Spurion analysis: pedagogical example I wanna to understand how Spurion analysis works. Physicist widely use this (From statistical field theory to quantum mechanics problems, as I understand from Google), but I don't know foundations behind this analysis and some clear and effective examples.
Could somebody present simple intuitive examples? 
 A: In this answer I summarize the concept of effective field theory and in particular that of the pion Lagrangian for QCD. Perhaps, however, a bit of prior knowledge is needed.
The prime example of a spurion appears in the context of effective theories for QCD. A widely studied approximation for low energy QCD is that where a certain number $N_f$ of flavors is considered massless and the remaining flavors are considered infinitely heavy (sometimes people say that they are "integrated out" in the path integral). The QCD Lagrangian in this limit reads
$$
\mathcal{L} = - \frac14 F^a_{\mu\nu}\,F^{a\,\mu\nu} + \sum_{j=1}^{N_f} i \,\overline{\psi}_j\gamma^\mu D_\mu \psi_j\,.
$$
Since there is no mass term, left and right Weyl spinors do not mix and the (classical) symmetry group is
$$
G = \mathrm{U}(N_f)_L \times \mathrm{U}(N_f)_R\,,
$$
where the subscripts $L$ and $R$ refer to the component on which they act. Namely
$$
\psi_{L(R), i} = \frac{1-(+)\gamma^5}{2}\psi_i\,,
$$
and $U(N_f)_{L(R)}$ acts as a left matrix multiplication on $\psi_{L(R), i}$.
\begin{equation}
\begin{aligned}
\psi_L \to V_L \psi_L\,,\\
\psi_R \to V_R \psi_R\,.\\
\end{aligned}
\label{psitransf}\tag{1}
\end{equation}
Then this symmetry is spontaneously broken down to $\mathrm{U}(N_f)_V \times \mathrm{U}(1)_A$, where $\mathrm{U}(N_f)_V$ is the diagonal subgroup that acts with the same matrix on both components and $\mathrm{U}(1)_A$ is the axial symmetry that acts as opposite phases on the two components but which is further broken by an anomaly. So the actual final symmetry of the quantum theory is $\mathrm{U}(N_f)_V$.
Long story short, we can write down an effective field theory that realizes this symmetry breaking pattern as a theory of massless Nambu-Goldstone bosons (the pions), associated to each of the broken generators. One common way to parametrize the pions is as
$$
U = \exp\left(\frac{i}{f_\pi}\sum_{a=1}^{N_f^2-1}\pi_a t^a\right)\,,
$$
where $t^a$ are the generators of $\mathrm{SU}(N_f)$. $U$ transforms under the full group as a left-right action
\begin{equation}
U \to V_L\, U\, V_R^\dagger\,. \label{utransf}\tag{2}
\end{equation}
Consistently with the breaking pattern, the diagonal subgroup $V_L= V_R$ is realized linearly, the $\pi_a$ transforming as the adjoint representation
$$
\pi_a \to V \pi_a V^\dagger\,.
$$
This is because $VV^\dagger = \mathbb{1}$ and we can bring the $V$'s to the exponent.
We can systematically write a Lagrangian by putting all terms that respect the above symmetry, in an expansion in the order of derivatives (this is because the effective Lagrangian is an expansion at low energy). One then has
$$
\mathcal{L}_{\mathrm{eff}} = \frac{f_\pi^2}{4}\,\mathrm{tr}\,\left(\partial_\mu U^\dagger \,\partial^\mu U\right) + O(\partial^4)\,.
$$
It is easy to see by expanding at first order that the $\pi_a$'s are massless particles. They better be because they are Nambu-Goldstone bosons. But if we want to introduce the effect of a non-zero mass of the quarks we need to break the chiral symmetry $\mathrm{U}(N_f)_L\times \mathrm{U}(N_f)_R$ explicitly. This will likely induce a non-zero mass on the pions too.
The way to break spontaneously a symmetry and carrying over the effects to the effective Lagrangian is to not break the symmetry at all! Namely we introduce a new field, the spurion, that leaves the Lagrangian invariant, but would break the symmetry if it were to be made a constant. This spurion, when turned into a constant, is the mass term. The new Lagrangian reads
$$
\mathcal{L}_m = - \frac14 F^a_{\mu\nu}\,F^{a\,\mu\nu} + \sum_{j=1}^{N_f}  i\,\overline{\psi}_ji\gamma^\mu D_\mu \psi_j + \sum_{j,k=1}^{N_f}\overline{\psi}_{R,j}\, S_{jk}\,\psi_{L,k} + \overline{\psi}_{L,j}\, S^\dagger_{jk}\,\psi_{R,k}\,.
$$
This Lagrangian is still invariant under the same group if $S$ (the spurion) transforms as
\begin{equation}
\begin{aligned}
S &\to V_R S V_L^\dagger \,,\\
S^\dagger &\to V_L S^\dagger V_R^\dagger\,,
\end{aligned}
\label{stransf}\tag{3}
\end{equation}
where the second line is put just for clarity but is indeed redundant and the $\psi$'s transform as before \eqref{psitransf}. Also note that $S_{ij} = m_{ij}$, this is indeed a fermion mass term.
Now let's go back to the pion Lagrangian. The term with lowest amount of derivatives that respects the transformation properties of the spurion is the following
$$
\mathcal{L}_{\mathrm{eff},m} = \frac{f_\pi^2}{4}\,\mathrm{tr}\,\left(\partial_\mu U^\dagger \,\partial^\mu U\right) + \frac{f_\pi^2 B}2 \,\mathrm{tr}\,\left(US + S^\dagger U^\dagger\right)\,.
$$
with $B$ an additional dimensionful constant. It can be checked that this term is invariant using \eqref{stransf} and \eqref{utransf}. Now we obtained a Lagrangian with a new field and the same symmetry as before. But let's remember: if $S$ is let to $m$ the effect it has is precisely that of a mass term! So we can, in a sense, condensate the spurion and make the Lagrangian
$$
\mathcal{L}_{\mathrm{eff},m} = \frac{f_\pi^2}{4}\,\mathrm{tr}\,\left(\partial_\mu U^\dagger \,\partial^\mu U\right) + \frac{f_\pi^2 B}2 \,\mathrm{tr}\,\left[m\left(U+ U^\dagger\right)\right]\,,
$$
where I assumed the mass term to be real. Now it is easy to see that the pions do acquire masses! The mass matrix is
$$
\mathcal{M}_{ab} = 2 B \,\,\mathrm{tr}\,\left(m\{t_a,t_b\}\right)\,,
$$
and the masses squared are the eigenvalues of that matrix. They scale as $m_{\pi_a}^2 \sim B m$.
So, to reiterate, the spurion is a trick to carry an explicit symmetry breaking down to the effective Lagrangian formulation. It consists in promoting an explicit symmetry breaking to a field that transforms in the right way to undo the breaking. The precise extent to which symmetry is broken is then encoded in the transformation property of $S$. When $S$ is restored to be a constant one obtains the desired theory.
