How Did Feynman Determine Specific Arrow Sizes and Angles in QED?

  • Thread starter Thread starter BiBi
  • Start date Start date
  • Tags Tags
    Turning
AI Thread Summary
Feynman determines specific arrow sizes and angles in QED through the principles of complex arithmetic, which he simplifies by referring to amplitudes as "little arrows." The lengths of these arrows represent probabilities, and their angles correspond to phase information. When multiplying two complex numbers, the resulting arrow's length is the product of the original lengths, while the angle is the sum of the original angles. This results in a shorter product since the initial lengths are less than one. The discussion highlights the mathematical foundation behind Feynman's visual representation of quantum mechanics, emphasizing the relationship between complex numbers and the behavior of photons.
BiBi
Messages
4
Reaction score
0
In Richard Feynman's book QED, when writing about multiplying arrows of photons (on page 61 if you have the book handy)he shrinks and turns them at specific numbers, and though I understand why and how it's effective, I don't understand how he determined to shrink and turn the arrows by those specific amounts. If anyone knows, when shrinking and turning arrows, how you come up with those amounts and numbers I would be grateful.
 
Physics news on Phys.org
The shrinking and turning of the arrows is a feature of the complex arithmetic that Feynman didn't want to get into, which is why he chose to speak of the amplitudes as "little arrows" rather than complex numbers. If you have two complex numbers they can be written u cos \theta + iu sin \theta and v cos \phi + iv sin \phi, where u and v give the lengths and the angles are the ones the two vectors make with the real axis.
If you multiply them you get after simplification uv cos(\theta + \phi) + iuv sin (\theta + \phi). So the product has length equal to the products of the two factors and its angle is the sum of theirs. Since the lengths were less than 1 to begin with, the product uv is smaller still, therefore the product is shorter than the factors. And the sum of the angles gives the rotation.
 
thank you so much! I've been asking my friends for almost a week now, but no one had done QED for a while, if at all. thanks again.
 
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
Back
Top