I have a question about the path integral formulation used in QED etc.

The path integral formulation implies to me that index of refraction (and reflection or anything also calculated using this formulation) are dependent on the macro geometry of the object. In the case of a refractive block of glass, the index of refraction should change based on the size of the block, and in the case of a mirror, the actual reflected image of my face would be different a 1m^2 mirror than it would be in a 10m^2 mirror... This is non-intuitive, as the phenomena is unobserved - My face looks the same in a 1m^2 mirror as it looks in a 10m^2 mirror...

Have I missed something? Or is it just that the coefficients in the integral have dropped off to such small numbers it makes no practical difference?

If that is the case, would an infinitely large mirror give a different looking reflection than a 1m^2 mirror? Would it stop reflecting or get better? The integral would be including infinitely many infinitely small terms... Would it slow light down to 0?

  • $\begingroup$ Can you explain how you arrived at that implication? $\endgroup$ – probably_someone Oct 8 at 15:49
  • $\begingroup$ Basically, it's only the paths which are "close" to the classical path matter in the path integral. This is because, near the classical path, the exponent is at a local maximum/minimum, and all the tiny phases "add up" there. Away from a local maximum/minimum, the nearby phases all basically cancel each other out. So making the mirror larger has basically no effect on the probability amplitudes. Light will still basically just hit it and bounce off such that the angle of incidence equals the angle of reflection. $\endgroup$ – user1379857 Oct 8 at 15:50
  • $\begingroup$ @probably_someone I watched some Feynman lectures and videos about refraction on YouTube... $\endgroup$ – Fries of Doom Oct 8 at 15:55
  • $\begingroup$ @user1379857 Your explanation satisfies me wrt. reflection, but the way refraction was explained in youtube.com/watch?v=YW8KuMtVpug seems somehow incompatible with your explanation. But that is to be expected from a YouTube video, which is why I wanted some clarification. Do you know an example of someone going through the math for this? $\endgroup$ – Fries of Doom Oct 8 at 15:57
  • $\begingroup$ That explanation goes for any classical path. So the same argument would go for a beam of light passing through an transparent material. Only the paths surrounding the classical path would matter. $\endgroup$ – user1379857 Oct 8 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.