Let us consider the $\phi^4$ theory, where $\phi$ is a real scalar field, such that the physical mass vanishes.
Is it true that the bare mass also vanishes?
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1$\begingroup$ Related: physics.stackexchange.com/q/821753 $\endgroup$– hftCommented Jul 18 at 19:47
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1$\begingroup$ I can't answer definitively, but note that: (1) At first order in perturbation theory the physical mass squared is proportional to the running mass squared; (2) at first order, the running mass squared is proportional to the bare mass squared, but the first order correction in dimensional regularization also depends on $\frac{1}{d-4}$, where $d$ is the space-time dimension. Thus, although it seems like these proportionalities would make the bare mass vanish when the physical mass does, I do not know and I am worried about the infinities. $\endgroup$– hftCommented Jul 18 at 20:58
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$\begingroup$ (Reference: Lowell Brown QFT Section 4.3) $\endgroup$– hftCommented Jul 18 at 20:58
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$\begingroup$ I'm not really sure what this question is asking - in the counterterm formulation of renormalization we have $m_B = m_R + m_C$, B/R/C stand for bare/renormalized/counterterm. The counterterm, and hence also the bare, quantities are formally infinite, the whole point of renormalization is to render the renormalized ("physical") part finite. In what sense can you talk about a finite value of a bare quantity? $\endgroup$– ACuriousMind ♦Commented Jul 19 at 14:42
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Assume spacetime dimension $2 \leq d \leq 4$. The $\phi^4$ coupling constant $\lambda$ has mass dimension $[\lambda] = 4-d$. The one-loop mass correction is linear in $\lambda$. Given the UV cutoff $\Lambda$, simply by dimensional analysis we can conclude that $m^2$ receives UV-divergent correction $$ \lambda(\Lambda^{d-2} + m^2 \Lambda^{d-4}). $$ Note that $\lim_{\epsilon \rightarrow 0} \Lambda^{\epsilon} \sim \log \Lambda$.
So, nope, generally the bare mass will contain the first term, which is independent of the renormalized mass $m^2$. You need to fine-tune the bare theory to get near the massless critical region.
Dimensional regularization is well-known to only pick up logarithmic divergence and completely ignores e.g. the quadratic $\Lambda^2$ term in $d=4$ in this case. It is true that in this case only the subleading term is important for the RG flow in $2 < d < 4$, and people often just merrily dim. reg. away. But do keep a record what you are sweeping under the carpet.
