Spontaneous symmetry breaking in the standard model

synoe
Messages
23
Reaction score
0
In the standard model, the Lagrangian contains scalar and spinor and vector fields. But when we consider spontaneous symmetry breaking, we only account for the terms contain only scalar fields, " the scalar potential", in the Lagrangian. And if the scalar fields have vacuum expectation value, then we recognize the symmetry is broken spontaneously. Why don't we have to contain spinors and vectors? I think it's related to Lorentz invariance.
 
Physics news on Phys.org
synoe said:
Why don't we have to contain spinors and vectors? I think it's related to Lorentz invariance.
If a spinor or vector had a nonzero expectation value, it would single out a preferred direction. The vacuum state must be rotationally invariant (Lorentz invariant too) so this is not possible.
 
Thank you, Bill_K.
Could you explan more formally or mathematically by using language of quantum theory?
I would like to proof \langle 0|\psi_\alpha|0\rangle=0 from Lorentz invariance of the vacuum. Your explanation seems that the vacuum expectation vales is supposed to be Lorentz invariant but I think Lorentz invariance means U^\dagger(\Lambda)|0\rangle=|0\rangle.

And another question occurred. Your explanation seems that the propagator, the vacuum expectation value of two point function also vanishes. Why propagators don't vanish even if they have Lorentz indices or spinor indices?
 
synoe said:
Thank you, Bill_K.
Could you explan more formally or mathematically by using language of quantum theory?
I would like to proof \langle 0|\psi_\alpha|0\rangle=0 from Lorentz invariance of the vacuum. Your explanation seems that the vacuum expectation vales is supposed to be Lorentz invariant but I think Lorentz invariance means U^\dagger(\Lambda)|0\rangle=|0\rangle.

And another question occurred. Your explanation seems that the propagator, the vacuum expectation value of two point function also vanishes. Why propagators don't vanish even if they have Lorentz indices or spinor indices?

See post#35 in
www.physicsforums.com/showthread.php?t=172461

Sam
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top