Timeline for Why does general irreducible representation $(A, A)$ of a quantum field correspond to traceless symmetric tensor of rank $2A$? [duplicate]

10 events

when toggle format	what		by	license	comment
Jul 7, 2022 at 12:07	comment	added	Qmechanic♦		The why is essentially the content of eq. (10) in my answer.
Jul 7, 2022 at 0:55	comment	added	physicsbootcamp		Thank you for the link, but seems like the link doesn't seem to answer my question, I was asking why is (A, A) representation symmetric traceless tensor. I think it is because as Weinberg says all terms are integer spins, which have symmetric traceless tensor representations. For example, for spin-2, we can use symmetric traceless tensor in 3-d which has 5 independent components as expected.
Jun 12, 2022 at 14:05	history	closed	Qmechanic♦		Duplicate of Lorentz representation of totally symmetric tensors?
Jun 12, 2022 at 14:05	comment	added	Qmechanic♦		The answer to the last sentence (v5) is Yes. Possible duplicate: physics.stackexchange.com/q/545943/2451
Jun 12, 2022 at 14:03	history	edited	Qmechanic♦	CC BY-SA 4.0	added 29 characters in body; edited tags
Jun 10, 2022 at 22:16	history	edited	Qmechanic♦	CC BY-SA 4.0	added 26 characters in body; edited tags; edited title; edited tags
Jun 10, 2022 at 21:09	history	edited	physicsbootcamp	CC BY-SA 4.0	added 111 characters in body
Jun 10, 2022 at 21:08	history	edited	physicsbootcamp	CC BY-SA 4.0	edited title
S Jun 10, 2022 at 21:07	review	First questions
Jun 10, 2022 at 21:36
S Jun 10, 2022 at 21:07	history	asked	physicsbootcamp	CC BY-SA 4.0