## Isospin: how serious must I take it? Superposition of proton and neutron?

Not to be smart-***-y, but how would you realize e.g. $|1,1 \rangle$ in nature?

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Science Advisor
 Quote by tom.stoer Then how shall $$|1,1\rangle\,+\,|2,2\rangle$$ be realized in nature? A simple definition of a superselection rule are states |1> and |2> for which <1|A|2>=0 holds for all observables A. I think this is satisfied for the above mention angular momentum eigenstates in QM.
As a chemist I would call this a pd hybrid. In a hydrogen atom you could prepare it by a pi/4 pulse of light starting from L=1, L_y=1. However, now I suspect that this is meant to be an artificial example and you only consider elements of the SO(3) algebra as observables?

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Science Advisor
 Quote by mr. vodka Not to be smart-***-y, but how would you realize e.g. $|1,1 \rangle$ in nature?
a massive spin-1 particle with z-component = +1, e.g. a rho meson

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Science Advisor
 Quote by mr. vodka I didn't even know there were any forbidden superpositions when it comes to regular spin...
That is the easiest superselection rule, called univalence: There are no superpositions possible between states with integer and half integer spin. Under a rotation by two pi (i.e. the identity), the interference term $\langle 1/2 ;A ;1\rangle$ gets $- \langle 1/2; A; 1\rangle$ and therefore cannot be observable. (sorry I don't find the pipe on the strange keyboard I am using right now. Use a ";" instead).
 Recognitions: Science Advisor I agree, $$|1,1\rangle\,+\,|1/2,1/2\rangle$$ is a better example

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Science Advisor
 Quote by tom.stoer yes exactly; remember the $|n\rangle + |p\rangle$ superposition; it is forbidden in QCD but can be realized when taking other interactions (and observables) into account, e.g. as $|n\rangle + |p\,e^-\,\bar{\nu}\rangle$ superposition
I am not sure about that. Mr_Vodka is completely right that the whole point of considering isospin is taking into account the possibility of superpositions of two states spanning a doublet. However in pure QCD, there are no electoweak charges whence superpositions of protons and neutrons are possible.
 Recognitions: Science Advisor you are right; we already discussed that; my fault; I deleted the post
 Recognitions: Science Advisor Spinning this line of thought further: left handed u and d quarks also form a doublet with respect to electroweak interactions as long as this symmetry is not broken. Hence superpositions are then really possible.

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Science Advisor
 Quote by tom.stoer I guess strangerep wants to discuss the states [...]
Not necessarily. The previous answers seemed to be missing the mark somehow,
so I wanted check Mr Vodka's prior knowledge of Lie groups, QM, and their use in
the classification of elementary particles.

 Quote by mr. vodka I'm familiar with the derivation yes. However, I don't recall SO(3) being mentioned (but I'm quite familiar with group theory, so no need to hold back). I took a quick look at Ballentine (happened to be my lying on my desk) and I don't see SO(3) being mentioned in the derivation either.
Ballentine covers the rotation group SO(3) much earlier, so he assumes the
reader already knows about how those noncommuting ##J_x,J_y,J_z## operators
form the Lie algebra of so(3) (or su(2), but let's skip that for the moment).
He also assumes that by section 7 the reader knows about the relationship between
a Lie group like SO(3) and the Lie algebra so(3) which "generates" the group.

From what you said above, I get the impression you're not yet clear on the intimate
connection between angular momentum and the rotation group SO(3) ?

 It seems the derivation in Ballentine is similar to the one I saw in my QM class: we basically define the angular momentum operator J as something that satisfies the well-known commutation relations, and then it was stated that Jē and J_z form a CSCO (that it's a SCO is clear, but we didn't see an argument for the "Complete" part [which, I suppose, depends on the context of the angular momentum]; maybe this is related to your SO(3) reference).
No, it's "complete" because there's only one "Casimir" invariant operator for
the SO(3). (A Casimir invariant is an operator which commutes with all the
group operators, other than the trivial identity operator.) Since this
derivation focuses only with angular momentum and nothing else, he
doesn't need to consider other stuff at this point.

 One defines the classifying quantum numbers j and m such that [...] With some similar arguments one can also argue that $-j \leq m \leq j$.
OK, that's close enough for present purposes.

 But I don't think my confusion stems from me misunderstanding (regular) spin (?).
The feeling I got reading through earlier parts of this thread is that you
were having trouble transferring the principles in the quantum theory of
angular momentum over to more general cases of classification of elementary
particles.

But... to ensure none of us waste our time..., maybe you should restate your
question(s) as clearly as you can now?
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