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I'm having trouble with a few concepts dealing with slater determinants and many electron systems. In particular, when dealing with 3 electron systems, I know that

$$ \psi_0 = \frac{1}{\sqrt{6}} \det \begin{bmatrix} 1s(1)\alpha(1) & 1s(2)\alpha(2) & 1s(3)\alpha(3) \\ 1s(1)\beta(1) & 1s(2)\beta(2) & 1s(3)\beta(3) \\ 2s(1)\alpha(1) & 2s(2)\alpha(2) & 2s(3)\alpha(3) \end{bmatrix}, $$

where the $1s$ and $2s$ are spatial functions and $\alpha$, $\beta$ are spin functions.

Performing the determinant expansion yields 6 terms, and it turns out that $\psi_0$ is an eigenfunction for the unperturbed Hamiltonian (in atomic units) $$ \hat{H}_0 = -\frac{1}{2} \nabla^{2}_{1} -\frac{3}{r_1} -\frac{1}{2}\nabla^{2}_{2} -\frac{3}{r_2} -\frac{1}{2}\nabla^{2}_{3} -\frac{3}{r_3} $$ and that the resulting eigenenergy is $$E^{0}=E_{1s}+E_{1s}+E_{2s}.$$ How does one show this?

My confusion mainly stems from the difficulty in acting the Hamiltonian on the determinant expression above. For example, how does one compute (in bra-ket notation) $$ \frac{1}{\sqrt{6}} \hat{H}_0 | 1s(1)\alpha(1) 1s(2)\beta(2) 2s(3)\alpha(3)\rangle? $$ This is only the first term in the expansion, and once I have some intuition on how to do this properly, then the general pattern should emerge quite clearly. Any help and guidance would be greatly appreciated!

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    $\begingroup$ 2 hints: First notice that the Hamiltonian has same terms for all 3 electrons, so we can just stick with the first electron ($\hat h_i$), and multiply its result by 3 in the end. Second, an intermediate result of the Hamiltonian acting on the ket vector is not needed. Directly consider the expectation value $\langle \hat h_i\rangle$. Notice which electronic coordinates the operator is acting on, and which not. Also keep in mind that the orbitals are form an orthonormal basis. $\endgroup$
    – Feodoran
    Commented Feb 16, 2019 at 16:46
  • $\begingroup$ So breaking the Hamiltonian into 3 components and using the first term in the expansion as an example, only $\hat{h}_{1}$ acts on the first term since it leads with $1s(1)$? Sorry, I'm having trouble in interpreting and understanding your second hint. $\endgroup$
    – Buttfor
    Commented Feb 17, 2019 at 0:14
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    $\begingroup$ Of course the whole Hamiltonian acts on all terms. But we can break it into the smaller parts ($\hat h_1+\hat h_2+\hat h_3$) and apply the sum rule of integration, i.e. consider the terms separately. Next the different terms of the Slater determinant can be considered separate as well. So the first term is $\langle 1s(1)\alpha(1)1s(2)\beta(2)2s(3)\alpha(3)|\hat h_1|1s(1)\alpha(1)1s(2)\beta(2)2s(3)\alpha(3)\rangle$, which can be rearanged to $\langle 1s(1)\alpha(1)|\hat h_1|1s(1)\alpha(1)\rangle\langle 1s(2)\beta(2)|1s(2)\beta(2)\rangle\langle 2s(3)\alpha(3)|2s(3)\alpha(3)\rangle$. $\endgroup$
    – Feodoran
    Commented Feb 17, 2019 at 10:26
  • $\begingroup$ By the orthonormality of spin orbitals, that term evaluates to $1s(1)$, correct? So going about this way, I'm going to end up dealing with 36 terms (many of which go to $0$). $\endgroup$
    – Buttfor
    Commented Feb 20, 2019 at 0:31
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    $\begingroup$ Yes, there should be 6 surviving terms, and 30 terms being 0 (I think). $\endgroup$
    – Feodoran
    Commented Feb 20, 2019 at 7:21

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