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I can remember while reading Ryder's book on quantum field theory, i a chapter intro, that Dyson remembered Feynman saying that an electron does anything it likes. It moves at any speed, in any direction, forward and backward in time, and if you add the amplitudes gives you the wavefunction.

Thus it seems that the wavefunction is a kind of cross section in time of all paths taken by a particle. Is this how the wavefunction is connected with QFT? Can we connect the states in Fock space with a wavefunction?

MatterGauge
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1 Answers1

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You may be asking 3.7 questions in an XY problem setup.

  1. To start with, a path integral is just a (powerful) gimmick to propagate an oscillator, and as you see from the diffusion kernel, it sums over an infinity of paths (unphysical, if you will). This is all in NR one particle QM. So, given the initial conditions, a wavefunction is just these, properly propagated thusly.

  2. Now, a one-particle oscillator wavefunction connects to the superior matrix/operator mechanics like this. You could redo the path integral propagation of the $a, a^\dagger$ operators to all times in path integral language, if you so desire.

  3. To go to QFT, you repackage an infinity of such oscillators $a_k, a_k^\dagger$ by decoupling them in momentum space: the famous normal mode reduction. You enforce Lorentz-invariance, much more easily than in any other way, and upload all to the path integral, if so desired.

It is a bad idea, for problem solving, to squeeze NR one particle wavefunctions out of this, but it could be done.... but why? This defocuses you from the wonderful reverse transition to QFT.

Your "cross section of paths" imagery is present in the first paragraph, (1), where the propagator is actually an integral transform giving preference to paths that maximize the contributions of the kernel $K(x',x;t)$ propagating the wavefunction from the initial state to all places and times. The outstanding book by Feynman and Hibbs on the subject is still a go-to classic.

Response to last comment/question

  • By and large, yes. Quantum fields are that mattress, pervading space. Its infinite coupled springs decouple to oscillator normal modes which represent free particles. This is the essence of QFT. Your text should detail it better than my refs. So, in my link to propagators, the propagator for the oscillator represents these notional springs when detached, and that of the free particle, integrating over decoupled amplitudes if you wish, amount to the free particle modes, for each and every E and p... "Real"? Notional? It's a philosopher's dream to tell the difference.
Cosmas Zachos
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  • Thanks for the answer. I just wanted to know if the wavefunctions used in qft, although there being defined as functionals with operators working in Fock space, can, for one particle, can be seen as a cross section of all these paths. For the non-interacting particle. Though some interaction is always needed. All particles would be lost over space. – MatterGauge Mar 02 '22 at 19:41
  • I'm not sure I fully identify your term... do you mean the kets in Fock space? or the wavefunctionals of quantum fields obeying a functional Schroedinger equation in the freak Hamiltonian formulation of qft? Or these guys? – Cosmas Zachos Mar 02 '22 at 20:13
  • ..or you might be talking about this object? $\langle x|\phi(x)|0\rangle $ ? They all have different properties and peculiarities... – Cosmas Zachos Mar 02 '22 at 20:21
  • I mean the free particle states in a Hilbert space. With fixed p and E. "To start with, a path integral is just a (powerful) gimmick to propagate an oscillator" ??? Oscillators being propagated? Sounds even more virtual than virtual particles. Is there really a spring matrass pervading space? – MatterGauge Mar 03 '22 at 06:50