## Number of PAW one-center entries in CHGCAR

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yuansheng_zhao
Newbie
Posts: 2
Joined: Mon Oct 16, 2023 8:02 am

### Number of PAW one-center entries in CHGCAR

Dear Admin and VASP users,

I am trying to interpret the one-center terms in CHGCAR.
These terms should represent Eq.7 of PhysRevB 59, 1758:
rho_ij = sum_n f_n <psi_n|p_i> <p_j|psi_n>
Thus, for element with a total of n lm-decomposed projectors, there should be n(n+1)/2 entries.
However, this is not the case.

For elements with 2s + 2p projectors such as O, n = 2*1 + 2*3 = 8, so I am expecting 8*9/2 = 36 entries while there are only 33 in CHGCAR.
Likely, for elements with 2s + 2p + 1d (such as Mg) or 2s + 2p + 2d (such as Zn), I am expecting 13*14/2 = 91 and 18*19/2 = 171 entries instead of 78 and 138 in CHGCAR, respectively.

The same also goes to augmentation part in POT file.

I am wondering what I am missing. Any help is appreciated.

Best regards,
YZ

alexey.tal
Global Moderator
Posts: 288
Joined: Mon Sep 13, 2021 12:45 pm

### Re: Number of PAW one-center entries in CHGCAR

Dear Younsheng,

If you would like to understand how the number of augmentation occupanices is calculated you should look at the routine RETRIEVE_RHOLM in paw_base.F
In the essence, we have two projectors for each angular momentum that gives us 2*n*(2*n'+1)/2. Then for each pair of angular momenta l and l' we add all occupancy channels J=2L+1, where L=|l-l'|...l+l'.
And certain elements of this array will be zero due to the sum rule.

yuansheng_zhao
Newbie
Posts: 2
Joined: Mon Oct 16, 2023 8:02 am

Dear Alexey