# What's the difference between divergences that can be corrected and those that can't

I'm confused by renormalization . If a lagrangian has a term with negative mass dimension , why can't the divergences be absorbed into lagrangian coefficients? What's the difference between divergences that can be corrected and those that can't

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You may also want to read my popular explanation here. –  Vladimir Kalitvianski Jan 19 '13 at 16:42

Divergences arising from nonrenormalizable terms in the Lagrangian can't be absorbed into coefficients in the Lagrangian because there were no terms in the Lagrangian of this sort to start with!

If one starts with a term $${\mathcal L}_{\rm nonren} = c\cdot {\mathcal O}_1$$ where $c$ has a negative mass dimension, then the loop divergences produce interactions (with infinite coefficients) of the form $$c^n \cdot {\mathcal O_2}$$ where $n\geq 2$ counts the number of vertices of the ${\mathcal O}_1$ type in the Feynman diagram. But then the dimension of the coefficient of this term is even more negative (the dimension of $c^n$ is $n$ times larger than that of $c$ itself) and the positive mass dimension of ${\mathcal O}_2$ is even higher (more positive), so one clearly created a term of a different type than we started with. To absorb the divergence, you have to include a new term in the original Lagrangian $$d\cdot {\mathcal O}_2$$ so that $c^2\times \infty$ may be absorbed to $d$. Here, $d$ has an even more negative dimension than $c$. I could repeat the same story with $d,e,\dots$ and get infinitely many new terms that would have to be present in the initial Lagrangian if the Lagrangian were supposed to be able to absorb all the divergences. These terms would have undetermined coefficients – an infinite number of undetermined coefficients – so one would need an infinite number of measurements to determine all of them. This is not really possible – the theory is, to a large extent, arbitrary and, more precisely, unpredictive, and this is the real problem with non-renormalizable theories.

For renormalizable theories, one only obtains divergent terms to the coefficients in front of a finite number of operators that were present from the beginning such as $e\cdot A_\mu \cdot \bar\Psi\gamma^\mu \Psi$. In this case, it's because $e$ is dimensionless (at least classically; let me neglect running and anomalous dimensions which are small if the perturbative expansion is any good) and $e^n$ has therefore the same dimension. We don't produce increasingly more non-renormalizable (i.e. those with very strongly negative-mass-dimension coefficients) operators with divergent coefficients.

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The basic idea is this. Take a $\phi^6$ theory in 4 dimensions. This is a nonrenormalizable theory. There is a divergent one loop graph contribution to the 8-point function due to two 6-point vertices (aside: is there an accepted way to include Feynman diagrams here?): take two six-point vertices and connect two legs of one to two legs of the other for a contribution to the 6+6-2-2=8 point function. To define the theory you need to add a $\phi^8$ counter-term to absorb this divergence, but now there is an 8 point vertex which contributes to a one-loop divergence of the 12 point function (there is also a divergent 10 point function due to a loop with 6 and 8 point vertices). These new divergences require new counterterms $\phi^{10},\ \phi^{12},\ldots$. Each counterterm you add creates more primitive divergences. You eventually end up with an infinitude of counterterms each with an adjustable coupling constant so the theory is no longer predictive.