0
$\begingroup$

In the Standard Model, the Higgs field is $\Phi=\left(\begin{array}{c}\phi^+\\\phi^0\\\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{c}\phi_1+i\phi_2\\\phi_3+i\phi_4\\\end{array}\right)$.

The component with isospin $+1/2$ is positively charged. The component with isospin $-1/2$ is neutrally charged.

We have 3 Goldstone bosons : $\phi_1$, $\phi_2$, $\phi_3$, and the Higgs boson $H$ is a subpart of $\phi_4$ : $\phi_4=v+H$.

  • Since the Higgs doublet has only a positive component, with hands, how could the negative W boson eat the positive Goldstone boson $\phi^+$?

  • How could we have $\phi^-$ for the $W^-$?

For example, in this web page Who ate the Higgs?

(at a moment where the figure was still readable), they introduce a $H^+$ and a $H^-$: but where are there coming from? As seen above, there is only one $\phi^+$, not a $\phi^+$ and a $\phi^-$.

  • Is $\phi_1$ and $\phi_2$ both a Goldstone boson, or is the combination $\phi_1+i\phi_2$ a Goldstone boson?
$\endgroup$
3
  • 2
    $\begingroup$ ? Since $\phi^+=(\phi^1+i\phi^2)/\sqrt{2}$, a goldston, it follows that $\phi^-=(\phi^1-i\phi^2)/\sqrt{2}$, also a goldston. The third goldston is $\phi^3$. The action is hermitian, so for every $\phi^+$ there is a term with a $\phi^-$. $\endgroup$ Apr 3 at 18:06
  • $\begingroup$ @Cosmas Zachos : Are both $\phi_1$, $\phi_2$ Goldstone bosons ? Is combination of $\phi_1$ and $\phi_2$ a Goldstone boson ? $\endgroup$ Apr 3 at 18:11
  • $\begingroup$ @Cosmas Zachos : where in the Lagrangian there is the $\phi^-$ term ? $\endgroup$ Apr 3 at 18:12

1 Answer 1

4
$\begingroup$

All three $\phi^1,\phi^2, \phi^3$ are massless Goldstone bosons. It turns out we may rearrange them in complex charged states, $$ \phi^3,\qquad \phi^{\pm}=(\phi^1\pm i \phi^2)/\sqrt{2}~. $$ You may (must!) check that all three combinations transform under the symmetries by infinitesimal pieces whose v.e.v.s don't vanish (the litmus test of Goldstone's realization).

With malice aforethought, we further define, contragrediently, $$ W^{\pm}_\mu= (W^{ 1}_\mu \mp i W^{ 2}_\mu)/\sqrt{2}~. $$

You then look at the kinetic term for the Higgs doublet, $$ \frac{1}{2} \partial_\mu {\boldsymbol \Phi}^* \cdot \partial^\mu {\boldsymbol \Phi}= \frac{1}{2} ( \partial_\mu \phi ^3 \partial^\mu \phi^{3~*}+ \partial_\mu h \partial^\mu h)+ \partial_\mu \phi^+ \partial^\mu \phi^ - ~. $$

I'll be cavalier/inconsistent with normalizations in the trail map that follows, since it is only a trail map―you are meant to flesh out the details in your basic SM course. They are all correct in Schwartz's text's (29.4) for the masses. All you need is write down the relevant terms that contain the goldstons in the covariant version of the above, $$ (D^\mu {\boldsymbol \Phi})^\dagger D^\mu {\boldsymbol \Phi}=( D^\mu {\boldsymbol \Phi})^* \cdot D^\mu {\boldsymbol \Phi}~.\tag{$\natural$} $$

The covariant derivative is completed to $$ D^\mu {\boldsymbol \Phi}\propto \partial^\mu \begin{pmatrix}\phi^+\\\phi^0 \end{pmatrix} +ig \begin{pmatrix} A_\mu /\cos\theta_w & W_\mu^+ \sqrt{2}\\ W^-_\mu \sqrt{2} & -Z_\mu/\sin\theta_W \end{pmatrix} \begin{pmatrix}\phi^+\\\phi^0 \end{pmatrix} ~~~\leadsto \\ D^\mu {\boldsymbol \Phi}^*\propto \partial^\mu \begin{pmatrix}\phi^-\\\phi^{0~*} \end{pmatrix} -ig \begin{pmatrix} A_\mu /\cos\theta_w & W_\mu^- \sqrt{2}\\ W^+_\mu\sqrt{2} & -Z_\mu/\sin\theta_W \end{pmatrix} \begin{pmatrix}\phi^-\\\phi^{0~*} \end{pmatrix} . $$ Note how each rung of the two-vectors has the same charge!

If we give $\langle \phi_4\rangle = v$, the hermitian term quadratic in the matrix in the ($\natural$) term above will provide the mass terms for the $W^\pm$ and the Z, namely $~~2W^+ W^-+Z^2/\cos ^2 \theta_W$, with no contribution for the photon, as expected.

But, pertaining to your question, you can also see in ($\natural$), among others, the emergent bilinear combinations $(\partial_\mu \phi^+ -gW^+_\mu v)(\partial^\mu \phi^- -gW^{-~^\mu } v)+ ...$, where the $W_\mu^\pm $ are poised to gauge-absorb the respective $\phi^{\pm}$ of the same charge, and net the mass term $g^2v^2W^+_\mu W^{-~\mu}$ .

There are more elegant and compact ways to present this, but I opted for a brutishly explicit trail map here, as I assumed that's where the problem was... Often, students are surprised by $W^1-iW^2$ eating $\phi^1+i\phi^2$, and crosswise for the $W^-$, but this is what the charge conservation dictates on you, which is why I wrote the covariant derivatives in terms of the physical fields.

$\endgroup$
5
  • $\begingroup$ thank you. I digest your kind useful explanations. $\endgroup$ Apr 4 at 11:37
  • $\begingroup$ Thank you. From your answer, do I understand that there are two Higgs fields : $\Phi=\left(\begin{array}{c}\phi^+\\\phi^0\\\end{array}\right)$, but also $\Phi=\left(\begin{array}{c}\phi^-\\\phi^{0*}\\\end{array}\right)$ ? $\endgroup$ Apr 6 at 21:23
  • $\begingroup$ From your derivation of $D^\mu {\boldsymbol \Phi}$, I see that there is the term $W^-_\mu \phi^+$, so how could the $W^-_\mu$ eat the $\phi^-$ since I don't see term $W^-_\mu \phi^-$ : I see term $W^-_\mu \phi^+$ $\endgroup$ Apr 6 at 21:25
  • 1
    $\begingroup$ The doublet containing the $\phi^-$ is $\Phi^*$, not $\Phi$. It is the bland complex conjugate thereof. When you expand the covariant derivative term I wrote down, you see the "emergent bilinear combination" (...)(...) I wrote down, where, e.g., $(W_\mu^- -\partial_\mu \phi^-/gv)$ gauge-transforms to $W_\mu^-$; this is the "eating". $\endgroup$ Apr 6 at 21:52
  • 1
    $\begingroup$ Thanks a lot for your generous help. $\endgroup$ Apr 7 at 13:25

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.