How does a fermion, like an electron, get its mass through the Higgs-mechanism? Can someone explain me this with formulas (Lagrangian)?

I know that the Yukawa interaction has something to do with this, is that right?

Maybe when I'm right, there is a term:

$$g \bar{\Psi} \Phi \Psi?$$

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    $\begingroup$ You're right. Since this is very standard material, let me first refer you to wikipedia en.wikipedia.org/wiki/…, en.wikipedia.org/wiki/Yukawa_interaction so you get can a feeling for what happens. $\endgroup$ – Vibert Jan 15 '14 at 13:47
  • $\begingroup$ And if I'm right again first there is a term g bar psi phi psi. Through the spontaneous symmetry breaking the yukawa term become g phi_0 bar psi psi. And phi_0 is the vacuum expectation value of the higgs field. g and phi_0 are the mass m_f of the fermion, right? $\endgroup$ – user37415 Jan 15 '14 at 14:15
  • $\begingroup$ Roughly yes, but in the Standard Model the interaction couples different fields, so the total Yukawa coupling is of the form $\sum_{ij} c_{ij} \bar \psi_i \phi_0 \psi_j$ where $c_{ij}$ is a matrix of c-numbers. $\endgroup$ – Vibert Jan 15 '14 at 14:22
  • $\begingroup$ And can you please explain me, how exactly a fermion (for example a electron) get mass through the higgs-mechanism? $\endgroup$ – user37415 Jan 15 '14 at 14:31
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    $\begingroup$ Before the symmetry breaking the Yukawa term (coupling of the higgs field with the fermion field) looks g bar psi phi psi, right? And after the symmetry breaking there is g phi_0 bar psi psi==> this is the mass term, right? $\endgroup$ – user37415 Jan 15 '14 at 15:21

It is about the "the 5-th force."

As you said the Yukawa term introducing the interaction between scalar field $\Phi$ and fermion $\Psi$ field: $$g \bar{\Psi} \Phi \Psi$$

The Higgs mechanism causes the $\Phi$ field condense at a classical expectation value (v.e.v: vacuum expectation value), due to the Higgs potential $U(\Phi)$, so $\Phi$ tend to find a classical minimum, which causes:

$$\Phi(x,t) \to \langle \Phi \rangle=m$$

as a fix value $m$. You can imagine this process as originally $\Phi(x,t)$ is a field variable free to have any real/complex values at any spacetime $(x,t)$ point due to quantum fluctuation. However, the Higgs mechanism causes $\langle \Phi \rangle=m$ finding a (local) classical stable minimum value of the potential $U(\Phi)$.

The remarkable result is that $\Phi(x,t)$ semi-classically now have to take the fix value at $m$ at any spacetime point! (This is the remarkable fact of the 5-th force: Higgs field introduces mass to fermions i.e. quarks, leptons, in the Standard Model. Some people coin the name the 5-th force - a different mechanism from the 4 fundamental forces.)

Add: Some people like to think about (fermions,W$^{\pm}$,Z$^{0}$ bosons) particles moving in the ocean of Higgs fields, thus (fermion,W,Z) particles become massive due to the buoyancy force effects in the Higgs ocean.

The mass $M$ of fermion fields now can be read as

$$g \bar{\Psi} \Phi \Psi \to (g\cdot m) \bar{\Psi} \Psi=M \bar{\Psi} \Psi $$ with fermion mass $M=g\langle \Phi \rangle=g\cdot m$.

Note that now Fermion mass takes the fixed value at $g \langle \Phi \rangle$, BUT there is quantum fluctuation around the v.e.v. ($\langle \Phi \rangle+\delta \Phi $) to cause fermion field interacting with the Higgs fluctuation $\delta \Phi $. You can draw a Feynman diagram to compute its effect.

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    $\begingroup$ I have still one question. m is in your answer the vacuum expectation value of the higgs-field? And then g*m is the mass of the fermion? $\endgroup$ – user37415 Jan 16 '14 at 12:22
  • $\begingroup$ Yes, it is correct. You are right. ps. Some people like to think about (fermion,W,Z) particles moving in the ocean of Higgs fields, (fermion,W,Z) particles become massive due to the buoyancy force effects in the Higgs ocean. It is pretty cute, and which description is helpful and imaginative in some way. :-) $\endgroup$ – wonderich Jan 16 '14 at 17:37
  • $\begingroup$ Regarding the idea with the higgs ocean and the buoyancy. This is a imagninative description, but it is not the 100 % right description. The 100 % correct description is your answer with the Yukawa interaction term, right? $\endgroup$ – user37415 Jan 17 '14 at 8:40
  • $\begingroup$ Yes, I agree. Higgs ocean is just a hand-waving way to explain in the manner for popular science. $\endgroup$ – wonderich Jan 17 '14 at 17:19
  • $\begingroup$ And your answer with the Yukawa interaction term is the 100 % correct answer in particle science? $\endgroup$ – user37415 Jan 17 '14 at 17:21

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