Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Klein Gordon operator

  1. May 3, 2017 #1
    Hello! I read that the Klein-Gordon field can be viewed as an operator that in position space, when acted upon vacuum at position x creates a particle at position x: ##\phi(x) |0 \rangle \propto |x \rangle##. It make sense intuitively and the mathematical derivation is fine too, but I was wondering what does it physically means, as according to uncertainty principle, you can't have a localized particle. So what exactly does it mean to create a particle at position x?
     
  2. jcsd
  3. May 4, 2017 #2

    DrDu

    User Avatar
    Science Advisor

    You can localise a particle as much as you like. Uncertainity principle only tells you that your knowledge about the momentum of the particle will decrease the more you localise it.
     
  4. May 4, 2017 #3
    I know this. But based on this definition of Klein-Gordon operator, you know the position of particle with 0 error, which means that you should know the momentum with infinite error, which is, the particle can have any momentum from 0 to infinity. But this is not possible, as the particle can't travel faster than light and you know the mass of the particle as you know what particle you create, and the particle is not virtual, so its mass is known (of course with a certain uncertainty, but again a finite one). Is this logic wrong?
     
  5. May 4, 2017 #4

    DrDu

    User Avatar
    Science Advisor

    The momentum of a relativistic particle can get as large as you like, yet it's velocity is always smaller than c.
    Namely ## p=\frac{mv}{\sqrt{1-v^2/c^2}}##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted