Recent content by PineApple2

I get this decomposition: T^{ijk}=T^{(ijk)}+T^{[ijk]}+\frac{4}{3}T^{(i[j)k]}+\frac{4}{3}T^{[i(j]k)} is that right? because the last two terms don't match to your last two terms. For example \frac{4}{3}T^{(i[j)k]}=\frac{4}{3}\cdot\frac{1}{4}(T^{ijk}...

Suppose that in the tensor component ##T^a_b ## the upper index is the ## \bf{3}## component and the lower index is the ##\bf{\bar{3}} ## component. To be concrete, consider the decomposition u^iv_j= \left( u^iv_j-\frac{1}{3}\delta^i_j u^kv_k \right) +\frac{1}{3}\delta^i_j u^kv_k which...

Sure. I was not concentrated last night. Thanks!

I guess my question is pretty basic, and following a procedure in the textbook by Lahiri and Pal. You can see the relevant pages at https://books.google.com/books?id=_UmPP8Yr5mYC&pg=PA245&source=gbs_toc_r&cad=4#v=onepage&q&f=false On eqs. (12.84)-(12.86). I don't see how to get from (12.85) to...

Hi. Thank you for the detailed answer, that helped me understand. I have a few further questions please. You found the dimension of the ##\mathbf{8}## as the dimension of the traceless antisymmetric tensor ##\Gamma^{j}_{l}##. and I understand that the tensor ##\varepsilon^{ikl}## doesn't matter...

Here is my trial. Please let me know if there is a better method to use. In this case, we have only upper indices, which we should symmetrize / anti-symmetrize, and not require being traceless. So, T^{ijk}=u^iv^{jk}=T^{(ijk)}+T^{[ijk]} now T^{(ijk)} =...

Thanks! @samalkhaiat: Thank you very much for the detailed post, that is helpful. However, I don't understand a few things: how do you know in eqn. (2) that the coefficient is ##-\frac{1}{8}##? Also, how do you know that you should use the relation...

I see, thanks. So is it correct to say that the irreducible representations need to be traceless with respect to upper and lower indices, because otherwise the invariant tensor ##\delta^i_j## will contract them into a smaller dimensional representation, which would mean that they were not...

Hi. In Georgi's book page 143, eqn. (10.29) he gives an example of decomposing a tensor product into irreps: u^iv_k^j=\frac{1}{2} \left( u^iv_k^j+u^jv_k^i-\frac{1}{4}\delta_k^iu^\ell v_\ell^j-\frac{1}{4}\delta_k^ju^\ell v_\ell^i \right)\\ +\frac{1}{4} \varepsilon^{ij\ell} \left(...

Why is it important if the quark and anti-quark have the same flavor?

Does this discovery reveal new physics, besides the assembly of new particles made of more quarks? i.e. does it change or give new information about current physical models or theories?

I see, that makes sense. Thank you all for answering

Hi, this is a rather mathematical question. The inner product between generators of a Lie algebra is commonly defined as \mathrm{Tr}[T^a T^b]=k \delta^{ab} . However, I don't understand why this trace is orthogonal, i.e. why the trace of a multiplication of two different generators is always zero.

I am talking about the weak isospin. I am not sure how post #6, which says that symmetry breaking is not involved, is consistent with #8, which looks like the Higgs breaks this symmetry. If someone can write some math and explain this. About the hadronic isospin - since you mentioned that, I...

I am a bit confused by this post. Is the weak isospin conserved before symmetry breaking? if not, then what is the conserved charge associated with SU(2)_L ? and why is it a useful quantity? and if it is, then how does symmetry breaking violate that? and how is the electron not having a defined...