I'm reading Weinberg's new book on Quantum Mechanics, and in Chapter 8.7 "Time-Dependent Perturbation Theory" he derives the usual Dyson series for the $S$ matrix when the interaction Hamiltonian $V_I(t)$ (interaction picture) is the integral of a local density $V_I(t) = \int \mathrm{d}^3x\ \mathcal{H}(x,t)$:

$$ S_{\beta\alpha} = \sum_{n=0}^\infty \left[ -\frac{i}{\hbar} \right]^n \int \mathrm{d}^4 x_1\cdots\int \mathrm{d}^4 x_n\left(\Phi_\beta,\ T\left\{\mathcal{H}(x_1)\cdots\mathcal{H}(x_n)\right\}\Phi_\alpha\right) $$

with his notation $\left(u,v\right)$ for the Hilbert space inner product. He then discusses when this formula is Lorentz invariant. There is no problem defining the time ordering when the $x_i$s are inside the light-cone, but the time ordering is ambiguous outside the light-cone. So the usual argument leads to the condition:


if $ (x'-x)^2 \geq c^2 (t'-t)^2 $. So far so good - I've seen all this before. But then he gives this parenthetical:

(This is a sufficient, but not a necessary condition, for there are important theories in which non-vanishing terms in the commutators of $\mathcal{H}(x,t)$ with $\mathcal{H}(x',t')$ for $ (x'-x)^2 \geq c^2 (t'-t)^2 $ are canceled by terms in the Hamiltonian that can not be written as the integrals of scalars.)

There are no references for this, and as far as I can tell it is not clarified anywhere else in the book. If this is true it seems to contradict some of the arguments for local quantum field theories as being somehow a unique (apart from string theories) set of consistent relativistic quantum theories. Does anyone know the theories Weinberg is making reference to here?

(If it's string theory I guess I'll go listen to the Derpy song.)

  • $\begingroup$ I'm also thinking the light-cone condition should be a strict inequality ($>$ rather than $\geq$). I've written it as it appears in the book, so if it's wrong it's a typo in the Kindle edition of the book. $\endgroup$ – Michael Brown Jan 24 '13 at 7:26
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    $\begingroup$ What is the Derpy song? $\endgroup$ – Dilaton Jan 24 '13 at 12:44
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    $\begingroup$ @Dilaton Pure silly: youtube.com/watch?v=pCaSO7eqsjo Derp as in silly mistake and/or oversight. I'd say me forgetting about Coulomb gauge QED deserves a small derp - though only a small one since to be fair I haven't actually used it for anything. :) $\endgroup$ – Michael Brown Jan 24 '13 at 15:28
  • $\begingroup$ Ha ha, that is very funny :-D; I'll go to listen to this when I ask the next silly question ... $\endgroup$ – Dilaton Jan 24 '13 at 23:56

Probably, Weinberg means the instant Coulomb interaction term that appears in QED formulated in the Coulomb gauge (see his Quantum Theory of Fields, V. 1.)

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    $\begingroup$ Thank you. I don't have Weinberg QFT in front of me, and it's not in Peskin and Schroeder, but I've found a little bit about it in Srednicki chapter 55. I've just been working out the commutator in position space. It is (unless I made a mistake): $\left[A_i(x,t), \Pi_j(y,t)\right]=i\delta(x-y)\delta_{ij} + \frac{i}{4\pi}\left(3(x_i-y_i)(x_j-y_j)-r^2 \delta_{ij}\right)/r^5$, $r=\left|x-y\right|$. This shows explicitly the non-locality in the second term. If this is the full story then I don't understand the "cannot be written as the integrals of scalars" comment - it's just a bad gauge choice! $\endgroup$ – Michael Brown Jan 24 '13 at 15:20
  • $\begingroup$ @MichaelBrown: Many consider the Coulomb gauge (also known as a "radiation" gauge) as a "more relevant physically" because of transverse character of vector potential $\mathbf{A}$ and because of presence of lovely Coulomb interaction in the zeroth-order (non perturbed) Hamiltonian, especially useful in atomic problems. In my opinion, this transverse potential contains not only radiation (real photons), but also a near field (virtual photons), so it is not really as "physical" as many think. $\endgroup$ – Vladimir Kalitvianski Jan 24 '13 at 15:38
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    $\begingroup$ I think I erased a small qualification from that comment to fit character limits. It should read "bad gauge choice for demonstrating Lorentz invariance." Of course, different gauges have different merits. I had no intention of being polemical. This still leaves me confused about his "cannot be written as the integrals of scalars" comment. That seems to imply to me that it's something more than a gauge artifact, though maybe I'm reading too much into it. $\endgroup$ – Michael Brown Jan 24 '13 at 16:01
  • $\begingroup$ @MichaelBrown I agree, "important theories" seems to be something "bigger" than some choice of gauge issue to me too... $\endgroup$ – Dilaton Jan 25 '13 at 0:02

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