# Path integrals in QED

1. Jul 17, 2009

### RedX

When calculating a path integral, if the Lagrangian is quadratic in the field, then you can perform the path integral by just substituting in the classical solution for the field.

So if you have free-field Lagrangians for electrons and photons, and add the standard QED interaction term - which is linear in the electromagnetic potential $$j^\mu A_\mu$$ - then can you in principle solve the path integral by plugging in the classical solution for $$A_\mu$$? When you do this, all $$A_\mu$$'s disappear from your Z[J(x)] functional. You still have to perform the path integral for the fermion terms, but the photon terms are out of the path integral now.

2. Jul 17, 2009

### xepma

If you plug in the classical solutions for the electromagnetic field you are obviously not treating this field quantum mechanically. In that case, you deal with a linear perturbation, and yea, the system is quadratic again. But that's not really something you want.. right?

3. Jul 18, 2009

### RedX

The way that QED is taught is to solve the free-field source functional: $$Z_0[J]$$. Then any interaction such as the QED interaction can be gotten from: $$exp(i \int \mathcal L_I(\frac{\partial}{\partial J(x)})d^4x)Z_0[J]$$. The interactions end up being vertices, while the free-field stuff end up being propagators, and you have your usual Feynman rules.

However, if we integrate the photon field out completely by substituting the classical solution of the Lagrangian for the photon field (and this solution will be expressed in terms of the Dirac fields), then you no longer have vertices that involve photons in your interaction! As long as you agree that no external particles are photons (you can set the source J(x) for the photon field equal to zero), then you should be able to do QED without virtual photons at all.

Substituting the classical solution for the Lagrangian is valid quantum mechanically as long as the fields are quadratic.

4. Jul 18, 2009

### StatusX

That's true, though the resulting effective action isn't free, but has an interaction term for the fermions, which I'm pretty sure is non-local.

Last edited: Jul 18, 2009
5. Jul 18, 2009

### RedX

By effective action do you mean the Wilsonian effective action, which is just another way to regulate and renormalize a theory?

Substituting in the classical solution should produce exact results, and is not just an approximation.

When you do substitute in the classical solution, some of your operators might be greater than dimension 4, so the Wilsonian effective action scheme is required to calculate in the theory. Is this how the effective action comes up?

You can integrate out the W boson in the same way (by substituting the classical solution for the W boson) and get Fermi's 4-point interaction. However, in deriving Fermi's 4-point interaction from the Standard Model, approximations are made such as ignoring the kinetic term of W and boson-boson vertices - this is why I think Fermi's 4-point interaction is an approximation, and not exact. But if we choose to solve classically for the W boson while including the kinetic terms and boson-boson vertices, then the resulting theory of weak interactions between leptons should be exact, and not involve any virtual W particles?