4
$\begingroup$

(Using $Q=T_3+Y_W$)

The Higgs boson has weak isospin $T_3=-\cfrac12$ and weak hypercharge $Y_W=+\cfrac12$.

Since every particle has all opposite charges of its antiparticle, why isn't there an anti-higgs boson with $T_3=+\cfrac12$ and $Y_W=-\cfrac12$ ?

Question: Why does the Higgs boson have electroweak charges $~T_3~$ and $~Y_w~$ but does not have an antiparticle with opposite charges ?

$\endgroup$
6
$\begingroup$

It might be best to stare at your SM book. The Higgs field weak isospin doublet is $$ H = \begin{pmatrix} \phi^+ \\ \phi^0 \end{pmatrix}\equiv \frac{1}{\sqrt 2} \begin{pmatrix} \varphi_1-i\varphi_2 \\ h +i\chi \end{pmatrix}, $$ the Higgs particle being the real neutral piece h of it. The entire doublet has $Y_w=1/2$ in your conventions.

What happened to the conjugate doublet?, you ask. It is all around us, and kicking.$$ \tilde H =i\tau_2 H^*= \begin{pmatrix} \phi^{0~~*} \\ -\phi^- \end{pmatrix} = \frac{1}{\sqrt 2} \begin{pmatrix} h -i\chi \\ - \varphi_1-i\varphi_2 \end{pmatrix}, $$
with $Y_w=-1/2$ now, and the Higgs particle now reassigned to the upstairs ($T_3=1/2$) slot, exactly as you posited. It has the same 0 charge, since both its weak hypercharge and $T_3$ flipped in synchronized splendor. Same particle, same charge, different labels. (Warn your teacher to never use that as a trick question...)

But... you already know this! The Higgs particle is but the quantum excitation left-over from the redefinition of the Higgs doublet field through shifting by a real neutral v.e.v, v. This is the very v.e.v. that gives masses to both down quarks and up quarks, in EW invariant terms! That is, you couple H to the same conjugate quark left doublet, to give d a mass and $\tilde H$ to give u a mass--the 2nd and as important job of the Higgs. The very same h trails the very same v.

$\endgroup$
5
$\begingroup$

As emphasized in the answer by Cosmas Zachos, the key is to distinguish between the full Higgs field, which has two complex components, and the part corresponding to the Higgs particle. (I'm using the word "particle" a bit loosely here. "Resonance" might be a better name.)

Here's another way of thinking about that distinction. Intuitively, if the gauge fields were absent, we would have two different types of excitation associated with the two-component Higgs field:

  • The first type would be the Goldstone modes corresponding to oscillations of the Higgs field along the "flat" directions of the Higgs potential $V(\phi)$ — that is, along the "valley" in the graph of $V$ versus $\phi$. These modes are massless because those oscillations don't cost any energy: $V(\phi)$ depends only on the magnitude of $\phi$, not on where we are in the "valley."

  • The second type corresponds to oscillations in the "radial" direction, transverse to the "valley" so that the magnitude of the Higgs field is oscillating (about a non-zero vacuum expectation value). This mode is massive, because these oscillations do cost energy: $V(\phi)$ depends on the magnitude of $\phi$.

When the gauge field is present, the first type of excitation gets "eaten" by the Higgs mechanism. The second type of excitation corresponds to the Higgs particle. An $SU(2)_L\times U(1)_Y$ gauge transformation acts only along the "flat" directions of the Higgs potential. Those gauge transformations don't affect the radial direction, and this is the degree of freedom that corresponds to the Higgs particle. As a result, the Higgs particle is not charged with respect to the gauge group: it is its own antiparticle.

More formally, the two-component Higgs field $\phi$ can be written in the form $$ \phi = U\left[\begin{matrix} 0\\ \rho\end{matrix} \right] $$ where $U$ is a $2\times 2$ unitary matrix that is affected by $SU(2)_L$ and $U(1)_Y$ gauge transformations, and where $\rho$ is not (by definition). The non-zero values of $T_3$ and $Y_W$ are carried by the factor $U$, but the Higgs particle correpsonds to oscillations of $\rho$ (about its vacuum expectation value). This is why having non-zero values of $T_3$ and $Y_W$ for the full Higgs field is consistent with the fact that the Higgs particle is its own antiparticle.

Textbooks typically mention this under the heading unitary gauge. Section 6.1 in Quantum Fields on a Lattice (Montvay and Münster, 1994) describes it in a non-perturbative context.

$\endgroup$

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.