group result from the sum - therefore eg the $(0,0)$ rep has dimension one. This is called trivial representation, so it doesn’t change the object it is acting on. And if a Lorentz transformation doesn’t change anything (=trivial), it is a Lorentz scalar.

ersbygre1
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  • What dimension does $(\frac{1}{2},\frac{1}{2})$ have then in this case? I could tell it is 4D but how do i get that from $(2m+1)$? Is the correct equation then be Dimension= $(2m+1)(2n+1)$? – Gradient137 Oct 14 '18 at 05:22
  • (2m+1) is related to just one SL(2,C). You can either use the equation you wrote in your comment or see it like this: Two spin-1/2‘s can couple to spin-0 and spin-1. therefore you have 20+1=1 and 21+1=3, which together yields 4! – ersbygre1 Oct 14 '18 at 07:34
  • Also, I'm reading the book but it only states that that the $S0(3,1)$ breaks into $SU(2)\oplus SU(2)$. Is $SU(2)$ the same as $SL(2,C)$? – Gradient137 Oct 14 '18 at 22:12
  • As far as I know, the representations of sl(2,C) and su(2) are the same, so since we know su(2) pretty well from topic like angular momentum and spin, we use su(2). But this is also where my „casual explanation“ ends ;) Maybe this helps further: https://en.m.wikiversity.org/wiki/Representation_theory_of_the_Lorentz_group#Strategy (see 4th and 5th paragraph in „Strategy“) – ersbygre1 Oct 14 '18 at 23:41