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I have already asked questions about the Higgs mechanism. But what still interests me is the following: The vacuum expectation value of the Higgs field is responsible for the emergence of the elementary particle masses, together with the Yukawa coupling constant. What causes it comes to the spontaneous symmetry breaking? Is it right when I simply write phi = v + h ? When I write phi = v + h the symmetry is broken? And the vacuum expectation value is educated?

And am I right when I say that all fermion masses and gauge boson masses arises through the v (vacuum expectation value). Not through the higgs-boson.

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Can you reword this sentence: "How come now to the spontaneous symmetry breaking and thus the emergence of the vacuum expectation value?" I don't understand it. –  JeffDror Feb 2 at 10:00
    
I have edit my question. –  user37415 Feb 2 at 10:13

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Spontaneous symmetry breaking happens, if the vacuum of the theory lies not at $\phi = 0$, but at $\phi = v$, (i. e. $V'(0) \neq 0$, but $V'(v) = 0$).

In order to make sense of your QFT expansion, you want to expand fields around the vacuum. This justifies the plane-wave approach and the expansion of gauge transformation around unity.

Therefore, you re-write $\phi(x) = v + h(x)$ (sometimes with factors of $1/\sqrt{2}$ floating around). Expanding the new dynamical field $h(x)$ around $h = 0$ is an expansion around the vacuum of the theory and can tehrefore be dealt with perturbatively. The shift $v$ now appears in all couplings of the old higgs field $\phi$ and gives masses to the gauge bosons (through the gauge couplings) and the fermions (via Yukawa couplings). Furthermore, the new dynamical field $h$ obtains very much the same couplings that $\phi$ previousely had.

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"Spontaneous symmetry breaking happens, if the vacuum of the theory lies not at ϕ=0, but at ϕ=v, (i. e. V′(0)≠0, but V′(v)=0)." I always thought V'(0) = 0, but V''(0) < 0, hence the system is unstable at 0 and the new vacuum expectation value becomes ϕ = v which is stable. –  physicsGuy Mar 4 at 13:08
    
Exactly. I put $V'(v) = 0$, which is the condition for a Minimum at a field value $\Phi = v$. –  Neuneck Mar 4 at 22:17

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