46 votes
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Why are particles thought of as irreducible representations, in plain English?

As you probably know, the Lie group of physical transformations of a quantum system acts on the Hilbert space of states of the system by means of a (strongly-continuous projective-) unitary ...
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32 votes

What does it mean for particles to "be" the irreducible unitary representations of the Poincare group?

Irreducible representations of the Poincare group are the smallest subspaces that are closed under the action of the Poincare group, which includes boosts, rotations, and translations. The point is ...
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15 votes

Casimir Operators and the Poincare Group

Whew. This answer got super out of hand, so buckle in if you're not familiar with Lie groups, Lie algebras, and representation theory. I'll summarize the main points here, and then embark on a long ...
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11 votes
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Why does one study the representations of Lorentz group instead of studying only the representations of Poincare group?

It is not so much true that we separately study the representations of the Poincaré and the Lorentz group as that simply the two coincide in some cases: In the case where we are interested in the ...
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11 votes
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What is CPT, really?

As you have mentioned, when we Wick rotate to Euclidean signature, the four-component Lorentz group $O(d,1)$ becomes the two-component $O(d+1)$. Let's suppose we have infinitesimal Lorentz symmetry. ...
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10 votes

Why are one-particle states called irreducible representations of Poincaré group?

This is answered in depth in Weinberg's book on quantum field theory (Vol. I, Chapter 2). Relativistic invariance means translation invariance and Lorentz invariance, hence - obviously - Poincare ...
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Expansion coefficients of an arbitrary state in the Hilbert space of one-particle states

By construction, the $\lvert p,\sigma\rangle$ are a basis of your vector space. So all your states are linear combinations of these basis vectors. However, you are right to inquire about the nature of ...
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Can Poincare representations be embedded in non-standard Lorentz representations?

A field in the $(A,B)$ representation of the Lorentz Group has a propagator that scales as $|p|^{2(s-1)}$ for $|p|\to\infty$, where $s=A+B$ is the "spin" of the field (Ref.1 §12.1) . Therefore, the ...
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Why aren't infinite-dimensional representations of the Poincaré group classified by *two* half-integers?

I think most of the confusion is due to mixing up what the Lorentz and Poincare groups are typically acting on. When we talk about Lorentz irreps, we usually mean finite-dimensional non-unitary irreps ...
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Why does having a representation of the Poincaré algebra imply conservation of energy, momentum and angular momentum?

It does sound a bit weird, at first, that using the Poincare algebra alone can give nontrivial conservation laws. After all, that's not how other algebras work. Many physical system carry a ...
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9 votes
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Why is a nonzero VEV for a spinor field said to break Lorentz invariance?

The $v$ you write is itself a spinor, not a scalar. A non-zero spinor is obviously not invariant under Lorentz transformations, so a non-zero spinorial VEV breaks Lorentz invariance of the 1-point ...
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Are particles in curved spacetime still classified by irreducible representations of the Poincare group?

What is a particle? Before we worry about how to classify particles, we should try to be clear about what "particle" means. Ideally, we would like the definition to have these features: ...
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Is the boost operator conserved (in the context of QFT)?

Let us suppose that $G$ is a Lie group admitting some faithful unitary strongly continuous representation $U$ on a Hilbert space $\cal H$, so that we can interpret the elements $a$ of the Lie algebra ...
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8 votes

What is the difference between dynamical and geometric phases?

The total phase is a sum of the dynamical and geometric phases: $$\phi = \int_0^T E(t) dt + \oint A_{\mu}(R) dR^{\mu}$$ Where $R^{\mu}$ are the coordinates of the parameter space, $E(t)$ is the ...
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8 votes

GR as a gauge theory: there's a Lorentz-valued spin connection, but what about a translation-valued connection?

