<div dir="ltr">I have a few questions about default units. I am trying to find the default units for the density matrix and the core hamiltonian. I am assuming they are in e/(bohr)^3 and Hartrees respectively. I am also trying to determine which units are being used in my input file for the Electric field intensity as they change based on the periodicity of the system. Is it in W/cm^2 or in a.u? I have used the PERIODIC NONE keyword for the cell so I am thinking it is in a.u. <div><br></div><div>My input file is as follows:</div><div><br></div><div>&GLOBAL<br> PROJECT BENZENE-UNPERTURBED<br> RUN_TYPE ENERGY_FORCE<br> PRINT_LEVEL low<br>&END GLOBAL<br> <br>&FORCE_EVAL<br> METHOD QS<br><br> &SUBSYS<br> &KIND H<br> BASIS_SET DZVP-GTH-PADE<br> POTENTIAL GTH-PBE-q1<br> &END KIND<br> &KIND C<br> BASIS_SET DZVP-GTH-PADE<br> POTENTIAL GTH-PBE-q4<br> &END KIND<br> &KIND N<br> BASIS_SET DZVP-GTH-PADE<br> POTENTIAL GTH-PBE-q5<br> &END KIND<br> &CELL<br> ABC 30.0 30.0 30.0<br> PERIODIC NONE<br> &END CELL<br> &TOPOLOGY <br> COORD_FILE_NAME <a href="http://benzene.xyz">benzene.xyz</a> <br> COORD_FILE_FORMAT xyz <br> &CENTER_COORDINATES <br> &END CENTER_COORDINATES <br> &END TOPOLOGY <br> &END SUBSYS<br><br> &DFT<br> BASIS_SET_FILE_NAME ./BASIS_SET<br> POTENTIAL_FILE_NAME ./POTENTIALS<br> &QS<br> EPS_DEFAULT 1.0E-10<br> &END QS<br> &POISSON <br> PERIODIC NONE <br> POISSON_SOLVER wavelet<br> &WAVELET<br> SCF_TYPE 100<br> &END WAVELET <br> &END POISSON <br> &MGRID<br> CUTOFF 800<br> NGRIDS 6<br> <br> &END MGRID<br> &SCF<br> SCF_GUESS RESTART<br> EPS_SCF 1.0E-10<br> MAX_SCF 100<br> &DIAGONALIZATION T<br> ALGORITHM STANDARD<br> &END DIAGONALIZATION<br> &END SCF<br> &XC<br> &XC_FUNCTIONAL PBE<br> &END XC_FUNCTIONAL<br> &END XC<br><br> &REAL_TIME_PROPAGATION<br> INITIAL_WFN SCF_WFN<br> EPS_ITER 1E-9<br> MAX_ITER 10<br> MAT_EXP TAYLOR<br> DELTA_PULSE_DIRECTION 0 0 1<br> # DENSITY_PROPAGATION .TRUE.<br> &END REAL_TIME_PROPAGATION<br><br> &EFIELD<br> INTENSITY 1E10<br> POLARISATION 1.0000000000000000 0.0000000000000000 0.0000000000000000<br> WAVELENGTH [nm] 1E6<br> &CONSTANT_ENV<br> START_STEP 0<br> END_STEP 1<br> &END CONSTANT_ENV<br> &END EFIELD<br><br> &PRINT<br> &AO_MATRICES<br> ADD_LAST NUMERIC<br> DENSITY .TRUE.<br> #KOHN_SHAM_MATRIX .TRUE.<br> #KINETIC_ENERGY .TRUE.<br> #MATRIX_VXC .TRUE.<br> #POTENTIAL_ENERGY .TRUE.<br> CORE_HAMILTONIAN<br> NDIGITS 30<br> &END AO_MATRICES<br> &END PRINT<br><br> &END DFT<br>&END FORCE_EVAL<br></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Tue, Oct 6, 2020 at 5:53 PM Jason Gray <<a href="mailto:grayj...@gmail.com">grayj...@gmail.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div dir="ltr">Thank you for the clarification! That's exactly what I needed to know.</div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Fri, Oct 2, 2020 at 5:08 AM <<a href="mailto:hut...@chem.uzh.ch" target="_blank">hut...@chem.uzh.ch</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">Hi<br> <br> CP2K calculates the KS energy assuming PBC. The KS-Matrix is<br> <br> KS = T + Vpp + Vxc[rho] +Vh[rhot] with rhot = rho + core<br> <br> The output matrices are<br> <br> KOHN_SHAM_MATRIX : KS<br> KINETIC_ENERGY : T<br> CORE_HAMILTONIAN : T + Vpp<br> POTENTIAL_ENERGY : Vpp<br> MATRIX_VXC : Vxc<br> <br> There is no option to print Vh[rhot]. <br> As the standard program flow calculates Vxc+Vh together, one<br> could print Vh at the same point as Vxc (that has to be calculated<br> again for the printing), but this would require to add a couple of<br> lines of code.