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# Proof That the Two-Electron Ground State of a Spin-Independent Hamiltonian Is a Singlet(a) The mean energy of a two-electron system with Hamiltonian (32.3) in the s...

## Question

###### Proof That the Two-Electron Ground State of a Spin-Independent Hamiltonian Is a Singlet(a) The mean energy of a two-electron system with Hamiltonian (32.3) in the state $psi$ can be written (after an integration by parts in the kinetic energy term) in the form:$$E=int d mathbf{r}_{1} d mathbf{r}_{2}left[frac{h^{2}}{2 m}left{left|mathbf{abla}_{1} psiight|^{2}+left|abla_{2} psiight|^{2}ight}+Vleft(mathbf{r}_{1}, mathbf{r}_{2}ight)|psi|^{2}ight]$$Show that the lowest value ( $32.28$ ) assumes over

Proof That the Two-Electron Ground State of a Spin-Independent Hamiltonian Is a Singlet (a) The mean energy of a two-electron system with Hamiltonian (32.3) in the state $psi$ can be written (after an integration by parts in the kinetic energy term) in the form: $$E=int d mathbf{r}_{1} d mathbf{r}_{2}left[frac{h^{2}}{2 m}left{left|mathbf{ abla}_{1} psi ight|^{2}+left| abla_{2} psi ight|^{2} ight}+Vleft(mathbf{r}_{1}, mathbf{r}_{2} ight)|psi|^{2} ight]$$ Show that the lowest value ( $32.28$ ) assumes over all normalized antisymmetric differentiable wave functions $psi$ that vanish at infinity is the triplet ground-state energy $E_{t}$, and that when symmetric functions are used the lowest value is the singlet ground-state energy $E_{mathrm{s}}$ (b) Using (i) the result of (a), (ii) the fact that the triplet ground state $psi_{t}$, can be taken to be real when $V$ is real, and (iii) the fact that $left|psi_{2} ight|$ is symmetric, deduce that $E_{s} leqslant E_{T}$.

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