# “Dimensional analysis” arguments in quantum field theory

I'm uncomfortable with dimensional analysis arguments made in quantum field theory, particularly those related to renormalization. For example, in section III.2 of Zee's QFT book, it says:

Consider Fermi's theory of the weak interaction. Imagine calculating the amplitude $$\mathcal{M}$$ for a four-fermion interaction. At lowest order, $$\mathcal{M} \sim G$$. Let us try to write down the amplitude to the next order. [...] By dimensional analysis, the only possibility is $$\mathcal{M} \sim G + G^2 \Lambda^2.$$ Without a cutoff on the theory, or equivalently with $$\Lambda = \infty$$, theorists realized that the theory was sick: infinity was the predicted value for a physical quantity. Fermi's weak interaction theory was said to be nonrenormalizable.

This is already confusing, because UV divergences occur in renormalizable theories too. It's also confusing why the energy scale $$E$$ of the process can't appear. Then, one page later, it reads:

Here is another way of making the same point. Suppose we didn't know anything about cutoffs and all that. Then by high-school dimensional analysis we see that the scattering amplitude at center of mass energy $$E$$ has to go as $$\mathcal{M} \sim G + G^2 E^2 + \ldots.$$ When $$E$$ reaches the scale $$(1/G)^{1/2}$$ the amplitude reaches order unity and some new physics must take over.

These two passages seem blatantly contradictory; didn't we just say $$E$$ wasn't allowed in the dimensional analysis? How did the cutoff disappear, when it really should still be there?

In general there always seem to be three dimensionful scales, $$G$$, $$E$$, and $$\Lambda$$. What's the right way to make a dimensional analysis argument using them, and why are the two passages above not inconsistent?

• My two-cents: in the first case you assume that the theory is unitary, so $\mathcal M$ has to be bounded as a function of $E$. It cannot grow quadratically, so the only way to make the dimensions work is to use $f(E/m)G^2\Lambda^2$, with $f$ a dimensionless constant. In the second case, you forget about unitarity and proceed more naively. I'm not sure this makes much sense though. – AccidentalFourierTransform Dec 11 '17 at 15:47

A good way to do dimension analysis in computing amplitudes relies on a good power-counting of the action. Let me explain how it works before answering your question. For simplicity, in the following, I will consider a QFT in $d=4$.

Usually, it is very useful to distinguish between couplings, decay constant and mass scales. In order to do so, we restore the dimensions of $\hbar$. What I mean is that the action of the system has to have dimension of $\hbar$, namely

$$[\mathcal{S}] = [\hbar].\qquad\qquad\qquad (1)$$

Let us consider the very simple example of a One Coupling One Scales scalar theory (1C1S scalar theory), characterized by one coupling $g_*$ and one scale $\Lambda$ (that you interpret as a cut-off which may drive the derivative expansion of the EFT in powers of $\partial/\Lambda$).

Having restored the powers of $\hbar$, the dimensions of the fields are not just those of powers of energy. Indeed, from the kinetic term

$$[\hbar]=[\mathcal{S}] = [\int d^4x \, (\partial\phi)^2] = M^{-4} M^2 [\phi]^2$$

where $M$ just means "dimension of a mass". From this dimensional analysis we see that $[\phi] = \hbar^{1/2} M$, and analogously for spinorial fields $[\Psi] = \hbar^{1/2} M^{3/2}$.

We define then a coupling $g_*$ whose dimension is $[g_*] = [\hbar]^{-1/2}$. For the scalar theory, you can see this as an equivalent definition of $g_*$ as the square of the coefficient of the marginal $\phi^4$ interaction. Indeed, the $\phi^4$ interaction has the correct dimension as in Eq.(1)

$$\int d^4 x\, [g_*^2 \phi]^4 = M^{-4} [\hbar]^{-1} M^4 [\hbar]^{2} = [\hbar].$$

You see then the Lagrangian can be written as

$$\mathcal{L}(\phi,\partial\phi) = \frac{\Lambda^4}{g_*^2}\hat{\mathcal{L}}\left(g_* \frac{\phi}{\Lambda},\frac{\partial}{\Lambda}\right)$$ where $\hat{\mathcal{L}}$ is a dimensionless function of its dimensionless arguments. In general, any insertion of the field can carry a $O(1)$-coefficient that is not fixed by the power-counting (these coefficients will be called $a_1,a_2,a_3....$; see below). So, for example, you can write the most general terms compatible with the symmetries of the theory. In this case, we do not have any symmetry, and we can write

\begin{align} \mathcal{L} &\supset \frac{\Lambda^4}{g_*^2}\left( a_1 g_*^2 \frac{(\partial\phi)^2}{\Lambda^4} + a_2 \,g_*^2 \Lambda^{-2} \phi^2 + a_3\, g_*^3 \Lambda^{-3}\phi^3+a_4\, g_*^4 \Lambda^{-4} \phi^4 + a_5\, g_*^4 \Lambda^{-6}(\partial\phi)^2\phi^2\right)\\ &= a_1(\partial\phi)^2 + a_2 \, \Lambda^{2} \phi^2 + a_3\, g_* \Lambda\phi^3+a_4\, g_*^2 \phi^4 + a_5\, g_*^2 \frac{(\partial\phi)^2\phi^2}{\Lambda^2}\,.\qquad \qquad \qquad(2) \end{align}

To have a canonically normalized scalar field we can set $a_1=1/2$.

