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IS weak isospin conserved by all interactions?

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metroplex021
#1
Oct4-13, 10:16 AM
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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?!
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mfb
#2
Oct4-13, 10:19 AM
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Quote Quote by metroplex021 View Post
other days that sometimes weak isospin is *not* conserved.
Ask there for processes that violate the conservation?

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.
Bill_K
#3
Oct4-13, 11:43 AM
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Quoting from this Google book:

The weak isospin is not conserved by any interaction... The left-handed components of all elementary fermions are assigned weak isospin 1/2. The right-handed components have Tw = 0. The non-conservation of weak isospin is already apparent in the fact that, with the possible exception of the massless neutrino, there is no particle with a definite handedness and therefore with a defined weak isospin.

metroplex021
#4
Oct4-13, 12:00 PM
P: 119
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?
The_Duck
#5
Oct4-13, 12:10 PM
P: 847
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 double-well 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.
Bill_K
#6
Oct4-13, 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 Tw = 1 and Tw = 0.
dauto
#7
Oct4-13, 02:18 PM
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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.
Bill_K
#8
Oct4-13, 02:42 PM
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Quote Quote by dauto View Post
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.
The Higgs field is chosen to be an isospin doublet specifically to guarantee that the interaction term eR v EL = (singet) (doublet) (doublet) comes out to be invariant under isospin, and consequently conserves isospin. The vacuum state spontaneously breaks the symmetry, but that's unrelated to the effect of the interaction term with fermions.
The_Duck
#9
Oct6-13, 01:33 PM
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Quote Quote by Bill_K View Post
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.
So how do you reconcile this claim with the fact that if you write down the Standard Model Lagrangian before electroweak symmetry breaking, weak isospin is an exact symmetry and so must be exactly conserved?

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 three-particle interaction between the left-handed electron, the Higgs doublet, and the right-handed electron. So the electron can change from left-handed to right-handed 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 electron-Higgs interaction term that violates weak isospin. But if you add these terms together the sum conserves weak isospin.

Quote Quote by Bill_K View Post
Also, the electromagnetic interaction disconserves weak isospin, the photon being a mixture of Tw = 1 and Tw = 0.
I think the electromagnetic interaction definitely doesn't violate weak isospin. Electromagnetism is the unbroken part of weak isospin and weak hypercharge; how can it break weak isospin or weak hypercharge?

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.
Vanadium 50
#10
Oct6-13, 04:15 PM
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Bill_K had it in #6, but maybe the implication needs to be expanded. If you have, e.g., an electron in your hand, it is not in an eigenstate of weak isospin.


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