# Tag Info

15

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 ...

10

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 ...

10

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 ...

9

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. ...

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

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 ...

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 ...

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

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.

5

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 $\Lambda ... 5 When done properly, none of the problems exists and some of your assumptions are invalid. First, concerning the two questions, in topologically trivial but arbitrarily curved spacetimes, the Poincaré symmetry holds in the sense that it is a small subgroup of the infinite-dimensional group of all diffeomorphisms; general relativity and all of its extensions ... 4 The fact that combinations such as$(e^-+e^+)/\sqrt{2}$, (or other combinations such as a proton with an electron) are suppressed is a little mysterious. It falls outside of quantum mechanics. Any "theory of everything" needs to answer this sort of question. We can't form combinations of electrons with positrons because they have different electric charges ... 4 Not all irreducible representations (irrep's for short) of the Poincaré group lead to a Lagrangian. One example (see my comment to Julio Parra's answer) are the zero-mass, "continuous-helicity" (sometimes called "infinite-helicity") representations. There is, however, a way to begin from a positive energy irrep of the Poincaré group (i.e. a 1-particle ... 4 The Coleman-Mandula Theorem is a theorem about the infinitesimal symmetry generators of S-matrices. 1) It's only a theorem about Lie algebras. It doesn't see discrete symmetries like parity and it can't tell the difference between Spin(3,1) and SO(3,1). It also assumes that the symmetry generators form a lie algebra rather than a super Lie algebra. 2) ... 4 For a thorough treatment of that expression, consider this paper. 4 This is answered in depth in Weinberg's book on quantum field theory (Vol. I, Chapter 2). Relativistic invariance means translation invariance and Lorentz invariance, hence - obviously - Poincare invariance, so that one has a representation of the Poincare group. Because of relativistic invariance and unitarity, the Hilbert space of a QFT carries a unitary ... 3 The way irreducible representations of the proper Poincaré group are studied is through the machinery of induced representations. Since the Poincaré group (or more precisely its 2-cover) has a semidirect product structure by an Abelian group$N$, one can used a special case of the general theory of induced representations, known as Wigner's little group ... 3 The electromagnetic potential$A^\mu$is a four-vector, and hence transforms in the fundamental representation of$\mathrm{SO}(1,3)$, i.e.$A^\mu\mapsto \Lambda^\mu_\nu A^\nu$where$\Lambda$is the usual 4x4 matrix associated to a Lorentz transformation. Your question seems fundamentally confused about the difference between the field and the particle. The ... 3 (1) The operator$U(\Lambda,a)$is a unitary "rotation" in the Hilbert space corresponding to an inhomogeneous Lorentz transformation of the spacetime coordinates. When$U(\Lambda,a)=\exp(iH\tau)$, it is an operator that adjusts the clock forward by$\tau$. Conceptually this is not a physical time evolution of the system. (2) A unitary rotation$U$in the ... 3 Just one comment to the higgsss answer. Formally from the Wigner theorem we have that if there exist time shift symmetry, for which the scalar product of quantum mechanical rays is conserved, $$\tag 1 |\langle \psi (t)|\kappa (t)\rangle| = |\langle \psi{'}(t+\tau)| \kappa{'}(t+\tau) \rangle|,$$ then the symmetry transformation acts on$|\psi\rangle$as ... 3 Take away, for a moment, any prejudgment about the answer to "what is a particle?"; QM provides us a precise answer. We know that physics’ laws should be invariant under the Poincare group. Since this group defines how Poincare transformations act on classic objects (tensors), in order to find out how they act on quantum objects (vectors in Hilbert space), ... 3 It sounds to me like he is saying the appearance of the interaction term in the Hamiltonian is due to our choice of fields with which to express the Hamiltonian. If we choose fields which annihilate the bound states instead, clearly the 1 particle sector of the theory looks just like a free theory. It might seem there is a problem with doing this with ... 3 A subrepresentation is not the same as a subspace on which a representation exists (which exists on any subspace): it has to be the same representation (group action), restricted to that space. 3 Each separable infinite-dimensional Hilbert space carries both irreducible and reducible representations of any noncompact Lie groups you can name. But this information in itself is of little use. The Hilbert spaces in quantum mechanics always come with distinguished representations that give certain operators an interpretation as distinguished ... 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 ...

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

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 ...

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