# Tag Info

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

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. ...

4

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

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

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 ...

2

It is safe to ignore curvature at the length scales of particle physics, as in the relevant region of space-time one can approximate the manifold very well by its tangent space, which is a flat Minkowski space with Poincare symmetry. For the same reason, engineers do not use general relativity but work with special relativirty (or even Newton's laws). ...

2

Symmetry group of the space time$^1$ on which QFT is defined is usually required to have a representation on the space of states. Quantum mechanics is just QFT in one dimensions. The spacetime in this case is the time line $\mathbb R$. Fields are $X(t)$, and $P(t)$. Symmetry group is group of translations $t\rightarrow t+b$ of $\mathbb R$. Infinitesimal ...

2

In the answer of Argopulos, one must assume that the Taylor series converges. This is never done, and maybe physicists don't care. From a mathematical perspective, (2.20) is in fact the definition of the dynamics, and the commutator relation follows simply by differentiating both sides. No special assumption is needed to get this.

2

0) It's weird to denote the action by $Ad$; this is usually reserved for adjoint actions. I'm going to use $\rho$. 1) Your expression is correct. Note that $(\rho\Phi)(f)$ is defined to be $\Phi(\rho f)$. In the end, we're just translating and transforming the test functions. 2) It should. I'm not 100% sure. It really ought to be an inner ...

1

The formula $e^{-i\hat{P}a/\hbar}\hat\psi(x)e^{i\hat{P}a/\hbar}=\hat\psi(x+a)$ is just the finite transformation version of infinitesimal transformation version formula $[\hat{P},\hat{\psi}(x)]=i\hbar\partial_x\hat{\psi}(x)$ as mentioned by DaniH and Argopulos above. $e^{-i\hat{P}a/\hbar}$ is a finite transformation operator acting on the Hilbert space, ...

1

Taylor expand the right-hand side, and you will see that it matches, order by order in $x$, with the left-hand side. You just need to know the definition of $\hat{P}$, i.e. $[\hat{P},\hat{\psi}(x)]=i \hat{\psi}'(x)$, where the prime denotes derivative with respect to the coordinate associated to $\hat{P}$. Of course, you need to remember that the momenta all ...

1

In axiomatic quantum field theory, it is assumed that there is a unique Poincare invariant (projective) state. An arbitrary normalized representative $|0\rangle$ of this state is called the vacuum state. Poincare invariance implies that for all $x$, the state $e^{x\cdot P/\hbar}|0\rangle$ is a multiple of $|0\rangle$. This implies that ...

1

The formulation of this question assumes that it is impossible to have a vacuum state where $\langle0|H|0\rangle > 0$ without violating Lorentz invariance. This is not true. Generally, when there is energy density in the vacuum, you have the appropriate pressure to keep Lorentz invariance, because the stress tensor ends up proportional to $g_{\mu\nu}$ is ...

1

The problem with Poincaré group is in the fact that it is not compact. That's why this question is non-trivial. Though, properly formulated search gives few papers on this topic. Try to find the answer in this paper http://arxiv.org/abs/math-ph/0507056 . The paper itself may be not that interesting, but there is a nice introduction with a number of useful ...

1

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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Dear lurscher, in quantum mechanics - as demonstrated in quantum field theory - particles of the same species are identical so their wave functions are symmetric (for bosons) or antisymmetric (for fermions). If your new hypothetical antiparticle species were physically different, this symmetry or antisymmetry would have to be broken, and this would not be a ...

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