# LMF Tutorial: Molecular Statics in Se

This tutorial demonstrates how to carry out lattice relaxations in lmf (molecular statics).

### Command summary

LDA self-consistency (starting from init.fe)

$nano init.se$ blm --ctrl=ctrl --pbe --gmax=5.4 --nk=6 --wsitex=site --molstat init.se
$lmchk se --shell --angles$ lmfa se --basfile=basp
$lmf se$ lmf se -vminx=t --wpos=pos
$lmchk se --rpos=pos --angles  ### Introduction Se and Te crystallize in the same hexagonal structure, which has atoms per cell. The Se (Te) s state is very deep and participates only weaking in the bonding. Each atom has three neighbors at approximately 90o angles. This allows each of the three p orbital to form strong covalent bonds with each of its neighbors. Thus this system may be considered as a covalently bonded spn bonded semiconductor, with n→∞. The angle between Se atoms is not given by symmetry; it is approximately 90o, but not exactly so. Thus there a degree of freedom in the site positions not determined by symmetry. This tutorial uses molecular statics to find the minimum-energy value of this free parameter. ### Tutorial #### 1. Building the input file For the structure, copy in the contents of the box below into file init.se. Note the parameter u is the undetermined parameter the molecular statics algorithm will find by minimizing the energy. # Se and Te structures. From Wyckoff pg. 36. Space group D_3^4 (C3(_1)21). % const se=t % ifdef se # For SE % const a=8.234 cbya=1.136 u=.217 % char nam=Se % else # For Te % const a=8.406 cbya=1.330 u=.269 % char nam=Te % endif LATTICE ALAT={a} PLAT= 1 0 0 -0.5 sqrt(3)/2 0 0 0 {cbya} # This one is for Bi2Te3 SITE ATOM={nam} X= {u} 0.000 0 ATOM={nam} X= -{u} -{u} 1/3 ATOM={nam} X= 0.000 {u} 2/3  We will use the PBE functional. Enter the following command: $ blm --ctrl=ctrl --pbe --wsitex=site --molstat init.se


Switches have the following meaning:

• --ctrl=ctrl   Tells blm to write directly to the actual input file ctrl.se. By default it writes to actrl.ext so as not to overwrite the input file.
• --pbe     Use the PBE functional
• --wsitex=site  Write the site file with site positions expressed in units of PLAT
• --molstat   Set up a category for molecular statics; see below

There are two quantities you must supply yourself,  GMAX  and the k-point mesh. For the former, lmfa can determine GMAX  for you once the basis set is known. Since lmfa also supplies a basis set provided  HAM_AUTOBAS_LMTO is appropriately set, it will knows what the basis set is and tell you what GMAX  should be. Do the following

$lmfa se  At the end of the standard output you should find  FREEAT: estimate HAM_GMAX from RSMH: GMAX=5.4  This tells you what GMAX to use. For the k mesh, we will use 6 k divisions (216 points in the full Brillouin zone). Rerun both blm and lmfa, now specifying these numbers and also telling lmfa to write the basis set directly to the basp.se. By default it writes to basp0.ext so as not to overwrite whatever is already present. $ blm --ctrl=ctrl --pbe --gmax=5.4 --nk=6 --wsitex=site --molstat init.se
$lmfa se --basfile=basp  --molstat has the effect of adding the following category to the input file: % ifdef minx DYN MSTAT[MODE=6 HESS=t XTOL=.001 GTOL=0 STEP=.010 NKILL=0] NIT=10 % endif  This line will be parsed only if variable minx is defined with a nonzero value. ##### 1.1 bond lengths and angles To confirm the that bond angles are approximately 90o angles, run lmchk with the following switches $ lmchk se --shell --angles


You should see this table for bond pairs:

 Shell decomposition for class Se        class   1  z=34
shell   d     nsh csiz  class ...
1  0.000000   1    1  1:Se
2  0.533531   2    3  1:Se(2)
3  0.796025   4    7  1:Se(4)
4  0.845471   2    9  1:Se(2)
5  1.000000   6   15  1:Se(6)
...


This establishes that there are three NN bonds. For bond angles, this table should appear

 Bond angles for site 1, species 1:Se
shl1    d1    shl2    d2     cl1      cl2       angle(s) ...
1  0.533531   1  0.533531  Se       Se        104.81
1  0.533531   2  0.796025  Se       Se         95.52   95.52   99.86  159.67


Angles between the NN Se atoms is somwhat larger 105o, somewhat larger than the simple nearest-neighbor tight-binding model would suggest.

