# What exactly is field strength renormalization?

One thing I have not fully understood is what field strength renormalization is. In Peskin & Schroeder's book "An Introduction to Quantum Field Theory" (Section 7.1) they introduce it as as a term $$Z$$ that makes all the math work out and do not elaborate further. But what exactly are we renormalizing and how is this carried out? In contrast, when renormalize something like the mass it is intuitive what we are doing, with the renormalization carried out to "match" the physical mass.

I apologize if this question is extremely general. Unfortunately I do not know enough to specify it further.

Below, I will give a brief introduction to the difference between the interacting fields and the free field. This is discussed in the first few chapters of Srednicki.

The point is

1. A vacuum expectation value of field operator gives one particle state.
2. In general, one particle state of interacting system is different from that of free theory, but its difference is only the existence of wave-function (= field strength) renormalization constant $$Z$$ (see below). This is an effect of interaction.
3. In particular, condensed matter people calls such a one-particle state in an interacting system as a quasiparticle excitation.

Let me elaborate a bit more.

1st point is simple. If we consider the free theory, one particle state can be obtained by the following relation $$\phi_0(x)|0\rangle\sim a^\dagger|0\rangle.$$

Samely, in the interacting system, we can define the one particle state as a state that is created by field operator (more rigorous definition of one particle state is found in Srednicki or Duncan) $$\phi(x)|0\rangle=|\mathrm{one}\ \mathrm{particle}\ \mathrm{state}\rangle,$$ where both $$\phi$$ and $$|0\rangle$$ are interacting ones.

However, this interacting one particle state is not the same to the one particle state of free theory. We can easily understand this fact. Firstly, in general the difference between interacting field and free field will take the following form: $$\phi(x)=\phi_0(x)+g\delta \phi(x,g),$$ where $$g$$ is a coupling constant and $$\delta \phi(x)$$ is a function of $$g$$, but is regular in the limit of $$g\rightarrow 0$$.

Notice that not only the first term but the second term have a contribution to the one particle state.

Let’s consider one example: $$\phi^3$$ theory. In this case, EoM is given by $$(\Box+m_0^2)\phi(x)=-\frac{g}{3!}\phi(x)^3.$$ For simplicity, we ignore the mass renormalization and assume that renormalized mass is equal to that of free theory $$m_0^2=m^2$$. In real world, this relation is modified to $$m^0=m^2+g\delta m^2$$, but this term does not contribute to the wave-function renormalization.

By substituting third eq. to forth eq., at the first order of $$g$$, we will obtain $$(\Box+m_0^2)\delta\phi(x,g=0)=-\frac{1}{3!}\phi_0(x)^3.$$ The general solution of this equation is given by $$\delta\phi(x,g=0)=\alpha \phi_0(x) -\frac{1}{3!}\int_y iG(x-y)\phi_0^3(y),$$ where $$\alpha$$ is arbitrary constant. A generalization of this equation is called as the Yang-Feldman equation.

From the above, it is clear that $$\delta\phi$$ can have a contribution to free one particle state.

Thus, the general interacting field is written as the following form: $$\phi(x)=(1+\alpha g+\cdots)\phi_0(x)+g\delta \phi(x,g)_{n\geq 2},$$ where $$\delta \phi(x,g)_{n\geq 2}$$ denotes the quantum correction from more than 2 particle states.

The coefficient of the first term is exactly the wave-function renormalization: $$\sqrt{Z}:= (1+\alpha g+\cdots)$$.

From these discussion, we can understand the following formula: $$\phi(x)|0\rangle=\sqrt{Z}\phi_0(x)|0 \rangle.$$

That is, the change of normalization about one-particle state due to the interaction is $$Z$$. Moreover, because one particle can interact with infinite number of particles around him, this coefficient $$Z$$ is usually divergent in QFT.

I give a few final comments. These results are correct only if the interaction results in essentially no change in the particle depiction. Sometimes, the definition of “particle” is strongly modified by the interaction. (e.g. Tomonaga-Luttinger fluid)

In this case, we cannot find the simple formula like $$\phi(x)|0\rangle=\sqrt{Z}\phi_0(x)|0 \rangle$$ between the interacting field and the free field.

In this sense, the discussion above is applied only to the case that the effect of interaction essentially does not modify the definition of particle. This type of discussion is found in the topics around Landau-quasiparticle in condensed matter textbook.

