Abstract

The Schrodinger equation for a system of two mobile spinless nuclei with $$\mathcal{Z}=2$$ and four electrons ( $$^4$$ He $$_2$$ -dimer) has been solved with exact diagonalization approach. It is found that the ground state of the system ( $$\sim -~4.2$$ a.u.) is higher than the energy of two separate $$^4$$ He atoms ( $$\sim -~5.7$$ a.u.), and the average internuclear distance $$\langle R_0\rangle \sim 1.7~\AA $$ is a half smaller than the typical lattice constant in solid helium phases. This result corresponds to a metastable state which can be stabilized only at extremely high external pressure estimated as 27,000 GPa. It means that formation of condensed helium phases at saturated vapor pressure (SVP) is possible only due to many-particle interaction. Below 4.2 K at SVP, $$^4$$ He behaves as simple liquid without long-range ordering (He I phase), but below 2.17 K it undergoes the $$\lambda $$ -transformation into He II phase due to long-range antiferromagnetic ordering in the $$^4$$ He spin subsystem with exchange of order 10 K. As a result, the He II phase can be interpreted as ‘spin ice’ built as polytypic structure of close-packed 2D planes on triangular lattice collected in a stack with alternated ordering in the direction perpendicular to the basal planes, but without breaking the closest packing principle between neighboring planes.

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