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## Isospin ad conservation of J

In weak desintegrations, the isospin is not necessarily conserved. But is the total angular momentum J=L+S+I always conserved?

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 Quote by quasar987 In weak desintegrations, the isospin is not necessarily conserved. But is the total angular momentum J=L+S+I always conserved?
Yes. The total angular momentum conservation is twin to energy and linear momentum conservation, and that has never been observed to be broken.

 Recognitions: Gold Member Homework Help Science Advisor This is not the answer I was hoping for! I have this problem here that roughly says "a B particle disintegrate into a pi+ and a pi-". So I said "B has isospin 1/2, spin 0 and (there exists a ref. frame where B has) L=0. So that's J=1/2 for B. I know that the state ket for pipi must be symmetrical (2 indistinguishable bosons). And now I know that J total must be conserved. Can someone show me the reasoning behind how to extract the nature (symmetric or antisymmetric) of $|\pi^{+}\pi^{-}>$ given the above information.

## Isospin ad conservation of J

I'm sorry I did not notice that you include isospin in your angular momentum. I'm not sure what you are doing here. Possibly G-symmetry...

Usually, J=S+L is the total angular momentum. Isospin is analogous to spin but acts in internal space, unlike spin and angular momentum.

 Recognitions: Gold Member Homework Help Science Advisor So the isospin is an angular momentum in the sense that it obeys the commutation relations defining an angular momentum ($[I_i,I_j]=\hbar\epsilon_{ijk}I_k$) but for conservation of J to hold, we must not include it in the total angular momentum.

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 Quote by quasar987 So the isospin is an angular momentum in the sense
No, isospin isn't angular momentum in any sense. Yes, it obeys an algebra that is a carbon copy of that used for angular momentum. But a nucleon that is "isospin up" is a proton, and a nucleon that is "isospin down" is a neutron. This has got nothing to do with the z-component of its angular momentum.

I think you're getting 2 different things confused here. You quote the total angular momentum as J=L+S+I, and then call I the isospin. But most textbooks refer to I as the nuclear spin, which is an angular momentum. Isospin is usually represented by $\tau$.

 Recognitions: Gold Member Homework Help Science Advisor Cohen-Tanoudji defines angular momentum as any operator wich satisfies the commutation relation $[J_i,J_j]=\hbar\epsilon_{ijk}J_k$. This is why I called the isospin an angular momentum. But it doesn't add to L and S.

 Quote by quasar987 Cohen-Tanoudji defines angular momentum as any operator wich satisfies the commutation relation $[J_i,J_j]=\hbar\epsilon_{ijk}J_k$.
He does !?
That only defines a symmetry group, not to what the symmetry is applied (as Tom Mattson said).
I am quite bugged.

 But it doesn't add to L and S.
So, what you are doing has no link with G-symmetry ? You were mentionning weak interaction. It maximally violates parity, so combining parity and isospin reverse, you often get (almost) conserved quatities...

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 Quote by quasar987 This is not the answer I was hoping for! I have this problem here that roughly says "a B particle disintegrate into a pi+ and a pi-". So I said "B has isospin 1/2, spin 0 and (there exists a ref. frame where B has) L=0. So that's J=1/2 for B. I know that the state ket for pipi must be symmetrical (2 indistinguishable bosons). And now I know that J total must be conserved. Can someone show me the reasoning behind how to extract the nature (symmetric or antisymmetric) of $|\pi^{+}\pi^{-}>$ given the above information.
The J=1/2 is wrong. Ispin has nothing to do with angular momentum.
Just the math is similar.
The two pions have L=0, a symmetrical state.

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 Quote by humanino He does !?
First sentence of p.646: This is why we shall adopt a more general view and define and angular momentum $\mathbf{J}$ as any set of three observables which satisfies: $[J_i,J_j]=i\hbar\epsilon_{ijk}J_k$

 Quote by humanino So, what you are doing has no link with G-symmetry ? You were mentionning weak interaction. It maximally violates parity, so combining parity and isospin reverse, you often get (almost) conserved quatities...
I don't know what G-symmetry is; this exercise is in the context of the Wigner-Eckart theorem in an ordinary undergrad QM class.

 Quote by quasar987 First sentence of p.646: [i]This is why we shall adopt a more general view and define and angular momentum ...
what is the previous sentence... or what was he talking about

As Meir Achuz said, the pions are in a L=0 state, which must be symmetrical.

 Recognitions: Gold Member Homework Help Science Advisor The three operators associated with the components of an arbitrary classical angular momentum therefor satisfy the commutation relation $[L_i,L_j]=i\hbar\epsilon_{ijk}L_k$. It can be shown, moreover thatthe origin of these relations lies in the geometric properties of rotation in three-dimensionnal space. This is why...

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 Quote by quasar987 First sentence of p.646: This is why we shall adopt a more general view and define and angular momentum $\mathbf{J}$ as any set of three observables which satisfies: $[J_i,J_j]=i\hbar\epsilon_{ijk}J_k$
OK, fine: Isospin doesn't satisfy that algebra, if only because there is no $\hbar$ in there. The isospin matrices satisfy:
$$[\tau_i,\tau_j]=i\epsilon_{ijk}\tau_k$$
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus Actually, I've been thinking about my last post and I'm not satisfied with it. In natural units (where $\hbar=1$), the algebras are identical. So that's not why isospin is not an angular momentum. Angular momentum is the generator of rotations in the normal 3-space in which we all live. It is conserved in physical systems that are invariant under rotations in that space. Isospin, on the other hand, is the generator of rotations in a completely different space altogether, called isospin space. Isospinors are not elements of the eigenspace of $J$, and neither are spinors elements of isospin space. And there is no reason that conservation of $J$ should imply anything about conservation of $\tau$, and vice versa.