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Q1: My question is, in the context of dimensional regularisation(DREG, in dimension $d$), why do they mention the absence of $d=2$ pole in the gauge theory cases[1], whereas the $d=2$ pole is not discussed about in $\lambda\phi^4$ theory[2]?

For details, refer to Peskin & Schroeder's "An introduction to Quantum Field Theory":

in [1], page 251 (Section 7.5), in calculation of the vacuum polarisation diagram in QED,

$$...\:\sim\:\Gamma(2-\frac{d}{2})g^{\mu\nu}...$$ We would have expected a pole at $d=2$, since the quadratic divergence in $ dimensions becomes a logarithmic divergence in 2 dimensions. But the pole cancels. The Ward identity is working.

also page 525 (Section 16.5) , One-Loop divergences of Non-Abelian Gauge Theory--The gauge Boson Self-Energy:

Now we are ready to put these results together. In the sum of the three diagrams, The coefficient of $\Gamma(1-\frac{d}{2})g^{\mu\nu}...$ is

$$...=(1-\frac{d}{2})(d-2).$$ The first factor cancels the pole of the gamma function at $d=2$. Thus, the sum of the three diagrams has no quadratic divergence and no gauge boson mass renormalisation.

in [2]. However, in page 328 (Section 10.2), calculating the self-energy diagram in $\lambda\phi^4$ theory, there is the contribution:

$$\Gamma(1-\frac{d}{2})...$$

which has a pole at $d=2$, and no comment is made there.

Q2: I've always thought that in DREG, as long as the $d=4$ divergences are canceled, everything is fine, so what's the significance in talking about $d=2$ divergence?

Q3: Is there any serious problem to overlook the $d=2$ pole, for example, in $\lambda\phi^4$ theory?

Q4: By the way, is the "fine-tunning problem" related to the $d=2$ pole in DREG, if yes, how?

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  • $\begingroup$ I assume the Ward identity mentioned states that there should be no divergences somehow, and if this Ward identity is valid for $d=2$ as well, the cancellation of divergences in $d=2$ is a check on the result. $\lambda\phi^4$ has fewer identities restricting divergences. Q2: That's the case when only considering the theory at $d=4$. Sometimes it is useful to consider other dimensionalities to understand better what's going on. $\endgroup$ – David Vercauteren Dec 17 '14 at 7:07

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