Am I understanding correctly the argument that leads to the need for field and mass renormalization? I'm studying Quantum Field Theory from Weinberg's book, and I'm to the point where he introduces the concept of renormalization. I'd like to know if I'm getting the point that Weinberg makes when speaking about the necessity of renormalization in perturbative calculations. I do get that renormalization is necessary to tame the divergences due to Feynman diagrams containing one or more loops and to extract useful information even from those diagrams. However, in chapter 10, to be more specific in section 10.3, "Field and Mass Renormalization", Weinberg states that if we are to use the usual Feynman rules to calculate S-matrix elements, then we should first re-define the normalization of the fields $\psi_{l}(x)$ so that
$$
\langle {\rm VAC}\,|\,\psi_{l}(0)\,|\ q,\sigma\rangle=(2\pi)^{-3/2}\ u_{l}(q,\sigma)\qquad\quad(*)
$$
where $|q,\sigma\rangle$ is an eigenstate of the four-momentum of the complete theory with eigenvalue $q^{\mu}$, and $u_{l}(q,\sigma)$ is the coefficient function appearing in the field $\psi_{l}(x)$. It thus appears that renormalization is necessary in principle to do perturbative calculations, independently of the fact that it is a useful technique to tame the divergences. Now, from the very beginning of the book, Weinberg has been introducing formulae that don't make use of renormalized fields, and everything seemed to go well. The point at which the need for renormalized fields turns up is when he studies poles in momentum-space amplitudes for vacuum-vacuum transitions. Here we learn that the masses of the physical states of the theory, i.e. of the eigenstates of the full four-momentum operator, are the points at which those amplitudes have poles. On the other hand, when deriving the Feynman rules for perturbation theory, we find that the propagators associated to the fields have a pole at the point $p_{\mu}p^{\mu}$, the mass of the free field. Thus the value for the masses in the unperturbed hamiltonian (or in the unperturbed action, in the path-integral formalism) should be taken equal to the value of the physical masses given by the true eigenstates of the full four-momentum. This leads to the necessity of introducing mass renormalization counter-terms in the hamiltonian/action. Moreover, a randomly normalized field would give rise to a discrepancy between the Feynman rules derived through perturbation theory and the Feynman rules derived through the study of the poles of the propagator. Thus one must impose condition $(*)$, and to do so one must introduce field renormalization counter-terms in the hamiltonian/action. Did I get this straight?
 A: *

*Weinberg argues that renormalization of the mass and fields appearing in the original Lagrangian is necessary even if no infinities arose or all momentum integrals were convergent. I think it's best to go through this with the example Weinberg takes (closely following the book with my comments).


*If we start a theory by postulating a particle has some `bare' mass, $m_B$ and is described by the following Lagrangian:
\begin{equation}
    \mathcal{L} = -\frac{1}{2} \partial_\mu \Phi_{B} \partial^\mu \Phi_B - \frac{1}{2} m_B^2 \Phi_B^2 - V_B(\Phi_B),
\end{equation}
we will find that the $m_B$ is not really the experimentally observed mass, and the propagator will not be normalized correctly with this field.

*

*To be specific, I think the problem arises when we think that $m_B$ will be the actual observed mass of the particle in the full theory or that a pole in $q^2$ will occur at $-m_B^2$ based on section 10.2-10.3 of Weinberg. Only in the free theory, $m_B$ will be the mass of the particle of the full theory. We can't even expect that the field $\Phi_B$ satisfies the following condition $(VAC, \Phi(0) \Psi_{\mathbf{q}}) = (2 \pi)^{-3/2} u(\mathbf{q})$ (true for a ''in" and ''out" state particles which in some way are like free particles). You can think of $\Phi^\dagger(0)$ acting on interacting vacuum as creating an ''in" state particle.


