Hello. I've got a QFT final tomorrow, and one question is still bothering me.(adsbygoogle = window.adsbygoogle || []).push({});

Consider two lagrangians.

The first one is

[tex]L = \frac{1}{2} D_{\mu} \vec{\phi}.D^{\mu} \vec{\phi} + \frac{m^{2}}{2} \vec{\phi}.\vec{\phi} - \frac{\lambda}{4} \left(\vec{\phi}.\vec{\phi}\right)^{2} [/tex]

The second one

[tex]L = D_{\mu} \Phi^{\dagger} D^{\mu} \Phi + m^{2} \Phi^{\dagger}\Phi - \lambda \left(\Phi^{\dagger}\Phi\right)^2[/tex]

In the first lagrangian [tex]\vec{\phi}[/tex] is a vector with real components, in the second on it is a SU(2) doublet.

The first lagrangian obviously has a SO(3) symmetry. If we let the vacuum aquire an expectation value, the symmetry gets broken down to SO(2). The two lost degrees of freedom become Goldstone bosons, which then can give mass to two of the gauge bosons.

The second one has SU(2) symmetry. If I understand it correctly, the symmetry gets completely broken and the three gauge fields acquire mass. (like in the electroweak sector in the Standard Model, but without Weinberg mixing with the U(1) hypercharge gauge field.)

I've got two questions :

Assuming my treatment is correct, why does the first symmetry breaking give mass to two bosons, while the second one gets broken completely?

Is the first lagrangian invariant under SU(2) transformations in the vector representation? Does something need to be alter, other than going with covariant derivatives?

Does anything of this have to do with the fact that SO(3) is locally isomorphic with SU(2), but globally with SU(2) modulo 2?

Any help or insights would be grealty appreciated.

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# Urgent question on spontaneous symmetry breaking

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