5
$\begingroup$

My question is in the context of the Standard Model, not about Beyond Standard Model.

At the beginning of the Universe, before electroweak symmetry breaking, we have a complex Higgs isospin doublet. So could we say that at that period, there were 4 Higgs bosons?

After electroweak symmetry breaking, 3 Higgs fields (called Goldstone bosons) are eaten by the gauge bosons, and get mass, while the last Higgs field gets mass? So could we say that after electroweak symmetry breaking, there is one Higgs only?

But then, does it mean that the number of Higgs boson has varied between the Big Bang and the time at which the electroweak symmetry breaking occured?

$\endgroup$
4
$\begingroup$

Whenever we say Higgs boson we actually mean "the quantum of the Higgs field, post SSB$^\dagger$". In the same way the "photon" is usually associated with the quantum of the post-SSB $A^\mu$ field, and not the pre-SSB (mixed) $B^\mu$.

The particles ( the quanta of the fields) are just excitations of the fields. The Higgs field operator $\hat{\phi}_H(\mathbf{x})$ will create one Higgs boson at position $\mathbf{x}$. The field itself, however, is the more imporant entity, and it permeates the whole of space.

If the field has $>1$ degrees of freedom, its quanta have $>1$ internal configurations.
The photon field $A^\mu$ has $2$ degrees of freedom: this corresponds to two mutually orthogonally polarised photons.

The $4$ degrees of freedom of the pre-SSB Higgs field can be thought as "polarisations" of the pre-SSB Higgs boson. Though this has imaginary mass so a physical interpretation is somewhat challenging (but let me know if you come up with one).

I think they key problem with you is the Higgs field before and after SSB.

  • Before SSB:

    The Higgs field is a $SU(2)$ complex doublet, as you said, meaning it is a $2$-component quantity, each entry being complex: $$ \phi_H = \left ( \begin{array}{c} \phi^+ \\ \phi^0 \end{array} \right ) = \left ( \begin{array}{c} \phi^+_1 + i\phi^+_2 \\ \phi^0_1 + i \phi^0_2 \end{array} \right ),$$

    with $\phi^+, \phi^0 \in \mathbb{C}$ and $\phi^+_1, \phi^+_2, \phi^0_1, \phi^0_2 \in \mathbb{R}$.
    So there $4$ independent parameters, or "degrees of freedom" (d.o.f.).
    Each can be considered as a scalar field, yes -- scalar fields carry one degree of freedom each, so the maths checks out.

    The Lagrangian density $\mathcal{L}$ has an $SU(2)\times U(1)$ symmetry, which has $4$ generators (different $4$ from before).
    Upon requiring this symmetry to be local, you introduce $4$ gauge fields $W^1, W^2, W^3, B$.

    These have to be massless since bare mass terms for gauge bosons are not allowed in the Standard model, as they would violate gauge symmetry. A massless gauge field has $2$ degrees of freedom (two polarisations, think of a photon).

    So: $4$ massless gauge fields ($8$ d.o.f.) $+$ $1$ complex Higgs doublet ($4$ d.o.f.) = $12$ d.o.f.
    Symmetry: $SU(2)_L \times U(1)_Y$.

  • After SSB:

    The Higgs mechanism happens.
    It breaks $SU(2)_L \times U(1)_Y$ into $U(1)_{\mathrm{em}}$. The number of broken generators are $4-1$ = $3$.

    These would-be Goldstone bosons are "eaten up" by $3$ of the massless gauge fields, which now become massive. The Higgs mechanism also mixes these fields around, so now they are labelled $W^+, W^-, Z^0$ and $A$, the latter being the only one retaining its massless status (the photon field).

    Massive gauge fields have $3$ degrees of freedom, think of an atom with spin $1$: it has projections $-1, 0, 1$.

    In the process, the Higgs complex doublet reduces to a single real scalar field, what is usually referred as "the Higgs boson".

    So: $3$ massive gauge fields ($9$ d.o.f.) + $1$ massless gauge field ($2$ d.o.f.) + $1$ real scalar field ($1$ d.o.f.) = $12$ d.o.f.

The number of degree of freedom of the $SU(2)_L \times U(1)_Y$ theory (electro-weak) remains unchanged. All that happens is a "reshuffling" around.

The Higgs boson pre-SSB, i.e. the quantum of the pre-SSB Higgs field, is a single particle carrying $4$ degrees of freedom ("polarisations"). Though it has imaginary mass so physical interpretation is challenging.

