I was looking at the Lagrangian density:(adsbygoogle = window.adsbygoogle || []).push({});

L= [itex]-\frac{1}{4}[/itex] [itex]Α^{α}_{μν}[/itex][itex]Α^{μν}_{α}[/itex][itex]-\frac{1}{4}[/itex] [itex]Β^{}_{μν}[/itex][itex]Β^{μν}_{}[/itex][itex]+\bar{e}_{R}[/itex](i∂-g'B)[itex]e_{R}[/itex]

+[itex]\bar{L}_{e} (i\partial-\frac{g'}{2}B+\frac{g}{2}τ^{α}Α_{α})L_{e}-G_{e}\frac{f+p(x)}{\sqrt{2}}(\bar{e}_{R}e_{L}+{e}_{R} \bar{e}_{L})[/itex]

[itex]+\frac{1}{2}({\partial}_{μ} p(x))^{2} +\frac{1}{8} (f+p)^{2} [ (g' {B}_μ-g A_{μ}^{3})^{2} + g^{2} ({Α}_{μ}^{1} Α^{μ1} +{Α}_{μ}^{2} Α^{μ2}) ] - V( \frac{(f+p(x))^{2}}{2})[/itex]

Now I will try to explain what everything is (although i don't know if it has anything to do with the question I want to make):

Le is the doublet of Isospin

[itex]{L}_{e}=( {v}_{e} , {e}_{L} ) [/itex]

[itex]{e}_{L}=\frac{1}{2} (1-γ^{5}) {Ψ}_{e} [/itex] the left component of the electron spinor-wavefunction

with hypercharge -1

[itex]{e}_{R}=\frac{1}{2} (1+γ^{5}) {Ψ}_{e} [/itex] the right component.

with hypercharge -2

there are the 3 gauge fields [itex] {A}_{μ}^{α}, α=1,2,3 [/itex] for the SU(2) with coupling constant g.

the 1 gauge field [itex] {Β}_{μ} [/itex] for Uy(1) with coupling constant g'/2

and for the symmetry breaking I used 4 scalar fields (2 complex)

Φ=(Φ* Φ^{0})

with hypercharge +1

and the well known Potential

[itex] V(Φ*Φ)= {μ}^{2}(Φ*Φ)+λ{Φ*Φ)^{2} [/itex]

One thing more about my notation, everywhere you see A,B or the partial derivative, it is DASHED. I don't know how to make it...

then to get the above lagrangian I expanded my initial lagrangian over the field Φ around the value:

Φ(x)= ( 0 [f+p(x)]/√2 ) where f^{2}= -μ^{2}/λ >0

then you get the above baby monster....

Now my question

When you define the new fields Z,W,A:

[itex]W^{\pm}_{μ}=\frac{1}{\sqrt{2}}(Α^{1}_{μ}\mp i Α^{1}_{μ})[/itex]

[itex]Ζ_{μ}=(g^{2}+g'^{2})^{-\frac{1}{2}}(-gΑ^{3}_{μ}+ g' B_{μ})[/itex]

[itex]A'_{μ}=(g^{2}+g'^{2})^{-\frac{1}{2}}(g B_{μ}+ g' A^{3}_{μ})[/itex]

I see that W is written in respect to the initially defined gauge field of the SU(2) so it means it could couple the isospin doublets (i hope my term was correct) but it cannot couple charges (because it does not have the Uy(1) gauge field B in it). Something that is wrong because W is a charged particle and it COULD interact electromagneticallly...

On the other hand, Z seems to be able to interact electromagnetically because it has the B field...

And finally the Photon A' can couple the isospin doublets? Coz it has the 3rd field A^{3}_{μ}

What am I thinking wrong of it about?

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# Question on Z,W,photon fields

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