Physics, asked by anshulsood6032, 1 year ago

Normalising a massive propagator?

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Answered by vrrunda
0
If you take a propagator Greens function for a massless photon from the origin as

G(0,x)=1x2−t2+iεG(0,x)=1x2−t2+iε

then the normalisation factor at time t is:

N=∫|G(0,x)|2d3x=∫1(x2−t2)2+ε2d3x≈|t|εN=∫|G(0,x)|2d3x=∫1(x2−t2)2+ε2d3x≈|t|ε

(i.e. the probability for a particle being anywhere at time t should be 1). So dividing by the normalisation factor the propagator is 0 outside the light cone in the limit ε→0ε→0. So the probability density is:

P(x,t)∝limε→0ε((x2−t2)2+ε2)|t|.P(x,t)∝limε→0ε((x2−t2)2+ε2)|t|.

I am trying to find the same reasoning for the massive propagator, e.g. an electron to show an electron can't travel faster than light, but I am getting wrong results that an electron always will be on the light cone. Any idea how to prove it?

Edit: Actually, in fact if an electron was known to start at a single point then the momentum would be completely unknown and hence it could be any momentum from 0 to infinity. Thus the electron would with probability 100% be going at the speed of light and stay on the light cone.

Answered by Anonymous
0
does not depend on batch statistics for normalizing the in- ... Normalization Propagation (NormProp) ...... half the mass of Y is concentrated on R+ with Half-Normal ...
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