# Tag Info

14

The general idea. Let's restrict the discussion to matrix Lie Groups for simplicity. Determining the generators of a given Lie group $G$ simply means (by definition) determining a basis for its Lie algebra $\mathfrak g$. Here's a standard method for finding such a basis: Recall that the Lie algebra $\mathfrak g$ of a matrix Lie group $G$ is defined as ...

9

This just says that you can decompose any unitary representation of the Poincare group (= inhomogeneous Lorentz group) into irreducible representations. He suggests to identify the irreducible representations with elementary particles, as suggested by the analogy irreducible = no longer decomposable = elementary. He doesn't really explain why (but only ...

8

In general, quantum numbers are labels of irreducible representations of the relevant symmetry group, not primarily eigenvalues of an otherwise simply defined operator. But for every label that has a meaningful numerical value in every irreducible representation, one can define a Hermitian operator having it as an eigenvalue, simply by defining it as the ...

8

For Poincare algebra there are (as far as I know) two different approaches to find its representations. In the first approach one begins from a finite dimensional representation of (complexified) Lorentz algebra, and using it one constructs a representation on the space of some fields on Minkowski space. Representation so obtained is usually not irreducible ...

8

First, note that in physics, we consider unitary representations $U$ of the Poincare group acting on the Hilbert space $\mathcal H$ of the theory because we are interested in a precise formulation of the concept of Poincare transformations acting on the quantum mechanical states of the theory as symmetries (since the laws of physics should be inertial ...

7

Essentially by definition (due to Wigner), one-particle Hilbert spaces of elementary particles support unitary strongly continuous irreducible representations of Poincaré group. Conversely, any multi-particle Hilbert space, with either fixed or undefined number of particles either identical or distinguishable, cannot be irreducible under the action of the ...

7

It is important to distinguish between three group actions that are named "Galilean": -The Galilean transformation group of the Eucledian space (as an automorphism group). -The Galilean transformation group of the classical phase space (whose Lie algebra constitute a Lie subalgebra of the Poisson algebra of the phase space). This is the classical action. ...

6

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, representations of the Lorentz group are labelled by the eigenvalues of both $p^2$ and $W^2$. When $p^2 = -m^2$, we have the property $W^2 = -m^2 {\bf J}^2$. ...

6

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 a function of position. Writing down a Lagrangian requires an assumption about the laws of physics (or more precisely, an assumption about the dynamics). There ...

5

When done properly, none of the problems exists and some of your assumptions are invalid. First, concerning the two questions, in topologically trivial but arbitrarily curved spacetimes, the Poincaré symmetry holds in the sense that it is a small subgroup of the infinite-dimensional group of all diffeomorphisms; general relativity and all of its extensions ...

5

Because a constant vector (like the translation vector) is annihilated by the differential operator $\partial^\mu$.

5

With respect to the discussion of momentum-eigenstates and the following derivation in Weinberg's book, $\sigma$ is just a label that denotes any degree of freedom that is not momentum. Even though it can be identified with spin, its nature is not relevant for the discussion at hand.

4

The Coleman-Mandula Theorem is a theorem about the infinitesimal symmetry generators of S-matrices. 1) It's only a theorem about Lie algebras. It doesn't see discrete symmetries like parity and it can't tell the difference between Spin(3,1) and SO(3,1). It also assumes that the symmetry generators form a lie algebra rather than a super Lie algebra. 2) ...

