The $P_{cs}(4459)$ pentaquark from a combined effective field theory and phenomenological perspective

TL;DR

This paper investigates the nature of the $P_{cs}(4459)$ pentaquark using effective field theory and phenomenological models, proposing it as a molecular state with specific spin and channel coupling effects.

Contribution

It combines effective field theory with phenomenological analysis to predict the spin and channel interactions of the $P_{cs}(4459)$ pentaquark, providing insights into its structure and hyperfine splitting.

Findings

The $P_{cs}(4459)$ is likely a $ar{D}^* \Xi_c$ molecular state.

Hyperfine splitting between $J=3/2$ and $J=1/2$ states is estimated to be 5-15 MeV.

Spectroscopy and decay patterns favor a $J=3/2$ assignment.

Abstract

The observation of the $P_{cs}(4459)$ by the LHCb collaboration adds a new member to the set of known hidden-charm pentaquarks, which includes the $P_c(4312)$, $P_c(4440)$ and $P_c(4457)$. The $P_{cs}(4459)$ is expected to have the light-quark content of a $\Lambda$ baryon ($I=0$, $S=-1$), but its spin is unknown. Its closeness to the $\bar{D}^* \Xi_c$ threshold -- $4478\,{\rm MeV}$ in the isospin-symmetric limit -- suggests the molecular hypothesis as a plausible explanation for the $P_{cs}(4459)$. While in the absence of coupled-channel dynamics heavy-quark spin symmetry predicts the two spin-states of the $\bar{D}^* \Xi_c$ to be degenerate, power counting arguments indicate that the coupling with the nearby $\bar{D} \Xi_c'$ and $\bar{D} \Xi_c^*$ channels might be a leading order effect. This generates a hyperfine splitting in which the $J=\tfrac{3}{2}$ $\bar{D}^* \Xi_c$ pentaquark will be lighter than the $J=\tfrac{1}{2}$ configuration, which we estimate to be of the order of $5-15\,{\rm MeV}$. We also point out an accidental symmetry between the $P_{cs}(4459)$ and $P_c(4440/4457)$ potentials. Finally, we argue that the spectroscopy and the $J/\psi \Lambda$ decays of the $P_{cs}(4459)$ might suggest a marginal preference for $J = \tfrac{3}{2}$ over $J = \tfrac{1}{2}$.