|
| |
"http://wwwinformationdelightinfo/encyclopedia/entry/QFT" class="copylinks">QFT with a conserved Stress-energy Tensor which is Poincaré Covariant (and Gauge Invariant if there happens to be any Gauge Symmetry which hasn't been Gauge-fixed ) does not admit massless particles with helicity ''h'' > 1
|
| |
"http://wwwinformationdelightinfo/encyclopedia/entry/S-matrix" class="copylinks">S-matrix element (assuming the S-matrix exists, which is a nontrivial assumption in theories without a Mass Gap however, we're only interested in one particle to one particle states, which doesn't presuppose the existence of an S-matrix) <math>\langle p'J^\mu(0)p
angle</math> where ''p''> and ''p''′> are the one-particle states of the massless charged particle with helicity ''h'' and 4-momentum ''p'' and ''p''′ respectively Let's assume that the sign of the helicity of both particles are the same (after all, we can't rule out self-dual particles with only one sign for the helicity!) Of course ''p'' and ''p′'' lie on the boundary of the forward Light Cone Let's look at the case where <math>p - p'</math> isn't a Null Vector (ie Lightlike Momentum Transfer ), which means that the momentum transfer is Spacelike
|
| |
\langle p'Qp
angle
|
| |
\int d^3x\, \langle p'J^0( ec{x},0)p
angle
|
| |
rac{q}{(2\pi)^3}</math>
|
| |
e^{2ih heta}</math> under Rotation s by θ counterclockwise about the ''z''-axis whereas <p'J<sup>1</sup>(0)+iJ&2(0)p> and <p'J<sup>1</sup>(0)-iJ&2(0)p> change by the phase factors <math>e^{i(2h+1) heta}</math> and <math>e^{i(2h-1) heta}</math> respectively
|
| |
"http://wwwinformationdelightinfo/encyclopedia/entry/Thomas_precession" class="copylinks">Thomas Precession ), but the One Particle Hilbert Space is Lorentz-covariant and so it still makes sense to speak of the Lorentz invariance of the S-matrix So, if we make any arbitrary but fixed choice for the phases, then each of the matrix components in the previous paragraph has to be invariant under the rotations about the ''z''-axis So, unless ''h'' = 0 or 1/2, all of the components have to be zero
|
| |
\lim_{p'
ightarrow p}\langle p'J^0(0)p
angle</math>,
|
| |
rac{q}{(2\pi)^3}</math>, <math>\langle p'J^\mu(0)p
angle </math> can't be zero for all spacelike momentum transfers (This is because the case of zero momentum transfer is the limit of a sequence of spacelike momentum transfers) But this is only possible if <math>h=0, rac{1}{2}</math>
|
| |
rac{E}{(2\pi)^3}</math>
|
| |
0, rac{1}{2},1</math>
|
| |
"http://wwwinformationdelightinfo/encyclopedia/entry/Yang-Mills" class="copylinks">Yang-Mills theories in the Coulomb Phase don't violate this theorem Yang-Mills theories don't have any conserved 4-current associated with the Yang-Mills charges that are both Poincaré covariant and gauge invariant Noether's theorem gives a current which is conserved and Poincaré covariant, but not gauge invariant As ''p''> is really an element of the BRST cohomology, ie a Quotient Space , it is really an equivalence class of states As such, <math>\langle p'Jp
angle</math> is only well defined if J is BRST-closed But if ''J'' isn't gauge-invariant, then ''J'' isn't BRST-closed in general The current defined as <math>J^\mu(x)\equiv rac{\delta}{\delta A_\mu(x)}S_\mathrm{matter}</math> is not conserved because it satisfies <math>D_\mu J^\mu=0</math> instead of <math>\partial_\mu J^\mu=0</math> where D is the Covariant Derivative The current defined after a gauge-fixing like the Coulomb Gauge is conserved but isn't Lorentz covariant
|
| |
"http://wwwinformationdelightinfo/encyclopedia/entry/superselection_sector" class="copylinks">Superselection Sector s Let's say they have momenta ''p''′ and ''p'' respectively Then as ''J''<sup>μ</sup>(0) is a local neutral Operator , it does not map between different superselection sectors So, <math><p'J^\mu(0)p></math> is zero The only way ''p''′'> and ''p''> can belong in the same sector is if they have the same velocity, which means that they are proportional to each other, ie a null or zero momentum transfer, which isn't covered in the proof So, infraparticles violate the continuity assumption
|
| |
\lim_{p'
ightarrow p}\langle p'J^0(0)p
angle </math>
|
The stress-energy operator is defined as a
Vertex Operator corresponding to this infinitesimal change in the background metric.
Unfortunately (or fortunately), not all backgrounds are permissible. Superstrings have to have
Superconformal symmetry, which is a super generalization of
Weyl Symmetry , in order to be consistent but they are only superconformal when propagating over some special backgrounds (which satisfy the
Einstein Field Equation s plus some higher order corrections). Because of this, the effective action is only defined over these special backgrounds and the functional derivative is not well-defined. The vertex operator for the stress-energy tensor at a point also doesn't exist.