4

It is incorrect to say that $$U(\Lambda) \left|p,\sigma\right> \propto |\Lambda p, \sigma\rangle~~~~~~ \text{WRONG!!}$$ Here is the correct logic. Consider the state $U(\Lambda) |p,\sigma\rangle$. We have just shown that (in eq. 2.5.2) that this state has a momentum eigenvalue $\Lambda p$. Now, there are a whole bunch of states with momentum $\Lambda ... 4 Not all irreducible representations (irrep's for short) of the Poincaré group lead to a Lagrangian. One example (see my comment to Julio Parra's answer) are the zero-mass, "continuous-helicity" (sometimes called "infinite-helicity") representations. There is, however, a way to begin from a positive energy irrep of the Poincaré group (i.e. a 1-particle ... 4 The fact that combinations such as$(e^-+e^+)/\sqrt{2}$, (or other combinations such as a proton with an electron) are suppressed is a little mysterious. It falls outside of quantum mechanics. Any "theory of everything" needs to answer this sort of question. We can't form combinations of electrons with positrons because they have different electric charges ... 3 Superselection is about relative phases. In the particular case of$e+\bar{e}$, it concerns the U(1) symmetry. Both observables and states have to be invariant under symmetry transformations (for the symmetry to be a symmetry), but$e$and$\bar{e}$transform differently under U(1) transformations, so$e+\bar{e}$cannot be invariant because it transforms to ... 3 You are still asking way too many questions at once. So again, consider splitting them next time. I will address only the topology part here. As topological spaces we have $$SO(3) = {\mathbb R \mathbb P}^3 = S^3 /\sim ,$$ $$SO(4) = S^3 \times S^3 / \sim,$$ $$SO^+(1,3) = {\mathbb R}^3 \times S^3 / \sim$$ (in all of these cases$\sim$is an identification ... 3 I think it is correct. However, just as a pedantic remark, you should prove$U(\Lambda)a^{s\dagger}_\mathbf{p}U^{-1}(\Lambda)$and$\sqrt{\frac{E_{\Lambda\mathbf{p}}}{E_{\mathbf{p}}}} a^{s\dagger}_{\Lambda \mathbf{p}}$act on all the vectors in the same way, not just on the vacuum$|0\rangle$, you need a slight modification of your proof: $$\sqrt{2 ... 3 I'm not sure such a thing exists. Usually reps only helps you classify the kind of particles you have (i.e the quantum numbers that identify them) and how they transform under the corresponding group. I believe how to represent this particles mathematically and what is their dynamics is a different matter. The only thing similar I know about is that some ... 3 For a thorough treatment of that expression, consider this paper. 3 There is nothing special about the group of isometries being non-abelian: Already the simplest isometry you encounter in physics, that of rotations in ordinary Euclidean three space plus (or, rather, semi-direct product with) translations, is non-abelian, since the group of rotations \mathrm{SO}(3) alone is non-abelian - it is not the same to first rotate ... 2 Spinors and vectors aren't two "competing formalisms". They are two inequivalent representations of the rotational or Lorentzian group. For fermions such as electrons, one absolutely needs spinors and it would be extremely awkward, nearly impossible, to produce the same physics just with vectors and tensors constructed out of vectors. On the other hand, ... 2 Consider$$M_{31} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \text{ and } P_0 = -i \begin{pmatrix} 0 & 0 & 0 & 0 & 1 ... 2 The Poincare group is really just the Lorentz group + translations. If you can show that the expression is invariant under rotations, boosts and translations, you're done. This is fairly trivial in the case at hand since the appearance of contracted tensor indices only implies Lorentz invariance. In fact, that's one of the main reasons why we use tensor ... 2 The definition of the Pauli-Lubanski pseudo-vector is : $$\hat {W}^{\alpha} = \frac{1}{2}\varepsilon^{\alpha \beta \gamma \delta}\hat {J}_{\beta \gamma}\hat {P}_{\delta}.$$ Where${J}_{\beta \gamma}$and${P}_{\delta}$correspond to the generators of the Poincaré group. (respectively total angular moment, and momentum) So, yes,${J}_{\beta \gamma}$is of ... 2 In the inhomogenous Lorentz group$ISO(1,3)$, you have the space-time translation group, and the Lorentz group$SO(1,3)$. You begin to find a representation of the space-time translation group, by choosing a momentum$p$. So your representation must have a$p$indice : $$\psi_p$$ After this, you will have to get the full representation, by finding a ... 2 The definition is that a particle in Minkowski space is a unitary irreducible representation of the Poincare group. So one needs to see how various P.D.E.s are related to the classification of unitary irreducible representations of$iso(3,1)$or$iso(d-1,1)$in the case of$d$-dimensions instead of$4$. Note that these are all the Poincare-invariant ... 2 This condition is due to the fact that for a free massless particle the Pauli-Lubanski vector$W=*(M\wedge P)\$ must be proportional to the linear momentum (The proportionality factor being the helicity). Thus the condition must be valid to all free massless relativistic field theories.

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