Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
In explicit symmetry breaking, the equations of motion of a physical system are variant under the broken symmetry; by contrast, for spontaneous symmetry breaking (SSB), these equations are invariant, but the entire system is not because its vacuum (background) is non-invariant. Further use for the SSB characteristic nonlinear realizations (Goldstone mode), and the group theoretical patterns involved.
7
votes
1
answer
1k
views
Georgi-Glashow model and the VEV of the scalar field
Consider the Georgi-Glashow model, an $SU(2)$ gauge theory with a real scalar in the adjoint (thus a 3-vector in the colour space) $\phi$. The Lagrangian is
$$ L = -\frac{1}{4g^2} F_{\mu \nu}^{\, a} …
2
votes
Why does the Higgs field have less energy when it's non-zero than when it's zero?
From the form of the Higgs potential (which is quartic, the famous Mexican hat) you can see that for $\Re \phi =0$ as well as for $\Im \phi=0$ (the real and imaginary parts of the Higgs field), it is …