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Consider a theory $$\mathcal{L}=(\partial_\mu\Phi^\dagger)(\partial^\mu\Phi)-\mu^2(\Phi^\dagger\Phi)-\lambda(\Phi^\dagger\Phi)^2$$ where $\Phi=\begin{pmatrix}\phi_1+i\phi_2\\ \phi_0+i\phi_3\end{pmatrix}$ is a complex $SU(2)$ doublet. After symmetry breaking there is no residual symmetry and hence there are $(2^2-1)=3$ goldstone bosons. The same Lagrangian can also be written as $$\mathcal{L}=\frac{1}{2}\sum\limits_{i=0}^{3}(\partial_\mu\phi_i)^2-\mu^2(\sum\limits_{i=0}^{3}\phi_i^2)-\lambda(\sum\limits_{i=0}^{3}\phi_i^2)^2$$ which is nothing but the Lagrangian of linear sigma model. After symmetry breaking the symmetry of the Lagrangian reduces from $O(4)$ to $O(3)$. Therefore, there are $3$ goldstone bosons once again and the results match. However, I'm having a confusion with the following. Consider the theory $$\mathcal{L}=(\partial_\mu\xi^\dagger)(\partial^\mu\xi)-\mu^2(\xi^\dagger\xi)-\lambda(\xi^\dagger\xi)^2$$ where $\xi=\begin{pmatrix}\xi_1+i\xi_2\\ \xi_3+i\xi_4\\ \xi_0+i\xi_5\end{pmatrix}$ is a complex $SU(2)$ triplet. The Lagrangian is again $SU(2)$ invariant. Right? After SSB there is no residual symmetry and umber of goldstone boson is 3. However, if we write it as $$\mathcal{L}=\frac{1}{2}\sum\limits_{i=0}^{5}(\partial_\mu\xi_i)^2-\mu^2(\sum\limits_{i=0}^{5}\xi_i^2)-\lambda(\sum\limits_{i=0}^{5}\xi_i^2)^2$$ then $O(6)$ symmetry breaks down to $O(5)$ and number of Goldstone bosons is $=5$. So it doesn't match. Then where am I making the mistake? What is the correct number of Goldstone bosons in this case?

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    $\begingroup$ How did you calculate that there were 3 goldstone bosons in the second case? You are correct that there are 5. $\endgroup$
    – Akoben
    Commented Feb 11, 2015 at 16:09
  • $\begingroup$ @ Akoben- Second Lagragian is SU(2) also invariant like the first one. When the symmetry is broken no residual symmetry is left as in the former case. So I concluded that the number is still 3 corresponding to the breakdown of SU(2) symmetry $\endgroup$
    – SRS
    Commented Feb 11, 2015 at 17:22

2 Answers 2

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Your first two theories, Φ in the spinor rep of SU(2), and φ in the vector rep of O(4), are dealt with correctly, with 3 generators broken in both cases, so 3 goldstons and one massive field.

You have completely messed up the counting and symmetry structure of your latter theory. The first form, with a complex triplet ξ, is SU(3)-, not just SU(2), invariant, and this SU(3) breaks down to the residual SU(2) by the v.e.v., so 8-3=5 broken generators, and thus 5 goldstons, and one residual massive field, just as in the language of your O(4)/O(3) vector representation model.

I am unclear as to how you concluded, erroneously, that "there is no residual SU(2)". There is: it mixes up the components not involving the v.e.v. So, for example, if the v.e.v. is dialed to the 3rd component, the SU(2) subgroup mixing up the upper two components ($\lambda_1, \lambda_2,\lambda_3$ Gell-Mann matrices) is unbroken. You ought to brush up on the standard elementary SSB counting arguments, which your teacher must have assigned to you, Ling-Fong Li (1974)

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  • $\begingroup$ Dear Professor, I am also troubled for this question. How about the Lagrangian $\mathcal{L}=(D_\mu\xi^\dagger)(D^\mu\xi)-\mu^2(\xi^\dagger\xi)-\lambda(\xi^\dagger\xi)^2$ with $D_\mu=\partial_\mu-igA_\mu^a T^a$, where $T^a$ is the SU(2) generator under triplet representation. In other words, the original symmetry is still SU(2). Would we will have three goldstone bosons? However, there are still five massless particles in $\xi$. $\endgroup$
    – Daren
    Commented Oct 26, 2023 at 11:53
  • $\begingroup$ @Daren In your construction, you have only gauged an O(3) subgroup of the global SSBroken O(6), which has 5 Goldstone bosons before breaking. The adjoint of SU(2)~O(3) is a real representation, so the real and imaginary parts of the complex SU(3) do not mix! I have not chased which of these 5 goldstons have been absorbed by gauge bosons, which you should do in a private exercise in a real representation. $\endgroup$ Commented Oct 26, 2023 at 13:14
  • $\begingroup$ PS You might find out only 2, not 3 (!!!), of the 5 goldstons are absorbed by gauge bosons--the O(3) is not completely broken, as there is a surviving O(2)... $\endgroup$ Commented Oct 26, 2023 at 14:29
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    $\begingroup$ No, the SU(3) global symmetry is there for vanishing gauge couplings only! The gauge couplings break it explicitly to SU(2)… but, as suggested, this SU(2) is not completely SSbroken, now…. $\endgroup$ Commented Oct 27, 2023 at 2:08
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    $\begingroup$ Hi professor, could please check this: physics.stackexchange.com/questions/785999/… $\endgroup$
    – Daren
    Commented Oct 27, 2023 at 9:33
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You are right that that the symmetry breaking breaks all three symmetries of $SU(2)$. Thus the $SU(2)$ generators give you three goldstone bosons in the theory with broken symmetry.

However, we have not yet considered all of the symmetries of the original theory. We know that the full symmetry group has six generators and that five of them must be broken. Thus there must be two additional generators of the symmetry group of the original theory (besides the three $SU(2)$ generators we have already counted) which get broken.

Once we include these two additional broken generators we get $5$ goldstone bosons.

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