# Hypercharge of the Higgs field

I am puzzled by the hypercharge of the Higgs field. Under the entry "Higgs Mechanism" in Wikipedia, it is written:

Its (weak hypercharge) U(1) charge is 1

However, under the entry "Higgs Boson" in Wikipedia, it is written:

... while the field has charge +1/2 under the weak hypercharge U(1) symmetry

Moreover, on page 527 of Srednicki's textbook "Quantum Field theory", it is written:

... and the complex scalar field $$\varphi$$, known as the Higgs field, in the representation $$(2, -\frac{1}{2})$$

here the hypercharge becomes $$-\frac{1}{2}$$. How do these different hypercharge values come about? And, generally, how is the hypercharge of a field determined?

• – innisfree Feb 28 '17 at 5:23
• – Qmechanic Feb 28 '17 at 19:30

The historical convention defines it as $$Y_{\rm W} = 2(Q - T_3)$$, as in the Gell-Mann—Nishijima formula of the strong interactions——for a conserved quantity. There, it was frequently used for strange particles, so the hypercharge could get to be 2, —2, etc... and a normalization like this one was warranted. In the weak interactions, thus, the weak hypercharge is defined as twice the average charge of a weak isomultiplet (where the average $$T_3$$ vanishes).
However, the more practical younger generation use $$Y_{\rm W} = (Q - T_3)$$, instead, so the average charge of the isomultiplet, so, e.g., for right-handed fermions, weak isosinglets, the hypercharge is the charge, without daffy gratuitous 2s in the way. But it is only a matter of convention, and references such as the one you quote also specify the convention, as they should.
• As I know, the Higgs field is $\Phi = \left( \begin{array}{cc} \varphi_{1} + i\varphi_{2} \\ \varphi_{3} + i\varphi_{4} \end{array} \right)$, with $\varphi_{1}$ and $\varphi_{2}$ carrying positive charge +1, while $\varphi_{3}$ and $\varphi_{4}$ are electrically neutral. Then their hypercharge is $+1/2$, using $Q = T_{3} + Y$. But why is Sredinicki's version of hypercharge is $-1/2$? Is the hypercharge also notation dependent? – Shen Feb 26 '19 at 6:50
• Not quite: $\varphi_1 +i\varphi_2$ has charge 1 and its complex conjugate has charge -1, so the real Lagrangian has charge 0. Here, it is the lower component, neutral, that picks up the v.e.v. The hypercharge , like the charge, depends on whether you are considering this $\Phi$ or its hermitian conjugate! – Cosmas Zachos Feb 26 '19 at 12:08