IS weak isospin conserved by all interactions?by metroplex021 Tags: conservation law, conserved, interactions, isospin, weak, weak isospin 

#1
Oct413, 10:16 AM

P: 115

Hi people: I keep reading one day that weak isospin is exactly conserved by all interactions; other days that sometimes weak isospin is *not* conserved. Can anyone clear this one up?!




#2
Oct413, 10:19 AM

Mentor
P: 10,853

Maybe some BSM models violate it, no idea, the standard model does not as far as I know. Edit: Okay, this is wrong, see Bill_K. 



#3
Oct413, 11:43 AM

Sci Advisor
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#4
Oct413, 12:00 PM

P: 115

IS weak isospin conserved by all interactions?
Aargh! I don't understand... why would we have even introduced isospin symmetry if it were not conserved by any interaction?! Isn't the above phenomenon more indicative of parity violation than anything else?




#5
Oct413, 12:10 PM

P: 796

Surely the issue is a little more subtle? If we write down the Standard Model Lagrangian before electroweak symmetry breaking, there is an exact SU(2)_isospin symmetry. Any process described by this Lagrangian should conserve weak isospin, period.
Yes, the gauge symmetry is then "spontaneously broken," but isn't it true that this is something of a misnomer and symmetries can't actually be "broken", only "hidden"? After spontaneous symmetry breaking, the vacuum is not invariant under weak isospin rotations, and so the particle spectrum does not consist of particles with definite weak isospin. But as far as I know that doesn't mean that weak isospin isn't conserved, contrary to what Bill_K's quote implies. For example, consider a 1D doublewell potential with an ##x \to x## parity symmetry, and let the barrier between the two wells be essentially infinite. This system will exhibit "spontaneous breaking" of the parity symmetry: you can construct energy eigenstates that localized to one well only, and so are not parity eigenstates. Nevertheless parity is still conserved in this system. 



#6
Oct413, 12:25 PM

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Spontaneous symmetry breaking is not what's involved. As the book points out, any interaction that fails to preserve handedness, fails to conserve weak isospin.
Also, the electromagnetic interaction disconserves weak isospin, the photon being a mixture of T_{w} = 1 and T_{w} = 0. 



#7
Oct413, 02:18 PM

P: 1,273

Isospin is exactly conserved before symmetry breaking but it is not conserved after symmetry breaking because the Higgs field is not an isospin singlet and it forms a condensate. Particles are moving through this condensate at all times and interacting with it. Why then introduce the weak isospin at all as a symmetry, someone asked?
That's the only way to renormalize the weak interaction correctly. 



#8
Oct413, 02:42 PM

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#9
Oct613, 01:33 PM

P: 796

Surely spontaneous symmetry breaking is at the heart of the matter here. Before electroweak symmetry breaking, the mass term for the electron (say) is actually a threeparticle interaction between the lefthanded electron, the Higgs doublet, and the righthanded electron. So the electron can change from lefthanded to righthanded as long as it emits a Higgs boson, which carries away the conserved weak isospin. Here is an interaction that changes the handedness of an electron but manifestly conserves isospin. If you write down the Lagrangian after spontaneous symmetry breaking things are more confusing, to me. Then there is an electron mass term that seems to violate weak isospin. There is also an electronHiggs interaction term that violates weak isospin. But if you add these terms together the sum conserves weak isospin. It's true that the photon isn't an *eigenstate* of weak isospin. But despite the claims of the book you cited, I don't see how this proves anything about whether weak isospin is conserved. A spontaneously broken symmetry isn't manifest in the particle spectrum, but the corresponding current is still conserved. 


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