Hi, and thanks in advance for reading this. I've been bashing my head on my desk for days on this now.
My problem is the first sentence of the paragraph after equation 32.1.5 in Weinberg III:
"... note that for a given $Z_{nm}$ to be non-zero, since it is a scalar all of the $\sigma$s in...
However, kinsemc and sniffer's original question was concerning jobs for *theoretical* physicists such as people who did a PhD in string theory. If I delete from the list all teaching jobs, postdocs and lectureships, pretty much all adverts are for *experimental* physicists (although I did count...
Hi there,
I'm giving lectures on SUSY following Weinberg III. Here's my problem: Is (27.1.12) correct? I mean, shouldn't the \Omega dependent factors be swapped? Otherwise \Phi^\dagger \Gamma is not gauge covariant!
My understanding is that the extended gauge transformations of (27.1.11)...
Well, hard photoproduction is my subject, so I'm bound to give a biased opinion! The advantage of colliding photons in order to study QCD, as opposed to protons, is that photons are much "cleaner" (after collision, the smashed protons produce a lot more junk, or "background"). Theoretically...
Photon - photon scattering in QED is indeed well established. Perhaps you might like to study the more unestablished QCD description of inclusive photon-photon scattering. Basically, one of the photons emits a quark or gluon at low energy, which interacts with the other photon at high energy...
I was put off Weinberg for a long time because his notation was different to others. But once I got familiar with it, I found in fact that his notation was simpler and more general.
One of my problems is I'm not sure! Let me just state what I understand: We introduce the principle that there exist "free" states, where "free" means that acting on them with &exp;[-iHt] gives the same result as acting on them with &exp;[iH0 t], where H0 is called the "free" particle...
If I understand you correctly, yes. If you have Weinberg(?), look at equation 6.2.1 for the propagator. This definition is unique, and comes from the commutation relations of the creation and annihilation operators. The iε and the choice of sign of ε is introduced in the...
I'm not sure I understand, but I will have a go.
If your problem is the apparent arbitrary insertion of the iε in the denominator: This correctly reproduces the position space propagator.
If your problem is a deeper one concerning scattering theory, I had the same problem when...
Definitely. Yet how easy it is to understand depends mostly on how well explained it is.
I've been trying to write a companion to Weinberg, to fill in the more technical steps and analyse in more depth the statements that Weinberg thinks is obvious. But it's a little messy.
Great - how much have you read, if any? How did you find it? What textbooks have you found helpful?
By the way, chapter one is a historical account of QFT, not necessary for understanding the rest of the book.
Sorry, I was referring only to those people who had never read Weinberg, and who want to learn / re-learn the basics. I think you're the first person I've met who has! What did you think, was it the best approach for you or is there another author you prefer?
By the way, Weinberg discusses...
Although this is not really his reason for making it so. He gives a more formal argument - that it's needed for Lorentz invariance.
Yes, I didn't mention LI + causality for brevity. CDP + LI + causality (+ anything else?) leads to fields.
Actually here I felt was one of Weinberg's...
My reason for liking Weinberg: His approach feels much more "pure" than other books, particularly in that all axioms are simple and physically inuitive, and are introduced only when needed. Thus your knowledge doesn't get entangled, the various theorems become more powerful since they can then...
I hope you and others will join. I would hope that those who don't have Weinberg Volume I but who want to join this thread can find it in a library or borrow from someone, although I understand this is a hassle.
My reason for choosing Weinberg: His approach feels much more "pure" than other...
Thanks for the quick reply!
However, I'm still confused. I understand that W(ℜ,p) must be SO(3) like ℜ but why does that mean W(ℜ,p)=ℜ? Is it implied by
W(ℜ[sub]1[\sub],p) W(ℜ[sub]2[\sub],p)=
W(ℜ[sub]1[\sub]ℜ[sub]2[\sub],p),
and if so, why?
Purpose
Actually the question was not the most important part of my last posting (although I still need the answer, if anyone has it!), I just wanted to get the ball rolling.
Of course everybody has a favourite textbook. What I want is a thread for those people whose favourite is Weinberg's...
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