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This is a follow up or better an edit to my previous question that was marked as a duplicate of this other question. I think I failed to emphasize what I really wanted. The tittle of my question was, Why are constant terms irrelevant in Lagrangians in quantum field theory? In the question whose duplicate my question was categorized, there is one answer in whose comments a guy points out that a constant term can be ignored because you can factor $$ e^{iS_{const}} $$ in functional integrals and since correlation functions are quotiens of functional integrals these terms will cancel. I get that. That was not what I wanted. My point was, constant terms in Lagrangians lead to infinite terms because after integrating $$ \int d^dxCONST=\infty $$ so what you would be factoring is infinite! even after dimensional regularization. So what I really ask is, how can we neglect this infinite constants? I suspect that the most likely answer I will get is gonna be, "you can just absorb them in the normalization of the functional integral". Is that all there is to it? Can we just add any infinite constant in the action and get away with it? always?

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  • $\begingroup$ I may be failing to see exactly what you are asking here, but if you have a term like $e^{i \int d^dx \text{const}}$ in the generating function then it is just going to cancel when you take quotients. I cannot see what your problem is? $\endgroup$ – Quantum spaghettification Jun 30 '18 at 11:23
  • $\begingroup$ @Quantumspaghettification My problem is about ignoring infinite terms. Surely that deserves some attention. You cannot just simplify infinite/infinite=1 and live happily ever after. $\endgroup$ – Yossarian Jun 30 '18 at 12:44
  • $\begingroup$ It depends on what kind of infinities you have. If your infinities are coming from asking your theory too much, such as integrating over very high energy scales or very low distances, it is not wrong to insert some cutoff to regulate such results. The statement ''infinity/infinity=1'' is then a well defined statement that is regulated to the same statement ''large number/same large number=1''. The same holds for integrating over very large distances. Similar statements of "infinity/infinity=1" are the regulated statements "(1/small number)/(1/small number)=1". $\endgroup$ – Panos C. Jun 30 '18 at 13:03
  • $\begingroup$ @PanosC. Agreed. But this infinity is not regulated. It is just there and it is ignored. $\endgroup$ – Yossarian Jun 30 '18 at 13:32
  • $\begingroup$ It is not regulated because it is obvious that any regularization procedure will lead to the same answer. Call your infinite integral $I$ and suppose that you regularized it in such a way that by introducing a dependence on some parameter $\alpha$ the quantity $I[\alpha]$ is finite for almost every $\alpha$, with $I[\alpha_{0}]=I$. Then $I[\alpha]/I[\alpha]=1$ for any value of $\alpha$, so that $\lim_{\alpha\to \alpha_{0}} I[\alpha]/I[\alpha]=1$. Which is the same as saying "large number/same large number=1" (cit. Panos). $\endgroup$ – Giorgio Comitini Jun 30 '18 at 15:53

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