[CP2K-user] [CP2K:17854] Query regarding output of DFT-D3

[Abhishek Sharma]

Dear CP2K users,

I was trying to check the effect of D3 interactions on the energy, forces,
and stress. So, I ran two CP2K(version 8.2) jobs (RUN_TYPE  ENERGY_FORCE)
with and without D3 correction and printed stress and forces using,
  &PRINT
    &FORCES             ON
      FILENAME          =forces
    &END FORCES
    &STRESS_TENSOR      ON
FILENAME        =stress
    &END STRESS_TENSOR
  &END PRINT
  STRESS_TENSOR         ANALYTICAL

and printed D3 output using,
      &vDW_POTENTIAL
        POTENTIAL_TYPE  PAIR_POTENTIAL
        &PAIR_POTENTIAL
          TYPE          DFTD3
          PARAMETER_FILE_NAME dftd3.dat
          REFERENCE_FUNCTIONAL PBE
          VERBOSE_OUTPUT  .TRUE.
          &PRINT_DFTD
                FILENAME =VDWForces
          &END PRINT_DFTD
        &END PAIR_POTENTIAL
      &END vDW_POTENTIAL

I obtained following energy values,
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
with D3:
  Dispersion energy:
-0.43391295044042
    Total energy:
-1822.59757123461372
without D3:
Total energy:
-1822.16365828416315
<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
It shows that E_DFT+D3 = E_DFT + E_D3. Since forces are negative gradients
of energy, so my expectation was F_DFT+D3 = F_DFT + F_D3.

I checked the files with forces and found following values:
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
with D3:
  Dispersion Forces
  Atom   Kind                            Forces
    1      1   -0.00064625376145   -0.00178929821343    0.00063408272599
    2      1   -0.00010368151520    0.00098624399705   -0.00068002896134
    3      1   -0.00015900750143    0.00000043929734    0.00016150708715
    4      1   -0.00176730962009    0.00065208064208   -0.00062160173245
----------------------------------------------------
 ATOMIC FORCES in [a.u.]

 # Atom   Kind   Element          X              Y              Z
      1      1      C           0.00005291     0.00012528    -0.00005636
      2      1      C          -0.00000900    -0.00001802     0.00002341
      3      1      C           0.00001580     0.00007087    -0.00002165
      4      1      C          -0.00011523     0.00002219    -0.00005222

without D3:
 ATOMIC FORCES in [a.u.]

 # Atom   Kind   Element          X              Y              Z
      1      1      C          -0.00059335    -0.00166402     0.00057772
      2      1      C          -0.00011269     0.00096823    -0.00065662
      3      1      C          -0.00014321     0.00007131     0.00013985
      4      1      C          -0.00188254     0.00067427    -0.00067382
<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

>From the printed forces, I found that (assuming printed dispersion forces
are in a.u.) that F_DFT+D3 =  F_DFT - F_D3 (e.g. F1x: 0.00005291  =
-0.00059335  - (-0.00064625376145) ), which should not be the case and it
should be F_DFT+D3 =  F_DFT + F_D3.

Similarly, I checked the stress and got following values,
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
with D3:
 Stress Tensor (dispersion)
 -0.302617856781     -0.301607753891E-04  0.414417744693E-03
 -0.301607753891E-04 -0.301666150572      0.101523675983E-03
  0.414417744693E-03  0.101523675983E-03 -0.302011925451
   Tr(P)/3 :   -0.30209864426787675
----------------------------------------------------
 STRESS| Analytical stress tensor [GPa]
 STRESS|                        x                   y                   z
 STRESS|      x        1.06013861421E-03  -2.63120964024E-05
 -3.93224150664E-03
 STRESS|      y       -2.63120964024E-05   1.25227540891E-02
1.45254611477E-04
 STRESS|      z       -3.93224150664E-03   1.45254611477E-04
7.18280089554E-03
 STRESS| 1/3 Trace
6.92189786629E-03

without D3:
 STRESS| Analytical stress tensor [GPa]
 STRESS|                        x                   y                   z
 STRESS|      x        2.67537379581E-01   2.45931708730E-07
 -4.29716685938E-03
 STRESS|      y        2.45931708730E-07   2.78161945635E-01
5.58511123164E-05
 STRESS|      z       -4.29716685938E-03   5.58511123164E-05
2.73126455701E-01
 STRESS| 1/3 Trace
2.72941926972E-01
<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
Assuming the dispersion stress in a.u., I got the following relation
Stress_DFT+D3 (GPa) = Stress_DFT (GPa) + Stress_D3 (a.u.)*4359.74393 /
volume (Angstrom^3), which seems fine.

So from output from CP2K, I got following relations:
E_DFT+D3 (a.u.)           =  E_DFT (a.u.)          +  E_D3 (a.u.)
F_DFT+D3 (a.u.)           =  F_DFT  (a.u.)          -  F_D3 (a.u.)
Stress_DFT+D3 (GPa)  =  Stress_DFT (GPa) +  Stress_D3 (a.u.)*4359.74393 /
volume (Ang^3)

Could you please help me to understand why there is subtraction of the
forces to obtain total (DFT+D3) forces?

Many thanks ,
Abhishek

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