• Thank you for this wonderful answer! I think I understand now. So the constant $Z$ accounts for contributions made by the interacting term to the one particle state? Commented Apr 6, 2023 at 19:00
• @Siam So, you say that $Z$ depends on an arbitrary constant $\alpha$? So $Z$ is arbitrary? By the way, you also say "the general solution is given by..". I do not quite see why it is the general solution. Say, one can add $\phi_0(x-a)$ to it with any $a$. It is still a solution. Commented Apr 6, 2023 at 19:36
• @Yoma 1. In this formal discussion, Z remains as an arbitrary constant; to get more precise information about Z, we need to apply the Z obtained here to the spectral representation of the interacting field. This allows us to put the well-known restriction $0\leq Z<1$, but the coefficient $\alpha$ given above cannot be determined and must be calculated by a perturbation or other appropriate procedure. 2. “General solution” is a littler bit too strong argument. I will correct it later.(But if we impose appropriate condition like locality, I feel this may be the unique solution. I’m not sure.)
– Siam
Commented Apr 6, 2023 at 23:36
• @CBBAM That's the way I understand it. If you want to know more, I suggest you study the Yang-Feldman equation and spectral representation about the interacting field. For this purpose, old-style textbooks like Itzykson-Zuber may be helpful.
– Siam
Commented Apr 6, 2023 at 23:45

I would say what explains what $$Z$$ is the best is its definition. Namely $$$$Z\equiv |\langle \Omega|\phi(0)| p\rangle|^2, \qquad (1)$$$$ where $$\Omega$$ is the interacting theory vacuum and $$|p\rangle$$ is the single-particle state of the interacting theory.

Let me elaborate a bit more on this. The notion of free particles in free theory is different from that in the interacting theory: these are completely different states. The former are much easier to construct, while the latter is what we actually need to define the $$S$$-matrix. $$Z$$ is one of the ingredients that allows us to access the free particles in the interacting theory.

In the free theory $$\phi(x)|0\rangle$$ already gives one-particle states. In the interacting theory, first of all, we need to take into account that the new vacuum $$|\Omega\rangle$$ is different from the free one $$|0\rangle$$. Besides that unlike in free theory $$\phi(x)|\Omega\rangle$$ does not give single-particle states of the interacting theory. In reality, it is a superposition of all multi-particle states. Still, as (1) shows, $$\phi(x)|\Omega\rangle$$ has a non-vanishing projection to single-particle states of the interacting theory. Roughly speaking, $$Z$$ is a relative normalization factor that one needs to take into account when replacing $$|p\rangle$$ with $$\int dx e^{-ipx}\phi(x)|\Omega\rangle$$ in computations. More rigorously this statement is given by the LSZ reduction formula.

Let me also answer it in a general way.

Suppose you want to define a continuum path integral $$\int\mathcal{D}\phi\,\exp\{-S[\phi,g]\}$$ where $$g$$ represents assorted couplings. We all know that $$\mathcal{D}\phi$$, an infinite-dimensional analog of the Lesbuege measure, does not exist at all. So we need to think about a well-defined situation first (regularization) and then take a clever limit to define a continuum theory (renormalization).

Suppose you now have a well-defined regularized path-integral $$\int\mathcal{D}\phi(\mu)\,\exp\{-S[\phi,g]\}$$ where the regularized measure $$\mathcal{D}\phi(\mu)$$ depends on some regularization scale $$\mu$$, such as an inverse lattice pace. If we naively take the limit $$\mu\to\infty$$, we just reproduce the same problem of the nonexistence of $$\mathcal{D}\phi(\infty)$$. But perhaps, the following limit has a chance to exist despite the nonexistence of $$\mathcal{D}\phi(\infty)$$, $$\lim_{\mu\to\infty}\int\mathcal{D}\phi(\mu)\,\exp\bigl\{-S[\phi(\mu),g(\mu)]\,+ N(\mu)\bigr\}\,,$$ after we suitably choose some functions $$\phi(\mu)$$, $$g(\mu)$$, and $$N(\mu)$$. Of course, only some theories can be defined in this way and they are called renormalizable.

Then, the function $$\phi(\mu)$$, or its asymptotic behavior, is the field renormalization.

• So just as how, for example, the mass term changes as the energy scale changes (mass renormalization), the field strength renormalization tells us how the field itself changes as the energy scale changes? Commented Apr 6, 2023 at 18:47
• Yes. The analog of running coupling here is anomalous dimension. You'll meet both of them in the Callan–Symanzik equation, which I think may help your understanding.
– Leo
Commented Apr 6, 2023 at 18:59
• Thank you for all your help! Commented Apr 6, 2023 at 19:00

Briefly speaking, the field-strength renormalization/wavefunction renormalization of the field $$\phi$$ is the introduction of a $$Z_{\phi}$$-factor between the bare field $$\phi_0=\sqrt{Z_{\phi}}\phi$$ and the renormalized field $$\phi$$.

• It seems the two (field-strength and wavefunction renormalizations) are very similar. What is the difference between the two? Commented Apr 6, 2023 at 18:49
• They are the same. Commented Apr 6, 2023 at 18:52
• Thank you. Is there any particular reason the two are sometimes referred to separately? Commented Apr 6, 2023 at 18:53