What is the path integral exactly?

Question

I asked a question here about path integrals and QFT. I just want to confirm something. Is the path integral in quantum field theory a mathematical tool only? I thought the path integral meant that particles literally do take all paths as stated by Victor Stenger and Stephen Hawking in their interpretation of what happens in reality? If particles are field excitations, as QFT says, does that mean Stenger and Hawking's interpretation is right and in reality particles do take all paths (sum over histories)?

Answer 1

I would say that it's only a poetic description of a mathematical device, not a literal description of physical reality. We don't actually get to observe what the particle is doing when we aren't looking at it, so it's not really meaningful to say that it's doing this or that.

If you're reading about path integrals at the popular science level, you should strongly consider reading Feynman's little book QED: The Strange Theory of Light & Matter. It contains a remarkably accurate and BS free popular explanation of how path integrals work. (A small caveat: Feynman's description of renormalization in the last chapter is a bit antiquated. Unfortunately, as far as I know there's not any good explanations of renormalization in the popular science literature. The best thing I can recommend is Wilson's Nobel lecture.)

Answer 2

I do not like the idea that the path integral is 'only a mathematical device', since the path integral is not well-defined for anything but a few special cases. The path integral is the fundamental connection between the classical and quantum worlds; it has real physical meaning in the same way that the action has real physical meaning, although we may not be able to go out and measure it directly.

I think it's a fair interpretation to say that particles and fields take all possible paths (or all possible field configurations) to contribute to the probability of some event occuring. The classical one contributes the most (as we expect), with the quantum corrections coming in as powers of $\hbar$. Once we observe the event, we have selected a specific state $a~la$ quantum mechanics intuition. To me, this is more then a 'mathematical device', since it's not even mathematically well-defined.

Answer 3

Perhaps one of the easiest ways to understand this problem is in terms of the plane wave solutions to the Klein Gordan equation and the process of wave diffraction in the $n$-slit experiments.

As discussed in the opening chapters of A. Zee Quatum Field Theory in a Nutshell, if one can imagine a situation of a particle, whose probability amplitude travels as a plane wave, encounters a panel with $n$-slits, then the diffraction pattern associated with the probability amplitude as it passes through the slits represents the probability of detecting the particle in some location at some time after passing through the slits. The sum of each amplitude associated with each point will give us the total probability of detecting the particle.

If the panel has an infinite number of slits, or more importantly an uncountably infinite number of slits, one would see that the plane wave represents the total probability of finding the particle at any point in space and time.

Plane waves are of course infinite in extent along one axis. We of course will never directly observe a particle traveling faster than light, however we need to consider the amplitudes of "off-shell" virtual particles must be considered and the vacuum itself is allowed to use such particles as long as they obey the uncertainty principle. Since virtual particles are always paired, this suggests the probabilities associated with seeing a particle traveling faster than light effectively cancel each other out.

This of course is enshrined by enforcing the commutation relationship:

$$[\phi(x),\phi(y)]=0$$ for fields outside each other's light cones.

This enforced by the combined use of the dirac delta function $\delta$ and the theta function $\theta$ inside the equation for propagators which restricts us to only consider values that are on the upper light cone and on-shell

$$\Delta_c(z)_{z_0 = 0} =\dfrac{1}{(2\pi)^3} \int d^4q \{e^{i(\vec{q}\vec{z})} - e^{-i(\vec{q}\vec{z})}\} \theta(q_0)\delta(q^2+m^2) $$

In the case of the general solution to the Klein-Gordon equation for single particle

$$\phi(x) = \sum_{\vec{p}} \dfrac{1}{\sqrt{2p_0V}}\{a(\vec{p})e^{i(px)} + a^{\dagger}(\vec{p})e^{-i(px)}\}$$ where $$p_0 = \sqrt{\vec{p}^2+m^2}$$

where $a(\vec{k})$ is the annihilation operator and $a^{\dagger}(\vec{k})$ is the creation operator, and because the equation for $p_0$ enforces the cancellations.

Effectively, one is accounting for the paths of particles along any path, while simultaneously showing that those paths will cancel each other out.

Answer 4

You have two descriptions which are equivalent (so long as the theory has certain nice/convenient properties). Both descriptions are equally good and can in-principle be used to solve QFT problems. In practive, one approach might be more convenient for some special cases. So which description is more fundamental? You tell me.

I don't think that's a question we can answer based on the physics we know today. They really seem to be pretty much equivalent and you can't distinguish between the two. So what's "fundamental" or "real" depends on what your notion of fundamental is. For some, it's the notion of particles and operators and states. For me, it's the democracy encoded in path integrals which is elegant and simple.