Infinities and short distances in QFT

[geoduck]

Infinities in QFT come from high momenta. I sometimes hear that is equivalent to coming from short distances, but I'm not sure I see the connection.

The free propagator G(x-y) which I think goes like 1/|x-y|^2 is singular for short distances (when x=y). In momentum space G(x)= ∫d^4k exp[ikx] /(k^2+m^2). If x=0, then clearly ∫d^4k exp[0] /(k^2+m^2) diverges for large momenta k. However, if x is not equal to zero, then the expression is perfectly finite even when k is allowed to be really large in the integral. So it seems distance is even more important than momenta in determining the divergence of the free propagator.

Also, for a phi^4 theory it's true that the self-energy at one loop is G(z-z)=G(0) where z is the vertex coordinate that is integrated over. This leads to infinity. So you can say short distances leads to infinities.

However, for phi^3 theory, it is this term which leads to infinities: G(y-z)G(y-z) where y and z are the vertex coordinates that are integrated over. You have an infinity here that's caused by repetition of the Green's functions rather than y=z.

So for the phi^3 case, the infinities aren't caused by short distances between the vertices.

So are infinities really explained by short distances?

 

[andrien]

high energy and short distance behaviour are same in meaning,not infinity.you can find theories which will behave badly in high energy limit.there will be infinities associated with it.they are said to be divergent at high energy or short distance.If a theory is renormalizable i.e. have coupling strength dimension as positive powers of mass(or zero) then they may not diverge badly at high energy behaviour or short distances.

 

[king vitamin]

Infinities in QFT come from high momenta. I sometimes hear that is equivalent to coming from short distances, but I'm not sure I see the connection.

The free propagator G(x-y) which I think goes like 1/|x-y|^2 is singular for short distances (when x=y). In momentum space G(x)= ∫d^4k exp[ikx] /(k^2+m^2). If x=0, then clearly ∫d^4k exp[0] /(k^2+m^2) diverges for large momenta k. However, if x is not equal to zero, then the expression is perfectly finite even when k is allowed to be really large in the integral. So it seems distance is even more important than momenta in determining the divergence of the free propagator.

Your argument for the Green's function being finite for infinite momenta is still being formulated in position space though. Since 1/|x-y|^2 ~ ∫d^4k exp[ik(x-y)] /(k^2+m^2), your statement that both functions diverge when x=y are the same.

When one refers to loop corrections being infinite in a QFT before renormalization, one can do so either purely in position space or momentum space. Consider the one-loop self-energy correction in phi^3 theory. Using position-space Feynman rules, we have ∫d^4xd^4y G(x-y)G(x-y) being infinite in the region where x=y. Using momentum-space Feynman rules, we get ∫d^4k G(k)G(p+k) being infinite where k goes to infinity. These are the two equivalent statements that field theories give infinities for short distance/large momenta.