#### 2. Self-consistency

Make the system self-consistent with:

$lmf se  If you have compiled it in parallel mode you can do it a bit faster, e.g. $ mpirun -n 6 lmf se


lmf should converge to self-consistency in 9 iterations.

You should see this table at the end:

 Forces, with eigenvalue correction
ib           estatic                  eigval                    total
1  970.90    0.00    0.00  -917.24    0.00    0.00    53.67    0.00    0.00
2 -485.45 -840.83    0.00   458.62  794.35    0.00   -26.83  -46.47    0.00
3 -485.45  840.83    0.00   458.62 -794.35    0.00   -26.83   46.47    0.00
Maximum Harris force = 53.7 mRy/au (site 2)  Max eval correction = 5.8


The forces are all equal in magnitude, but rotated by 120o angles.

Note: Unlike the total energy forces are not variational in the density; thus they are more difficult to converge.

That is, the forces do not converge to their self-consistent value as $O(n-n^*)^2$, where $n$ and $n^*$ are the current and self-consistent densities. Rather it converges as $O(n{-}n^*)$. However, an estimate for the correction can be performed. One way is to find a contribution to the force that would arise if the free-atomic density were dragged along with the nucleus. That is the ansatz used here (this ansatz is used when HAM_FORCE=1). An alternative is to use HAM_FORCE=12.

Inspect the output from the first iteration. You should find this table

 Harris correction to forces: shift in free-atom density
ib         delta-n dVes             delta-n dVxc               total
1 -846.44    0.00    0.00   -30.21    0.00    0.00   876.65    0.00    0.00
2  423.22  733.04    0.00    15.11   26.16    0.00  -438.32 -759.20    0.00
3  423.22 -733.04    0.00    15.11  -26.16    0.00  -438.32  759.20    0.00
shift forces to make zero average correction:            0.00    0.00    0.00


It says that the correction term to the force is about 880 mRy/au. This is a very large term!

The correction table is soon followed by

 Forces, with eigenvalue correction
ib           estatic                  eigval                    total
1  152.09    0.00    0.00  -118.67    0.00    0.00    33.42    0.00    0.00
2  -76.05 -131.72    0.00    59.34  102.77    0.00   -16.71  -28.94    0.00
3  -76.05  131.72    0.00    59.34 -102.78    0.00   -16.71   28.94    0.00
Maximum Harris force = 33.4 mRy/au (site 1)  Max eval correction = 876.6


Without the correction term, the would be 33.42−876.65 mRy/au. But with this ansatz, the force is only 33.42 mRy/a.u.. So the correction term is huge — much larger than the force itself. The corrected force, even from the initial [Mattheis construction]/tutorial/lmf/si_calculation/#self-consistency), is not so far from the self-consistent one (see below).

Now look at the equivalent output from the last iteration. You should find

 Harris correction to forces: shift in free-atom density
ib         delta-n dVes             delta-n dVxc               total
1   -5.64    0.00    0.00    -0.14    0.00    0.00     5.78    0.00    0.00
2    2.82    4.88    0.00     0.07    0.12    0.00    -2.89   -5.00    0.00
3    2.82   -4.88    0.00     0.07   -0.12    0.00    -2.89    5.00    0.00
shift forces to make zero average correction:            0.00    0.00    0.00


The correction term has become very small; indeed the tighter the tolerance the density is converged, the smaller this term becomes.

Look for the table shown below in the output of the last iteration, when self-consistency has been reached. It should read:

 Forces, with eigenvalue correction
ib           estatic                  eigval                    total
1  970.90    0.00    0.00  -917.24    0.00    0.00    53.67    0.00    0.00
2 -485.45 -840.83    0.00   458.62  794.35    0.00   -26.83  -46.47    0.00
3 -485.45  840.83    0.00   458.62 -794.35    0.00   -26.83   46.47    0.00
Maximum Harris force = 53.7 mRy/au (site 2)  Max eval correction = 5.8


#### 3. Molecular statics

lmf can relax the site positions to their equilibrium point where the forces vanish (molecular statics).

The parameters that control how this relaxation proceeds are given in as tokens in the DYN_MSTAT tag. Notice the last lines of the input file. To see whether the proprocessor includes this line or not, compare the output of the following two commands.

$lmf se --showp$ lmf se -vminx=t --showp


In the second command the preprocessor includes the DYN category in the input file.

