Goldstone bosons

The 4>2 particle is known as a Goldstone boson. The important point is that this phenomenon is general. The Goldstone theorem [48] which states that massless scalars occur whenever a continuous symmetry of a physical system is spontaneously broken (or, more accurately, is not apparent in the ground state) will be accepted without further proof, however, compare [49],... 

The particle spectrum consists of a massless Goldstone boson 2, a massive scalar i, and more crucially a massive vector A. The Goldstone boson can be eliminated by gauge transformation. For infinitesimal gauge factor a(x),... 

In order to test this experimental finding we are comparing with the eigenvalues of a Hamiltonian for a realistic quark model, namely the Goldstone-boson-exchange (GBE) constituent quark model (Glozman et al, 1998). It includes the kinetic energy in relativistic form... 

Figure 8. Histograms of the nearest-neighbor spacing distribution for the nucleon (left plots) and the delta (right plots). The data is for Goldstone-boson exchange and for one-gluon exchange compared to a pure linear confinement potential of the same strength. Curves represent the Poisson and the GOE-Wigner distributions.

This constraint implicitly defines the matrix, K (4>,gL,gR) Here, we wish to examine the CFL spectrum of massive states using the technique of integrating in/out at the level of the effective Lagrangian. is the Goldstone boson decay... 

This must be contrasted with the situation at zero chemical potential, where the coefficient of the four-derivative term is always a pure number before quantum corrections are taken into account. In vacuum, the tree-level Lagrangian which simultaneously describes vector mesons, Goldstone bosons, and their interactions is ... 

By expanding the effective Lagrangian with the respect to the Goldstone boson fields, one sees that g is also connected to the vector meson coupling to two pions, through the relation... 

The validity of the t Hooft anomaly conditions at high matter density have been investigated in [32, 33], A delicate part of the proof presented in [33] is linked necessarily to the infrared behavior of the anomalous three point function. In particular one has to show the emergence of a singularity (i.e. a pole structure). This pole is then interpreted as due to a Goldstone boson when chiral symmetry is spontaneously broken. 

Alternatively, if we had taken rns to be finite for fixed regulator d (so that, as /j, —> oo, eventually rns < m ), the inequality in (60) could be applied to exclude a Nambu-Goldstone boson, but we would find ourselves in the phase without a kaon condensate. 

Note that this phase breaks no global symmetry at all. This is an important point in connection with the identification of the lowest lying excitation modes which in turn strongly influence the thermodynamic properties of the system. In the 2SC phase we do not find, even in the chiral limit, any real Goldstone bosons, although there are some low-lying (pseudo)-Goldstones related to the axial 17,4(1) [15]. 

The third term has the form of a mass term ( rn2r)2) for the 77 field. Thus, the 77-mass is mv = y/—2fj.2, as before. The first term in L represents the kinetic energy of the -field, but there is no mass term for . This means that the field is carried by massless particles, known as Goldstone bosons. 

According to the Goldstone theorem [79] the three real fields Pi x) would introduce three massless Goldstone bosons to the theory. These can, however, be gauged away if one takes advantage of the local SU 2)i, gauge symmetry given by... 

The same calculation as before shows that still only one field acquires a genuine mass whereas the other n — 1 scalar fields remain massless. These massless bosons are usually called Goldstone bosons . 

Physically, the various equivalent ground states differ according to the number of Goldstone bosons of zero energy and momentum that they contain. It can be shown that the physical scattering amplitudes do not show any zero-mass pole terms. 

What has happened is that in the spontaneously broken symmetry the gauge boson has acquired mass at the expense of the would-be Goldstone boson, which simply disappears. For each vector gauge field that gets massive we need one complex scalar field, one piece of which becomes unphysical and disappears (it reappears as the longitudinal mode of the vector field) leaving one real scalar physical field, the Higgs boson. 

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