# Do we really need a cutoff in QFT

by geoduck
Tags: cutoff
 P: 231 Does the cutoff really have to go to infinity in QFT? It seems that once we replace bare parameters by experimental (i.e. physical) parameters, the cutoff vanishes from the expressions for physical quantities, so it didn't matter what the value of the cutoff was, whether it's 0, 10000000000, or infinity. So do you really have to take the step of setting the cutoff to infinity after replacing bare parameters with physical parameters? It doesn't do anything, right, since the final expression doesn't involve the cutoff at all, so the cutoff can be any value and that wouldn't change the value of physical quantities. Also, say the cutoff is 10000000000, and you conduct an experiment at the same energy (10000000000). Does the bare parameter then equal the experimental parameter?
 Sci Advisor P: 2,470 In practice, nobody actually sets cutoff to infinity. They set it to a value that's high enough for results to converge.
P: 231
 Quote by K^2 In practice, nobody actually sets cutoff to infinity. They set it to a value that's high enough for results to converge.
Do you work with a cutoff, or a mass scale μ?

I thought usually the cutoff was traded for a mass scale μ, and this occurs when you replace bare parameters with physical parameters. You can vary the mass scale μ to get your results to converge, but the cutoff should vanish once μ appears in your equations.

 Sci Advisor P: 2,470 Do we really need a cutoff in QFT Mass scale, of course. But that's basically what you use to get away with using a finite cutoff. Is there another way of doing that? I wasn't aware of one.
P: 231
 Quote by K^2 Mass scale, of course. But that's basically what you use to get away with using a finite cutoff. Is there another way of doing that? I wasn't aware of one.
The mass scale remains even when the cutoff goes to infinity.

If you can integrate out all fields with momentum greater than the cutoff in the path integral for the action, then that can be viewed as a renormalization group transformation that produces a new action with new coefficients and even new, unrenormalizeable interactions. Using this new action, you only need to integrate to a finite cutoff. I'm not sure if a mass scale is every introduced in this scheme, but rather you rely on the finite cutoff, and figure out how the coefficients change with cutoff.

Since the new action is numerically equal to the old action, the physics remains the same, only now you can't calculate correlators of momentum greater than the cutoff since you coarse-grained everything. But for low energy calculations this is not a problem.

Taking the cutoff to infinity implies the distance scale goes to zero which implies the correlation length goes to infinity which implies all theories where the cutoff goes to infinity are based on fixed/critical points. I'm not exactly sure what the point is of interpreting renormalization this way. If you don't force the cutoff to infinity, then your theory doesn't have to be based on a fixed point, but in any case this doesn't help you at all with the calculations.