3
$\begingroup$

In the SM we can not add fermionic mass terms like $m \overline{e}_R e_L$ to the Lagrangian since these terms are not invariant under $SU(2)\times U(1)_Y$.

After introducing the Higgs in the unitary gauge $\phi=\left(\begin{array}{c} 0\\ v + H(x)\\ \end{array}\right)$ into the system we break the symmetry and are able to put mass terms into the Lagrangian which look like $$y \overline{\Psi}_L\phi e_R$$ with $\Psi_L$ the doublet left handed fermion field and $e_R$ the singlet right-handed fermion and the Yukawa coupling $y$. They will provide us mass terms like $m_e \overline{e}_L e_R$ with $m_e$ dependent on the vev and the Yukawa coupling.

Why do we need these $SU(2)\times U(1)_Y$-invariant terms if our symmetry is already broken? How is this method better than just putting the mass terms $m \overline{e}_R e_L$ in our system by hand neglecting the $SU(2)\times U(1)_Y$ symmetry?

$\endgroup$
  • $\begingroup$ You are braking symmetry without breaking the symmetry! $\endgroup$ – AMS Jun 4 '16 at 16:58
3
$\begingroup$

Because "spontaneous symmetry breaking" does not actually break any symmetries. This is a pretty important principle that is not always adequately taught.

In spontaneous symmetry breaking the symmetry in question is always a full symmetry of the theory. The difference between a spontaneously broken symmetry and an unbroken symmetry is just in how the symmetry is realized. In the spontaneously broken case, the symmetry has a non-trivial action on the vacuum state of the system, and often has a non-linear action on the fields.

It's incredibly important to the standard model that the $SU(2)\times U(1)$ symmetry is not broken because it is a gauge symmetry. A theory of interacting spin-1 vector bosons requires a fully unbroken gauge symmetry or the theory will not be unitary. A sensible quantum theory can be a lot of crazy things but it must be unitary, and hence the Standard Model Lagrangian must only contain terms that do not break gauge invariance.

The Higgs mechanism is important because theorists finally figured out how to square the theoretical requirement that the theory have an unbroken gauge symmetry and the experimental requirement that the gauge bosons / fermions have mass. The Higgs mechanism tells you how to start with a theory with terms that don't break gauge invariance and turn it into a theory with an apparently broken symmetry without violating unitarity.

$\endgroup$
  • $\begingroup$ So the spontaneous symmetry breaking is not really a breaking of the physical symmetry, but more breaking of the mathematical description? $\endgroup$ – Statics Jun 5 '16 at 20:38
  • $\begingroup$ @Statics Kind of, but there is more to it than that. The mathematical description of the symmetry becomes more complicated, but it's still there. And SSB does have physical consquences, namely Goldstone bosons that appear as massless modes or longitudinal modes for gauge bosons. $\endgroup$ – Luke Pritchett Jun 6 '16 at 16:02
  • $\begingroup$ @Statics Your above comment is quite disturbing, it is not the case that "SSB is not a breaking of a physical symmetry; And when we say SSB, we really mean "Spontenous breaking of the symmetry of the vaccum". Which is definitely a breaking of a physical symmetry. You could understand SSB in better way through a simpler group like U(1). Also, you should not call this as a "breaking of the mathematical description". $\endgroup$ – Aman pawar Feb 21 at 0:17
1
$\begingroup$

The gauge symmetry group associated to the SM is $SU\left(3\right)_{c}\times SU\left(2\right)_{L}\times U_{Y}\left(1\right)$. Then we can not build the lagrangian of the SM with terms of the form $m\bar{\psi}\psi$ because they are not gauge invariant. A term of this kind mix the right and left handed parts, which transforms differently. In order to give mass to the electroweak bosons and fermions, a scalar doublet is introduced. After developing a non zero VEV $\left(\left\langle \phi\right\rangle _{0}\right)$, we say that the $SU\left(2\right)_{L}\times U\left(1\right)_{Y}$ is spontaneously broken. That means that the vacuum has not the symmetry of the lagrangian. But note that the lagrangian still has the gauge symmetry required. Once the VEV is adquired, we parametrize oscillations around the VEV introducing some goldstone bosons and a scalar field (the Higgs). In the unitary gauge, that means that we choose a gauge, we remove the unphysical degrees of freedom (goldstone bosons) that are eaten by the gauge bosons associated to the broken generators, acquiring mass.

In the case of the fermions, the terms we can construct with this doublet are of the kind you are written. After developing a VEV and choosing a gauge, they take the form you say, but that's the key, you have choose a gauge, the term is not gauge invariant anymore. However, the theory still is gauge invariant.

$\endgroup$
0
$\begingroup$

In a few words, as Luke Pritchett writes, the Higgs mechanism provides us description of particles mass without breaking of the unitarity, i.e., breaking gauge symmetry explicitly. It is interesting fact that even if You start from electroweak theory in the broken phase and don't know about Higgs boson, $W/Z$-boson and existense of hidden gauge invariance (i.e., from Fermi theory), then the requirement of tree unitarity will lead You to SM largangian which exactly looks like, with Higgs boson (see Horejsi, "Introduction to electroweak unification...").

But more correct is to say that due to Higgs mechanism the physical states - i.e., particles, - doesn't form the representation of the full gauge group below the EW crossover scale (see 't Hooft, Conceptual basis of QCD, chapter 4). However, without fixation of the gauge the gauge invariance of the lagrangian isn't broken even after shifting the Higgs doublet on the given VEV (You could clarify this in the simple case of Higgs mechanism for $U(1)$ gauge theory), i.e., the full theory is gauge invariant. That is the great difference between Higgs mechanism and spontaneous symmetry breaking, where even below SSB scale physical states form representation of the "broken" group, like mesons form non-linear representation of $SU_{L}(3)\times SU_{R}(3)$, which is spontantously broken in QCD. For example, three scalar fields of Higgs doublet, being physical bosons upper from the electroweak crossover scale, become unphysical ghosts below it. It is clearly seen in the unitary gauge, which provides only physical particles in the spectrum.

Moreover, it is possible to look on the SM below EW crossover in the gauge invariant way, see an article.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.