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Wightman axiom 2: what kind of representation?

Physics Asked on March 27, 2021

Both in Wikipedia and on page 98 of Streater, Wightman, PCT, Spin and Statistics and all that, the second axiom postulates that a field must transform according to a representation of the Poincaré group.

I am a mathematician and I wonder if there are implicit assumptions there. Is any representation of the Poincaré group acceptable? Should such a representation be real-valued (that is, if $rho$ is such a representation, for any element $g$ of the Poincaré group, is $rho(g)$ a real-valued matrix)? Should it be orthogonal, unitary (that is, should the aforementioned matrices be orthogonal, unitary)?

EDIT: Let me rephrase my question.

As far as I understand, axiomatically,

  • a QFT should come with a (strictly speaking, projective, but I’m not sure it’s relevant here) unitary representation $U$, that is, a continuous morphism from the Poincaré group to the group of unitaries of the Hilbert space;
  • a $n$-dimensional vector-valued field is described by an $n$-tuple of maps $phi := (phi_1,cdots,phi_n)$ from the Minkowski space to the set of operators on the Hilbert space (I also knew that, strictly speaking, we should consider distributions instead of maps but I don’t think it is relevant here);
  • now, under the action of a symmetry (that is, under conjugation by a unitary -from the unitary representation $U$) each coordinate of $phi$ becomes a linear combination of all the coordinates, and the coefficients are stored in a matrix that Streater and Wightman call $S$ (equation 3-4 in Streater-Wightman, page 99).

This $S$ is a morphism from the Poincaré group to the group of square invertible complex matrices of size $n$, so, as a mathematician, I also call $S$ a (finite-dimensional) representation of the Poincaré group.

My question is: is there any implicit assumption on $S$?

I think my question is motivated by my fear of coordinates (I don’t like the idea that a field $phi$ should be implemented as a tuple; it looks that we are making an arbitrary choice of coordinates).

2 Answers

The value of the classical field at a point in spacetime $phi(x)$ may transform under any finite-dimensional representation, not necessarily unitary or orthogonal etc.

But the quantum field as an operator-valued distribution transforms under an infinite-dimensional unitary representation that acts on the Hilbert space of the QFT.

Wightman axioms relate the two representations, postulating that

$$ U(Lambda) phi(f) U(Lambda)^{dagger} = P(Lambda) phi (S(Lambda^{-1}) f). $$

Here $U(Lambda)$ is the infinite-dimensional unitary representation, $P(Lambda)$ is the finite-dimensional representation that acts on the classical field's value at a point, and $S(Lambda)$ is the natural representation that acts on test functions over spacetime.

Answered by Prof. Legolasov on March 27, 2021

Well the answer here is that the author is talking about Irreducible representations (Irreps). But @Plop you have asked me a great question in the comments, "why physicists like so much irreducible representations?". So I done my best to answer it.

TLDR: Physicists primarily deal with Lie Groups (Poincare group is a Lie Group) and there is a theorem which states that "If the Lie group representation isn't already irreducible than it can be "completely reduced" into a collection of irreducible representations." So by the nature of our math tools we MUST be using Irreps

$$----------------- text{Long Answer} -----------------$$

I studied the Burau representation in undergrad (https://en.wikipedia.org/wiki/Burau_representation) and then quantuum chromodynamics in grad school, so I'll try to connect express how I have connected those two experiences.

