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The equivalence is shown using 4-D delta functions, so it works in relativistic QFT.Only in nonrelativistic quantum field theory.
The equivalence is shown using 4-D delta functions, so it works in relativistic QFT.Only in nonrelativistic quantum field theory.
But if you take 4D Fourier transforms then the N-point Green's functions get N times, and can for N>2 no longer be interpreted in terms of inner products of multiparticle states (where there is one time only but multiple space coordinates).Feynman diagrams depicting S-matrix elements lead to the same result in both the space-time and the energy-momentum representation. The corresponding (amputated) time-ordered ##N##-point Green's functions are Fourier transforms of each other.
Expand it all out in terms of propagators and point vertices and the equivalence becomes trivial.But if you take 4D Fourier transforms then the N-point Green's functions get N times, and can for N>2 no longer be interpreted in terms of inner products of multiparticle states (where there is one time only but multiple space coordinates).
I don't complain about the equivalence but about the interpretation that you give in post #38.Expand it all out in terms of propagators and point vertices and the equivalence becomes trivial.
Representing creation at X and destruction at Y. Or vice versa for the antiparticle.I don't complain about the equivalence but about the interpretation that you give in post #38.
Applying two field operators gives a state with two time variables.
And what does ##\phi(x)\phi(y)\phi(z)|vac\rangle## with 4D ##x,y,z## represent? Surely not an ordinary wave function ket, which would be ##|x,y,z\rangle## with 3D position vectors ##x,y,z##.Representing creation at X and destruction at Y. Or vice versa for the antiparticle.
Is the whole idea of particle positions muddled by the lost simultaneity in relativistic quantum mechanics? As in, how do we make sense of unitary time evolution when particle positions defined at the same times are boosted to different times?And what does ##\phi(x)\phi(y)\phi(z)|vac\rangle## with 4D ##x,y,z## represent? Surely not an ordinary wave function ket, which would be ##|x,y,z\rangle## with 3D position vectors ##x,y,z##.
Not only that of particle positions but that of more than one particle.Is the whole idea of particle positions muddled by the lost simultaneity in relativistic quantum mechanics?
Wasn’t there a recent textbook on classical electrodynamics where the problem of charged point particles interacting via dynamical fields was given a satisfactory treatment?Not only that of particle positions but that of more than one particle.
Even in classical relativity, there is no good relativistic multiparticle theory: see
- Currie, Jordan and Sudarshan, Relativistic Invariance and Hamiltonian Theories of Interacting Particles, Reviews of Modern Physics 35 (1963), 350.
To my knowledge only the problem of a single point particle in a field is tractable, not that of several....Wasn’t there a recent textbook on classical electrodynamics where the problem of charged point particles interacting via dynamical fields was given a satisfactory treatment?
There have been already a lot of answers, I only want to quote the introduction of "quantum electrodynamics" by Beretetskii-Lifshitz-PitaevskiiHe said that this was down to the fact that we should be considering multi-particle states in relativistic situation, before introducing Fock-space states.
Of course not. The QFT formalism automatically takes care of the Bose/Fermi (anti-)symmetrization.But if you take 4D Fourier transforms then the N-point Green's functions get N times, and can for N>2 no longer be interpreted in terms of inner products of multiparticle states (where there is one time only but multiple space coordinates).