Transition from QFT to Classical Physics using Path Integrals?

In summary: But just the fact that the quantum lagrangian is the same as the classical lagrangian means that, in suitable limits, the physics is the same.
  • #1
RedX
970
3
I have a question in Srednicki's book regarding path integrals, but first I'll set it up so that no familiarity of the book is required to answer the question.

The vacuum to vacuum transition amplitude for the photon field in the presence of a source is given by: [tex]<0|0>_J=\int \mathcal D A e^{iS+\int J_\mu A^\mu}[/tex] where [tex]S=\int -\frac{1}{4}F_{\mu \nu}F^{\mu \nu} d^4x[/tex]. The result is [tex]e^{i\int \int J^\mu(x)\Delta(x-y)_{\mu \nu} J^\nu(y)} [/tex]. If you rewrite J, the current density, in terms of the line integral [tex] e \oint dx^\mu [/tex], and choose a path for the line integral, then this vacuum to vacuum transition amplitude, being equal to [tex]e^{-iE_{0}t [/tex], should give you the energy of the situation described by your line integral. In this way by choosing the path where an electron and anti-electron are just sitting a small distance R apart for a long time T, you actually recover the Coloumb force from QFT, [tex]V(R)=-\frac{\alpha}{R}[/tex]. This is equation (82.22) Srednicki.

My question is what does it mean to have the vacuum to vacuum amplitude in the presence of a source? If there is a source then is it really vacuum? Or does the vacuum mean it's a vacuum with respect to only photons, but you are allowed to have a fermion source [obviously J(x)]? If this is the case, then could you in principle use this same method but use the Dirac Lagrangian in place of the photon Lagrangian, and have the source J(x) be a photon source, and from this derive the energy between two photons? If one wants to derive Maxwell's equations (and not just Coloumb's law) from the path integral approach, how would you do this?

Also, this is the first time I've ever seen QFT used to describe something in position space. Is this the standard way to check if QFT gives the same results as classical physics - vacuum expectation values in presence of sources?

The coupling constant in QED gets big if the energy is high, and high energy is equivalent to short distances. Is this reflected in Coloumb's law with the 1/r^2 dependence? The beta function and renormalization seemed to have disappeared from this example, so this must mean that the 1/r^2 dependence is not due to changing coupling constants with energy?
 
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  • #2
RedX said:
If there is a source then is it really vacuum?
Not when the source is "on". The idea is that the source is "off" (equal to zero) before some time t1 and after some later time t2, and that the initial state (before t1) is the vacuum. Then the computed amplitude is the amplitude to find the system in the vacuum state after time t2.
RedX said:
could you in principle use this same method but use the Dirac Lagrangian in place of the photon Lagrangian, and have the source J(x) be a photon source, and from this derive the energy between two photons?
No, for the same trick to work the source has to couple just to the field, so you would need a source like Srednicki's eta that couples to the Dirac field psi.
RedX said:
Also, this is the first time I've ever seen QFT used to describe something in position space. Is this the standard way to check if QFT gives the same results as classical physics - vacuum expectation values in presence of sources?
Of course, QFT does not give the same results as classical physics; it's quantum! But just the fact that the quantum lagrangian is the same as the classical lagrangian means that, in suitable limits, the physics is the same.
RedX said:
The coupling constant in QED gets big if the energy is high, and high energy is equivalent to short distances. Is this reflected in Coloumb's law with the 1/r^2 dependence?
No, the 1/r^2 is a classical effect. Quantum effects effectively make the electron charge a function of r, and e^2 grows logarithmically with r (though this language is somewhat misleading; a more accurate statement is that there are quantum corrections to Couloumb scattering amplitudes that can be summarized by taking e^2 to be a function of r).
 
  • #3
Avodyne said:
No, for the same trick to work the source has to couple just to the field, so you would need a source like Srednicki's eta that couples to the Dirac field psi.

I'm guessing there's no easy way to have [tex]\eta (x)[/tex] describe a photon, whereas it's easy to have [tex]J(x)[/tex] describe a fermion.

I was hoping one could calculate the energy between two photons because I heard somewhere that QFT says that this vacuum energy should be infinite, which is troublesome because cosmology says it ought to be zero (the cosmological constant or something - it's been awhile since I looked at general relativity).

Of course, QFT does not give the same results as classical physics; it's quantum! But just the fact that the quantum lagrangian is the same as the classical lagrangian means that, in suitable limits, the physics is the same.

At first I was amazed that Srednicki, in deriving (82.22), seem to have got a value for the self-energy. And indeed that term does come from integrating over the same line in the rectangular Wilson loop. But then I thought this was sweeping the self-energy under a rug, because the reason that this term is not infinite is because we assumed space-time is discrete via a lattice spacing, so of course there is no self-energy because a continuum doesn't exist! So in a way that's pretty cheap.

I'm unconvinced of equation (82.18), or the "perimeter law". So that's basically saying that if you keep the aspect ratio of a certain Wilson loop the same (to maintain shape or c-tilda), then the value of vacuum expectation value will vary as the perimeter P. Just looking at this statement seems wrong. If you imagine two points infinitismally close together, and connect them with a rigid rod (so that their distance is constant), and move them along the loop, then I can see that the integral is proportional to the perimeter, because if you travel twice the distance on a path then that's twice the distance you integrate over a function whose value you insert the length of the rod. But clearly this doesn't work for a rod of finite length. It seems this type of problem would come up in self-inductance in classical E&M, but I can't find a book that shows this. I should probably check Jackson's book.
 

1. What is the concept of Path Integrals in Quantum Field Theory (QFT)?

Path Integrals are a mathematical tool used in QFT to calculate the probability of a particle moving from one point to another in a given amount of time. It involves summing over all possible paths that the particle could take, taking into account the different interactions and forces acting on it.

2. How do Path Integrals relate to Classical Physics?

In the classical limit, where the physical systems are large and the quantum effects are negligible, Path Integrals reduce to the classical action, which is the integral of the Lagrangian over time. This means that the classical equations of motion can be derived from the quantum theory using Path Integrals.

3. What is the significance of the transition from QFT to Classical Physics using Path Integrals?

The transition from QFT to Classical Physics using Path Integrals allows us to connect the descriptions of physical systems at the quantum and classical levels. This is important in understanding the behavior of physical systems and making predictions about their behavior.

4. Can Path Integrals be applied to all physical systems?

Yes, Path Integrals can be applied to any physical system, as long as it can be described by a Lagrangian or Hamiltonian. This includes both quantum and classical systems.

5. Are there any limitations to using Path Integrals in the transition from QFT to Classical Physics?

One limitation is that Path Integrals can become computationally challenging for systems with a large number of particles or interactions. In these cases, alternative methods may be used to study the transition from QFT to Classical Physics.

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