Introduction to conformal field theory, Lecture 3
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- čas přidán 2. 05. 2018
- In this seminar I will, over some 10 lectures, introduce the basics of conformal field theory. The emphasis will be on the physical content, however, there will be reference to mathematical formulations throughout.
The course is based on a mixture of Ginsparg's "Applied Conformal Field Theory", hep-th/9108028 and Schottenloher's "A mathematical introduction to conformal field theory".
Prerequisites for the course comprise: advanced QM, QFT, advanced QFT, and some familiarity with symplectic methods, which you can cover by watching my previous videos.
In this third lecture I discuss the classical field representations of the conformal group.
6:49, 13:05, 20:20, 24:57, 31:46, 40:03, 48:47, 55:20, 1:03:01, 1:11:55, 1:17:43, 1:21:02, 1:24:16 oddly satisfying! Amazing lecture. 41:59 "Even here life is complicated so ugggghhhhhh" Totally felt that!
Thanks for sharing!
Dear Prof. Osborne,
Thanks for another great lecture!
I particularly liked the picture you gave to explain the active transformation of the field and how the representations of the group arise to describe the transformation of the field indices.
At 1:16:10, I didn't quite understand what the relation between lambda and epsilon was, could you specify it please?
Thanks!
The idea is that If epsilon = 1/n then epsilon = (log\lambda)/n. I hope this helps; sincerely,
Tobias Osborne
pure treasure
Hi @Tobias Osborne, thanks for the lectures! I have a question regarding 1:14:13... Initially you write down the finite transformation on the field Phi(0) as exp(-*delta), which makes sense to me. But then you erase it and replace the exponential with lambda.
I am not sure why that is justifiable? the limit as n goes to infinity of (1+x/n)^n is indeed exp(x), so why is it okay to replace exp with lambda?
Thanks for your clarifications!
Noam
Did you get the answer ... I am confused too in this
Hi Dr. Osborne,
At the end of the lecture, you argued with BCH formula and take the result into the representation. However, in the representation, the dilation is trivial and if we take it into the Lie algebra between dilation and translation, should we have the translation in this representation be 0? If so, will this argument still be valid?
Thank you!
Thankyou for your comment. I don't think the representation of the dilation is trivial. At least, I could not find the point where this is said. So I don't think there is a problem in this case?
Sincerely,
Tobias Osborne
@@tobiasjosborne Hi Dr. Osborne,
Thank you very much for you reply! By saying trivial, I mean it is a constant. At 1:09:34, with Schur's lemma, we have the representation of dilation proportional to identity matrix. The identity matrix should commute with any matrix in the space. Thus, with the Lie algebra between dilation and the translation, [D,P_mu]=iP_mu, will our only choice of translation be 0?
@@hongyao6885 At this point in the lecture I am focussing on the subgroup of the Conformal group which leaves the origin invariant (so we can isolate/study the action on the spin degrees of freedom). In this case we don't have any translation generators (as these would take us out of the subgroup), and hence the commutators don't appear. I hope this helps. Sincerely,
Tobias Osborne
At 1:27:52 I saw that you wrote
\phi'_a (x) = \lambda ^{-\Delta} \phi_a (\Lambda x)
but in the second line you wrote
\phi'_a (x) = \lambda ^{-\Delta} \phi_a (\lambda x)
why can you change \Lambda to a \lambda?
Thank you for answering.
The big Lambda is the notation for a general transformation. In the particular case of scaling it is defined to be equaivalent to the small lambda transformation.
I hope that helps; sincerely,
Tobias Osborne
I don't get why you say, at 58:00, that special conformal transformation keeps the origin fixed? It comes with a translation b^{\mu}
oops. Thankyou for spotting the mistake; many thanks!
Sincerely,
Tobias Osborne
@Haowu Duan, the claim in the video is correct. You can convince yourself by considering SCT = inversion o translation o inversion. The inversion moves the origin to the point at infinity. The translation (b) leaves the point at infinity (and only this point) invariant. Then the second inversion moves the point at infinity back to the origin.