# Why does a cutoff break gauge invariance?

It's been stated repeatedly that introducing a sharp momentum cutoff $\Lambda$ into a gauge theory breaks gauge invariance. Apparently, this is because momentum modes directly at the cutoff cannot be subjected to a gauge transformation.

Is this the only problem and could someone point to some resource where this calculation is done explicitly?

• "It's been stated repeatedly........due to the partial derivative the $\partial_\mu \alpha$ adds a slight increase in momentum" I have never heard this. Can you provide a reference? Sep 9, 2016 at 17:40
• Actually, the Fourier-transformed analogue of the gauge transformation is $A_{\mu}(p) \rightarrow A_{\mu}(p) + p_{\mu} \alpha(p)$. This can be valid for any $p$ within the cutoff range, so it seems that the cut-off is gauge-invariant though not Lorentz invariant. +1 for providing a reference. Sep 9, 2016 at 18:14
• Because it cuts off Euclidean momentum not Minkowski square of momentum, Lorentz is broken right away, isn't it? Sep 9, 2016 at 19:01
• @innisfree ofcourse, that is exactly what I meant. Sep 9, 2016 at 21:37
• @Prahar I was quite sure I had read this somewhere. I looked everywhere but couldn't find it, until at last I remembered that it was in a transcript of an oral exam. I found the original passage now but it's in german and an exam transcript probably wouldn't count as a valid reference anyway. So I went ahead and removed the statement. It might be a good thing to leave the question more open-ended anyway. Sep 11, 2016 at 9:06

A part of the problem with this question is what is meant by gauge invariance. I think the issue with gauge fields is covariance. The momentum operator is made gauge covariant by $\hat P~=~\hat p~+~ie\hat A$ then for $\hat A~\rightarrow~\hat A~+~\partial\chi$ these transformations are incomplete if we demand that $|\hat P|~<~\Lambda$. The whole set of possible transformations on the principal bundle are not permitted.
This carries over to the fields as well. The transforms gauge connections in general are $$A'~=~g^{-1}Ag~+~g^{-1}\partial g.$$ The gauge transformation here is seen from letting $g~=~e^{\chi}$ $=\simeq~1~+~\chi$ for $\chi$ small so that $$A'~=~A~+~[A,~\chi]~+~\partial\chi~+~O(\chi^2),$$ where the commutator is zero in the caseabove. The fields can be seen from the covariant momentum according to differential forms $${\bf F}~=~{\bf P}\wedge {\bf P}~=~d\wedge {\bf A}~+~{\bf A}\wedge {\bf A}.$$ The transformed fields are then $F'~=~g^{-1}Fg$ that gives $$F'~=~F~+~[F,~\chi].$$ A cut off again means that the range of transformations is restricted.
What is invariant is the Lagrangian. For ${\cal L}~=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$, with indices restored, or ${\cal L}~=~-\frac{1}{4}F\cdot F$, is such that ${\cal L}'~=~-\frac{1}{4}g^{-1}Fg\cdot g^{-1}Fg$ is just a transformation of a scalar so ${\cal L}'~=~{\cal L}$. Consequently the cut-off does not appear to influence the gauge invariance of the Lagrangian.
• Any QFT book will do. If you want something really mathematical the book by Donaldson and Khronheimer works the deep mathematics of 4-manifolds and gauge theory. The mass of the photon will be zero. The propagators for the photon and electron have the form $1/(k^2~-~m^2~+~i\epsilon)$ and $1/(\gamma\cdot k~-~m + i\epsilon)$ and there is no mass to perform mass renormlization on with the photon. Sep 11, 2016 at 12:50