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Do we really need a cutoff in QFTby geoduck
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#1
Mar2712, 07:30 PM

P: 229

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? 


#2
Mar2712, 07:35 PM

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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.



#3
Mar2712, 08:23 PM

P: 229

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. 


#4
Mar2712, 11:18 PM

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.



#5
Mar2812, 01:05 AM

P: 229

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 coarsegrained 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. 


#6
Mar2812, 01:39 AM

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#7
Mar3012, 09:05 PM

P: 229

I've had time to do a little more reading, and evidently it's all understandable from Wilson's renormalization group flow.
The idea is that you have a cutoff Lambda and some coupling constants g_i. If you change Lambda, the renormalization group dictates you must change the coupling constants g_i to give you the same physics, so that g_i=g_i(Lambda). Now imagine a multidimensional phase space, with the g_i's as the axes. You begin your system with some cutoff Lambda_init and coupling constants g_i_init. If you alter the parameter Lambda_init to Lambda', your point g_i_init in phase space flows to a new point g'_i(Lambda'). Now there are two ways you change Lambda_init: you can go up or down. Going up is what you're doing when you take the cutoff to infinity. You go down by integrating out higher momentum fields. Once you integrate out higher momentum fields, in all your loop integrals you don't need to integrate up to that high of a momentum since you integrated it directly out of the path integral in the beginning to give you new effective Lagrangian. The mass scale μ corresponds to integrating out momenta from Lambda to Lambda/b, where b>1: μ=Lambda/b What you do in renormalization is that you set μ constant and take the cutoff Lambda to infinity. That is, your renormalized couplings don't change even though you take the cutoff to infinity. Why is this so? It's because what you do is take Lambda to infinity, but then you go back down by increasing b in such a way that: μ=Lambda/b is constant. So when you increase Lambda, you're moving up the trajectory in phase space, but after you do that, you go back down by integrating out more modes that are less than Lambda but greater than Lambda/b. So the cutoff can have any arbitrarily high value which will move you up the RG trajectory, but your parameters won't change because you just move back down by integrating out more modes. All this is only possible if your theory is based on an ultraviolet fixed point so that if your cutoff is exactly equal to infinity, you arrive at the fixed point in the opposite direction of integrating out momenta. If the theory is based on an IR fixed point then as you take your cutoff to infinity you actually move towards the same direction as integrating out momentum so you can't undo taking the cutoff to infinity. Note also that when μ=cutoff, your wavefunction renormalization Z_phi goes to 1 as all your twopoint loop integrals go to zero. This is good because it is demanded by invariance of physics under renormalization group transformation. μ=cutoff corresponds to b=1, or just the bare theory. Because in the fundamental (bare) theory, there is no such thing as Z_phi. Anyways, I might have some things wrong, but this is how I understand it so far from reading some beginning books and papers on the renormalization group (from the viewpoint of phase transitions and critical phenomena rather than particle physics though). I have a presentation to give on this in 3 weeks, so I'll be studying this further. 


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