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1

I not quiet sure that I understand your statement: For ordinary spontaneously broken symmetries, you can demonstrate relations between S-matrix elements with a soft goldstone emission and another S-matrix element without the emission. Are you referring to soft emission theorem where the soft goldstone boson is always emitted from an external leg and as a ...


1

In the standard model the mass for the quarks and other elementary particles comes from from the Higgs mechanism. In the case of the quarks, the masses given in the table of the link are calculated using convoluted theoretical models and data input from scattering experiments.4 At the moment chiral symmetry breaking does not contribute to the quark masses ...


2

It is indeed a an example of symmetry breaking. Basically, the problem (the equactions, and the fields, aven the solution) is symmetric, however because the solution is unstable, in a realistic case where noise is present will break the symmetry into a solution that is not symmetric. The solution "choose" one specific direction to get to a state of lower ...


3

The Higgs mass does not stem from eating Goldstone bosons, since the Higgs is not a gauge field. Since we are breaking an $\mathrm{SU}(2) \subset \mathrm{SU}(2)_L \times \mathrm{U}(1)_Y$ completely, we have three Goldstone bosons, which are eaten by three of the four electroweak gauge bosons to form the massive $W^\pm,Z$ with the photon remaining massless. ...


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I think its easiest to understand this if one has a minimal understanding of QFT. I'm not sure about your background knowledge but hopefully this isn't gibberish to you. The QCD Lagrangian for massless quarks is given by, \begin{equation} {\cal L} = - g \sum_i \bar{\psi} _i A _\mu \gamma ^\mu \psi _i - \frac{1}{4} F _{ \mu \nu } F ^{ \mu \nu } ...


0

A simple (and quite accurate) answer is that quantum fluctuations are the fluctuations that exist at zero temperature. What it means is that even at zero temperature, there might be fluctuations in the measurements of observables, which does not happen for classical systems at zero temperature, due to the non-commutativity of the dynamical and potential ...



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