True, during acceleration I am not an inertial observer. But after I released the accelerator pedal, I move with a constant speed and I may regard myself as an inertial observer, which is boosted with respect to my previous state.
I agree that this is not a perfect analogy for boost, but not a...
By definition, boost is a transformation that changes velocity of reference frame.
When I press accelerator pedal, velocity of my car changes. Then the inertial reference frame associated with me and my car experiences a boost.
Eugene.
Imagine yourself in a car. You press accelerator pedal (this is boost), then you wait for some time (=time translation). Obviously, after these two transformations you'll find yourself far from the place you've started from.
In reverse order: You wait for some time (=time translation), then you...
Somewhat related to this discussion. Decay law of a moving unstable particle was calculated in a number of articles. Few examples:
E. V. Stefanovich, Quantum effects in relativistic decays. Int. J. Theor. Phys. 35 (1996), 2539
M. Shirokov, Decay law of moving unstable particle. Int. J. Theor...
S-matrix is a mapping from (free) states in the remote past to (free) states in the distant future. So, basically S-matrix can be regarded as a result of integrating the time evolution in an infinite time interval.
The Hamiltonian of renormalized QED contains divergent counterterms. So...
I was talking about renormalization in the context of removing UV divergences from the S-matrix of QED as explained by Tomonaga, Schwinger, Feynman and Dyson.
Eugene.
@LittleSchwinger: Thank you for the kind words about my work.
The authors of this work used the "standard" Lagrangian of QED in equation right after eq. (2.56). I have two major issues with their approach:
1. This is the Lagrangian before renormalization. At page 295 the authors expressed a...
There is an approach called "dressed particle" QFT or "clothed particle" QFT, which succeeds in formulating a well-defined finite Hamiltonian in the Fock space. With this Hamiltonian one can
1. Reproduce S-matrix of the traditional renormalized QFT in all perturbation orders.
2...
Are you saying that interaction changes particle observables/operators?
Then our definitions of "interaction" are quite different. In my view, relativistic interaction should be introduced in a theory by adding operators V and W to generators of the non-interacting representation of the...
How did your reach this conclusion?
Theoretically, Newton-Wigner position operators of particles are defined separately in each N-particle sector of the Fock space. So, by definition, NW position commutes with all particle number operators. They are measurable simultaneously. In theory.
Eugene.
The principle of relativity does not say that all observers must see the same things. If the particle looks localized for one observer and delocalized for another observer, that's OK with the principle of relativity. This principle only insists that descriptions of different observers should be...
Rules of quantum mechanics say: if position commutes with the number of electrons, then there should exist states with sharp values of position and a definite number of electrons (e.g., N=1). QM does not specify "physical mechanisms" of measurements.
Eugene.
I would like to disagree with Landau on these points.
1. One may have troubles with the "negative frequency" solutions, if one regards one-particle states as solutions of wave equations (such as the Dirac equation). But this kind of thinking and confusion was characteristic for early days of...