Register to reply 
Counting degrees of freedom for Goldstone bosons 
Share this thread: 
#1
Apr614, 04:46 AM

P: 86

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(N1), the # of Goldstone bosons is [itex] (N^21)((N1)^21)=2N1[/itex] . The Goldstone modes are from some scalar fields. Take the electroweak theory in SM as an example, the [itex]SU(2)_L*U(1)[/itex] is broken to [itex]U(1)[/itex] 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 [itex] 2N[/itex] degrees of freedom. If all [itex] N^21[/itex] symmetries in the gauge group are broken, how can we have [itex] N^21[/itex] Goldstone modes since there are only [itex] 2N[/itex] scalars in total? 


#2
Apr614, 02:25 PM

P: 842




#3
Apr614, 03:39 PM

Sci Advisor
P: 892

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


#4
Apr614, 03:42 PM

Sci Advisor
P: 892

Counting degrees of freedom for Goldstone bosons



#5
Apr614, 05:04 PM

P: 842




#6
Apr614, 06:34 PM

P: 86

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. 


#7
Apr714, 09:46 AM

P: 1,020




#8
Apr714, 02:12 PM

P: 842




#9
Apr714, 06:45 PM

Sci Advisor
P: 892

In general, the pattern of a symmetry breaking [itex]G \rightarrow H[/itex] or (which is the same thing) the nature of the unbroken subgroup [itex]H[/itex] depends on [itex]G[/itex], on the representations chosen for the scalar fields [itex]( \phi_{ i } , \chi_{ i } , … )[/itex], and on the form of the invariant potential. OK, here is an exercise for you: Take [itex]G = SU(3) \times SU(3)[/itex] and let the field [itex]\Phi[/itex] be a [itex]3 \times 3[/itex] matrix in the [itex]( 3 , \bar{ 3 })[/itex] representation. Write down the most general invariant potential for the matrix [itex]\Phi[/itex]. Minimize your potential and show that, for hermitian [itex]\langle \Phi \rangle[/itex], the unbroken subgroup [itex]H[/itex] contains at least [itex]SU(2) \times U(1)[/itex]. Sam 


#10
Apr714, 06:57 PM

Sci Advisor
P: 892

One can show this using Cartan decomposition of the algebra of [itex]G[/itex], 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 [itex]G[/itex] be the symmetry group of order [itex]r[/itex], and let [itex]\phi_{ i } ( x )[/itex] be a multiplet of scalar fields transforming according to some ndimensional representation. The Ginvariant action is of the form: [tex]S = \int d^{ 4 } x \ \left( \frac{ 1 }{ 2 } \partial_{ \mu } \vec{ \phi } \cdot \partial^{ \mu } \vec{ \phi } + U ( \phi ) \right) . \ \ \ (1)[/tex] Now, let us assume that at the classical level the symmetry is spontaneously broken, i.e. [itex]U( \phi )[/itex] has degenerate minima. We choose one of them, [itex]\vec{ V }[/itex], around which we expand perturbation theory, and call [itex]H[/itex] the little group (of order s) of the vector [itex]\vec{ V }[/itex], i.e. the subgroup of [itex]G[/itex] which leaves [itex]\vec{ V }[/itex] invariant. We know that when [itex]G \rightarrow H[/itex] a number of field components of [itex]\vec{ \phi }[/itex] 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 [itex]\vec{ \phi } ( x )[/itex] can be parametrized in terms of a matrix representation of G, [itex]R ( g(x) )[/itex], acting on [itex]\vec{ V }[/itex]: [tex]\phi_{ i } ( x ) = R_{ i j } ( g(x) ) \ V_{ j } , \ \ g(x) \in G . \ \ \ (2)[/tex] Now, if we multiply [itex]g(x)[/itex] on the right by an element [itex]h(x) \in H[/itex] and use the fact that [itex][H][/itex] leaves [itex]V_{ j }[/itex] invariant, we find that [itex]\phi_{ i } ( x )[/itex] does not get modified: [tex] 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 ) . [/tex] This shows that [itex]\phi_{ i } ( x )[/itex] is a function of the elements of the coset space [itex]G / H[/itex]. We can now divide the set of generators of the Lie algebra of [itex]G[/itex] into the set of generators corresponding to the Lie algebra of [itex]H[/itex]: [tex]\mathcal{ L } ( H ) : \ \ T^{ a }_{ i j } \ V_{ j } = 0 , \ \ ( a = 1 , 2 , … , s) ,[/tex] and the remaining set which generates the coset space [itex]G / H[/itex]. It is such that [tex]\sum_{ a = s + 1 }^{ r } T^{ a }_{ i j } \ V_{ j } \lambda_{ a } = 0 \ \Rightarrow \ \lambda_{ a } = 0, \ \ \forall a .[/tex] Thus, for [itex]a > s[/itex], the vectors [itex]( V^{ a } )_{ i } \equiv T^{ a }_{ i j } \ V_{ j }[/itex] are linearly independent. Therefore, the matrix [itex]R_{ i j }[/itex] can be parametrized in terms of a set of local coordinates, [itex]\chi_{ a } ( x )[/itex], on the coset space [itex]G / H[/itex] as: [tex] R ( \chi ( x ) ) = \exp \left( \sum_{ a = s + 1 }^{ r } T^{ a } \chi_{ a } ( x ) \right) . \ \ \ (3) [/tex] To finish the job, we need to verify that the fields [itex]\chi_{ a } ( x )[/itex] are massless. Since [itex]U ( \phi )[/itex] is group invariant and derivativefree, it does not depend on [itex]g(x)[/itex] 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: [tex]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 } .[/tex] Expanding this action near the identity shows that all the remaining fields are indeed massless. I hope that helps. Sam 


#11
Apr714, 10:22 PM

P: 86




Register to reply 
Related Discussions  
Difference between gapless excitations and Goldstone bosons  Atomic, Solid State, Comp. Physics  5  
Question about counting degrees of freedom  General Math  5  
What is meant by saying that the Goldstonebosons are eaten by gauge bosons?  High Energy, Nuclear, Particle Physics  3  
Goldstone bosons in Models with global symmetry, broken by Orbifolding  Beyond the Standard Model  4  
Goldstone bosons 1st order phase transition  Quantum Physics  13 