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2 votes
1 answer
80 views

Why does $S$-matrix theory end up being a covariant formalism when it is not obvious that it is?

A principle of QFT that is frequently invoked, repeated, and potentially subject to rigorous verification is that the theory in question must exhibit Lorentz covariance and be invariant under the ...
Davius's user avatar
  • 1,640
2 votes
1 answer
127 views

Why do the eigenvalues of the 4-momentum operator organize themselves into hyperboloids?

Specifically I'm asking for the motivation behind figure 7.1 in page 213 of the QFT textbook by Peskin and Schroeder. In that section they just consider eigenstates of the 4-momentum operator $P^\mu=(...
Function's user avatar
  • 151
5 votes
1 answer
219 views

How is Lorentz invariance of $S$-matrix related to vanishing of Hamiltonian density commutator at spacelike separations?

In Section 5.1 of the book, 'Quantum Theory of fields Vol-1' by Steven Weinberg, he says that if the Hamiltonian density commutes with itself at spacelike separation then the $S$-Matrix satisfies ...
Gagan's user avatar
  • 195
1 vote
0 answers
133 views

If the scattering amplitudes are Lorentz scalars, why is S-matrix Lorentz covariant?

All observers should agree on the probabilities: $\mathcal{P}(\mathcal{R}_1 \rightarrow \mathcal{R}_2)$ in an inertial frame $\mathcal{O}$ = $\mathcal{P}(\mathcal{R}_1' \rightarrow \mathcal{R}_2')$ ...
physicsbootcamp's user avatar
2 votes
2 answers
419 views

Usefulness of $| {\rm in}\rangle$ and $| {\rm out}\rangle$ states in S-matrix description of QFT

I am currently reading Niklas Beisert's lecture notes on QFT, Chapter 10, on the scattering matrix $S$.$^1$ My main confusion lies in the construction of $\vert \rm in \rangle$ and $\vert \rm out \...
Sito's user avatar
  • 1,215
4 votes
1 answer
185 views

Weinberg's Coleman-Mandula theorem proof sufficient condition for isomorphism?

In Weinberg's QFT Volume 3 book on Supersymmetry, he presents his own proof of the Coleman-Mandula theorem. As part of the proof, he proves that the only possible internal symmetry generators must ...
fewfew4's user avatar
  • 3,514
3 votes
1 answer
600 views

Confusion over assumptions made in the LSZ reduction formula

I've been reading through a derivation of the LSZ reduction formula (http://www2.ph.ed.ac.uk/~egardi/MQFT_2013/, lecture 2, pages 2-3) and I'm slightly confused about the arguments made about the ...
user35305's user avatar
  • 3,207
2 votes
1 answer
949 views

Scattering Amplitude Not Invariant under Little Group?

I am trying to make sense of scattering amplitude recently. In some literature people say that if some number of massless particles collide together, one can theoretically express the scattering ...
Daps's user avatar
  • 261
2 votes
3 answers
2k views

Derivation of the full generator of the Lorentz transformations

Let us study the subgroup of the Poincare group that leaves the point $x=0$ invariant, that is the Lorentz group. The action of an infinitesimal Lorentz transformation on a field $\Phi(0)$ is $L_{\mu \...
CAF's user avatar
  • 3,569
6 votes
1 answer
1k views

S-operator lorentz invariance

How to show that $\hat {S}$-operator must be lorentz-invariant operator? $$ |\Psi (t)\rangle = \hat {S} | \Psi (0) \rangle , \quad \hat {S} = \hat {T}e^{-i\int \hat {H}_{I}d^{4}x}. $$ I have read ...
Andrew McAddams's user avatar