<div>Dear all,</div><div> </div><div>I have a few questions about computing the third order coupling tensor implemented by d3.x code. I have gone through all previous messages about this similar content, as well as the suggested papers. Basically, from example14, I have known how to use pw.x, ph.x and d3.x to get the result, but the output is still confusing to me. </div>
<div> </div><div>Question1: Take the example14 for example, I think si. anh_G file contains the third order derivative, in the example output it looks like this</div><div>------------------------------------------------------------------------------</div>
<div>Third derivative in Cartesian axes</div><div> q = ( 0.000000000 0.000000000 0.000000000 )</div><div><br> modo: 1</div><div> 1 1<br> 0.000000000000E+00 0.000000000000E+00 0.000000000000E+00<br>
0.000000000000E+00 -0.138777878078E-16 0.000000000000E+00<br> 0.416333634234E-16 0.000000000000E+00 0.277555756156E-16<br> 0.000000000000E+00 0.383235688562E+00 0.000000000000E+00<br>
-0.693889390391E-16 0.000000000000E+00 0.383235688562E+00<br> 0.000000000000E+00 -0.693889390391E-16 0.000000000000E+00<br> 1 2<br> 0.555111512313E-16 0.000000000000E+00 -0.138777878078E-16<br>
0.000000000000E+00 0.000000000000E+00 0.000000000000E+00<br> 0.416333634234E-16 0.000000000000E+00 -0.693889390391E-16<br> 0.000000000000E+00 -0.385638438801E+00 0.000000000000E+00<br>
0.138777878078E-16 0.000000000000E+00 -0.385638438801E+00<br> 0.000000000000E+00 0.000000000000E+00 0.000000000000E+00<br> 2 1<br> -0.832667268469E-16 0.000000000000E+00 -0.416333634234E-16<br>
0.000000000000E+00 0.277555756156E-16 0.000000000000E+00<br> -0.277555756156E-16 0.000000000000E+00 0.138777878078E-16<br> 0.000000000000E+00 -0.385638438801E+00 0.000000000000E+00<br>
-0.971445146547E-16 0.000000000000E+00 -0.385638438801E+00<br> 0.000000000000E+00 0.277555756156E-16 0.000000000000E+00<br> 2 2<br> -0.138777878078E-16 0.000000000000E+00 0.000000000000E+00<br>
0.000000000000E+00 -0.138777878078E-16 0.000000000000E+00<br> -0.277555756156E-16 0.000000000000E+00 -0.277555756156E-16<br> 0.000000000000E+00 0.385638438801E+00 0.000000000000E+00<br>
0.138777878078E-16 0.000000000000E+00 0.385638438801E+00<br> 0.000000000000E+00 -0.138777878078E-16 0.000000000000E+00</div><div> modo: 2<br>........................</div><div>......................</div>
<div>---------------------------------------------------------------</div><div> </div><div>According to the theory, the anharmonic tensor should be something like C_xyz(q1=0,j1;q2=0,j2;q3=0,j3|s1,s2,s3) in this particular case, where xyz is cartesian indices, j's branch number, s1,s2,s3 atom basis index in a cell. </div>
<div> </div><div>I guess modo1 means j1=1, i.e, the first mode for q1=0, and every q2 and q3 should each have 6 modes. Then why in the above matrices, their entries are 6 by 3? Also how do I understand the 1 1, 1 2, 2 1, 2 2 indices on top of these matrices, I bet they correspond to the basis index of a unit cell. </div>
<div> </div><div>Question2: This question is similar to the first one. I read the README in example14 as below, I am not sure about the arguments of C's, such as C_{x,y,z} (0,X,-X|1,1,1), does 1 1 1 means atom basis index? </div>
<div> </div><div>----------------------------------------------------------</div><div>By displacing one atom, one can get also these coefficients by a<br>finite-difference mathod. We give first the values obtained by<br>the 2n+1 method, then the values by the finite-differences.<br>
All units are in Ryd/(a_b)^3.</div><div> tensor | 2n+1 | fin. dif.<br>------------------------------------------------------------<br>C_{x,y,z} (0,0,0|1,1,1) | 0.38314 | 0.38446<br>------------------------------------------------------------<br>
C_{x,y,z} (0,X,-X|1,1,1) | 0.34043 | 0.34109<br>C_{x,x,z} (0,X,-X|1,1,2) | -0.25316 | -0.25296<br>C_{z,x,y} (0,X,-X|1,1,1) | 0.35781 | 0.35767<br>C_{z,x,x} (0,X,-X|1,1,2) | -0.25706 | -0.25491<br>
C_{z,z,z} (0,X,-X|1,1,2) | -0.13133 | -0.12813<br></div><div>---------------------------------------------------------- <br clear="all"></div><div>Question3: using DFPT, we can get the third order derivative of energy w.r.t phonon displacement u(q), but I think in order to compute the lifetime, the polarization vectors e1, e2, e3 must be multiplied, as mentioned in many papers including Dr. Debernardi's thesis, Eq 2.9. However, I don't see any comment in the code. So have those polarization vectors already been added in the d3.x output?</div>
<div> </div><div>In all, I am confused at this point, because I cannot match those output to my general understanding of phonon anharmonic coupling. I know this is a specialized topic, and I only understand the minimum, hopefully I can get some help from some professionals. Thank you in advance !</div>
<div><br>-- <br>Lan, Tian<br>
Ph.D. Candidate, Department of Applied Physics and Materials Science<br>
California Institute of Technology,<br>
Caltech M/C 138-78, Pasadena, CA, 91125<br>
</div>