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Spontaneous symmetry breaking in the standard model

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synoe
#1
Feb25-14, 10:53 AM
P: 5
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.
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Bill_K
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Feb25-14, 01:52 PM
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Quote Quote by synoe View Post
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.
synoe
#3
Feb25-14, 03:17 PM
P: 5
Thank you, Bill_K.
Could you explan more formally or mathematically by using language of quantum theory?
I would like to proof [itex] \langle 0|\psi_\alpha|0\rangle=0[/itex] 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 [itex] U^\dagger(\Lambda)|0\rangle=|0\rangle[/itex].

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?

samalkhaiat
#4
Feb25-14, 05:24 PM
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Spontaneous symmetry breaking in the standard model

Quote Quote by synoe View Post
Thank you, Bill_K.
Could you explan more formally or mathematically by using language of quantum theory?
I would like to proof [itex] \langle 0|\psi_\alpha|0\rangle=0[/itex] 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 [itex] U^\dagger(\Lambda)|0\rangle=|0\rangle[/itex].

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
Bill_K
#5
Feb25-14, 05:38 PM
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Quote Quote by samalkhaiat View Post
Which directly is: http://www.physicsforums.com/showthr...96#post1398896


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