<br> <br> or<br> <br> Vh = KS - CORE_HAMILTONIAN - Vxc<br> <br> regards<br> <br> Juerg Hutter<br> --------------------------------------------------------------<br> Juerg Hutter Phone : ++41 44 635 4491<br> Institut für Chemie C FAX : ++41 44 635 6838<br> Universität Zürich E-mail: <a href="mailto:hut...@chem.uzh.ch" target="_blank">hut...@chem.uzh.ch</a><br> Winterthurerstrasse 190<br> CH-8057 Zürich, Switzerland<br> ---------------------------------------------------------------<br> <br> -----<a href="mailto:cp...@googlegroups.com" target="_blank">cp...@googlegroups.com</a> wrote: -----<br> To: <a href="mailto:cp...@googlegroups.com" target="_blank">cp...@googlegroups.com</a><br> From: "Jason Gray" <br> Sent by: <a href="mailto:cp...@googlegroups.com" target="_blank">cp...@googlegroups.com</a><br> Date: 10/02/2020 02:42AM<br> Subject: [CP2K:13992] Hartree Potential in Terms of AO Matrix<br> <br> Hello, <br> <br> I need to determine the external potential in the atomic orbital basis. I know that V_Kohn_Sham = V_Hartree + V_XC + V_External, and I see that I can print the Kohn Sham matrix and the XC matrix in the atomic orbital basis with AO_MATRICES. Is it possible to do the same with the Hartree potential, so that I can simply subtract away from V_Kohn_Sham to get V_External? I see that I can print the "POTENTIAL_ENERGY" in the AO_MATRICES section, but is this the Hartree energy?<br> <br> All the best,<br> Jason Gray<br> <br> -- <br> 440-465-4608<br> B.S. Physics from Virginia Tech<br> Physics PhD. Student at Rensselaer Polytechnic Institute <br> -- <br> You received this message because you are subscribed to the Google Groups "cp2k" group.<br> To unsubscribe from this group and stop receiving emails from it, send an email to <a href="mailto:cp2k%2Bun...@googlegroups.com" target="_blank">cp...@googlegroups.com</a>.<br> To view this discussion on the web visit <a href="https://groups.google.com/d/msgid/cp2k/CAMyXkYmx6QOVKf_BWJGmjNpp_pnnuANvTWtyoWApZVw2G3qdUg%40mail.gmail.com" rel="noreferrer" target="_blank">https://groups.google.com/d/msgid/cp2k/CAMyXkYmx6QOVKf_BWJGmjNpp_pnnuANvTWtyoWApZVw2G3qdUg%40mail.gmail.com</a>.<br> <br> <br> -- <br> You received this message because you are subscribed to the Google Groups "cp2k" group.<br> To unsubscribe from this group and stop receiving emails from it, send an email to <a href="mailto:cp2k%2Bun...@googlegroups.com" target="_blank">cp...@googlegroups.com</a>.<br> To view this discussion on the web visit <a href="https://groups.google.com/d/msgid/cp2k/OFE45D959D.4F35E3E1-ONC12585F5.00323BDE-C12585F5.00323BDF%40lotus.uzh.ch" rel="noreferrer" target="_blank">https://groups.google.com/d/msgid/cp2k/OFE45D959D.4F35E3E1-ONC12585F5.00323BDE-C12585F5.00323BDF%40lotus.uzh.ch</a>.<br> </blockquote></div><br clear="all"><div><br></div>-- <br><div dir="ltr"><div dir="ltr"><div><div dir="ltr"><div dir="ltr"><div style="font-size:small">440-465-4608</div><div style="font-size:small">B.S. Physics from Virginia Tech</div><div style="font-size:small">Physics PhD. Student at Rensselaer Polytechnic Institute</div></div></div></div></div></div> </blockquote></div><br clear="all"><div><br></div>-- <br><div dir="ltr" class="gmail_signature"><div dir="ltr"><div><div dir="ltr"><div dir="ltr"><div style="font-size:small">440-465-4608</div><div style="font-size:small">B.S. Physics from Virginia Tech</div><div style="font-size:small">Physics PhD. Student at Rensselaer Polytechnic Institute</div></div></div></div></div></div>