What is an example of a massless field that acquires a mass from quantum corrections? It's all in the title. In eq. (7.75) of Peskin they say the full photon propagator takes the form
$$\frac{-i\ g_{\mu\nu}}{q^2 (1- \Pi(q^2))}\, \tag{1}$$
and that the photon remains massless as long as $\Pi (q^2)$ is regular at $q=0$. Is there an example of a theory (if possible something simple, like a scalar field), say in $4$d, where a massless field becomes massive at the quantum level? Can it be made visible with a one-loop computation already?
(Note, as a bonus, that I am particularly interested in seeing that effect in position space.)
 A: The separation between the masses on the tree and loop level is actually non-physical. So your question, as it is posed now, is actually non-answerable.
E.g. consider two scalar fields $\phi$ and $\chi$. Let us assume that $\chi$ has some nonzero mass $M$ and there is an interaction,
\begin{equation}
\frac{g}{4}\phi^2\chi^2
\end{equation}
Then the one-loop correction to $\phi$ self-energy comes from the diagram with $\chi$-loop and the mass counterterm $\delta_m^2$ diagram,

From the usual $\phi^4$ result employing the dimensional regularization we get that at the one-loop level the pole mass (defined as the position of the pole in the full propagator),
\begin{equation}
(m_\phi^2)^{(pole)}=(m_\phi^2)^{(tree)}+\frac{g}{32\pi^2}M^2\Big(\frac{2}{\epsilon}-\gamma+\ln\frac{4\pi\mu^2}{M^2}\Big)+\delta_m^2+\mathcal{O}(\epsilon)
\end{equation}
Now, the counterterm must cancel the divergent part. However the finite part of $\delta_m^2$ may be chosen according to our wishes. These choices necessary for the complete definition of the theory are known as the renormalization conditions.
E.g. in the on-shell renormalization scheme you require $(m_\phi^2)^{(pole)}=(m_\phi^2)^{(tree)}$. So at each order of $g$ the counterterm $\delta_m^2$ must cancel all the contributions to the mass. Then if you start with the massless $\phi$ it will remain to be massless.
In the minimal substraction scheme ($\text{MS}$) and the modified minimal substraction scheme ($\overline{\text{MS}}$) you set,
\begin{equation}
\text{MS}:\quad \delta_m^2=-\frac{g}{32\pi^2}M^2\cdot\frac{2}{\epsilon},\quad\quad \overline{\text{MS}}:\quad \delta_m^2=-\frac{g}{32\pi^2}M^2\cdot\Big(\frac{2}{\epsilon}-\gamma+\ln{4\pi}\Big).
\end{equation}
Then the difference between the pole and the tree masses is generally speaking nonzero. You may notice that it depends on the choice of the renormalization point $\mu$. So you may find such $\mu$ that the tree mass coincide with the pole mass. In case of the $\overline{\text{MS}}$ scheme this corresponds to $\mu=M$ at one loop. But for different choices of $\mu$ there will be a difference so you may have the situation when the particle is massless on a tree level but gains mass through the radiative correction.
As the pole mass corresponds to the masses of the asymptotic states that propagate almost freely and hit the detector (which is why it is often called a physical mass) you may be compelled to use only the on-shell renormalization scheme. However it is not always a good idea. E.g. for the light quarks you usually can't find the situation where they make sense as asymptotic states. So their pole mass is rarely discussed (as it is rarely makes sense to consider such a pole) and usually tables list their $\overline{\text{MS}}$ masses for $\mu=2\,\text{GeV}$ or $\mu=1\,\text{Gev}$. In contrast $t$-quark is usually so short-lived that it does not escape the asymptotic freedom regime and it is its pole mass that often is given.
So the proper question would be: when it may be useful for you to choose $(m^2)^{(tree)}=0$ even if $m^2\neq 0$?
For example you may consider a massless particle (e.g. a photon) that is travelling through a  medium. Inside the medium it may gain an effective mass (though usually in a Lorentz-symmetry breaking fashion). If the medium is thin you may find it natural to consider this effective mass as a radiative correction.
You may also consider a symmetric model that experience the symmetry breaking associated with the generation of mass, like in case of the Higgs mechanism. But in that case the mass shift comes not from the loop diagrams but from the tree processes with the nonzero Higgs v.e.v. Even when the Higgs potential is produced radiatively (as in Coleman-Weinberg case) you don't really represent its v.e.v. through the Feynman diagram expansion (which is evident from the expressions for its v.e.v. that are not decomposable into a Taylor series of the perturbation parameter)
You may also consider the situation when the classical theory has the symmetry that is ruined by the anomaly in the quantum case. E.g. the Goldstone bosons of the spontaneously broken symmetry may gain mass because of its explicit breaking by the anomaly. However in this case I can't think of any model where such mass shift would be described as a perturbative radiative contribution. The classic example would be a $\eta'$ gaining mass due to the gluon triangle anomaly for the axial current but while the anomaly may be derived in the perturbative QCD, $\eta'$ itself lies in the non-perturbative realm. And in $\frac{1}{N_c}$ expansion I would call it a "tree" process.
