# How does the Higgs boson give mass to other elementary particles like electrons? [duplicate]

So, the nucleus of an atom can be broken to protons and neutrons, and those can be broken down to quarks.

Electrons however are a different story, they can't be broken down since they are elementary particles but they do have mass.

So the higgs boson itself does not give mass to particles but the interaction between the Higgs field and the Higgs boson does? How does the electron for example interact with the Higgs field?

• "How does the electron for example interact with the higgs field?" - See Yukawa interaction: "The Yukawa interaction is also used in the Standard Model to describe the coupling between the Higgs field and massless quark and lepton fields (i.e., the fundamental fermion particles). Through spontaneous symmetry breaking, these fermions acquire a mass proportional to the vacuum expectation value of the Higgs field." – Hal Hollis Jun 2 '17 at 22:18
• Possible duplicates: physics.stackexchange.com/q/17944/2451 , physics.stackexchange.com/q/6450/2451 and links therein. – Qmechanic Jun 3 '17 at 4:37
• This one is very relevant too: physics.stackexchange.com/q/95419/154997 – user154997 Jun 3 '17 at 9:19

In field theory, the dynamics of a system is described by a Lagrangian. For instance, for a free scalar field (spin 0) field, the Lagrangian is $${\cal L} = - \frac{1}{2} ( \partial \phi)^2 - \frac{1}{2} m^2 \phi^2 ~.$$ Then, the mass of the particle that corresponds to this field is determined as follows. Recall that in field theory a particle is understood as fluctuation of a field. So a field like $\phi$ may have a base value $h$ and a particle is represented as a fluctuation $f$ on top of this base value. So writing $\phi(x) = h + f(x)$ and substituting this in the Lagrangian and expanding to quadratic order, we find $${\cal L} = - \frac{1}{2} ( \partial f)^2 - \frac{1}{2} m^2 f^2 - m^2 f h - \frac{1}{2} m^2 h^2$$ We should think of this Lagrangian as describing the fluctuation $f$. Then, the mass is read off from the quadratic term as being $m$. Note that for this simple case the would-be "mass" of $\phi$ read off from the first form of ${\cal L}$ is the same as the mass of $f$. This is not always true. For instance, consider a scalar field that is self-interacting with an interaction term $\phi^3$ so that the action is $${\cal L} = - \frac{1}{2} ( \partial \phi)^2 - \frac{1}{2} m^2 \phi^2 -g \phi^3 ~.$$ Then, writing as before $\phi(x) = h + f(x)$ we find $${\cal L} = - \frac{1}{2} ( \partial f)^2 - \frac{1}{2} [ m^2 + 6 g h ] \,f^2 + \cdots$$ In this case, we find that the mass of the particle $f$ is modified to $\sqrt{m^2 + 6 g h}$ so it depends on the "base" value of the field $h$ as well as the coupling constant $g$. Now, whether a field can have a base value or not depends on other factors that I will not explain here.

In this way, we see that adding interactions modifies the mass of the particle if the field happens to have some sort of base value. It is in this way that the Higgs field gives mass to the electron. Let us see this in some more detail.

We work with a simplified version of the Higgs field described by a real scalar field $\phi$ (as opposed to the real one which is a fundamental of the SU(2)). It interacts with itself and the electron field $\psi$ according to the following Lagrangian $${\cal L} = - \frac{1}{2} ( \partial \phi)^2 + \frac{1}{2} m^2 \phi^2 - \frac{1}{4} \lambda \phi^4 - i {\bar \psi} \gamma^\mu \partial_\mu \psi - g \phi {\bar \psi} \psi ~.$$ The last term is known as the Yukawa interaction term. Now, $\psi$ is a fermionic field and is not allowed to have a base value. For the Higgs field, we write $\phi(x) = h + f(x)$ and we find $${\cal L} = - \frac{1}{2} ( \partial f)^2 - \frac{1}{2} [ 3 \lambda h^2 - m^2 ] f^2 - i {\bar \psi} \gamma^\mu \partial_\mu \psi - g h {\bar \psi} \psi + \cdots ~.$$ Thus, we find that if the Higgs field has a base value $h$, then the Higgs boson (described by $f$) has mass $m_{\text{Higgs}} = \sqrt{ 3 \lambda h^2 - m^2 }$ and the electron (described by $\psi$) has mass $m_{el} = gh$. This process is the Higgs mechanism. If $h=0$, the electron is massless and this was the case a long time ago. At some point, the Higgs field attained a vev (base value) and $h$ became nonzero and the electron was now massive!

