# Relation between isospin invariance and charge independence

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 Lipkin'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, Lipkin tries to show why the isospin formalism implies that the strong interaction must be charge independent 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. 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 $$W:=V_{1/2}\otimes V_{1/2}=\operatorname{Sym}^2V_{1/2}\oplus\Lambda^2 V_{1/2}.$$

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 $$\operatorname{Sym}^2 V_{1/2}=V_1$$ (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?

• Interactions

The way I understand it, mathematically, an interaction which is isospin invariant is an element of $$\operatorname{Hom}(W,W)$$.

By Schur lemma, $$\operatorname{Hom}(W,W)=\operatorname{Hom}(V_1\oplus V_0,V_1\oplus V_0)=\operatorname{Hom}(V_1,V_1)\oplus \operatorname{Hom}(V_0,V_0)=\mathbb{C}\oplus\mathbb{C}$$.

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 independence follows from this ?! The two numbers are independent of each other, aren't they ?
• I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. May 27, 2020 at 11:11

You are in the right forum, as runaway hyper-mathematicized expressions, applied inappositely, are all but guaranteed to obscure the point rather than enlighten it. If the pedestrian language of Harry's expression is obscure, further formalization will only drive one barking up the wrong tree, as you are evidently rushing to do here.

Q 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 ?!

You are missing the common background assumed of the reader: the generalized Pauli principle, antisymmetrizing two fermions (your nucleons). A space-spin antisymmetric wave function requires (for over-all antisymmetry) a symmetric isospin wavefunction, T=1, the one you wrote and is a triplet representation of isospin (su(2)).

By contrast, a space-spin symmetric wave function requires (for over-all antisymmetry) an antisymmetric isospin wavefunction, T=0, the one you wrote and is a singlet representation of isospin (su(2)).

1. Is my description accurate ?

Invariant interactions $$[H,\vec T]=0$$, mean that all members of an isomultiplet in question will be treated the same by the interactions (su(2) invariant). So all members of the triplet will be treated the same, even though they have different eigenvalues ("charges") for $$T_3$$. The singlet state does not connect to another state by $$T_3$$, and of course, will have a different Casimir invariant $$T(T+1)$$ than any of the triplet states. The Hamiltonian, and thus interactions, may depend on the Casimir, so you cannot connect different isomultiplets, and nobody suggested you could.

1. How does charge independence follow from this ?! The two numbers are independent of each other, aren't they ?

If you have not appreciated the Gell-Mann–Nishijima_formula, $$Q = T_3 + \frac{1}{2} (B+S),$$ underlying the formal structure, maybe you could go earlier in the text. The last term on the r.h.s. for dinucleons amounts to just one. The charge is a linear function of $$T_3$$, so isospin is a proxy for charge dependence.

• Thank you for your detailed answer. A follow-up, if i may, for the last point. How does charge independence translate here ? I would have thought it meant $[H,Q]=0$ but this can't be true because it would imply charge independence any time the Hamiltonian and $T_3$ commute, which isn't true (as in the case of pions for example) . May 27, 2020 at 16:35
• I'm not sure what you mean by the case for pions--I don't have that book. Pions decay electromagnetically and weakly, both interactions violating isospin; while the weak interactions further may violate S. Tell me what the precise question is. May 27, 2020 at 16:51
• Ok, then, let's forget about the pions and consider this : "in general, what does charge independence translate mathematically ? Does it mean [H,Q]=0 ?" May 27, 2020 at 17:15
• No, $[H,Q]=0$ means charge conservation: charge will not emerge or disappear. Charge independence means, for instance, your interaction will not care if your charge is 1 or 0, or 2 (there are several such hadrons...). Isospin invariance can ensure that, including several charges in the same isomultiplet. For instance, the baryon decuplet for SU(3) flavor symmetry ensures the $\Delta ^{++}$ and the $\Omega^-$ are treated the same by the strong interaction! May 27, 2020 at 17:44
• "A symmetry generator (such as $T_3$) includes the charge operator and has multidimensional representations": the important point here being that $T_+$ for instance may raise your $T_3$ eigenvalue, and hence charge, within the same isomultiplet, all of whose members are treated the same by the strong interactions. You need a nonabelian group for that. May 27, 2020 at 17:57