One can make a Poincare-Lie-Algebra-valued Cartan Connection by setting $$ \eta = \tau_a {\bf e}^{*a} + \frac 12 \sigma^{ab} \omega_{ab} $$ where $\tau_a$ and $\sigma^{ab}$ are the Lie algebra ...
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Lorentz transformation of vector field

Precisely because of the reason you have put forward, the Lorentz transformation for a scalar field $\phi(x)$ is usually written as: $$ \phi(x) \rightarrow \phi(x)' = \phi(\Lambda^{-1}x), $$ and for a ...
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Poincaré group and the Bargmann theorem

Yes, indeed that statement has been already corrected in the second edition of my book. Poincare group is a semi direct product of $SL(2,C)$ and the abelian Lie group $R^4$. The former is a (semi) ...
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7 votes

Klein-Gordon inner product

The Klein-Gordon inner product is a natural construction for functions defined on the mass hyperboloid $k^2=m^2$, because if you write your function in momentum space, $$ \phi(x)\sim \int\widetilde{\...
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7 votes

Local translations in curved spacetime

Ref. 1 defines a local translation on spacetime $M$ as a diffeomorphism. Note that the words local and global in this physics context mean point-dependent (=$x$-dependent) and point-independent (=$x$-...
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What is the matrix representation of the momentum operator (generator of translations) that is used in the commutators of the Poincaré Group?

The boost generators are First of all, note that You've chosen specific representation of Lie algebra generators of Poincare group, which is vector-like matrix representation. There are many ...
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Inonu-Wigner Group Contraction

In physical, non-covariant language, (WP conventions), the Poincaré algebra presents as $$ [P_0,P_i]=0,\\ [P_i,P_j]=0, \\ [J_i,P_0] = 0 ~, \\ [K_i,P_k] =- i \delta_{ik} P_0 ~,\\ [K_i,P_0] = -i ...
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Inonu-Wigner Group Contraction

Cosmas Zachos has already given a correct answer. The main point is that The natural non-relativistic Lie algebra in Newtonian mechanics is the Bargmann algebra, not the Galilean algebra! The ...
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Lie group compactness from generators

As Cosmas Zachos hints at in the comments, a non-Abelian Lie algebra belongs to a compact Lie group if its Killing form $K(X;Y) = \mathrm{tr}(\mathrm{ad}_X\circ \mathrm{ad}_Y)$ is negative-definite, ...
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How is it possible for quantum fields in different irreps of the Poincaré group to interact?

There are several confusions here. First, you seem to be mixing up particles (described by infinite-dimensional unitary irreps of the Poincare group) with fields (described by finite-dimensional, ...
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6 votes

Poincare non-invariance in real world and field theory

The state of the universe is not homogeneous and isotropic, but the laws of physics are. For example, the speed of light propagation is the same in all directions, and the mass of the electron is not ...
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Misuse of $\mathbf J^2$ in classifying Poincare reps

The Poincare group has two Casimir Invariants - namely $p^2$ and $W^2$ where $$ W_\mu = \frac{1}{2} \epsilon_{\mu\nu\rho\sigma} J^{\nu\rho} p^\sigma $$ is the Pauli-Lubanski pseudo-vector. Thus, ...
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6 votes

How to arrive at the Dirac Equation from Poincaré Algebra?

First, note that on the Hilbert space of physical states is realized the projective representations of the Poincare group, precisely, the representations "up to the sign". This can be incorporated ...
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6 votes

Is the boost operator conserved (in the context of QFT)?

You are correct that, in the abstract Poincaré algebra, the "Hamiltonian" $P^0$ does not commute with the boosts $K^i = J^{0i}$. But, crucially, "conservation" has no meaning in the abstract setting. ...
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6 votes

Is local Lorentz + diffeomorphism invariance equivalent to full local Poincaré invariance?

I cannot give you a full answer, that honor has to pass on to someone else, but nontheless I can give you some info you might value. It is also known (but there's lot of confusion out there about ...
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6 votes

Is there a type of supersymmetry where supercharges have spin 3/2?

Under general assumptions, the answer is no. Supersymmmetry generators belong only to the $\left (1/2, 0 \right ), \left( 0, 1/2\right)$ representation of the homogeneous Lorentz group (so they can ...
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