The Higgs boson post-SSB, i.e. the quantum of the post-SSB Higgs field, is a single particle carrying $1$ degree of freedom. This has a real mass term, and is what we associate with the experimentally observed particle.


$\dagger$ = Spontaneous Symmetry Breaking.

$\endgroup$
  • 1
    $\begingroup$ I'm not so sure that the canonical interpretation of the 4 dof should be one field with 4 dof. After all the upper and lower components of the doublet have different charges and CP (though charged defined after EWSB). Consider the Yukawas. The neutral components couples e.g. $H^0 \bar e_L e_R$ whereas the upper components couple e.g. $H^+ \bar e_L \nu_R$. $\endgroup$ – innisfree Sep 26 at 1:12
  • $\begingroup$ I think a more natural description might be a charged scalar (2 dof) and 2 neutral scalars, one CP odd, one CP even. $\endgroup$ – innisfree Sep 26 at 1:12
  • $\begingroup$ Do you have a reference for where the pre-SSB Higgs doublet is treated in more detail? $\endgroup$ – SuperCiocia Sep 26 at 7:51
  • $\begingroup$ Dear SuperCiocia, what is your opinion about the comment from innisfree : before EWSB, should we say (academically speaking) "one field with 4 dof", or "4 fields with one dof each" ? $\endgroup$ – Mathieu Krisztian Sep 27 at 21:31
  • $\begingroup$ I used to think of it as one field with 2 complex degrees of freedom. If you do 4 real scalars then you imply an $O(n)$ symmetry which is technically not the same thing as an $SU(n-1)$ symmetry... but I never heard of this CP labelling. I’d be interested to learn more. $\endgroup$ – SuperCiocia Sep 28 at 1:11
0
$\begingroup$

There is a confusion here, there is one Higgs boson in the standard model, which appears after symmetry breaking. Before symmetry breaking there is the Higgs field :

In the Standard Model, the Higgs particle is a boson with spin zero, no electric charge and no colour charge. It is also very unstable, decaying into other particles almost immediately. The Higgs field is a scalar field, with two neutral and two electrically charged components that form a complex doublet of the weak isospin SU(2) symmetry. bold mine.

So the particles with masses come after symmetry breaking.

There is a complete distinction between particles and fields. There is everywhere a Higgs field on which creation and annihilation operators produce and annihilate the Higgs particle. Similar as for all particles in the table, for example: there is everywhere an electron field on which creation and annihilation operators produce and annihilate the electron. There is one electron.

Though there is a difference on the effect of symmetry breaking in the model between fermions and bosons. In the standard model table, the fermions exist before symmetry breaking with mass zero. The vector bosons are all within the multiplet of the unbroken gauge "particle". At symmetry breaking energy in the cosmic models, the fermions acquire mass through the higgs field, and the gauge bosons acquire mass and separate identities , as seen in the table, and separate fields on which creation and annihilation operators will work. The Higgs boson is a scalar particle , and it also gets its mass from the Higgs field afaik . See also this question.

$\endgroup$
  • $\begingroup$ I'm not sure this gets to the nub of the question: how do we count degrees of freedom prior to EWSB/choosing the unitary gauge? Do we say there are Higgs particles corresponding to every (real) dof in the complex Higgs doublet? $\endgroup$ – innisfree Sep 25 at 7:11
  • $\begingroup$ Hello anna v. There are 4 scalars field. Scalar field is synonym to boson. Are you really sure that Higgs boson and Higgs fields are not synonyms ? $\endgroup$ – Mathieu Krisztian Sep 25 at 7:53
  • $\begingroup$ I am absolutely sure that the fields are different than the particles generated by creation operators on the fields. Before symmetry breaking there exists a Higgs field and the gauge boson are diffent than this field. Before symmetry breaking there are four gauge bosons also massles some of which through the Higgs mechanism also acquire a mass and become the Z W+W- and the photon. The higgs particle is an extra bonus of the higgs field which is responsible at the energy of symmetry breaking. see en.wikipedia.org/wiki/Electroweak_interaction and also here $\endgroup$ – anna v Sep 25 at 8:07
  • $\begingroup$ physics.stackexchange.com/questions/341411/… $\endgroup$ – anna v Sep 25 at 8:07
  • $\begingroup$ @innisfree in the standard model there is one higgs particle. it is the field that has a complex doublet. $\endgroup$ – anna v Sep 25 at 8:09

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.