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Is it true that $\left(H^\dagger H\right)^2$ is invariant under $U\left(1\right) \times SU\left(2\right)$ where $H$ is the Higgs field $(1,2,1/2)$?

Does this invariance imply that its hypercharge is invariant under $U\left(1\right)$ and its spin is invariant under $SU\left(2\right)$? .

$$H = [H_+, H_0]$$

$$H_+ = [H_-]$$

but

$$H_0 = [?]$$

$$H^\dagger H = [H_-][H_+] + [?][H_0]$$

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Yes, $(H^\dagger H)^2$ is invariant under $SU(2)\times U(1)$ because even without the second power, $H^\dagger H$ is invariant under it. By that, we mean $$\sum_{i=1}^2 H_i^* H_i$$ Of course that it's invariant under $U(1)$ because the $U(1)$ charges of $H^\dagger$ and $H$ are opposite in sign and add up to zero. Note that the transformation of a charge-$Q$ field is $F\to F\exp(iQ\lambda)$.

The invariance under $SU(2)$ is also self-evident because $\sum_i z^*_i z_i$ is exactly the bilinear (with one asterisk) invariant that defines the unitary groups.

No, the invariance of $H^\dagger H$ or its square does not mean that the hypercharge of $H$ itself is zero or isospin is zero. Moreover, the OP seems to confuse the spin and the isospin.

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  • $\begingroup$ Thank you, Motl. I appreciate your help... so the sum of the hypercharge must equal 0 to be invariant under U(1), is that right? $\endgroup$ – curiousGeorge119 Jul 18 '13 at 13:55
  • $\begingroup$ Yup, this condition is equivalent to the conservation of the charge in the interaction vertices. It's no coincidence - symmetries are linked to conservation laws. $\endgroup$ – Luboš Motl Jul 18 '13 at 14:42

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