# A Do the weak isospins of the w1 and w2 combine?

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1. Jul 5, 2016

### QuantumForumUser

Do the weak isospins of the w1 and w2 bosons combine as their fields combine?

2. Jul 11, 2016

### vanhees71

I don't know what you mean by that. You should look at some textbook how quantum flavor dynamics (the Glashow-Salam-Weinberg model of the electromagnetic and weak interactions) are constructed from the corresponding chiral gauge group $\mathrm{SU}(2)_{\mathrm{wiso}} \times \mathrm{U}(1)_{\text{Y}}$ and "Higgsed" to the $\mathrm{U}(1)_{\mathrm{em}}$ to give the particles and some of the gauge fields (the $W^{\pm}$ and $Z$ bosons) masse without violating the vital chiral local gauge symmetry of the model. A very good book on that is

https://www.amazon.de/Elementary-Particle-Physics-Theoretical-Mathematical/dp/3540504966

3. Jul 12, 2016

### QuantumForumUser

What I meant was when the w1 and w2 bosons combine into the w+ and w- bosons through w+ or w- = (w1 -or+ iw2)/sqrt(2). The w+ has weak isospin +1 and the w- has weak isospin -1. So does that mean the weak isospins of the w1 and w2 can be found by rewriting the previously stated transformation (+or- 1= x -or+ iy/sqrt(2))?

4. Jul 13, 2016

### ChrisVer

Does weak isospin make sense after you do the combination?

5. Jul 16, 2016

### QuantumForumUser

Weak isospin (according to Wikipedia: https://en.wikipedia.org/wiki/Weak_isospin) is a conserved quantity. This means that the weak isospin values don't change whether symmetry breaking happens or not.

6. Jul 16, 2016

### vanhees71

True, quantum flavor dynamics bases on the local chiral gauge symmetry $\mathrm{SU}(2)_{\text{wiso}} \times \mathrm{U}(1)_{\text{Y}}$, and this symmetry must not be broken explicitly (and can also not be broken spontaneously):

https://en.wikipedia.org/wiki/Elitzur's_theorem

7. Jul 16, 2016

### ChrisVer

not the point which i tried to make, but that the W+- are not eigenstates of weak isospin and so they don't have definite eigenvalues..
Try to get what the operator thatgave you the +1 or -1 eigenvalue for the W_i does on them... it transforms the one into the other,

8. Jul 16, 2016

### vanhees71

Hint: The SU(2) x U(1) are defined on the "original fields", before introducing the non-vanishing VEV of the Higgs field explicitly. To express the transformation in terms of the "physical" fields (W's, Z, and $\gamma$) is awful!

9. Aug 14, 2016

### QuantumForumUser

According to my book on Electroweak Physics, the w1, w2, and w3 bosons form a weak isospin triplet. This probably means the w1 boson has weak isospin -1, the w2 has weak isospin 1, and the w3 has weak isospin 0. Otherwise, the weak isospins of the w1, w2, and w3 could be eigenvalues of the su(2) Pauli matrices.

10. Aug 15, 2016

### nikkkom

I'm not even sure that in unbroken symmetry, weak isospin *can be represented as a scalar* similar to the electric charge. Since it's related to SU(2) symmetry, shouldn't it have two charges? SU(3) has three charges, "colors". SU(2) should have two, no?

Last edited: Aug 15, 2016
11. Aug 15, 2016

### nikkkom

I think this is not true. Weak isospin is conserved in interactions. But one of the interactions is with Higgs field. When Higgs field VEV is nonzero, it means that particles interact with it all the time. This changes weak isospin and weak hypercharge of the particles. Only their combination which we call "electric charge" is conserved.

For example, a free electron changes between T3=−1/2,Y=−1 ("left electron") and T3=0,Y=−2 ("right electron"). Only Q=T3+Y/2 stays unchanged.

Last edited: Aug 15, 2016
12. Aug 22, 2016

### ofirg

Actually, w$\pm$, w3 are the eigenstates of the weak isospin you are refering to ( which is the third component of the weak isospin - T3)
The usefullness of the w1,2,3 basis is that it transforms as a vector in a three dimensional space.

13. Aug 22, 2016

### ChrisVer

are you sure that the W± are eigenstates of the weak isospin?

14. Aug 22, 2016

### ofirg

Unless someone knows otherwise, they are eigenstates of the third component of the weak isospin - T3
For example, the electric charge Q=T3+Y ( Y is the hypercharge)
Since Y=0 in this case, Q=T3. So the states with well defined electric charge also have well defined and equal T3.

15. Aug 22, 2016

### ChrisVer

if the $W^{1,2,3}$ are eigenstates of $T^3$ with eigenvalues $1,-1,0$ respectively, then:
$T^3 W^\pm =\pm W^\mp$
(in particular the W+/- are more like the ladder operators that you had for the spin).