Note:: you should consider values for tokens in DYN_MSTAT. Some default values were chosen, but it is very difficult to make a general algorithm that relaxes atoms. In this case it uses a Broyden scheme (a kind of Newton-Raphson algorithm), with an initial step size of 0.01. This a good choice in the present context, but in other contexts it may not be.

To relax the structure, do

$lmf se -vminx=t --wpos=pos  lmf iterates the density to self-consistency; when it is reached it shifts the nuclear positions using the Broyden scheme. At the end of the first self-consistency cycle you should see  RELAX: (warning) failed to read Hessian; set to unity gradzr: begin Broyden xtol=1e-3 dxmx=0.01 isw=211 brmin: start xtol=|1e-3| isw=15 |g|=0.0931 brmin: max shift = 0.05376 is larger than dxmx. Scale by 0.1860 brmin: iter 1 max shift=0.01 |g|=0.09312 max g=0.05376 Updated x: 0.227000 0.000000 0.000000 -0.113500 -0.196588 0.378667 -0.113500 0.196588 -0.378667 RELAX: completed Broyden iter 1; max shift=0.01000 |g|=0.0931 Gradients: -0.054 -0.000 -0.000 0.027 0.047 0.000 0.027 -0.047 0.000 Diagonal inverse Hessian: 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 Updated atom positions: Site Class Position(relaxed) 1 Se 0.22700000(T) 0.00000000(T) 0.00000000(T) 2 Se -0.11350000(T) -0.19658777(T) 0.37866663(T) 3 Se -0.11350000(T) 0.19658777(T) -0.37866663(T)  The x coordinate of the first Se atom changed by 0.01 (this was controlled by STEP=.010, which limits the largest allowed change in any component). Next follows this table:  smshft: add shifted free atom densities site old pos new pos shift 1:Se 0.21700 0.00000 0.00000 0.22700 0.00000 0.00000 0.010000 2:Se -0.10850 -0.18793 0.37867 -0.11350 -0.19659 0.37867 0.010000 3:Se -0.10850 0.18793 -0.37867 -0.11350 0.19659 -0.37867 0.010000  Rather than build the density from a Mattheis construction, it takes the self-consistent density, adds a Mattheis construction calculated from the new positions and subtracts the same construction built from the starting positions. This provides a reasonable ansatz for the density at the new positions. When self-consistency is reached with these positions the forces read:  Forces, with eigenvalue correction ib estatic eigval total 1 899.40 0.00 0.00 -879.01 0.00 0.00 20.39 0.00 0.00 2 -449.70 -778.90 0.00 439.50 761.24 0.00 -10.20 -17.66 0.00 3 -449.70 778.90 0.00 439.50 -761.24 0.00 -10.20 17.66 0.00 Maximum Harris force = 20.4 mRy/au (site 2) Max eval correction = 1.4  They are about half as large as the starting force. At the close of the fourth relaxation step the forces read  Forces, with eigenvalue correction ib estatic eigval total 1 835.61 0.00 0.00 -835.12 0.00 0.00 0.48 0.00 0.00 2 -417.80 -723.65 0.00 417.56 723.24 0.00 -0.24 -0.42 0.00 3 -417.80 723.65 0.00 417.56 -723.24 0.00 -0.24 0.42 0.00 Maximum Harris force = 0.483 mRy/au (site 2) Max eval correction = 1.1  This is indeed a very small force. The final site positions are printed to stdout and also to file pos.se (because of --wpos=pos). The x coordinate of the first atom in pos.se should be 0.2346. Note that the coordinates last output by lmf are slightly different: these correspond to what would be the updated coordinates if the molecular statics would proceed another step. It is interesting to see how the total energy changes with each relaxation step. Do the following: $ grep ^c save.se


You should see

c minx=1 ehf=-14580.5132479 ehk=-14580.5132302
c minx=1 ehf=-14580.5221027 ehk=-14580.5220784
c minx=1 ehf=-14580.523951 ehk=-14580.5239234
c minx=1 ehf=-14580.524057 ehk=-14580.5240287


The energy drops monotonically, with a large change after the first relaxation step, and almost no change at the last step. This is because the energy stops changing as the forces go to zero.

You can see how the Se-Se bond length and angles change on relaxation:

\$ lmchk se --rpos=pos --angles


You should find that the bond angle drops to 101o, and the bond length increases a few percent.

Note also the bandgap

 VBmax = 0.136522  CBmin = 0.201860  gap = 0.065338 Ry = 0.88860 eV


The experimental gap is about 1.8 eV.