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$\left(H^\dagger H\right)^2$ is invariant under $U(1)\times SU(2)$?

+ 2 like - 0 dislike
1054 views

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

$$H_0 = [?]$$

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

This post imported from StackExchange Physics at 2014-03-24 03:49 (UCT), posted by SE-user curiousGeorge119
asked Jul 18, 2013 in Theoretical Physics by curiousGeorge119 (10 points) [ no revision ]
retagged Apr 19, 2014 by dimension10

2 Answers

+ 3 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-03-24 03:49 (UCT), posted by SE-user Luboš Motl
answered Jul 18, 2013 by Luboš Motl (10,178 points) [ no revision ]
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?

This post imported from StackExchange Physics at 2014-03-24 03:49 (UCT), posted by SE-user curiousGeorge119
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.

This post imported from StackExchange Physics at 2014-03-24 03:49 (UCT), posted by SE-user Luboš Motl
Wow... thanks! Thank you for mentioning that!

This post imported from StackExchange Physics at 2014-03-24 03:49 (UCT), posted by SE-user curiousGeorge119
+ 0 like - 0 dislike

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

Yes, clearly. The Higgs field $\phi $ is invariant under the transformation mentioned. To quote wikipedia:

In the standard model, the Higgs field is an $SU(2)$ doublet, a complex spinor with four real components (or equivalently with two complex components). Its weak hypercharge ($U(1)$ charge) is 1. That means that it transforms as a spinor under $SU(2)$. Under $U(1)$ rotations, it is multiplied by a phase, which thus mixes the real and imaginary parts of the complex spinor into each other—so this is ''not the same'' as two complex spinors mixing under $U(1)$ (which would have eight real components between them), but instead is the spinor representation of the group $U(2)$ .

And the conclusion follows.

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

Yup, see the quote from wikipedia above-mentioned. Almost by' definition.

Make separate questions for the rest.

P.S. You may find this link useful for learning QFT. It is more of QCD, but it covers other things too.

answered Jul 18, 2013 by dimension10 (1,950 points) [ revision history ]
Thanks you for your answer, Dimension 10, and for the link... I have been wondering if the top quark was heavier than the Higgs, and the answer is right there on the second page! Thanks!

This post imported from StackExchange Physics at 2014-03-24 03:49 (UCT), posted by SE-user curiousGeorge119
@curiousGeorge119: For questions regarding such factual things, like masses of particles, there is always a [Wikipedia article. Click "Show" for the parameters of the standard model.](en.wikipedia.org/wiki/…

This post imported from StackExchange Physics at 2014-03-24 03:49 (UCT), posted by SE-user Dimensio1n0
Ahh, on the SM diagram... yes, I overlooked that! Thank you.

This post imported from StackExchange Physics at 2014-03-24 03:49 (UCT), posted by SE-user curiousGeorge119
[Wikipedia article. Click "Show" for the parameters of the standard model.](en.wikipedia.org/wiki/Standard_Model#Construction of the Standard Model Lagrangian) Links are not working? , .

This post imported from StackExchange Physics at 2014-03-24 03:49 (UCT), posted by SE-user Dimensio1n0
@curiousGeorge119: No, not on the diagram. I meant here

This post imported from StackExchange Physics at 2014-03-24 03:49 (UCT), posted by SE-user Dimensio1n0
Thanks for your help! One more question?? :) Does this result prove translational and rotational invariance?

This post imported from StackExchange Physics at 2014-03-24 03:49 (UCT), posted by SE-user curiousGeorge119
@curiousGeorge119: Translational and Rotational invariance of what? It just means that the Higgs field is gauge invariant under $U(1)\otimes SU(2)$. ,

This post imported from StackExchange Physics at 2014-03-24 03:49 (UCT), posted by SE-user Dimensio1n0

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