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I do not understand the distinction between configuration state functions and Slater determinants. Is not every Slater determinant a CSF ?

Please explain the difference and the use/need of configuration state functions.

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Slater determinants are not eigenfunctions of the $\hat S^2$ operator, but CSFs are.

The Hamiltonian commutes with the operators for total and projected spin \begin{align} [\hat H, \hat S^2] &= 0 \\ [\hat H, \hat S_z] &= 0 \end{align} Therefore, a set of common eigenfunctions to all three operators exists. As Slater determinants are eigenfunctions of $\hat S_z$, but not $\hat S^2$, using them as a basis for the electronic wave function (=eigenstate of $\hat H$), will not guarantee the (approximate) solution to be an eigenfunction of $\hat S^2$. The spin multiplicity of the found solution may not be a pure Singlet (or Doublet, Triplet, etc.).

As an example, consider a 2 electron system, with both electrons occupying different spatial orbitals. The situation with same spin for both electrons can be represented as the Slater determinant $|\alpha\alpha\rangle=\frac{1}{\sqrt{2}}[\alpha(1)\beta(2)-\alpha(2)\beta(1)]$. The spatial part is omitted here, to keep the notation short. The corresponding eigenvalue equations are then \begin{align} \hat S_z |\alpha\alpha\rangle &= 1 |\alpha\alpha\rangle \\ \hat S^2 |\alpha\alpha\rangle &= 2 |\alpha\alpha\rangle \end{align} Thus we have $S=1$, from $S(S+1)=2$, and the spin multiplicity is $2S+1=3$, i.e. a Triplet. Here the Slater determinant directly corresponds to a CSF.

For the case of opposite spin, we have \begin{align} \hat S_z |\alpha\beta\rangle &= 0 |\alpha\beta\rangle \\ \hat S^2 |\alpha\beta\rangle &= |\alpha\beta\rangle + |\beta\alpha\rangle \end{align} and therefore $|\alpha\beta\rangle$ is not an eigenfunction of $\hat S^2$. The expectation value for total spin would yield $\langle\hat S^2\rangle=1$, which is neither a Singlet nor a Triplet (not even a Doublet where $S(S+1)=0.75$).

This can be fixed by taking suitable linear combinations of Slater determinants. In this example we have 2 options: \begin{align} |^1\Psi\rangle &= \frac{1}{\sqrt{2}} \left( |\alpha\beta\rangle - |\beta\alpha\rangle \right) \\ |^3\Psi\rangle &= \frac{1}{\sqrt{2}} \left( |\alpha\beta\rangle + |\beta\alpha\rangle \right) \end{align} This yields the eigenvalue equations \begin{align} \hat S^2 |^1\Psi\rangle &= 0 |^1\Psi\rangle \\ \hat S^2 |^3\Psi\rangle &= 2 |^3\Psi\rangle \end{align} which correspond to a Singlet and a Triplet state respectively.

Overall, this yields 1 Singlet component $|^1\Psi\rangle = \frac{1}{\sqrt{2}} \left( |\alpha\beta\rangle - |\beta\alpha\rangle \right)$, and the 3 Triplet components $|^3\Psi\rangle = \frac{1}{\sqrt{2}} \left( |\alpha\beta\rangle + |\beta\alpha\rangle \right)$, $|\alpha\alpha\rangle$ and $|\beta\beta\rangle$.

For the arithmetic on how to apply the $\hat S_z$ and $\hat S^2$ operators to multi-electron systems, see Chapter 2.5 in Modern Quantum Chemistry by A. Szabo and N. Ostlund.

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  • $\begingroup$ The kets are shorthand for determinants i assume ? So |ab> = |a(1)b(2)> - |a(2)b(1)>. And are alpha and beta spinorbitals or do you assume some factorization of spin and spatial part ? Sorry for the poor format of my comment but how can do line breaks in comments and can i do mathjax in the comments ? $\endgroup$ – Hans Wurst Apr 1 at 16:19
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    $\begingroup$ @HansWurst Mathjax can be used in comments, e.g $|ab\rangle$ yields $|ab\rangle$. Linebreaks are unavailable, however you can do display math, e.g. $$|ab\rangle$$ yields $$|ab\rangle$$ $\endgroup$ – Tyberius Apr 1 at 16:32
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    $\begingroup$ @HansWurst yes, the ket is a Slater determinant here. $\alpha$ and $\beta$ refers to the spin of one orbital each, so spin orbitals. The wave function can be factorized into spatial and spin part, however I omitted the spatial part. $\endgroup$ – Feodoran Apr 1 at 16:39
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    $\begingroup$ Do the spatial orbitals alone also form a determinant or do they factorize as a simple product ? Like $$ |^1 \Psi \rangle = \phi_1(r1) \phi_1(r2) \cdot (|\alpha \beta \rangle - |\beta \alpha \rangle) $$ or does it take some other form ? $\endgroup$ – Hans Wurst Apr 1 at 17:09
  • $\begingroup$ @HansWurst its always a Slater determinant, otherwise you get problem with the Pauli principle. $\endgroup$ – Feodoran Apr 1 at 17:10

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