*But from experiments we know the mass of single-particle states, so let's posit that by this process of so-called re-normalization, we can fix this issue, such that the non-interacting part of the theory will have $m$ as the mass expected for one-particle ''in" state, and we get the correct normalization on the propagator of the field as well. Weinberg says it very clearly: ''A renormalized field is one whose propagator has the same behavior near its pole (which comes from our condition on matrix element) as for the free field, and the renormalized mass is defined by the position of the pole (which is saying that we want $q^2$ to be $-m^2$ at pole as is measured in the experiments)". Therefore, we introduce renormalized field and mass:
\begin{align*}
    & \Phi \equiv Z^{-1/2} \Phi_B\\
    & m^2 \equiv m_B^2 + \delta m^2.
\end{align*}
Again, what we are saying is that if I define my fields and masses like this and demand a normalization like $(VAC, \Phi(0) \Psi_{\mathbf{q}}) = (2 \pi)^{-3/2} u(\mathbf{q})$ from the new field and a pole at $-m^2$, I can just find conditions on $Z$ and $\delta m^2$ from experimental observations. In a way, we are trying to get to the correct renormalized terms in the Lagrangian for the phenomenon we are trying to describe from what we initially posited, which would not agree with experiments (because pole doesn't occur at $m_B^2$).
\item So, we get:
\begin{align*}
    & \mathcal{L} = \mathcal{L}_0 + \mathcal{L}_1\\
    & \mathcal{L}_0 = -\frac{1}{2} \partial_\mu \Phi \partial^\mu \Phi - \frac{1}{2} m^2 \Phi^2\\
    & \mathcal{L}_1 = -\frac{1}{2}(Z-1) (\partial_\mu \Phi \partial^\mu \Phi + m^2 \Phi^2) + \frac{1}{2} Z \delta m^2 \Phi^2 - V_B(\sqrt{Z}\Phi)
\end{align*}


*We know now how to actually calculate the propagator $<\Phi(x_1) \Phi(x_2)>_0$ for the interacting field by just summing up the contribution at all orders. We do it using so called one particle irreducible graphs, which I will not go into. Ultimately, in momentum space, we get the full propagator $\Delta'(q)$ by:
\begin{equation*}
    \Delta'(q) = \frac{1}{q^2+m^2-\Pi^*(q^2) - i\epsilon}
\end{equation*}


*Let's reiterate the conditions we wanted from this whole process, we want the propagator to have a pole at $-m^2$ and the correct normalization (this means residue 1) because one-particle ''in" state is the single-particle state for our theory. These conditions can be written as:
\begin{align*}
    &\Pi^*(-m^2) = 0\\
    &\left[\frac{d}{dq^2}\Pi^*(q^2)\right]_{q^2 = -m^2} = 0
\end{align*}


*Now, you can find $\Pi^*(q^2)$ from the Feynman diagrams, and clearly from the interaction terms we know it will have contributions from vertices we introduced, and we obviously have loops.
\begin{equation*}
    \Pi^*(q^2) = -(Z-1)[q^2+m^2] + Z \delta m^2+ \Pi^*_{LOOP}(q^2). 
\end{equation*}


*From our conditions on $\Pi^*(q^2)$, we can evaluate the variables we introduced in the problem, i.e. $Z$ and $\delta m^2$. In the equation above, it turns out what we are subtracting from $\Pi^*_{LOOP}(q^2)$ are infinities, but it is just a consequence rather than the purpose from Weinberg's perspective (which is mostly to touch base with what we observe). The exact conditions we get are:
\begin{align*}
    & Z \delta m^2= -\Pi^*_{LOOP}(-m^2)\\
    & Z = 1+ \left[\frac{d}{dq^2}\Pi^*_{LOOP}(q^2) \right]_{q^2 = -m^2}.
\end{align*}


*Ultimately, we started with some factor $m_B^2$ in the posited Lagrangian for so called 'bare' or unrenormalized field $\Phi_B$, but single particles states in "in" or "out" state or just theory when particles are just far apart and not interacting don't have that mass, or more rigorously, the propagator doesn't have a pole at $m_B^2$ but at $m^2$, and the propagator doesn't have the correct residue. We do this process to define our new renormalized fields with $m$ in unperturbed Lagrangian and the parameters we used in this process are defined by the experimental conditions. So, renormalization process has kind of replaced our initially posited variables like mass with the experimentally observed ones.


*With this understanding, we can quote Weinberg: ``the renormalization of masses and fields has nothing directly to do with the presence of infinities, and would be necessary even in a theory in which all momentum space integrals were convergent."


*One amazing that happens now is that for external lines on mass shell, you don't need radiative corrections, because "on-shell" all the perturbative correction are zero based on our condition $\Pi^*(-m^2) = 0$, and other condition gives the right residue.
