Dear Stefano, Thanks for your reply. I still have a couple of unresolved issues.
Silicon has an inversion center about (a/8, a/8, a/8) right? So shouldn't the force constants also have an inversion symmetry? i.e., Phi(na=1,nb=1)_xy(R=a1) == Phi(nb=1,na=1)_xy(R=-a1) apart from Phi(na=1,nb=1)_yx(R=a1) == Phi(nb=1,na=1)_xy(R=-a1). So Phi(na=1,nb=1)_yx(R=a1) == Phi(nb=1,na=1)_xy(R=a1)? I also found another problem with my computations. The force constants matrix between atoms at (0, 0, 0) and (a/4, a/4, a/4) and also the other 3 equivalent points around (0, 0, 0) which make the first nearest neighbour shell should be of the form Phi_xx = a; Phi_xy = +/- b; Phi_xz = +/- b; Phi_yx = +/- b; Phi_yy = a; Phi_yz = +/- b; Phi_zx = +/- b; Phi_zy = +/- b; Phi_zz = a; right? In the force constants file I had sent to the forum earlier, this is not the case. In fact, there are 4 matrices of this type but are associated with other atomic positions. I have this problem only with the second atom in the basis. The force constants in connecting the first atoms in the basis Phi(nb=1,na=1) seem to have the expected symmetries. For brevity, I am attaching a force constants file with 3X3X3 grid. To point out one such example, please have a look at the following section of the force constants file (Format : Line number, tau_1, tau_2, tau_3, Phi_xx(na=1, nb=2) ): * 47 1 1 1 -6.84986974074E-02* 48 2 1 1 1.06531878102E-04 49 3 1 1 1.11977775153E-03 * 50 1 2 1 -6.84986969382E-02* 51 2 2 1 -6.67371571926E-04 *52 3 2 1 -6.84986979674E-02* 53 1 3 1 -7.21441565439E-03 54 2 3 1 -6.67372403004E-04 55 3 3 1 1.11977897933E-03 56 1 1 2 1.06531878102E-04 57 2 1 2 6.18314184023E-04 58 3 1 2 1.11977814815E-03 59 1 2 2 -6.67371571926E-04 60 2 2 2 -7.21441521994E-03 61 3 2 2 1.11977888383E-03 62 1 3 2 -6.67372403004E-04 63 2 3 2 6.18311460290E-04 64 3 3 2 6.18312968022E-04 65 1 1 3 1.11977775153E-03 66 2 1 3 1.11977814815E-03 67 3 1 3 6.72231001162E-04 *68 1 2 3 -6.84986979674E-02* 69 2 2 3 1.11977888383E-03 70 3 2 3 1.06531502673E-04 71 1 3 3 1.11977897933E-03 72 2 3 3 6.18312968022E-04 73 3 3 3 1.06533738456E-04 Line numbers 47, 50, 52 and 68 (highlighted in bold font) seem to have the same value (a). However, the lattice points corresponding to these lines are not at the locations where the first nearest neighbours are located for diamond lattice (Si). From my understanding, these lattice vectors correspond to (0 0 0), (0 a/2 a/2), (-a/2 0 a/2) and (-a/2 a/2 0), but they should have been (0, 0, 0), (-a/2, -a/2, 0), (0, -a/2, -a/2) and (-a/2, 0, a/2). I have checked this with 3X3X3 to 8X8X8 grid, the problem persists. It is quite possible that I have misunderstood the syntax. I would appreciate it if you or someone else in the forum could clarify my questions. Thanks for your time! --Navaneeth On Tue, Nov 11, 2014 at 2:53 AM, stefano de gironcoli <[email protected]> wrote: > dear Navaneeth, > as noted previously IFC should be symmetric w.r.t exchange of the two > atoms. this include the vector R that joins them. > in your case IFC Phi(na=1,nb=1)_yx(R=a1) == > Phi(nb=1,na=1)_xy(R=-a1) because when you exchange the two atoms you need > to invert the connecting vector as well. > indeed line 280 is equal to line 802 and line 282 is equal to 800 as > they should. > > stefano > > > > On 11/11/2014 08:23 AM, Navaneetha Krishnan wrote: > > Hi Stefano, > > Thanks for your reply. I am particularly looking to address an anomaly in > my simulation. I've attached my scf input, ph input and q2r input files > along with the generated force constant file. > > The force constants are not symmetric (in fact anti-symmetric, for > certain direction pairs) with respect to the displacement directions. For > eg., if you look at line number 280 and line 800, the fcs are supposed to > be equal as they are Phi_{XY} and Phi_{YX} for the same pair of atoms. I > have tried this with several combinations of k-sampling and Q-sampling and > I get the same anti-symetry (except when no. of Q's is very small, like > Q=2). I also used a large ecutwfc so that there is no convergence issue. > This problem was not there with Al lattice. The force constants were > symmetric in this sense. Is there bug in the code or am I making a mistake > in one of the input files? These are not negligible IFCs. Please have a > look. > > Thanks in advance! > > On Mon, Nov 10, 2014 at 2:44 AM, stefano de gironcoli <[email protected]> > wrote: > >> the IFC should be symmetric w.r.t. exchange of the two atoms and >> displacement directions as a consequence of being second order derivatives. >> other symmetries may apply in specific cases. >> difficult to comment more without more info on what you are looking for. >> stefano >> >> >> >> On 11/10/2014 09:16 AM, Navaneetha Krishnan wrote: >> >> Hello QE users, >> >> I was investigating the IFCs generated by q2r.x after scf and ph >> calculations. It looks like the force constant matrix for certain pairs of >> atoms is not symmetric (with respect to the cartesian coordinates). For >> silicon, the IFC matrices have to be symmetric with respect to the >> cartesian coordinates - right? For aluminum, the output is symmetric, as >> expected. Any thoughts? >> >> Thanks a lot! >> --Navaneeth >> Caltech. >> >> >> _______________________________________________ >> Pw_forum mailing >> [email protected]http://pwscf.org/mailman/listinfo/pw_forum >> >> >> >> _______________________________________________ >> Pw_forum mailing list >> [email protected] >> http://pwscf.org/mailman/listinfo/pw_forum >> > > > > -- > Navaneetha Krishnan Ravichandran, > Graduate Student, > Mechanical Engineering, > Caltech. > > > _______________________________________________ > Pw_forum mailing > [email protected]http://pwscf.org/mailman/listinfo/pw_forum > > > > _______________________________________________ > Pw_forum mailing list > [email protected] > http://pwscf.org/mailman/listinfo/pw_forum > -- Navaneetha Krishnan Ravichandran, Graduate Student, Mechanical Engineering, Caltech.
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