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In p.8 of Michio Kaku book Introduction to Superstrings and M-Theory-Springer (1998), he said

The gravitational force. Gravity research was totally uncoupled from research in the other interactions. Classical relativists continued to find more and more classical solutions in isolation from particle research. Attempts to canonically quantize the theory were frustrated by the presence of the tremendous redundancy of the theory. There was also the discouraging realization that even if the theory could be successfully quantized, it WOULD ... still be nonrenormalizable.

My question is that

  1. what does Kaku mean for nonrenormalizable but quantizable theory?

what are the criteria to be nonrenormalizable?

what are the criteria to be quantizable?

  1. What are examples of nonrenormalizable but quantizable theories?

Fermi weak interaction theory? Gravity? and how come?

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    $\begingroup$ A general theorem: A theory is non-renormalizable if it contains any interaction whose coupling $g_i$ has negative dimension, $[g_i] < 0$. (From Mandl & Shaw) $\endgroup$ – SG8 Jun 13 at 21:32
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The most basic definition of the terms (and I think the one Kaku is using) is

Quantizable theory - A classical theory that can be quantized in a way where the UV divergences are all cancelled by introducing counterterms.

Renormalizable theory - A quantizable theory which needs a finite number of counterterms.

Non-renormalizable theory - A quantizable theory which needs an infinite number of counterterms.

A theory is renormalizable as long as the classical coupling constants have non-negative mass dimension. For instance,

  1. QED has coupling constant $e$ (electric charge) which has mass dimension 0 so it is renormalizable.

  2. Fermi theory has coupling constant $G_F$ (Fermi coupling constant) which has mass dimension $-2$ so it is non-renormalizable.

  3. Gravity has coupling $G$ (Newton's constant) which has dimension $-2$ so it is non-renormalizable.

Mass dimension of a quantity is the total power of mass dimension in a quantity when working in natural units where $[M]=[L]^{-1}=[T]^{-1}$. For instance, the dimension of Newton's constant are $[G] = [L]^3 [M]^{-1} [T]^{-2}$ (because $G = 6.674 \times 10^{-11} m^3 kg^{-1} s^{-2}$). In natural units, we have $[G] = [M]^{-2}$. This implies that $G$ has mass dimension $-2$.

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  • $\begingroup$ This is the usual textbook answer, but I do like to mention that there are theories which are perturbatively non-renormalizable yet still perfectly UV finite because they are asymptotically safe. In these theories, even though the interaction couplings have negative mass dimension, there is a UV fixed point beyond perturbation theory, and if one considers the IR of the theory as a relevant deformation from that fixed point then no UV divergences appear. $\endgroup$ – Seth Whitsitt Jun 15 at 19:42

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