# Counting degrees of freedom for Goldstone bosons

1. ### karlzr

95
I mean Goldstone bosons in the title. Sorry I don't know how to edit the title.

Goldstone's Theorem says that there is a massless Goldstone mode for each breaking symmetry. For instance symmetry of a theory is broken from SU(N) to SU(N-1), the # of Goldstone bosons is $(N^2-1)-((N-1)^2-1)=2N-1$ . The Goldstone modes are from some scalar fields. Take the electroweak theory in SM as an example, the $SU(2)_L*U(1)$ is broken to $U(1)$ where three degrees of freedom from the Higgs scalar H become Goldstone modes.

Then comes my question. Let's consider that the scalar fields live in the fundamental representation of SU(N) and thus there are $2N$ degrees of freedom. If all $N^2-1$ symmetries in the gauge group are broken, how can we have $N^2-1$ Goldstone modes since there are only $2N$ scalars in total?

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3. ### The_Duck

915
A scalar in the fundamental representation of SU(N) can't break all ##N^2-1## symmetries. It can at most break SU(N) down to SU(N-1), breaking ##2N-1## symmetries.

4. ### samalkhaiat

1,022
Here you are doing ok. But then, you come up with this!
This means that you did not understand the OK part of your post.

Goldstone bosons need a space to live in. If a symmetry group $G$ breaks down spontaneously to one of its subgroup $H$, then Goldstone bosons appear as the parameters of the coset space $(G / H)$, i.e., there will be $(\mbox{ dim } G - \mbox{ dim } H)$ Goldstone bosons. So, if you (as you are suggesting) take $H = G \subset G$, then the “coset space” $G / G$ will have zero dimension, i.e., zero Goldstone bosons.

Sam

5. ### samalkhaiat

1,022
Why not down to $SU(N - M)$?

6. ### The_Duck

915
Let's take N = 3 for concreteness. Suppose we have a scalar in the fundamental representation of some SU(3) group which acquires a nonzero vacuum expectation value and therefore breaks some of the SU(3) symmetries. Without loss of generality we can take the VEV to be (0, 0, v) for some v, because any VEV can be rotated into the 3 direction by an SU(3) transformation. So this is the general case. This VEV breaks SU(3) down to SU(2). So in general a scalar field in the fundamental representation can only break SU(N) to SU(N-1).

7. ### karlzr

95
Thanks, that makes sense.

Thanks.
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.

8. ### andrien

Can not you break ##SU(3)## to ##U(1)× U(1)##.

9. ### The_Duck

915
How? Not by the VEV of a single scalar in the fundamental representation, I think, which is the scenario I was considering.

10. ### samalkhaiat

1,022
Yes, that is true for the simple case of one complex vector in $SU(N)$. However, for example, putting two complex N-vector $( \vec{ \phi } , \vec{ \chi } )$ requires that the $SU(N)$ invariant potential to depend not only on $\phi^{ \dagger }_{ i } \phi_{ i }$ and $\chi^{ \dagger }_{ i } \chi_{ i }$, but also on the invariant product $( \phi^{ \dagger }_{ i } \chi_{ i } + C.C. )$. The minimization of such potential shows that the two vectors $( \vec{ \phi } , \vec{ \chi } )$ can have non-zero entries in the last two components. These configurations are invariant under the $SU(N - 2)$ mixing of the first $( N - 2 )$ components.
In general, the pattern of a symmetry breaking $G \rightarrow H$ or (which is the same thing) the nature of the unbroken subgroup $H$ depends on $G$, on the representations chosen for the scalar fields $( \phi_{ i } , \chi_{ i } , … )$, and on the form of the invariant potential.

OK, here is an exercise for you: Take $G = SU(3) \times SU(3)$ and let the field $\Phi$ be a $3 \times 3$ matrix in the $( 3 , \bar{ 3 })$ representation. Write down the most general invariant potential for the matrix $\Phi$. Minimize your potential and show that, for hermitian $\langle \Phi \rangle$, the unbroken subgroup $H$ contains at least $SU(2) \times U(1)$.

Sam

11. ### samalkhaiat

1,022

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:
$$R_{ i j } \left( g(x) h(x) \right) \ V_{ j } = R_{ i k } \left( g(x) \right) \ R_{ k j } \left( h(x) \right) \ V_{ j } = R_{ i k } \left( g(x) \right) \ V_{ j } = \phi_{ i } ( x ) .$$
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:
$$R ( \chi ( x ) ) = \exp \left( \sum_{ a = s + 1 }^{ r } T^{ a } \chi_{ a } ( x ) \right) . \ \ \ (3)$$
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

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12. ### karlzr

95
##(3, \bar{3})## means ##\Phi \rightarrow M\Phi N^\dagger##, right? So the most general potential should look like ##U\big(Tr(\Phi^\dagger \Phi)^n\big)##. If ##\langle \Phi \rangle## is hermitian, can we rotate it so that only the (11) component is nonvanishing? In that case, the symmetry should be broken to ##SU(2)\otimes SU(2)##.