Field-strength renormalization problem (13.1) in Srednicki

Scott1137

Hi-

I've just completed problem 13.1 in Srednicki in which he tells us to relate the field-strength renormalization $Z_{\phi}$ to the spectral density $\rho(s)$ that appears in the Lehmann representation of the exact propagator. It seems straightforward-- I follow the hint, insert unity using the 0, 1, and multi-particle states of the interacting theory, and make use of canonical equal-time commutation relations to get

\begin{equation}

\frac{1}{Z_{\phi}} = 1+ \int_{4m^2}^{\infty} ds \rho(s)

\end{equation}

My question is this-- How do I reconcile the above result, which implies that $0 < Z < 1$, with the divergent expressions one obtains for $Z_{\phi}$ in standard perturbative calculations?

I'm almost positive I did the problem correctly (in many-body theory there are similar bounds one derives for the residue at the quasi-particle pole of the Green function). Would the bound on Z be restored if I could do a non-perturbative summation of diagrams?

Bill_K

Isn't it just that this Z is the reciprocal of the infinite one? So this Z goes to zero.

Scott1137

Sorry Bill_K, I don't follow. Let me rephrase my question in case the original wasn't clear. In problem 13.1, we use non-perturbative arguments to derive a sum rule that relates the field-strength renormalization $Z_{\phi}$ to the spectral density $\rho(s)$:

\begin{equation}
Z_{\phi} = \frac{1}{1+\int_{4m^2}^{\infty}ds\, \rho(s)}
\end{equation}

where $\rho(s)$ is defined in Eq. 13.11. Since the integral is positive-definite, this implies that $0 ≤ Z_{\phi} ≤ 1$.

Now look at section 14 where Srednicki calculates the 1-loop contribution to the self-energy in phi**3 theory. In \bar{MS}, he finds that the counterterm vertex $A=Z_{\phi}-1$ diverges as $1/\epsilon$ where $\epsilon = 6 -d$ (see Eq. 14.37).

My question is how do I reconcile the fact that the non-perturbative sum-rule implies $0 ≤ Z_{\phi} ≤ 1$, whereas any perturbative calculation gives a divergent $Z_{\phi}$? Is this just an artifact of perturbation theory that would disappear if I summed all diagrams to all orders?

Bill_K

Thanks, it was clear. It is Z^-1 that diverges, not Z. The fact that 0 ≤ Z ≤ 1 implies that the renormalized charge is less than the bare charge. In electrodynamics, and most other theories, Z^-1 diverges, so that the matrix element of the field operator φ between a one-particle state and the vacuum state <Ψ | φ | p> = Z^1/2/(2π)^3/2 vanishes.

One therefore defines a new, renormalized operator φ_R = Z^-1/2 φ which has the property that its matrix element between the vacuum and the one-particle state is finite.

Lapidus

Great question, I wondered about that before, too.

How is the renormalization constant in the exact two point function related to the renormalization constants in the perturbative loop calculations?

Why is the in the former case something between zero and one, and in the latter infinite??

If you do not own Srednicki book, take a look at Peskin and Schroeder equations (10.14) to (10.18.).

My uneducated guess is that since renormalization constants are unobservable, they are only required to fullfill some renormalization conditions and except from that can be whatever they want.

Scott1137

Ok, sorry to be so dense, but I don't understand your statement that it is 1/Z and not Z that diverges. This seems to be in conflict with what's in Srednicki and every other QFT book I've glanced at. To 1-loop order, in Eq. 14.37 Srednicki finds (in phi**3 theory in d= 6-ε spacetime)

\begin{equation}
Z_{\phi} = 1 - \frac{\alpha}{6}\,\bigl[1/\epsilon + \rm{log}(\mu/m) + 1/2 + \kappa_A\bigr]
\end{equation}

where

\begin{equation}
\alpha = \frac{g^3}{(4\pi)^3}\,.
\end{equation}

This diverges, no?

Bill_K

They mention this point in a Google Book. Go here and see the paragraph following Eq. (10.15.12).

Edit: I think the point is the distinction between perturbative and nonperturbative. You can have something which produces infinite terms in the perturbation expansion but is actually zero nonperturbatively. For example

Z = 1 - ∞ + ∞² - ... = (1 + ∞)^-1

Scott1137

Thanks. Nevertheless, this reasoning makes me very uneasy. This seems to say that the Z-factors are 1 for free field theory, but discontinuously jump to 0 with an interacting theory, even if the coupling in the interacting theory is arbitrarily small. This conflicts with my experience with interacting many-fermion systems, where the analogous quantity (the residue at poles in the Green function near the Fermi surface) passes continuously from 1 to 0 ≤Z ≤ 1 as on cranks up the strength of the interactions. There, moderate values of Z (i.e., not too close to zero) would indicate a system that qualitatively resembles the free theory albeit with renormalized couplings and masses (this is Landau's Fermi liquid theory), whereas Z~0 would indicate that the relevant degrees of freedom bear no resemblance at all to the free theory. I thought that much of the success of QFT relied on the fact that the interacting states bear a qualitative resemblance to the states of the zeroth-order free theory. I don't see how this can possibly be the case with Z = 0.

Lapidus

From the book that Bill K gave the link to: " Z is somewhere between 0 and unity, with Z=1 for a free field. Actually, for a interacting theory Z is a divergent function of a regularization parameter." (below 10.15.12)

So what now, is Z for an interacting theory smaller than unity or is it divergent??

Just for starters, what is the field renormalization factor anyway?

The probability to create one-particle states from the vacuum or a constant that relates physical fields to bare fields?

thanks