7
$\begingroup$

I was wondering how one can see that the hypercharge of the complex Higgs doublet is $Y=\frac{1}{2}$. Complex Higgs doublet. $$ \Phi(x) = \begin{pmatrix}\ \Phi^{+}(x)\\ \Phi^{0}(x) \end{pmatrix} $$ Lagrangian density: $$ L = \frac{1}{2}\partial_{\mu}\Phi^{\dagger}\partial_{\mu}\Phi + \frac{m^2}{2}\vert{\Phi(x)}\vert^2 + \frac{\lambda}{4!}\vert{\Phi(x)}\vert^4 $$ There is now a local $U(1)_Y$ symmetry called hypercharge, given by $\Phi'(x) = e^{-i\tfrac{1}{2}\varphi}\Phi(x)$. But how is the hypercharged deduced? I would really ask more specific questions, but I dont really get it and I am learning this stuff just for me, because I am interested how it works.

$\endgroup$
  • 1
    $\begingroup$ The $SU(2)_L \times U(1)_Y$ is a (local) gauge symmetry, not a global symmetry. Use the formula $Q= T_3 + \frac{Y}{2}$ to obtain the hypercharge. Here $Q$ is the electric charge and $T_3$ is the eigenvalue of the $SU(2)$ $T_3$ generator ($\frac{1}{2}$ for the first member of the doublet and $-\frac{1}{2}$ for the second member of the doublet ). Your doublet is not "correct" because you have a different value of $Y$ for the two members of the doublet. You have to choose, for instance $(\Phi_+,\Phi_0)$, corresponding to $Y = 1$ or $(\Phi_0,\Phi_-)$, corresponding to $Y = -1$ $\endgroup$ – Trimok Jul 17 '14 at 9:53
  • $\begingroup$ thanks for the remarks. I see that with $T_3 = \tfrac{1}{2}\sigma^3$ the eigenvalues are $\tfrac{1}{2}, -\tfrac{1}{2}$. Using this one gets Q=1 for $\Phi^{+}$ and Q = 0 for $\Phi^{0}$ with Y = 1. But why is there written $\tfrac{1}{2}$ and $-\tfrac{1}{2}$ ? $\endgroup$ – nerdizzle Jul 17 '14 at 10:59
  • $\begingroup$ Sorry, I don't understand clearly your question... $\endgroup$ – Trimok Jul 17 '14 at 11:19
  • $\begingroup$ is there a convention where one uses $Y_{\Phi}=\tfrac{1}{2}$ and $Y_{\tilde{\Phi}}=-\tfrac{1}{2}$? becuase that is what is written in my notes... $\endgroup$ – nerdizzle Jul 17 '14 at 11:31
  • $\begingroup$ OK. I think some authors describe the hypercharge by $Q= T_3+Y$ instead of $Q= T_3 + \frac{Y}{2}$. So you have to check the definition in your notes... $\endgroup$ – Trimok Jul 17 '14 at 11:41
3
$\begingroup$

The hypercharge of a doublet cannot be "deduced". When one builds a gauge theory, the first step is to define the particle content of your theory and to postulate the representation of all particle multiplets. In particular, if the gauge group is abelian, then we have to assign numbers usually called charges.

So, I reformulate your question: Why do we choose the hypercharge of the Higgs doublet to be $Y=1$? The idea is that we want to respect the Gell-Mann–Nishijima formula $Y=2(Q-T_3)$. Since we know the charges and the values of $T_3=\pm 1/2$ for the doublet, it is straightforward to find that $Y=1$.

There are several ways to see why the formula $Y=2(Q-T_3)$ must be respected:

  • One possibility is to compute the commutator of the generators $Q$ and $T_i$. You will notice that $[Q,T_i]=[T_3,T_i]\neq0$, but $[Q-T_3,T_i]=0$. This means that the global symmetries $SU(2)$ and $U(1)_{\rm em}$ cannot be simultaneously satisfied, but we can define the hypercharge $Y$ as new Abelian symmetry.
  • Other possibility is to try to assign Abelian charges for the fermionic multiplets of the Standard Model (knowing that the charge must be the same for all members of a multiplet). You will conclude that $Y=2(Q-T_3)$ is the only possible choice up to a multiplicative constant.
$\endgroup$

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.