Violation of non-interacting $\cal V$-representability of the exact solutions of the Schrödinger equation for a two-electron quantum dot in a homogeneous magnetic field

Details Violation of non-interacting $\cal V$-representability of the exact solutions of the Schrödinger equation for a two-electron quantum dot in a homogeneous magnetic field Violation of non-interacting $\cal V$-representability of the exact solutions of the Schrödinger equation for a two-electron quantum dot in a homogeneous magnetic field

2009-07-14

arxiv.org/abs/0907.2383v1

Physics

Condensed Matter

Mesoscale and Nanoscale Physics

18 pages, 4 figures

Scientific paper

We have shown by using the exact solutions for the two-electron system in a parabolic confinement and a homogeneous magnetic field [ M.Taut, J Phys.A{\bf 27}, 1045 (1994) ] that both exact densities (charge- and the paramagnetic current density) can be non-interacting $\cal V$-representable (NIVR) only in a few special cases, or equivalently, that an exact Kohn-Sham (KS) system does not always exist. All those states at non-zero $B$ can be NIVR, which are continuously connected to the singlet or triplet ground states at B=0. In more detail, for singlets (total orbital angular momentum $M_L$ is even) both densities can be NIVR if the vorticity of the exact solution vanishes. For $M_L=0$ this is trivially guaranteed because the paramagnetic current density vanishes. The vorticity based on the exact solutions for the higher $|M_L|$ does not vanish, in particular for small r. In the limit $r \to 0$ this can even be shown analytically. For triplets ($M_L$ is odd) and if we assume circular symmetry for the KS system (the same symmetry as the real system) then only the exact states with $|M_L|= 1$ can be NIVR with KS states having angular momenta $m_1=0$ and $|m_2|=1$. Without specification of the symmetry of the KS system the condition for NIVR is that the small-r-exponents of the KS states are 0 and 1.

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