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

nonequilibrium

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

DrDu

Quote by tom.stoer

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?

tom.stoer

Quote by mr. vodka

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

a massive spin-1 particle with z-component = +1, e.g. a rho meson

DrDu

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 [itex] \langle 1/2 ;A ;1\rangle [/itex] gets [itex]- \langle 1/2; A; 1\rangle [/itex] and therefore cannot be observable. (sorry I don't find the pipe on the strange keyboard I am using right now. Use a ";" instead).

tom.stoer

I agree,

[tex]|1,1\rangle\,+\,|1/2,1/2\rangle[/tex]

is a better example

DrDu

Quote by tom.stoer

yes exactly; remember the [itex]|n\rangle + |p\rangle[/itex] superposition; it is forbidden in QCD but can be realized when taking other interactions (and observables) into account, e.g. as [itex]|n\rangle + |p\,e^-\,\bar{\nu}\rangle[/itex] 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.

tom.stoer

you are right; we already discussed that; my fault; I deleted the post

DrDu

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.

strangerep

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 J² 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 [itex]-j \leq m \leq j[/itex].

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?

Similar Threads for: Isospin: how serious must I take it? Superposition of proton and neutron?
Thread	Forum	Replies
quarks but if a proton is uud and a neutron is udd	High Energy, Nuclear, Particle Physics	3
Neutron proton hit	High Energy, Nuclear, Particle Physics	2
Neutron to Proton	High Energy, Nuclear, Particle Physics	53
Is neutron-neutron fusion easier to facilitate than proton-proton	General Physics	5
proton and neutron	High Energy, Nuclear, Particle Physics	11

nonequilibrium	Isospin: how serious must I take it? Superposition of proton and neutron? Not to be smart-***-y, but how would you realize e.g. [itex]\|1,1 \rangle[/itex] in nature?

tom.stoer Recognitions: Science Advisor	I agree, [tex]\|1,1\rangle\,+\,\|1/2,1/2\rangle[/tex] is a better example

DrDu 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.