Why is cut-off regularization is not Lorentz invariant?

Why is it said that the cut-off regularization is not a Lorentz invariant regularization method?

• I believe it's simply because wavelength / frequency are not Lorentz invariant scalars, so that in principle one could detect uniform motion relative to the quantum ground state by measuring the cutoff. I should warn that my QFT knowledge ends pretty much with this book: amazon.com/Quantum-Electrodynamics-Advanced-Books-Classics/dp/… Jun 24, 2015 at 9:51

The UV cutoff procedure that is used to regulate Feynman diagrams must be chosen so that the region of integration involving large values of $p^\mu$ (all components individually) is cut-off. A Lorentz invariant cut-off such as $p^2 < \Lambda^2$ does not achieve this.

However, what we can do is to Wick rotate $p^0 \to - i p^0$ and then impose a cut-off on $p_E^2 < \Lambda^2$. The latter cut-off does exactly what we require it to do. In fact, this Wick rotation is convenient for other purposes. In any case, it is clear that such a cut-off is not Lorentz invariant.

• I think this is a +1, but a naive question from a non quantum field theorist: why should imaginary frequencies not transform like real ones? In systems theory, if I write a complex frequency $s = \sigma + i\,\omega$, then the real part describes damping whilst the imaginary describes oscillation: both phenomena are simply different kinds of evolution with time, and so therefore should be changed in exactly the same way by a Lorentz boost. PS: If this seems like a detailed question, I'll ask it on the main list if you like, although I may not be able to understand the answer. Jun 24, 2015 at 9:57
• @WetSavannaAnimalakaRodVance: Wick rotation is not really "imaginary frequency", it is the passage to a Euclidean theory instead of a Minkowskian one, where the convergence properties of the path integral are much nicer. Mar 3, 2016 at 17:44
• @ACuriousMind You know what? You've just taken a thorn of misunderstanding out from my side that's been there for years. That makes beautiful sense! Mar 5, 2016 at 9:34

Reason 1: Rotational Invariance Is Broken

Consider the following integral: $$\int d^4l\, F(l^2)\,l_\mu$$ in general, that is don't yet think about "cut-off" or anything. This integral is zero because $l_\mu$ is not rotationally invariant while everything else ($d^4l\, F(l^2)$) is. Now, when one imposes a "cut-off" on $l$ which is equivalent to moving from a continuous space-time to a discrete lattice. In the continuous space-time suppose $l$ is directed along a certain direction, you rotate the whole continuous space-time such a way that now in this new rotated frame the original vector $l$ becomes $-l$ and as $d^4l F(l^2)$ is invariant the sum of $l+(-l)$ gives zero. What happens in the discrete case is that you are not allowed to rotate by any angle you wish, only a finite discrete set of angles are allowed under which your $d^4l F(l^2)$ remains invariant, and so using this restricted set of angles you may not find a frame where the vector becomes $-l$ so that it cancels $l$ !

Reason 2: Translation Symmetry Is Broken Too

A translation $l\mapsto l+q$ for any $q$ is not possible, because you have fixed lattice axes and you are allowed to move only along them and by an integral multiple steps of the lattice spacing.

So the summary is the following:

A momentum cut-off $\Lambda$ let's say, is equivalent to a lattice which does not enjoy the Poincar`e symmetry.