karlzr said:
Could you explain a little more about "Goldstone bosons appear as the parameters of the coset space"? I only know they would appear in the exponent.
One can show this using Cartan decomposition of the algebra of G, but unless one has a sound knowledge in group theory, one can easily get lost in the details. This is (badly) explained in sec. 19.6 of Weinberg’s text QFT Vol.2.
However, we can simplify the subject considerably by some physical arguments. So, let G be the symmetry group of order r, and let \phi_{ i } ( x ) be a multiplet of scalar fields transforming according to some n-dimensional representation. The G-invariant action is of the form:
S = \int d^{ 4 } x \ \left( \frac{ 1 }{ 2 } \partial_{ \mu } \vec{ \phi } \cdot \partial^{ \mu } \vec{ \phi } + U ( \phi ) \right) . \ \ \ (1)
Now, let us assume that at the classical level the symmetry is spontaneously broken, i.e. U( \phi ) has degenerate minima. We choose one of them, \vec{ V }, around which we expand perturbation theory, and call H the little group (of order s) of the vector \vec{ V }, i.e. the subgroup of G which leaves \vec{ V } invariant. We know that when G \rightarrow H a number of field components of \vec{ \phi } are massless (Goldstone modes). Since we are only interested in these modes, i.e. in the long distance behaviour (IR limit) of correlation functions, the fluctuations of the massive fields in the functional integral can be neglected. In this limit, the remaining massless components of \vec{ \phi } ( x ) can be parametrized in terms of a matrix representation of G, R ( g(x) ), acting on \vec{ V }:
\phi_{ i } ( x ) = R_{ i j } ( g(x) ) \ V_{ j } , \ \ g(x) \in G . \ \ \ (2)
Now, if we multiply g(x) on the right by an element h(x) \in H and use the fact that [H] leaves V_{ j } invariant, we find that \phi_{ i } ( x ) does not get modified:
<br />
R_{ i j } \left( g(x) h(x) \right) \ V_{ j } = R_{ i k } \left( g(x) \right) \ R_{ k j } \left( h(x) \right) \ V_{ j } = R_{ i k } \left( g(x) \right) \ V_{ j } = \phi_{ i } ( x ) .<br />
This shows that \phi_{ i } ( x ) is a function of the elements of the coset space G / H.
We can now divide the set of generators of the Lie algebra of G into the set of generators corresponding to the Lie algebra of H:
\mathcal{ L } ( H ) : \ \ T^{ a }_{ i j } \ V_{ j } = 0 , \ \ ( a = 1 , 2 , … , s) ,
and the remaining set which generates the coset space G / H. It is such that
\sum_{ a = s + 1 }^{ r } T^{ a }_{ i j } \ V_{ j } \lambda_{ a } = 0 \ \Rightarrow \ \lambda_{ a } = 0, \ \ \forall a .
Thus, for a > s, the vectors ( V^{ a } )_{ i } \equiv T^{ a }_{ i j } \ V_{ j } are linearly independent. Therefore, the matrix R_{ i j } can be parametrized in terms of a set of local coordinates, \chi_{ a } ( x ), on the coset space G / H as:
<br />
R ( \chi ( x ) ) = \exp \left( \sum_{ a = s + 1 }^{ r } T^{ a } \chi_{ a } ( x ) \right) . \ \ \ (3)<br />
To finish the job, we need to verify that the fields \chi_{ a } ( x ) are massless. Since U ( \phi ) is group invariant and derivative-free, it does not depend on g(x) and therefore gives constant contribution to the action. So, it can be neglected. Inserting eq(2) and eq(3) in eq(1), we obtain the following for the action integral:
S = \frac{ 1 }{ 2 } \int d^{ 4 } x \ V_{ j } \ \partial_{ \mu } R^{ - 1 }_{ j k } ( x ) \ \partial^{ \mu } R_{ k i } ( x )\ V_{ i } .
Expanding this action near the identity shows that all the remaining fields are indeed massless.
I hope that helps.
Sam
