# Tag Info

1

The simplification follows from the theorem which states that if such operator is conserved in Heisenberg sense, $$\frac{d\hat{Q}}{dt} = \frac{\partial \hat{Q}}{\partial t} - \frac{i}{\hbar}[\hat{Q}, \hat{H}] = 0,$$ than it commutes with S-operator: $$[\hat{Q}, \hat{S}] = 0$$ So that these two operators can be diagonalized simulatenously: in ...

1

There is a long and formal way, and also an easy and dirty way. I will tell you the easy option. The algebra tells you that $[\delta_Q (\epsilon_1), \delta_Q (\epsilon_2)] = \delta_{P}(\xi^\mu_3)$ where $\epsilon$ is your SUSY parameter and $\xi^\mu_3 = \bar\epsilon_1 \gamma^\mu \epsilon_2$ is your translation parameter. Now, the only Lorentz vector that ...

0

Comments to the question (v3): The point group in question is the chiral tetrahedral symmetry group $T$ of order 12, i.e. the symmetry group of the tetrahedron. Problem 3.1(c) confusingly talks about a 2-dimensional irreducible representation $E$, which is in fact the reducible sum of a 1-dimensional representation and its complex conjugate ...

0

opinion based question, so it may be closed. The author Vincent (family name) has a very good introduction to group theory for molecules. I like this book as it has questions for you to answer as you go along so you really learn it as you read. If you are interested in solid state then you will have to go further to space groups with another text - this ...

1

The spin group $Spin(3,1)\cong SL(2,\mathbb{C})$ is the double cover of the restricted Lorentz group $SO^+(3,1)$, cf. e.g. this Phys.SE post and links therein.

0

You need a two box column for the anti fundamental-representation of SU(3) to accommodate the rules for filling in the boxes in the columns of the Young tableau and having the right number of anti-particles or anti-colors. There are three distinct states 1,2,3 in SU(3). A two box column has exactly 3 possible different configurations using these numbers. In ...

1

Short answers Apply the Young calculus (per ACuriousMind's suggestion in the comments). For finding the multiplicity of the trivial representation in a tensor product of representations of $SU(n)$, note that each irreducible representation $D$ of $SU(n)$ has a unique conjugate irreducible representation $\bar D$ such that the Young calculus allows ...

1

This is not my answer, it's one of the answers you can find here Is there a reason why the spin of particles is integer or half integer instead of even and odd? I just wrote here (and re-posted) the work of @Siva which I found a very good answer. However, follow the link to read more interesting useful answers The "spin" tells us how the wavefunction ...

Top 50 recent answers are included