Power counting with a cutoff

In Effective Field Theory video lectures found here, the professor explained power counting in effective field theories and the difficulties of power counting associated with loop diagrams. He then mentions that introducing a cutoff ($\Lambda_{UV}$) to regulate our divergences does not preserve power counting due to the new scale that we are introducing. To see this he uses four-fermi theory with the diagram,

$\hspace{6cm}$ We do our power counting (i.e., Taylor expansions) in powers of $m^2/M^2$ and then go on to consider the mass correction through,

$\hspace{6cm}$ Using a cutoff this gives a mass correction, \begin{align} a\frac{ m }{ M ^2 } \int _0^{\Lambda_{UV}}\frac{ \,d^4k _E }{ (2\pi)^4} \frac{1}{ k _E ^2 + m ^2 } & = a\frac{ m }{ ( 4 \pi ) ^2 } \left[ \frac{ \Lambda _{ UV } ^2 }{ M ^2 } + \frac{ m ^2 }{ M ^2 } \log \frac{ m ^2 }{ \Lambda _{ UV } ^2 } - \frac{ m ^4 }{ M ^2 \Lambda _{ UV } ^2 } + ... \right] \end{align}

If I understand correctly this breaks the power counting because even if $\Lambda_{UV} \sim M$, the first term is an order 1 correction since its not proportional to $m^2/M^2$. So far so good. However, then the professor says that you can still use power counting with a cutoff if you fix the power counting order by order and that this can be done by introducing an intermediate scale, $\Lambda$. But I don't how this fixes anything...

With an intermediate scale ($\Lambda$) we have, \begin{equation} a\frac{m}{M^2}\int _{ \Lambda } ^{ \Lambda _{ UV }} \frac{ \,d^4k _E }{ (2\pi)^4 } \frac{1}{ k _E ^2 + m ^2 } = \frac{a\,m}{ (4\pi)^2M^2 } \left\{ \left(\Lambda ^2 + m ^2 \log \frac{ m ^2 }{ \Lambda ^2 + m ^2 } + ... \right) + \left( \Lambda _{ UV } ^2 - \Lambda + m ^2 \log \frac{ \Lambda ^2 + m ^2 }{ \Lambda ^2 _{ UV }} \right) \right\} \end{equation} but how does this fix anything?

For more context see my lecture notes here under Effective Field Theory (starts around equation 4.6)

Before introducing any additional scales, assuming that the only relevant scale is given by $m^2$, power counting gives us the following relation for the integral in question:

$$\Delta m\sim\frac{am}{M^2}\int \frac{d^4k_E}{k_E^2+m^2}\sim \frac{am^3}{M^2},$$

which is a small mass correction.

Next, we see what happens when we introduce a regulator in the form of a UV-cutoff. In this case, the evaluated integral contains a term of the form

$$am\frac{\Lambda_{UV}^2}{M^2},$$

which is not a small correction as in the case without the regulator. Therefore, power counting fails as soon as we introduce a UV-cutoff.

Introducing an additonal scale $\Lambda$, which we assume to be small, solves this problem. Within the renormalization procedure we can absorb the $\Lambda_{UV}$-dependent terms into counterterms $\delta m(\Lambda,\Lambda_{UV})$ while still having some other terms left, i.e. something proportional to

$$am\frac{\Lambda^2}{M^2},$$

which is again a small correction. We have therefore "restored" the validity of power counting.

• Thanks for your response! But how come this fixes the power counting? Either way we absorb the large terms in the counterterms. So how does adding in this new scale make any difference? – JeffDror Mar 15 '14 at 16:51
• Without the new scale, there would be no small correction at all. Power counting does not give the right prediction, i.e. a small correction of the right dimension. – Frederic Brünner Mar 15 '14 at 18:50
• In this new method of adding an intermediate scale, we certainly have a small correction but we're just hiding the larger correction in $\delta m$. So power counting is still broken but we are just hiding it in $\delta m$? – JeffDror Mar 15 '14 at 21:21
• The intermediate scale is part of the renormalization procedure, in which counterterms arise naturally. So yes, one can think of it as hiding it in the counterterm. – Frederic Brünner Mar 15 '14 at 22:38