- #1

- 151

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter metroplex021
- Start date

- #1

- 151

- 0

- #2

mfb

Mentor

- 35,332

- 11,642

Ask there for processes that violate the conservation?other days that sometimes weak isospin is *not* conserved.

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.

Edit2: It is more complicated than a single yes/no answer, see later replies.

Last edited:

- #3

Bill_K

Science Advisor

- 4,155

- 201

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 T_{w}= 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.

- #4

- 151

- 0

- #5

- 1,006

- 105

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.

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.

Last edited:

- #6

Bill_K

Science Advisor

- 4,155

- 201

Also, the electromagnetic interaction disconserves weak isospin, the photon being a mixture of T

- #7

- 1,948

- 200

That's the only way to renormalize the weak interaction correctly.

- #8

Bill_K

Science Advisor

- 4,155

- 201

The Higgs field is chosen to be an isospin doublet specifically to guarantee that the interaction term eIsospin 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.

- #9

- 1,006

- 105

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.

Also, the electromagnetic interaction disconserves weak isospin, the photon being a mixture of T_{w}= 1 and T_{w}= 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.

- #10

Vanadium 50

Staff Emeritus

Science Advisor

Education Advisor

- 26,705

- 10,260

- #11

- 49

- 0

- #12

ChrisVer

Gold Member

- 3,379

- 460

What kind of isospin are we talking about? The SU(2) of the SM is broken by the Higgs mechanism...

The SU(2) isospin which appears in the strong sector is conserved by strong interactions (in the limit of massless quarks) and is broken (except for the 3rd component) by electromagnetic...

- #13

- 49

- 0

I am talking about the weak isospin. I am not sure how post #6, which says that symmetry breaking is not involved, is consistent with #8, which looks like the Higgs breaks this symmetry. If someone can write some math and explain this.

What kind of isospin are we talking about? The SU(2) of the SM is broken by the Higgs mechanism...

The SU(2) isospin which appears in the strong sector is conserved by strong interactions (in the limit of massless quarks) and is broken (except for the 3rd component) by electromagnetic...

About the hadronic isospin - since you mentioned that, I think the requirement is that the quark masses are equal, not necessary zero.

- #14

ChrisVer

Gold Member

- 3,379

- 460

I don't understand why that book says that the weak isospin is not conserved - it is not after the SSB and because of the Higgs vev (or in other words because the fermions are massive) . The force bosons (W,Z,gamma) appear only after the symmetry breaking and so you can say that those interactions (weak and em) don't preserve this symmetry...the strong interactions still do (since they don't couple left to rights)

- #15

samalkhaiat

Science Advisor

- 1,735

- 1,045

- #16

- 8

- 0

But...what I find in wikipedia(Weak isospin, the second paragraph) is thatQuoting from this Google book:

So I am confused again...

- #17

- 218

- 101

But...what I find in wikipedia(Weak isospin, the second paragraph) is thatThe. (Weak isospin is usually given the symbolweak isospin conservation lawrelates the conservation of T3; all weak interactions must preserve T3. It is also conserved by the other interactions and is therefore a conserved quantity in generalTorIwith the third component written asTz,T3,Iz orI3.[1]).

So I am confused again...

I think one needs to distinguish between the "weak isospin" (##T##, ##\vec{T}##, ##I##) and "third component of the weak isospin" (##T_3##, ##I_3##). The first one is apparently what this book is referring too, while the conservation law from wikipedia is about the latter one (note the next sentence after your quote: "For this reason

- #18

- 8

- 0

Sorry, I didn't see that difference. Thanks to point it out.I think one needs to distinguish between the "weak isospin" (##T##, ##\vec{T}##, ##I##) and "third component of the weak isospin" (##T_3##, ##I_3##). The first one is apparently what this book is referring too, while the conservation law from wikipedia is about the latter one (note the next sentence after your quote: "For this reasonT3 is more important thanTand often the term "weak isospin" refers to the "3rd component of weak isospin".").

- #19

- 2,075

- 398

But...what I find in wikipedia(Weak isospin, the second paragraph) is thatThe. (Weak isospin is usually given the symbolweak isospin conservation lawrelates the conservation of T3; all weak interactions must preserve T3. It is also conserved by the other interactions and is therefore a conserved quantity in generalTorIwith the third component written asTz,T3,Iz orI3.[1]).

So I am confused again...

As The_Duck said, "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" and "the vacuum is not invariant under weak isospin rotations".

IOW, our vacuum contains a Higgs condensate, which you may visualize as the ability to consume without a trace, or supply "invisible" Higgs bosons.

Share: