Recent content by jackson1

Thanks for the reply. Is it fair to say that the correlation functions are solely computed to get at the scattering amplitudes, or do they have any other significant relevance?

I've been reading Coleman's notes and the book on QFT by Ticciati. There they both place a lot of emphasis on computing the scattering matrix S. I can follow their computations (using Wick's theorem etc.) but I don't really have a good understanding of what S actually tells you. Ticciati even...

Thank you so much.

Ok, I see that later on (lecture 16) he shows that the total contribution of the tadpoles, possibly for a specified theory, is zero - just skimmed ahead and read a few sentences so I'm not sure of the proof, but I'm sure it's exactly as you mentioned. Just to make sure I'm understanding what...

@ Bill_K - Duh, thank you for pointing out the P.B. relations for the complex field. @Chopin - Thanks. Could you do the same for all odd-term interactions?

Hi, I'm currently going through Ticciati's book along with the notes from Sidney Coleman's course and I have a question pertaining to Wick diagrams/expansion of S. In their example (section 4.3 of Ticciati and lecture 9 in Coleman's notes) they never seem to contract the adjoint nucleon field...

Thanks so much for your help.

Ok, but now I have more questions :( . But first, let me just make sure I understand how to answer my original question. Skipping ahead, we have U(\Lambda)\alpha(k)^\dagger U(\Lambda)^\dagger U(\Lambda)|0\rangle = \alpha(\Lambda k)^\dagger |0\rangle which implies U(\Lambda)\alpha(k)^\dagger...

Ok, so then I can say U(\Lambda)|k\rangle = |\Lambda k\rangle implies U(\Lambda)\alpha(k)^\dagger |0\rangle = \alpha(\Lambda k)^\dagger |0\rangle and since the vacuum is invariant, U(\Lambda)^\dagger |0\rangle = |0\rangle , we have U(\Lambda)\alpha(k)^\dagger U(\Lambda)^\dagger |0\rangle...

I'm sorry for being so annoying but I still don't see it. When you say the vacuum is invariant do you mean U(\Lambda)|0\rangle = |0\rangle and similar for U(\Lambda)^\dagger ?

Thanks for the hint. However, I'm still stuck/puzzled. I have U(\Lambda)\alpha(k)^\dagger |0\rangle = U(\Lambda)|k\rangle = |\Lambda k\rangle = \alpha(\Lambda k)^\dagger |0\rangle , and so, I would think U(\Lambda)\alpha(k)^\dagger = \alpha(\Lambda k)^\dagger . I'm not sure how the...

Hi, I'm currently reading the book "Quantum Field Theory for Mathematicians" by Ticciati and in section 2.3 he mentions that the Lorentz action on the free scalar field creation operators \alpha(k)^\dagger is given by U(\Lambda)\alpha(k)^\dagger U(\Lambda)^\dagger = \alpha(\Lambda...