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I am following Peskin and Schroeder chap 13.3 on the 2D NLSM. At page 459, eq (13.95), we have to consider the correlator $$\langle\partial^\mu\phi_a(0) \partial_\mu \phi_b(0)\rangle.\tag{13.95}$$ Because of the renormalisation scheme, this correlator is to be computed for a free theory. This gives us the contribution (it is just a loop in the vacuum) up to some constant which are not relevant for this discussion:

$$ \int_{\Lambda'<k<\Lambda}d^2k\frac{1}{k^2}k^2\propto (\Lambda^2-\Lambda'^2) $$

Where the $\frac{1}{k^2}$ comes from the propagator and the $k^2$ factor from the derivatives $\partial^\mu\partial_\mu$. Now, in the renormalisation we say that we ditch this term, compared to the $\langle\phi_a\phi_b\rangle \propto \log(\frac{\Lambda}{\Lambda'})$ contribution. I am not sure why is that so. My guess is that because, after many renormalisation steps, since the contribution is $\propto \Lambda^2$ it is going to be negligible in the IR, compared to the contribution in $\ln(\Lambda)$ which is actually going to grow. I am really not convinced by that explanation however.

In the Peskin and Schroeder it is somehow justified at (13.83), but I am not sure of what he means in the explanation and how the dependence in the IR cutoff is important. (Note that in this part he does the renormalisation in another way)

I am really not sure this is a valid justification, could you confirm if this is a sensible reasoning, or is there something else at play?

Note that this derivation is made also in the paper https://doi.org/10.1016/0370-2693(75)90161-6

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    $\begingroup$ Related. $\endgroup$
    – Cosmas Zachos
    Commented Mar 23, 2018 at 15:09
  • $\begingroup$ Thank you for bringing my attention to this thread. Indeed, he seems to have stumbled upon the same problem, at one level lower in the computation, with terms like $<\phi_a>$. I didn't pay attention to those because I thought they vanished at first level in perturbations, but indeed I might have counted the powers of the coupling a bit hastily. This seems also highly related, although I have a bit of trouble following the answer. $\endgroup$
    – Frotaur
    Commented Mar 23, 2018 at 15:24
  • $\begingroup$ Well, in (13.82), you are invited to scale out $\mu^d$, no, so the expression vanishes as you take $\mu \to 0$ for d=2, provided the integral itself is finite for d=2. $\endgroup$
    – Cosmas Zachos
    Commented Mar 27, 2018 at 19:22

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Cosmas Zachos has already answered your question. But I will spell it out a bit more.

The key observation is that the term $\frac{\Gamma(-d/2)}{(\mu^2)^{-d/2}}$ in (13.82) is equal to $\Gamma(-1+\epsilon/2) (\mu^2)^{1-\epsilon/2}$. The pole of the $\Gamma$ function at -1 does not matter, because $(\mu^2)^{1-\epsilon/2}$ is well defined (i.e. zero) when we let $\mu \rightarrow 0$

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