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.


1 Answer 1


I am not sure if you are asking about the Taylor expansion of the Higgs field or of the potential. In order to describe the particle interactions and masses through the Higgs mechanism the two-term potential is enough. No other terms are required.

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.

  • $\begingroup$ 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. $\endgroup$
    – Eweler
    Aug 29, 2016 at 3:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.