# I Position in QFT

#### Michael Price

Only in nonrelativistic quantum field theory.
The equivalence is shown using 4-D delta functions, so it works in relativistic QFT.

#### A. Neumaier

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.
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).

#### Michael Price

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).
Expand it all out in terms of propagators and point vertices and the equivalence becomes trivial.

#### A. Neumaier

Expand it all out in terms of propagators and point vertices and the equivalence becomes trivial.
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.

#### Michael Price

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.
Representing creation at X and destruction at Y. Or vice versa for the antiparticle.

#### A. Neumaier

Representing creation at X and destruction at Y. Or vice versa for the antiparticle.
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 $x,y,z$.

#### HomogenousCow

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 $x,y,z$.
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?

#### A. Neumaier

Is the whole idea of particle positions muddled by the lost simultaneity in relativistic quantum mechanics?
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.

#### HomogenousCow

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.
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?

#### A. Neumaier

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?
To my knowledge only the problem of a single point particle in a field is tractable, not that of several....

#### andresB

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.
There have been already a lot of answers, I only want to quote the introduction of "quantum electrodynamics" by Beretetskii-Lifshitz-Pitaevskii

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