Hi again,<br /><br />Let me please answer at least party to myself as I just figured out how to get the dimensionless wavefunction and someone might need this at some point.<br /><br />Considering:<br />1. the volume V of a voxel<br />2. the DFT wavefunction Psi<br /> 2. that the sum of the "actual" or dimensionless wavefunction Psi_A squared over the normalization volume V equals 1<br />we can say: <br />V * sum( Psi_A² ) = 1, where the sum goes over all voxels.<br />as well as:<br />sum( Psi² ) = F .... F is just some factor that appears due to the normalization<br /><br />we can then get the dimensionless probability density by<br />Psi_A²=Psi²/(V*F).<br /><br />So far so good. Now I am not sure how to proceed when calcuating the hyperfine Hamiltonian S A I, with S is the spin (here 1/2) and I is the nuclear spin.<br /><div>A is given in MHz but do i multiply this with hbar or h to get the energy?</div><div><br /></div><div>Lukas<br /></div><br /><div class="gmail_quote"><div dir="auto" class="gmail_attr">Lukas C schrieb am Donnerstag, 26. Oktober 2023 um 13:10:21 UTC+9:<br/></div><blockquote class="gmail_quote" style="margin: 0 0 0 0.8ex; border-left: 1px solid rgb(204, 204, 204); padding-left: 1ex;">hi everyone!<br><br>I am using the cp2k implementation to calculate hyperfine tensors. no problems here so far, i got some results using an all-electron basis set.<br><br>now, unfortunately, i cannot find any information on how this is done internally and also i cannot assign the units. e.g. in which units is |Psi(x,y,z)|^2 ? I know that it has to be normalized, but to which volume?<br><br>one starting point was the vasp manual, in which they clearly say how the calculations works. <a href="https://www.vasp.at/wiki/index.php/LHYPERFINE" target="_blank" rel="nofollow" data-saferedirecturl="https://www.google.com/url?hl=de&q=https://www.vasp.at/wiki/index.php/LHYPERFINE&source=gmail&ust=1698387196954000&usg=AOvVaw01dfMsnS96ShQGMbHOT2Na">https://www.vasp.at/wiki/index.php/LHYPERFINE</a><br><br>does cp2k work similarly?<br><br>Thank you!<br><br>Best,<br>Lukas<br></blockquote></div>
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