Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to find some names or models of a particle interpretation of quantum field theory which isn't a literal path integral approach? Are there any particle interpretations of quantum field theory which don't use path integrals?

share|cite|improve this question
Please leave the body of the question intact. You may still find helpful answers - maybe even find out that the answer you found was incorrect. It may also be helpful to people in the future. – Michael Brown Mar 14 '13 at 0:08
John, you'll note that you edit has been rolled back. That is because we hope and expect that questions on Physics.SE will not only help the person who asked it, but also be a resource to help other in the future. Toward that end you are encouraged---if you find the answer yourself---to answer your own question so that what you have learned will be available to the next person with the same question. – dmckee Mar 14 '13 at 0:09
The particle intepretation is independent of the calculational tool. You get the same physical result using any formalism: path integral (particles are excited field configurations), canonical (particles are created by field operators acting on Fock space), Schwinger-Dyson equations (particles are poles in the Green functions), ... – Michael Brown Mar 14 '13 at 0:09
I'm currently writing a book on the interpretation of quantum field theory (which you could read online) where you would find, in the first chapter, a formulation of quantum field theory based on the Hamiltonian formalism, so without any path integral. I think it could be interesting to you, but I don't know if you would call it a "particle interpretation"? What do you mean exactly? – Sébastien Fauvel Jun 25 '13 at 7:56

Quantum mechanics has a path integral interpretation, but it also has a description in terms of operators acting on a Hilbert space.

QFT is pretty much (special)relativistic quantum mechanics, so it turns out that particle number is not conserved and you can create/annihilate particles. So it's not enough to have a Hilbert space (fixed number of particles) but a Fock space (space of states of the theory) where you can have an arbitrary number of particles. $\mathcal{F} = \mathcal{H}_{1-particle} \oplus \mathcal{H}_{2-particles} \oplus \mathcal{H}_{3-particles} \oplus \ldots$

QFT can be described as a bunch of operators acting on this Fock space. Just like interacting QM, we can go to the interaction picture (where the Fock space corresponds to the space of states of the free theory, since we don't know the states of the full interacting theory). Here, the time evolution operator can be written in terms of the Dyson series. Each term in the Dyson series is the time-ordered product of a bunch of operators corresponding to the interacting part of the Hamiltonian (since the free part acts trivially on the Fock space). So, you can see that the Dyson series is inherently perturbative as each term corresponds to a higher order in the interaction coupling. One can then define (say) scattering amplitudes as the the Dyson series (representing time evolution) sandwiched between states of the Fock space. That's how one does QFT in the operator picture, without any mention of path integrals.

Let's now go over to the path integral description. In the path integrals picture, when we try do do the integral perturbatively, note that the perturbation series we get is similar to the Dyson series. That's because we've solved the free part (which is an easy Gaussian integral) and trying to solve for the interaction (by expanding out the exponential). That formulation is very similar to what we do in the interaction picture! Essentially, paths which interfere constructively and contribute to the amplitude correspond to events (alternatively) modeled as a bunch of interaction operators acting on a state in the Fock space. So I hope I've also managed to motivate how the two descriptions are connected.

Update: I recently came across this relevant and useful blog post by Lubos.

share|cite|improve this answer

In some sense 2nd quantization is equivalent to path integral approach when introduce quantized fields. But the point is, even if you use 2nd quantization you still need to calculate things like cross sections, which is more related to generating functional of path integral. Also, many things of more advanced QFT are based on path integral. I think you would like Ron's comment on path integral which is more professional: What is the fundamental probabilistic interpretation of Quantum Fields?.

share|cite|improve this answer

Actually, Path Integrals are useful as a method to get physical sense out of a theory. From just the Lagrangian Density of the Standard model:

$${\mathcal L} = - {1 \over 4}{F^{\mu\nu }}{F_{\mu \nu }} + i\overline \psi \not \nabla \psi + \overline \psi \phi \psi + \mbox{h.c.} + {\left| {\nabla \phi } \right|^2} - V(\phi )$$

We can find the Phase, the Kernel, the Wavefunction, etc. of anything excluding gravitation using Path integrals, since:

$$S=\int\int\int\int\mathcal L \mbox{ d}x^4$$ $$\phi=\operatorname{exp}{\frac{iS}{\hbar}}$$ $$K(a,r)=\int\phi\mbox{ }\mathcal D r$$ $$\Psi(r)=\int K\Psi(a) \mbox{ d} a$$ Here, a and r have both spatial and temporal coordinates. $\phi$ is the phase of the path, $K$ is the Kernel, $\Psi$ is the wavefunction and $S$ is the Action.

Of course, if you want to incorportate general relativity in, then, $$S=\int\int\int\int\mathcal L \sqrt{-\det g_{\mu\nu}}\mbox{ d}x^4$$ But, as far as the standard model is concerned, general relativity is not included. That's why the spacetime metric is not found in the Standard Model Lagrangian Density and that's also why the Dirac Equation, Klein-Gordon Equation and Wave Equation used are always taken to be on a Minkowski spacetime.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.