Eqs. (5) & (6) here show how leptons couple to the Higgs field, giving the former mass. They are terms in the Lagrangian, viz.

\begin{aligned} \mathcal{L}_{Y} = &-\lambda_u^{ij}\frac{\phi^0-i\phi^3}{\sqrt{2}}\overline u_L^i u_R^j +\lambda_u^{ij}\frac{\phi^1-i\phi^2}{\sqrt{2}}\overline d_L^i u_R^j\\ &-\lambda_d^{ij}\frac{\phi^0+i\phi^3}{\sqrt{2}}\overline d_L^i d_R^j -\lambda_d^{ij}\frac{\phi^1+i\phi^2}{\sqrt{2}}\overline u_L^i d_R^j\\ &-\lambda_e^{ij}\frac{\phi^0+i\phi^3}{\sqrt{2}}\overline e_L^i e_R^j -\lambda_e^{ij}\frac{\phi^1+i\phi^2}{\sqrt{2}}\overline \nu_L^i e_R^j + \textrm{h.c.} \end{aligned}\tag5

\begin{aligned} \mathcal{L}_{m} = -m_u^i\overline u_L^i u_R^i -m_d^i\overline d_L^i d_R^i -m_e^i\overline e_L^i e_R^i+ \textrm{h.c.} \end{aligned}\tag6

Here each term with an $_L$ or $_R$ subscript is a fermion of left or right chirality, while coefficients such as $m_u^i$ are effective masses that follow from the $\lambda$s by setting the Higgs field in (5) to its vacuum expectation value, viz. $\phi^0=\frac{v}{\sqrt{2}},\,\phi^1=\phi^2=\phi^3=0$.

Gauge bosons require a different treatment, viz. Eqs. (1)-(4) in the same source. Fermions can in theory be massive without violating gauge invariance even without a gauge boson. For example, the electron's electromagnetic Dirac Lagrangian $\overline{\psi}\left(i\gamma^\mu\left(\partial_\mu+qA_\mu\right)-m_e\right)\psi$ allows this. (Having said that, whether this works depends on the gauge group.)

By contrast, the photon $A_\mu$ can't just gain a mass term like that, because adding an $m^2A_\mu A^\mu$ term to $\left(\partial_\mu-iqA_\mu\right)\phi^\ast\left(\partial^\mu+iqA^\mu\right)\phi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ would break the gauge invariance. In fact, the photon is massless. The problem is the W and Z bosons are not, and giving them a gauge-preserving effective mass requires terms of the form $q^2|\phi|^2B_\mu B^\mu$. As with leptons, the mass is proportional to the Higgs vacuum amplitude.

• Sorry, but the OP seems to be a non-expert, and it is hard to see much from this wall of formulas without connecting terms in Lagrangians to physical behavior. It might help to explain what "having mass" means and how interactions with the Higgs field make that happen. – Conifold Jun 2 '17 at 21:17
• @Conifold I originally only linked to the equations, since Wikipedia's article provides explanation at all levels of expertise. AccidentalFourierTransform added the equations, probably due to the long-term link rot risk. I'll edit my post in case the OP needs more exposition. – J.G. Jun 2 '17 at 21:19
• @Conifold I just want to point out that the correctness/usefulness of an answer does not depend on the level of expertness of OP. Also, the Higgs mechanism is a rather technical concept, so I feel that it deserves a technical answer. I'm not really sure I understand what OP is asking, but it seems to me that any possible answer must contain, to some extent, the Lagrangian terms that describe the fundamental interaction. Otherwise the answers would be too superficial IMHO. In other words, I don't see anything particularly bad about this answer (J.G. could extend it a bit though). – AccidentalFourierTransform Jun 2 '17 at 21:59
• @AccidentalFourierTransform Whether the answer is useful to the OP is up to the OP, of course, but other users might also benefit from more context. The problem is not with having Lagrangian terms, but with having nothing but Lagrangian terms. – Conifold Jun 2 '17 at 22:21
• @J.G. In SM, Dirac fermionic mass terms would not be invariant under SU(2)$_L$. So we need the Higgs mechanism there too, don't we? Fair enough, such terms would be invariant under U(1) and that would be enough for QED. But not for GSW theory. – user154997 Jun 3 '17 at 9:06