In chapter 6.2 of Weinberg's QFT Vol.1, he gave the general form of Wick contractions of all possible fields(scalar, spinor, vector, etc.), he showed

$$\Delta_{lm}(x,y)=\theta(x-y)P^{(L)}_{lm}\left(-i\frac{\partial}{\partial x}\right)\Delta_+(x-y) +\theta(y-x)P^{(L)}_{lm}\left(-i\frac{\partial}{\partial x}\right)\Delta_+(y-x)\tag{6.2.8}$$

where $$\Delta_+(x)=\int d^3p(2p^0)^{-1}e^{ip\cdot x},$$ and $P^{(L)}_{lm}$ is a covariant polynomial if its argument is a on-shell 4-momentum(i.e. $P^{(L)}_{lm}(\sqrt{\mathbf{p^2}+m^2},\mathbf{p})$), but may not be covariant for off-shell 4-momentum.

Then by a series of mathematical identities, he was able to show $\Delta_{mn}(x,y)$ can be written in a 4-momentum integral:

$$\Delta_{lm}(x,y)=\int d^4q\frac{P^{(L)}_{lm}(q)\exp(iq\cdot (x-y))}{q^2+m^2-i\epsilon}\tag{6.2.18}$$

From here he argued since the 4-momentum $q$ in $(6.2.18)$ is not always on-shell, $P^{(L)}_{lm}$ may not be covariant, and in turn the $\Delta_{mn}(x,y)$ may not be covariant. It's indeed not for a vector field as what he showed immediately after, and that's why we need to add a non-covariant term in the Hamiltonian to make the propagtor covariant ect.etc.

I did follow the steps he showed, but something strange occured to me when I looked back at the expression $(6.2.8)$: $(6.2.8)$ looks completely covariant, since $\theta$ is invariant, $\Delta_+$ is invariant and is written in 3-momentum integral so that $P^{(L)}_{lm}(-i\frac{\partial}{\partial x})\Delta_+(x-y)$ must be covariant, yet $(6.2.18)$-which is supposed to be equivalent to $(6.2.8)$- is not covariant. I am wondering if there's any subtlety I missed.

Also there's another mystery to me: in chapter 3.5 (page 144), he gave a perturbative proof of Lorentz invariance based on Dyson series, the invariance of Hamiltonian density and micro-causality condition, and he did mention Lorentz invariance could be disturbed by the reasonings given in chapter 6.2, but the proof in chap 3.5 is completely formal and I can't see exactly how the reasonings in 6.2 can jeopardize it.

(PS: I wish I could have formulated my question in a more self-contained way, but I couldn't unless I copied some whole pages of Weinberg, so I apologize for the potential vagueness in advance)

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    $\begingroup$ The expression 6.2.18 is written for a generally off-shell value of $P_{lm}$ but if you trace it, it's clear that the original expression didn't depend on $P_{lm}$ at off-shell momenta so the final (6.2.18) can't, either. It's a speedy derivation with a cheating but the final expression is correct if you also incorporate the comments below (6.2.18) which actually tell you how the $P_{lm}$ should be defined away from the mass shell so that it works... Indeed, the proof in 3.5 is "formal" which is the same as "naive" and when trying to make it explicit, there are sometimes subtleties. $\endgroup$ Feb 21, 2013 at 11:14
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    $\begingroup$ @LubošMotl: Could you elaborate a bit more about "It's a speedy derivation with a cheating but the final expression is correct if you also incorporate ..."? If after the integration (6.2.18) actually doesn't receive contribution from off-shell momenta, shouldn't (6.2.18) just be covariant(though not manifestly)? I don't quite see how the comment below (6.2.18) can make it non-covariant. Asfor the proof in chap 3.5, I totally agree with what you said, but just want to see exactly at which step this formal and naive derivation fails. $\endgroup$
    – Jia Yiyang
    Feb 22, 2013 at 12:29

1 Answer 1


I don't know if the poster is still interested in this question. Anyway here are my two cents.

Without introducing $P^{(L)}_{\ell m}$, just consider the massive spin one case and use E.(6.2.6) into Eq.(6.2.8).

Eq.(6.2.6) can be written as $P_{\mu\nu} = \sum_{n=0}^2 P_{\mu\nu}^{(n)}$ where $P_{\mu\nu}^{(i)}$ is proportional to $q_0^n$.

Follow Weinberg steps and you'll get an equation similar to (6.2.12) with terms proportional to $P_{\mu\nu}$, $\delta(x_0-y_0) P_{\mu\nu}^{(1)}$ and $\delta^\prime (x_0-y_0) P_{\mu\nu}^{(2)}$.

The $\delta$ term vanishes but the $\delta^\prime$ doesn't and gives the last term in Eq.(6.2.21).

Bottom line: Eq.(6.2.8) is not covariant because $\theta(x_0-y_0) \partial_\mu f(x-y)$ is not covariant (the $x_0$ dependence in f results in $\delta^{(n)} terms).

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    $\begingroup$ Thank you for your answer, but I need to remember the details of my own question since it was two years ago, I'll get to it as soon as I get some free time. $\endgroup$
    – Jia Yiyang
    Sep 24, 2015 at 3:15

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