# What is the physical interpretation/justification of field renormalization?

In renormalization, we are forced to set several quantities in the Lagrangian to infinite values in order to account for physical results. In QED, for example, we start with a Lagrangian like this: $$\mathcal{L} = \bar{\psi}\left(i\gamma^\mu \partial_\mu -e \gamma^\mu A_\mu -m\right)\psi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ We then perform some calculations (say, finding the cross section for some scattering event) and realize that to get the right physical result, we need to set several constants in the Lagrangian to (some variety of) infinity. This is pretty strange, but relatively unproblematic. The constants in the Lagrangian are free parameters, and they can be whatever they value need to be for the theory to be predictive—if that value is infinite, so be it. This procedure amounts to charge and mass renormalization.

Where I'm really confused is what's going on in field renormalization. Naively, the normalization of the field has already been set by appropriately defining $$a$$ and $$a^\dagger$$ and mandating the canonical commutation relations. And unlike the parameters $$m$$ and $$e$$, the bare fields $$\psi$$ and $$A_\mu$$ seem like they already have physical meanings: we defined $$a^\dagger$$ as an operator creating a one-particle momentum state, then used them to build up quantum fields that on average satisfy classical equations of motion, which are provided by the original Lagrangian. In renormalization, however, we stipulate that the actual field we measure in experiments, and the one that creates one-particle states, is the original field divided by an infinite constant: $$\psi_R= \frac{\psi}{Z}$$ It can be pretty easily seen that the equations of motion $$\psi_R$$ satisfies include several "counteterms" that aren't in the equations for $$\psi$$. So the renormalized field $$\psi_R$$ really is a different beast from the original field.

My question is—how can this move be justified? What physical reason is there for interpreting anything other than $$\psi$$ as the field we measure in experiment, except that it differs from experiment by a factor of infinity? How is the normalization of the field in any sense a free parameter in the theory?

• Let me reverse this and ask you a question - What makes you think that $\psi$ is the field that we measure in an experiment? Apr 2 at 20:25

Here's my shot at an explanation—would love anyone's feedback. In the above, I argued that since we normalized the fields to create a single particle state in the free theory (and satisfy the canonical commutation relations), we should expect them to do so in the interacting theory as well. As far as I can tell, this is just wrong. It assumes that the transition from free theory to interacting theory is entirely continuous, and that the free theory operators retain their exact meaning throughout the transition. It's roughly analogous to claiming that since $$A_\mu$$ creates a one-photon state at all times in the free theory (Heisenberg picture), it must do so in the interacting theory as well—which is wrong, since $$A_\mu$$ evolves under the full Hamiltonian to include amplitudes for electron-positron creation as well.

The first hint that our field operator isn't creating a normalized one-particle state at any time comes when we consider the Green's function $$\langle \Omega|T\{\phi(x_1)\phi(x_2)\}|\Omega\rangle$$ (I'm using $$\phi$$ for simplicity; the argument works the same for $$A_\mu$$ and $$\psi$$). Roughly speaking, this is the probability that a particle at $$x_1$$ propagates to $$x_2$$ or vice versa; in the free theory, it is given by the Feynman propagator. Now, for this probability to have any sense, it should be finite, modulo a possible singularity at $$x_1 = x_2$$. But when we compute the function in the interacting theory, we get $$\langle \Omega|T\{\phi(x_1)\phi(x_2)\}|\Omega\rangle = \infty$$ for all $$x_1, x_2$$. What's going on? It turns out the normalization we thought we had was all wrong: instead of $$\phi(x_1)|\Omega\rangle = |\text{particle at } x_1\rangle$$ we instead have $$\phi(x_1)|\Omega\rangle = \infty|\text{particle at } x_1\rangle$$ This field operator isn't all that useful to us, so we define a new, renormalized operator that retains the standard one-particle interpretation: $$\phi_R(x_1) = \frac{1}{Z}\phi(x_1),\quad \phi_R(x_1)|\Omega\rangle = |\text{particle at } x_1\rangle$$ where we choose $$Z$$ to cancel out the infinity.