Let's come to your question. First of all, you have to identify the correct dimension of the scatting amplitude (and this depends on the definition of the S-matrix and the normalization of the states of the Hilbert space). I use the same relativistic normalization of states and definition of $S-$matrix as in the Schwartz's book of QFT. For a $2\rightarrow 2$ scattering in $d=4$, the amplitude has dimension

$$[\mathcal{A}_{2\rightarrow 2}] = [\hbar]^{-1} = [g_*]^2\,\qquad \qquad \qquad (3)$$ Indeed, my scattering amplitude contribution from $g_*^2\phi^4$ goes like $\sim g_*^2$

Please, note the dimension of the amplitude depends on the space-time dimensions and the number of scattered particles. In general $d$ dimension the $n-$point amplitude has dimension $d-n\,d/2+n$.

Eq.(3) means that terms like $g_*^2 (E^2/\Lambda^2 + E^4/\Lambda^4)$ are generically allowed if there are irrelevant operators which contribute at these orders in the energy; here $E$ is the energy of the center of mass $E= \sqrt{s}$ with $s$ the Mandelstam variable.

In the example of 4-Fermi theory, you can identify (in my normalization)

$$G\simeq \frac{g_*^2}{\Lambda^2}$$

If you have the operator $\frac{g_*^2}{\Lambda^2} (\bar{\psi}\gamma^\mu \psi)^2$, the only contribution at tree-level is $g_*^2\, s/\Lambda^2$. The contribution proportional to $G^2$ comes from a loop. This is because loops carry a power of $\hbar$ which cancels the dimension of the extra $G$ insertion. When dealing with loops and renormalization, you have to fix a renormalization scheme.

$$\text{loop contribution}\simeq \frac{g_*^4}{\Lambda^4} (\#)$$

What do we put in $(\#)$ ?

If you use cut-off regularization (i.e. you cut-off the momenta integral of the loop) you can have a combination of powers of $s$ and $\Lambda$; However, these contributions are local and can be eliminated by choosing counter-terms; they are not physical. The only physical contribution from the loop-integral is a logarithm, namely $s^2 log(s/\Lambda^2)$.

I believe that Zee is discussing two different scenarios in a very general way, without entering in details and without discussing renormalization of fields

• Very high cut-off (but you know the theory has a physical cut-off), small probing energies. The loop integral $I = \int d^4 k \frac{1}{k}\frac{1}{k+p}$ where $p$ is an external momenta, has dimension $2$. Then, the result can be $s$ or $\Lambda^2$. In the assumption $s\ll \Lambda$ you take $\Lambda^2$ possibly multiplying a log.

• You might believe your theory is UV-completed and that it can be probed at any energy; namely, we do not have a cut-off and the only dimensionful quantity is the center of mass energy (and maybe the renormalization scale or particle masses which are however assumed to be small); then, the result of the loop can be $s$ possibly multiplying a log.

Note that in a massive (UV-completed) QFT the amplitude is constrained by the Froissart-Bound to be bounded from above, namely $|\mathcal{A}(s)| < s\cdot log^2(s)$.

• just one minor remark: power law divergences are not necessarily unphysical. They just represent the contribution from unkwon UV physics compatibly with the symmetry of the Regularized theory, once you interpreted $\Lambda$ as a physical threshold rather than an artificial cutoff. They can be removed changing scheme because different UV physics may or may not generate that terms (one can eg tuning it away dialing the parameters of the UV theory). What Zee is presenting is both sides of the coin: the theory needs a physical cutoff, precisely because it grows with energy polynmially. – TwoBs Dec 17 '17 at 7:25

I agree that Zee's presentation is unsatisfactory, and like Srednicki's presentation (Chapter 18) better. Rather than just making hand-waving arguments based on dimensional analysis, Srednicki presents the dimensional analysis as just a useful shortcut for figuring out whether a given Feynman diagram contains a divergent loop integal, using some straightforward but non-obvious graph-theory arguments. You might also like Lecture 1 of these notes, which gives a rougher argument in the spirit of Zee's, but which I think is much clearer than Zee's.

There is nothing wrong with Zee's presentation. In the first case, he discusses the first quantum correction to $G$, in which he cuts the integral off at $\Lambda$. Then the only dimensionful quantities in the game are $G$ and $\Lambda$, and he finds that the corrections go like $G\,\Lambda^2$, and argues why that's sick. Of course, this is not the full story, e.g. corrections to scalar masses also go like $\Lambda^2$, and one can kill them by the corresponding counterterms. The real problem here is that you'll end up needing to introduce infinitely many counterterms, for instance at the $G^2$ level there will also be a 6-fermion interaction. The theory is hence no longer predictive.

In the second case, he discusses scattering experiments, and explains why the theory cannot be trusted if $E\gtrsim G^{-1/2}$. In contrast to the above case, he is discussing tree level amplitudes, and here the only dimensionful quantities are $E$ and $G$.

Both observations tell you that (point-particle) quantum field theories in 4D with couplings of negative mass dimension are not UV complete.