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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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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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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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