Difference in requesting LC-PBE0 and LC-wPBEh functionals?
Just Got Here
9:31:50 AM PDT - Thu, Aug 29th 2013
In the current documentation, it says that LC-PBE0 (also called CAM-PBE0) is requested by specifying:

xc xcampbe96 1.0 cpbe96 1.0 HFexch 1.0
cam 0.30 cam_alpha 0.25 cam_beta 0.75

which I interpret as alpha=25% short-range HF exchange, varying with an error function (scaling=cam) to alpha+beta=100% HFexch at long range. Conversely, we have 1-alpha =75% campbe96 exchange at short range, being damped to 1-(alpha+beta)=0% at long range range. The "1.0" after xcampbe96 and HFexch, I interpret as "global" scaling factors.

Now, to request LC-wPBEh, one should specfiy:

xc xwpbe 0.80 cpbe96 1.0 hfexch 1.00
cam 0.2 cam_alpha 0.20 cam_beta 0.80

And, here, I do not understand the first line. Why the 0.80 after xwpbe? If I have to take care that alpha and the (global?) scaling factor of xwpbe add up to 1.0, why is this not the case for CAM-PBE0? Does that mean I have 125% exchange in CAM-PBE0? Or does it mean that I have just 0.8*(1-alpha) short-range wpbe in LC-wPBEh? Or do the two short-range GGA exchange philosophies require fundamentally different input specification?

I would be very much obliged if someone could help me out in my limited understanding here.

Forum Regular
3:40:49 PM PDT - Wed, Sep 11th 2013
Hi,

I see your confusion. For the LC-PBE0 (or CAM-PBE0), The xcampbe96 functional is the LC or CAM modified short-range DFT (PBE exchange). We actually scale this by (1-alpha-beta*erf) in the code so that the total amount of exchange in the short and long-range are 1.0

We do the LC-wPBEh functional a bit differently. The scale factor is directly used from the first line.

Best,
-Niri

niri.govind@pnl.gov

Just Got Here
1:53:52 PM PDT - Wed, Apr 1st 2020
Hi,

For Range separated functional wPBEh:

As $r_{12} => 0$, the HF exchange fraction is $\alpha$ i.e., cam_alpha, while the DFT exchange fraction is $(1-\alpha)$ i.e., cam_beta. As $r_{12} => \infinity$, the HF exchange fraction approaches $\alpha$ + $\beta$ i.e., hfexch 1.0 and the DFT exchange fraction approaches $(1- \alpha - \beta)$ i.e., xwpbe=0.8. It is not very clear to me how the scale factor is used.

I am interested in summing up $\alpha+\beta = 0.5$. I know that I can set hfexch as 0.5. But, how do I set my xwpbe value so that it is equal to $(1-0.5)$.

I would be happy if someone could help me to understand how the xwpbe value is set?

Forum Regular
2:22:54 PM PDT - Fri, Apr 24th 2020
Hi Chandra,

For the LRC-wPBEh functional, which is defined as follows:

xc xwpbe 0.80 cpbe96 1.0 hfexch 1.00
cam 0.2 cam_alpha 0.20 cam_beta 0.80

The short-range part has 80% xwpbe and 20% (cam_alpha = 0.2) HF, while the long-range (r->infinity) is fully HF (cam_alpha + cam_beta = 1.0). The transition from short to long-range is controlled by the attenuation parameter cam.

Hope this helps.

Best,
-Niri
niri.govind@pnnl.gov

Just Got Here
8:38:05 AM PDT - Wed, May 6th 2020
Hi Niri,

Thank you very much for the reply. I have couple of doubts with the implementation of this functional in the code. Is xwpbe and cam_beta both corresponds to SR DFT exchange?

And another thing is that how can I add LR DFT exchange term (1-alpha-beta) along with SR and LR HF parts. Generally, this is zero for alpha+beta=1.0. We cannot rule out this term for alpha+beta<1.



Best,
Chandra


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