Describing abelian symmetry breaking in his book on gauge theories, after favouring a vacuum (whose expectation value is $v$) from the symmetric continuum, Quigg parametrize the complex scalar as

$$ \phi=e^{i\frac{\zeta}{v}}\frac{v+\eta}{\sqrt{2}} \tag{1} $$

He then proceeds to use the gauge freedom to eliminate the Goldstone (the gauge boson is irrelevant to my doubt here) with a transformation whose gauge parameter is $\alpha=-\frac{\zeta}{v}$.

My problem is the non-abelian case. Suppose a $SU(2)$ model in which the abelian subgroups corresponding to the first two generators $T_1$ and $T_2$ were broken. Then, analogously to (1), what is done is writing the field as

$$ \phi=e^{\frac{i}{v}(\zeta_1T_1+\zeta_2T_2)} \left( \begin{matrix} 0 \\ 0 \\ v +\eta\end{matrix} \right) \tag{2} $$

after which we go to the U-gauge by means of a gauge trasformation with parameter $\vec{\zeta}=-\frac{1}{v}(\zeta_1, \zeta_2, 0)$.

I do not understand equation (2). It is clear that what we do after shifting the vacuum component around its expectation value is to write the field as the resulting vector acted on by the exponential of a arbitrary linear combination of the broken generators, but I don't know why that is the case. At first, equation (1) looked to me as just writing the field in a polar form, by convenience, since it is absolutely and obviously general. I would then expect that (2) is again just a general way of expressing a 3-component complex arbitrary vector, but I can't see why this is a general fact. That is, the third component is taken care of by the 'arbitrary' $\eta$, but I don't know if the first two are taken care of by the exponential.

Of course, my way of thinking and of interpreting (1) might just be plain wrong. So, why does the exponential in (2) carry only the broken generators?

(Edit) Example: In his book, Frampton deals with a $SO(N)$ model: breaking the (associated subspace of the) first $N-1$ generators, he chooses the vacuum

$$ \phi_0=\left(\begin{matrix} 0 \\ 0 \\ ... \\ v+\eta \end{matrix}\right) $$

and using the explicit generators $(L_{ij})_{kl}=-i(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk})$ he shows that we can parametrize

$$ \phi=\exp\left({\frac{i}{v}\sum_{i=1}^{N-1}\xi_{i}T_i}\right)\left(\begin{matrix} 0 \\ 0 \\ ... \\ v+\eta \end{matrix}\right) = \left(\begin{matrix} \xi_1 \\ ... \\ \xi_{N-1} \\ v+\eta \end{matrix}\right) $$

where $T_i \equiv L_{iN}$. After this, he goes to the U-gauge by the standard procedure. The point is that in this example what happens is exactly what I was under the impression of being the case: the (imaginary argument) exponential of a linear combination of the broken generators just happens to, when acting on the vacuum, generate a general (modulo the fact that the $\xi_i$ look to be phases) vector in the $N-1$ first entries. How general is this and why?

(Edit2): A long time ago I had accepted the answer below (which is very thoughtful and greatly appreciated), but rereading it now I realize I have actually never understood it. I will try to identify some specific confusions of mine regarding it (of course, you can also just ignore this part of the question and answer its core, above, if you think the answer below is inadequate).

(i) Take the passage "Fields transforming under G furnish a representation and they are elements of G. Then, you can make the identification ϕ=g and apply the decomposition Eq. (1).": How are fields transforming under some representation identifiable with the elements of the representation itself (to the point of formally equating to the transformations' form)?

(ii) Equation (1) seems to be a quite general and powerful result. What is its validity (meaning, are there other conditions than the ones stated needed for it to work) and how would one prove it?

TL;DR Question: To state briefly the question: the operational beauty with the higgs formalism is that it is easy to go to the unitary gauge where the Goldstones manifestly decouple by means of a gauge transformation, but how and why can one always parametrize a general expansion of the field around the favored vacuum as something like

$$\text{exp}\left\{i\sum_{\text{broken}} \xi_i T^i\right\}\left(\begin{matrix} ... \\ v+H \end{matrix}\right),$$

where $T_i$ are the generators and $H$ is the 'Higgs', so that a gauge transformation may cancel the exponential and the spectrum is obvious.


1 Answer 1


Consider a Lie group $G$ spontaneously broken to $H$. Then, using the commutation relations between broken and unbroken generators, you can show that any element of $g \in G$ can be decomposed in the following way

$$ g = G\cdot h \equiv \text{exp}\left(i \xi_a \hat{T}^a \right) \cdot \text{exp}\left(i\chi_a T^a\right)\,,\qquad (1) $$

where $\hat{T}^a$ and $T^a$ are the broken and unbroken generators respectively.

Fields transforming under $G$ furnish a representation and they are elements of $G$. Then, you can make the identification $\phi = g$ and apply the decomposition Eq. (1). In this case, the parameters $\xi_a,\chi_a$ takes a space-time coordinate dependence.

The element $h(x)$ now parametrize your vacuum. However, as you make a transformation along the broken generators, you end up in a new vacuum. You can interpret the Goldstone bosons as "coordinates " on a vacuum manifold.

In particular, for the element $h(x)$ you can choose a non-linear or a linear representation, which are related through field redefinitions. Usually, the decomposition is taken to be

$$ \phi(x) = \text{exp}\left(i \xi_a(x)\ \hat{T}^a \right) \cdot \text{exp}\left(i\chi_a(x) T^a\right) = \text{exp}\left(i \xi_a(x)\ \hat{T}^a \right) \left(\,\langle\phi(x)\rangle + \varphi(x)\,\right) $$ where $\varphi(x)$ and $\chi_a(x)$ are related by a non-linear field redefinition.

In the case of Electroweak $SU(2)_L$, you can choose $\varphi(x) = (0,h(x))^T$ where $h(x)$ is the physical Higgs field.

This is a very specific case of CCWZ construction. See the references for more details.


G.Panico and A.Wulzer, https://arxiv.org/abs/1506.01961

  • $\begingroup$ Thanks a lot, it was very helpful! I still have some doubts: (i) To show equation (1) would I use only the CBH formula (besides the comm. relations)? And (ii): altought it is intuitive, I could never fully grasp the concept of how we see the space where some representation of the group acts as a representation itself, which is the main argument of your answer (in the second paragraph). Can you give me a direction? $\endgroup$
    – GaloisFan
    Sep 1, 2018 at 22:27
  • 1
    $\begingroup$ (I) you need to use CBH for sure. However, you should show the solution of the decomposition exists and it is unique. (ii) this is just terminology. In particle physics, we identify the two concepts having in mind their inequivalent mathematical meaning. Rigorously, the representation is a map from the group to a vector space. Here, the field is an element of the group and you give to this element a representation. Is it clearer? $\endgroup$
    – apt45
    Sep 3, 2018 at 11:23
  • $\begingroup$ I'm still a little confuse. Isn't a representation a map from the group to the space of linear transformations on the vector space (where the field lives)? How is the field an element of the group? $\endgroup$
    – GaloisFan
    Sep 4, 2018 at 22:20
  • $\begingroup$ Sorry, I was not clear. By vector space I meant GL(V), which has a structure of vector space. So, a representation of G on the vector space V is the map R: G -> GL(V). The field is an element of the group which you can write as exponential of the generators. The transformation properties of the field under the action of the group G, depends on the representation of the generators. I know the answer is not satisfactory. It would need more space, so I suggest you to open another question about representations and fields. $\endgroup$
    – apt45
    Sep 4, 2018 at 22:50

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