Npsi for LF calculations

Dear Saki,

First the short answer: Npsi should be so large that all many-body eigenstates in the temperature range are included. For a $d^0$ configuration this is (normally) reached when Npsi=1. You can test this by taking Npsi=2 and calculating the eigenenergies of the states. If the gap between state $i$ and $i+1$ is larger than $\pm$ 0.1 eV you can take Npsi=$i$.

Now the longer answer.

In order to describe a many electron quantum system we can start by one particle states $\phi_{\tau}(\vec{r})$. These can also be written as states $|\phi_{\tau}\rangle$ or in second quantisation using creation operators as $a^{\dagger}_{\tau}|0\rangle = |\phi_{\tau}\rangle$. The quantum number $\tau$ defines the spin-orbital in an atom and the atomic site in a molecule. We define a one particle basis with NF one particle states such that $\tau \in [0,...,\mathrm{NF}-1]$.

With these one particle states we can define an $n$ electron basis as single Slater determinants. If $D_i^{(n)}$ is a set of $n$ one particle states selected from the $NF$ basis states, then the single Slater determinants are $\psi_i = \prod_{\tau \in D_i{(n)}} a^{\dagger}_{\tau}|0\rangle$.

General many particle states and especially many particle eigenstates are superpositions of single Slater determinants $\Psi_j = \sum_i \alpha_{i}^{(j)} |\psi_i\rangle$.

Eigenstates in Quanty are superposition of Slater determinants, i.e. the $|\Psi\rangle$ states in the previous paragraphs. This is completely independent of Npsi. So independent of Npsi the eigenstates can be superpositions of different configurations. The parameter Npsi tells you how many of these many particle states that are superpositions of different one particle state fillings are calculated.

Hope this helped, best wishes, Maurits

Dear Maurits,

Thank you for your quick reply with the detailed explanation!

Best regards, Saki