Relation between isospin invariance and charge independance
I don't know if this question is better suited for this forum or math.stackexchange.com.
I come from a mathematical background and I'm struggling to understand a passage in <PERSON>'s book "Lie groups for pedestrians" (.
More precisely, I would like to understand what the following passage means in "more mathematical" terms. In this passage, <PERSON> tries to show why the isospin formalism implies that the strong interaction must be charge independant for nucleons but not for other particles such as pions.
Here's the passage (emphasis is mine) (Chapter 2, section 3.):
Consider the interaction between two nucleons in a state which is antisymmetric in space and spin. Such a space-spin state has three possible charges : it can be a two-proton state, a two-neutron state or a proton-neutron state. The isospin formalism says that these three states form an isospin multiplet with $T= 1$. If the interactions are invariant under isospin transformations; i.e., they commute with the isospin operators, then the interaction must be the same in every state of the multiplet. 617
The proton-proton, proton-neutron and neutron-neutron interactions are thus all the same in states which are antisymmetric in space and spin. For states which are symmetric in space and spin there is no argument since such states can only be neutron-proton states with no possibility of other charges.
Here's how I try to understand and formulate it.
* States.
The state of a nucleon is described by a vector in "the" 2-dimensionnal representation $V_{1/2}$ of $\mathbb{su}(2)$.
The state of a pair of nucleons is therefore described by a vector in $W:=V_{1/2}\otimes V_{1/2}=Sym^2V_{1/2}\oplus\Lambda^2 V_{1/2}.$
On the other hand, $V_{1/2}\otimes V_{1/2}=V_{1}\oplus V_0$.
I'm pretty sure, for simple dimensional reasons, that $Sym^2 V_{1/2}=V_1$ (which is 3-dim) and $\Lambda^2 V_{1/2}=V_0$ (which is 1-dim).
Question :
1. I would have argued that symmetric states have three possible charges ($p\otimes p$, $p\otimes n+n\otimes p$, $n\otimes n$ for example) and antisymmetric states only one ($p\wedge n=p\otimes n -n \otimes p$). Why is it the contrary ?! what am i missing ?!
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* Interactions
The way I understand it, mathematically, an interaction which is isospin invariant is an element of $Hom(W,W)$.
By <PERSON> lemma, $Hom(W,W)=Hom(V_1\oplus V_0,V_1\oplus V_0)=Hom(V_1,V_1)\oplus Hom(V_0,V_0)=\mathbb{C}\oplus\mathbb{C}$.
This implies that such an interaction is specified by two numbers (one for the symmetric part, one for the antisymmetric part).
Questions :
1. Is my description accurate ?
2. How does charge independance follows from this ?! The two numbers are independant of each other, aren't they ? 656