Michael Price
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The equivalence is shown using 4-D delta functions, so it works in relativistic QFT.A. Neumaier said:Only in nonrelativistic quantum field theory.
The equivalence is shown using 4-D delta functions, so it works in relativistic QFT.A. Neumaier said: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).vanhees71 said: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.A. Neumaier said: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.Michael Price said: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.A. Neumaier said: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##.Michael Price said:Representing creation at X and destruction at Y. Or vice versa for the antiparticle.
A. Neumaier said: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.HomogenousCow said:Is the whole idea of particle positions muddled by the lost simultaneity in relativistic quantum mechanics?
A. Neumaier said: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...HomogenousCow said: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?
WWCY said:He 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.A. Neumaier said: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).