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?
[Physics] Problem with determining number of Goldstone bosons
group-theoryhiggsquantum-field-theorysymmetrysymmetry-breaking
Related Solutions
I understand the statement in the following way:
Pions, which are pseudo-goldstone bosons of chiral symmetry breaking, are described by the introduction of a unitary matrix $U(x)$, defined as
$$U(x)=\text{exp}\left(2i\pi^a(x)T^af_\pi^{-1}\right),$$
where $\pi^a$ is the pion field, $f_\pi$ is the pion decay constant and $T^a$ are the generators of the broken symmetry, i.e. the coset space. The pion Lagrangian can be written down in terms of $U(x)$
$$\mathcal{L}=-\frac14 f_\pi^2\text{Tr}\partial^\mu U^\dagger\partial_\mu U,$$
which by expanding the exponential form results in
$$\mathcal{L}=-\frac12\partial^\mu \pi^a\partial_\mu \pi^a+\dots,$$
where dots denote higher order terms. Thus, the statement that goldstone bosons live in the coset space can be related to the fact that the fields themselves are linked to the generators of the coset.
This can be understood in terms of Goldstone's theorem: if the original Lagrangian exhibits a continuous symmetry, the number of goldstone bosons is equal to the number of generators of the broken symmetry. Take for example the linear sigma model: if your original theory is $O(N)$-symmetric, it has $N(N-1)/2$ symmetries. If the symmetry is broken spontaneously, you end up with $O(N-1)$, leaving you with $(N-1)(N-2)/2$ symmetries. The amount of broken symmetries is the difference, i.e. $N-1$. But this is precisely the number of pions you have in your theory. We can conclude that the pions are linked directly to the broken symmetries, i.e. the coset space.
First, note that, strictly speaking, there is no such thing as spontaneous symmetry breaking in Higgs mechanism. I mean, that below and under the Higgs scale (i.e., at scale, at which non-zero Higgs VEV appears) the lagrangian can be rewritten in a gauge invariant way. How is it possible? The answer is that there are different physical states (i.e., eigenstates of hamiltonian) below and under the Higgs scale. Under the Higgs scale eigenstates of hamiltonian form also representations of the gauge group. But below it eigenstates of hamiltonian don't form representations of the gauge group. These eigenstates are linear combination of transverse and longitudinal degrees of freedom.
So, in some sence, nothing happens with three scalar fields in Higgs doublet. They just don't appear as physical states below the Higgs scale.
Best Answer
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)