This is all well and good, but what about the dynamics? Our Lagrangian is defined in terms of the (infinite) field $$\phi(x)$$, not the finite, renormalized $$\phi_R(x)$$. No problem: if we write $$\phi = (1+\delta)\phi_R$$, where $$\delta$$ is some (formally infinite) counterterm, we can write the Lagrangian in terms of our renormalized fields. We expect any correlation functions we write down involving $$\phi_R$$ to come out finite, and lo and behold, this turns out to be the case, our $$\delta$$ term canceling out the infinities order by order in perturbation theory. Again, this is totally unsurprising: we simply defined $$\phi_R$$ to be whatever we needed it to be for the Green's function to correspond to a finite one-particle propagator, so this cancellation at each order is exactly what we expect.

• This is on the right track. One correction, though: in an interacting model (with a non-quadratic lagrangian), applying a field operator to the vacuum state generally does not produce a 1-particle state. Instead, it produces a state of the form $|\text{1 particle}\rangle+|\text{not 1 particle}\rangle$, and the 1-particle term is typically relatively small (I mean its coefficient in the superposition is small). In a correlation function, we can isolate the 1-particle contribution because it corresponds to a pole, and the field renormalization convention is often based on this correpondence. Jun 30 '20 at 2:05
• @ChiralAnomaly interesting, do you mean immediately? My impression was that the field operator initially creates a one particle state that after any interval evolves into a superposition of indefinite number (unless you're in Heisenberg picture). Jun 30 '20 at 2:59
• Right, I do mean immediately. Here's some intuition: The field operator can be written as a linear combination of "creation" and "annihilation" operators that satisfy the same commutation relations as in a free field theory, but the so-called "annihilation" operator does not annihilate the vacuum state. In other words, it does not annihilate the state with the minimum possible eigenvalue of the Hamiltonian. Once we recognize this, it's not surprising that applying the corresponding "creation" operator to the vacuum cannot create a pure 1-particle state. Jun 30 '20 at 3:06
• Ohhhh, I get it. The field operator is no longer creating eigenstates of the Hamiltonian, which (if you take quanta of energy to correspond to particles) indicates that the field operator is no longer creating a state of definite particles. Jun 30 '20 at 3:48
• Right, exactly. You said it much better than I did. :) Jun 30 '20 at 3:50

Interaction forces us to distinguish between free field (Asymptotic In/Out field in S-Matrix) and the interacting field, which is reflected in these relations:
$$\phi(x)=\sqrt{Z}\phi_{in}(x)+\int dy \Delta_{ret}(x-y) \frac{\partial L_{int}}{\partial \phi(y)}$$ $$\phi(x)=\sqrt{Z}\phi_{out}(x)+\int dy \Delta_{adv}(x-y) \frac{\partial L_{int}}{\partial \phi(y)}$$ Now a bit of calculation can be done to arrive: $$<0|\phi(x)|p>=\sqrt{Z}<0|\phi_{in/out}(x)|p>$$
In the L.H.S the In/Out bare fields only connects single particle states with the vacuum while on R.H.S the field $$\phi(x)$$ (field considering interaction) should connect the vacuum with multi particle states too. Thus a normalization factor would be required to make both the side equal. In a nutshell its the presence of Interaction and consideration of higher order diagrams (in pursuit of obtaining results valid in all order in perturbation theory) that forces to renormalize the fields.

I would like to draw a line here that the infinities in the theory neither triggers the field renormalization nor the renormalization is intended primarily to remove them; this removal is just a byproduct that comes free with renormalization sometimes. I am quoting 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 ."

References

1. https://en.wikipedia.org/wiki/LSZ_reduction_formula
2. Section 10.3, The Quantum Theory Of Fields, Volume (1), S. Weinberg