  1. What is representation theory? A represenation of a group $G$ is a homomorphic mapping of the group $G$ onto a non-singular group of $d times d$ matrjces $Gamma(T)$, where matrix multiplication is the group's multiplicative operation. (The group of matrices $Gamma(T)$ forms a $d$-dimensional representation $Gamma$ of group $G$)

Example I think that the unreduced Burau representation of the Braid Group $B_n$ (https://www.youtube.com/watch?v=uMMxD0Ak4lg) gives a great visual interpretation! This representation maps the act of crossing two strands of hair (left over right) $sigma_i$ onto the matrix, $$ begin{align} (0) && Gamma(sigma_i) = left[ begin{array}{c|cc|c} I_{i-1} & 0 & 0 & 0 hline 0 & 1-t & t & 0 0 & 1 & 0 & 0 hline 0 & 0 & 0 & I_{n-i-1} end{array} right] end{align} $$ By definition the representation connects to uncrossing two strands of hair (right over left) $sigma_i^{-1}$ with matrix $Gamma(sigma_i)^{-1}=Gamma(sigma^{-1})$.

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  1. What are (ir)reducible representations? Reducible representations have the form $$ begin{align} label{eq_reducible} (1) && Gamma(T) = begin{bmatrix} Gamma_{11}(T) & Gamma_{12}(T) 0 & Gamma_{22}(T) end{bmatrix} end{align} $$. Such that $$ Gamma(T_1) Gamma(T_2) = left[ begin{array}{c|c} Gamma_{11}(T_1) Gamma_{11}(T_2) & Gamma_{11}(T_1) Gamma_{12}(T_2) + Gamma_{12}(T_1) Gamma_{22}(T_2) hline 0 & Gamma_{22}(T_1) Gamma_{22}(T_2) end{array} right] $$ We have arrived at the homomorphic property which defined a representation. $$Gamma_{11}(T_1 T_2) = Gamma_{11}(T_1) Gamma_{11}(T_2) $$ $$Gamma_{22}(T_1 T_2) = Gamma_{22}(T_1) Gamma_{22}(T_2) $$ Therefore the representation $Gamma$ contains representations $Gamma_{11}$ and $Gamma_{22}$ which are of a smaller dimension than $Gamma$. Since these smaller representations exist within $Gamma$, gamma is reducible. An irreducible group contains no such smaller dimensional representations.

Example Fortunately for us, the Reducible Burau representation Eq.0 has this structure! Consider $Gamma(sigma_1)$ and $Gamma(sigma_3)$ in $B_4$ $$ Gamma(sigma_1) Gamma(sigma_3) = left[ begin{array}{cc|cc} 1-t & t & 0 & 0 1 & 0 & 0 & 0 hline 0 & 0 & 1 & 0 0 & 0 & 0 & 1 end{array} right] left[ begin{array}{cc|cc} 1 & 0 & 0 & 0 0 & 1 & 0 & 0 hline 0 & 0 & 1-t & t 0 & 0 & 1 & 0 end{array} right] = left[ begin{array}{cc|cc} 1-t & t & 0 & 0 1 & 0 & 0 & 0 hline 0 & 0 & 1-t & t 0 & 0 & 1 & 0 end{array} right] $$ This is great and it indicates that the Unreduced Burau representation can be reduced! (https://en.wikipedia.org/wiki/Burau_representation#Explicit_matrices)

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  1. How does the process of Reduction Happen?
    Suppose that $Gamma_{11}(T)$ is itself reducible and be expressed in the same for as Eq.1. Fortunately you can always similarity transform $Gamma(T)$ such that $$S^{-1} Gamma S = Gamma' = left[ begin{array}{cccc} Gamma'_{11}(T) & cdots & cdots & Gamma'_{1n}(T) 0 & ddots & cdots & Gamma'_{2n}(T_2) vdots & 0 & ddots & vdots 0 & cdots & 0 & Gamma'_{nn}(T) end{array} right]$$ If this upper triangular representation can be "completely reduced" to a block diagonal using another similarity transform, then $Gamma(T)$ is a completely reducible representation. There is a theorem which states that "if a Lie group is reducible it is completely reducible." - "Group Theory in Physics" Ch4-S4 J.F. Cornwell (1984).

$$text{CITATION: https://www.sciencedirect.com/topics/mathematics/reducible-representation}$$

Answered by ThomasTuna on March 27, 2021

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