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 in QFT

  1. Jun 27, 2017 #1
    Hello! So in the Klein-Gordon equation you have a field ##\phi## which becomes an operator in QFT and when you apply it on the vacuum state ##|0>## you get a particle at position x: ##\hat{\phi}(x)|0>=|x>##. So if you look at this particle (in a non interaction theory) the wave function of this particle is a delta function (as you know it's exact position where it was created)? Also if you let the system evolve for a long amount of time, will the wavefunction of this particle turn into a free wave, as there is no potential, so it is basically a free particle? And if you apply the KG operator at the same initial position as before, you don't have 2 particles at position x, you have 1 particle localized at x and the other one (the first one) spread in space, is this right? Also the particle created is a spin 0 particle? Thank you!
    Last edited: Jun 27, 2017
  2. jcsd
  3. Jun 29, 2017 #2


    User Avatar
    Science Advisor
    Gold Member

    I'll address the spin question first. The Klein-Gordon equation applies to any massive particle (and reverts to the wave equation for zero mass). It is agnostic about spin. Spin 1/2 particles which satisfy the Dirac equation also satisfy the Klein-Gordon equation.

    As to the field satisfying the Klein-Gordon equation, it is rather a class of fields which can be partitioned into creators and annihilators of particles but not necessarily of particles localized at the origin or at any point.

    In the mathematical construction of the field theory the textbooks will typically begin by interpreting the field as a field of operators acting at all points of space exciting the local field from the vacuum mode. But you should understand that this field of operators is not effecting an actualized operation. They rather act sort of like a position vector giving relative connection so that one may describe a given quantum actuality in terms of its difference from the vacuum actuality. Further these operators are utilized in combination and in superposition to decompose the physical actions which are actualized... e.g. the propagation of a physical field which can decomposed into a superposition of series of coupled creation-annihilation actions.

    However, within all of that you can construct a creation operator for a particle localized as the origin and said particle should then over time propagate outward as dictated by the KG equation (and we can express this by the equivalence of the initial creation operator to the action of a later non-localized creation operator which creates from the vacuum that same spread out particle).

    And then, yes a subsequent application of the creation operator may effect a situation with two particles however there's more to the story than just the KG equation and even the spin. There is a determination of the particle's statistics which the KG equation does not specify. The statistics of the particle field will slightly modify your description. For example if we assume the particle is a Fermion, no two of which can occupy the same quantum state, then attempting to create a second one at the origin will not yield a 100% 2 particle mode. The earlier particle may still have a finite probability of being at the origin thus, its wave function has a component which is not compatible with creating the 2nd particle. You will get a reduced amplitude indicating a less than 100% chance the result is a 2 particle system.

    It's a bit hard to speak in this way though because again, you don't apply creators in isolation and so the interpretation is subject to inconsistency from the start.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Threads for Klein Gordon
I Klein-Gordon propagator
I Klein Gordon propagator