Abstract

The ground-state energy of infinite symmetric nuclear matter is usually described by strongly interacting nucleons obeying the Pauli exclusion principle. We can imagine a unitary transformation which groups four nonidentical nucleons (i.e., with different spin and isospin) close in coordinate space. Those nucleons, being nonidentical, do not obey the Pauli principle, thus their relative momenta are negligibly small (just to fulfill the Heisenberg principle). Such a cluster can be identified with an $\ensuremath{\alpha}$ boson. But in dense nuclear matter, those $\ensuremath{\alpha}$ particles still obey the Pauli principle since are constituted of fermions. The ground state energy of nuclear matter $\ensuremath{\alpha}$ clusters is the same as for nucleons, thus it is degenerate. We could think of $\ensuremath{\alpha}$ particles as vortices which can now braid, for instance making $^{8}\mathrm{Be}$ which leave the ground state energy unchanged. Further braiding to heavier clusters ($^{12}\mathrm{C}$, $^{16}\mathrm{O}$$,\dots{}$) could give a different representation of the ground state at no energy cost. In contrast $d$-like clusters (i.e., $N=Z$ odd-odd nuclei, where $N$ and $Z$ are the neutron and proton number, respectively) cannot describe the ground state of nuclear matter and can be formed at high excitation energies (or temperatures) only. We show that even-even, $N=Z$, clusters could be classified as non-Abelian states of matter. As a consequence an $\ensuremath{\alpha}$ condensate in nuclear matter might be hindered by the Fermi motion, while it could be possible a condensate of $^{8}\mathrm{Be}$ or heavier clusters.

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