<div dir="ltr">Hi, Kai,<div><br></div><div>I think eclectic field with 10 Hartree/e/Bohr is 514 V/Angstrom. But I guess this is electric field can be balanced by the nearby eclectic field with opposite directions, so the system can be stable. </div><div><br></div><div>For water, when you apply an external periodic electric field where the direction is fixed, so under such electric field, water will become more ordered like ice structure because electric field forces to be ordered, and since the electric field is along the same direction, this can result in a break of bond when the force is large enough because there is a persistent force along the same direction. So far this is what I understand. </div><div><br></div><div>Thanks!</div><div><br></div><div>Fangyong</div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Tue, May 10, 2022 at 12:40 PM Fangyong Yan <<a href="mailto:fyyan2019@gmail.com">fyyan2019@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">Hi, Kai,<div><br></div><div>Regards the strength of eclectic field, an external applied periodic electric field of 0.30 V/Angstrom is able to dissociate water. (see paper, G. Cassone et al., pccp, 2019, DOI: 10.1039/C9CP03101D (Paper) Phys. Chem. Chem. Phys., 2019, 21, 21205-21212)</div><div><br></div><div>Fangyong</div><div><br></div><div><br></div><div><br></div><div><br></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Tue, May 10, 2022 at 12:31 PM Fangyong Yan <<a href="mailto:fyyan2019@gmail.com" target="_blank">fyyan2019@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">Hi, Kai,<div><br></div><div>I think the electric field gradient unit is hartree/e/bohr. </div><div><br></div><div>10 hartree/e/bohr is a large electric field, may I ask what is the location for these large electric field in the cube file? </div><div><br></div><div>Thanks!</div><div><br></div><div>Fangyong</div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Wed, Apr 13, 2022 at 10:42 AM gary Washington <<a href="mailto:lindagary0@gmail.com" target="_blank">lindagary0@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">I find that we can test for the units of the Hartree potential by using a test molecule. I selected Li2 and applied an external electric field (Volts/Angstroms). First obtain the Hartree Potential cube file for a zero external potential<br><div><br></div><div># electric field along the z direction at efieldmag eV/Angstrom or efieldmag V/Angstrom<br> &EXTERNAL_POTENTIAL<br> FUNCTION (A/B)*Z<br> VALUES [eV] ${efieldmag} [angstrom] 1.0<br> PARAMETERS A B<br> &END EXTERNAL_POTENTIAL</div><div><br></div><div><br></div><div>Cutoff = 400, RelCutoff=80<br></div><div>so we set efieldmag = 0.0</div><div><br></div><div>Now using a box/cell **without periodic boundary conditions** we can optimize the geometry ( an energy calculation should also work ) now we use cubecruncher to obtain the potential along the z axis. We expect the initial slope, which is equal to the initial electric field magnitude in the z-direction. Now molecules will have an electric field around them due to their charge distribution of nuclei and electrons so even when efieldmag=0.0 the electric field around the molecule will be nonzero and will diminish as you move away from the molecule. If the box is large enough the influence of the molecule on the initial electric field will nearly vanish far from the molecule. I used {Lx, Ly, Lz} of {12,12,15} , {12,12,20}, {12,12,25} and the initial slopes are -0.019047, -0.00027563, -0.0000032. The initial slope is obtained on the left in the limit as z --> 0. We see the influence of the molecule can be greatly reduced in a sufficiently large box and this is expected to be true for any molecule in a sufficiently large box/cell. <br></div><div><br></div><div>Now we set efieldmag=0.005 Volts/Angstrom, using a box with dimensions of {12,12,15}, {12,12,20}, {12,12, 25} and the initial slopes are -0.01932522, -0.00008413067, -0.0001806136</div><div>Next we set Cutoff=600, relCutoff=80 and box size {12,12, 25} and the initial slope is -0.0001811421</div><div>Next we set Cutoff=800, relCutoff=80 and box size {12,12, 25} and the initial slope is -0.0001815272</div><div>Next we set Cutoff=800, relCutoff=80 and box size {12,12, 35} and the initial slope is -0.0001837458</div><div><br></div><div>* assumption 1: Hartree potential has the units of hartree/e = 27.211386 Volts / a.u.</div><div>*assumption 2: Electric Field Magnitude (Volts/Angstroms) = 27.211386 * (Hartree Potential at a point)</div><div><br></div><div>Check: Given an electric field magnitude of 0.005 V/Ang we expect that the observed Hartree potential sufficiently far from the molecule will yield a slope along the z axis, Ez (electric field in the z direction) of the same magnitude. We expect that the result should approach (0.005 V/Ang)/27.211386 = 0.000183747 a.u. ( the expected slope of the Hartree potential due only to the external potential)</div><div><br></div><div>We see that the initial slope is equal to efieldmag/27.211386 (Volts/a.u.) which shows that assumption 1 holds. <br></div><div><br></div><div><br></div><div><br></div><div><br></div><div><br></div><div class="gmail_quote"><div dir="auto" class="gmail_attr">On Monday, March 22, 2021 at 2:25:52 PM UTC-4 <a href="mailto:fyya...@gmail.com" target="_blank">fyya...@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">Hi, Kai,<div><br></div><div>50000 MV/cm = 50 V / (10E8) Angstrom = 5E-7 V/Angstrom, which is a very very small electric field. </div><div><br></div><div>Best,</div><div><br></div><div>Fangyong</div></div><br><div class="gmail_quote"></div><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Mon, Mar 22, 2021 at 1:07 PM Kaixuan Chen <<a rel="nofollow">kaixu...@gmail.com</a>> wrote:<br></div></div><div class="gmail_quote"><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><br><div>Dear all,</div><div>I have generated the cube files of hartree potential (<span><a href="https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT.html" rel="nofollow" target="_blank">CP2K_INPUT</a> /
<a href="https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT/FORCE_EVAL.html" rel="nofollow" target="_blank">FORCE_EVAL</a> /
<a href="https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT/FORCE_EVAL/DFT.html" rel="nofollow" target="_blank">DFT</a> /
<a href="https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT/FORCE_EVAL/DFT/PRINT.html" rel="nofollow" target="_blank">PRINT</a> /
<a href="https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT/FORCE_EVAL/DFT/PRINT/V_HARTREE_CUBE.html" rel="nofollow" target="_blank">V_HARTREE_CUBE</a></span>) and electric field (<span><a href="https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT.html" rel="nofollow" target="_blank">CP2K_INPUT</a> /
<a href="https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT/FORCE_EVAL.html" rel="nofollow" target="_blank">FORCE_EVAL</a> /
<a href="https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT/FORCE_EVAL/DFT.html" rel="nofollow" target="_blank">DFT</a> /
<a href="https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT/FORCE_EVAL/DFT/PRINT.html" rel="nofollow" target="_blank">PRINT</a> /
<a href="https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT/FORCE_EVAL/DFT/PRINT/EFIELD_CUBE.html" rel="nofollow" target="_blank">EFIELD_CUBE</a></span>) from cp2k. I don't see an explicit description on the units that are used in these cube files. If I take the a.u. as the default unit (hartree/e for potential, and hartree/e/bohr for electric field), the value seems pretty large. For example, I study the single water molecule system. The largest electric field at some density grid is 10~15 hartree/e/bohr, that is, ~50000 MV/cm. Please correct me if I am wrong, but the value seems unreasonable to me.</div><div>Any suggestion will be welcome, thanks in advance.</div><div>Best,</div><div>Kai<br></div>
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