Expanding about the VEV to find the spectrum of the theory

When attempting to find the spectrum of the theory after spontaneous symmetry breaking, of a gauge symmetry or otherwise, should we expand the potential in a Taylor series about the VEV or just substitute the shifted fields into the potential?

I am asking this question because I have seen both methods in various textbooks (eg Aitchison/Hey, Quigg).

For example, when attempting to showcase the existence of massless particles in the spectrum when we break a continuous symmetry, most books I have seen will give a potential, eg. $U(\phi) = \mu^2 |\phi|^2 + \lambda (|\phi|^2 )^2$, and expand in a Taylor series about the VEV.

Whereas in derivations for the Higgs mechanism with a Higgs potential $$\lambda \left[H^{\dagger} H - \frac{v^2}{2}\right]$$ In the unitary gauge, are left with a single real scalar field:

$$H(x) \rightarrow \begin{pmatrix} 0 \\ \bar{h}(x) \end{pmatrix}$$

we shift the Higgs field by its VEV: $$\bar{h}(x) = h(x) + v$$ and then substitute this directly into the potential. Why do we not simply undertake a Taylor expansion about the VEV for the Higgs field? Please excuse me if I am missing something obvious.

I feel I am mixing up concepts for Goldstone bosons emerging from spontaneous symmetry breaking of a global continuous symmetry and massive gauge bosons arising from the spontaneous symmetry breaking of a gauge theory.

About the Higgs field and Goldstone bosons: The general Higgs doublet would look like this: $$H(x) \rightarrow \begin{pmatrix} h_1 +ih_2 \\ h_3 + i h_4 \end{pmatrix}$$ with the goldstone bosons $h_2$ and $h_4$ which would not acquire any mass in the kinetic Lagrangian part. In the unitary gauge the Goldstone bosons components together with $h_1$ are being "set" to zero. The 3 degrees of freedom should later manifest in the gauge boson masses for $Z$-, $W^{\pm}$-bosons.
The Higgs potential $U(\phi) = \mu^2 |\phi|^2 + \lambda (|\phi|^2 )^2$ needs a negative $\mu^2$ to "break" the symmetry. This results in a non-zero vacuum i. e. $|\phi(0)|=v\ne 0$. Hence the component $h_3$ can be written as $$h_3= v + h(x).$$ This expression is exact. Furthermore, a Taylor expansion would result in more terms and other different interactions between Higgs and other particles in the Lagrangian which I believe have not been observed yet and are not required to describe the Standard model.
• Yes, I was confused because in descriptions of spontaneous symmetry breaking authors usually undertake a Taylor expansion about the potential, whereas in discussions about the Higgs the single scalar component was written as you said, $h(x) = \nu + h(x)$. I went to the trouble to undertaking a Taylor expansion about the VEV and got additional terms so your answer makes sense. – Eweler Aug 29 '16 at 3:09