The input file used here is tests/Arn2_test.inp in the Hibridon directory tree
Initiate execution of your code
% hib_arn2_251
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HIBRIDON SCATTERING CODE V 4.0 03/18/96 18:34:31 EST
AUTHORS: M. ALEXANDER, D. MANOLOPOULOS, H.-J. WERNER, B. FOLLMEG
CONTRIBUTORS: D. LEMOINE, P. VOHRALIK, G. COREY, R. JOHNSON, T. ORLIKOWSKI
A. BERNING, A. DEGLI-ESPOSTI, C. RIST, P. DAGDIGIAN, B. POUILLY
G. VAN DER SANDEN, M. YANG
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Hibridon> inp=arn2_test.inp
Hibridon> show
*** Parameters:
JTOT1 = 20 JTOT2 = 20 JTOTD = 5 JLPAR = 1
NERG = 1 NUMAX = 0 NUMIN = 0 NUD = 1
LSCREEN= 48 IPRINT = 0
FSTFAC = 15.00 RINCR = 3.000 RCUT = 30.00 RENDAI = 25.00
RENDLD = 8.000 RSTART = 5.600 SPAC = .1500 TOLAI = 1.150
XMU = 16.47
NOUT: 4; JOUT: 0 2 4 6
INDOUT: 0
*** 1-SIGMA system parameters:
NTERM = 1 VMIN = 0 VMAX = 0 JMIN = 0
JMAX = 4
BROT = 2.010 DROT = 0.0000E+00 HROT = 0.0000E+00 EVIB = 0.0000E+00
LAMMIN: 2
LAMMAX: 2
MPROJ: 0
*** Flags:
AIRYFL= T BASTST= F BATCH = F CHLIST= T CSFLAG= F FLAGHF= F
FLAGSU= F IHOMO = T IPOS = F LOGDFL= T NOPRIN= F NUCROS= F
PHOTOF= F PRAIRY= F PRLOGD= T PRPART= F PRSMAT= T PRT2 = T
PRXSEC= F READPT= F RSFLAG= F T2TEST= F TWOMOL= F WAVEFL= F
WRPART= F WRSMAT= F WRXSEC= F
** Maximum Channels: 251; Anisotropic Terms: 80
** Energies: 500.000000
** Label: N2-Ar CC PATTENGILL POTENTIAL
** Pot name: PATTENGILL-LABUDDE-BERNSTEIN AR-N2
** Input File: Arn2_test.inp
** Output file: Outpt
** Jobname: Job
Here we will try to optimize the parameter SPAC. A good initial
guess is 1/5 the deBroglie wavelength. To estimate this we run
the command DEBROGLI
Hibridon> debrog
de Broglie wavelength = .537 bohr / 5 = .107
wavevector = 11.696 Bohr^-1 = 22.101 Angstroms^-1
so we will an initial value of SPAC of 0.6, and keep reducing it by
multiplying by 0.8, with a minimum values of 0.02. Check the OPTIMIZE
help file for more information on the command syntax
Hibridon> opt,spac,.6,.02,.8
Flag WRSMAT set to .TRUE. for optimization
Flag NOPRIN set to .TRUE. for optimization
Optimization of SPAC for Jtot = 20
Start: .600 End: .020 Factor: .80 Increment: .000
Average error limit: 1.0%
Maximum error limit: 5.0%
Threshold for S-matrix elements: .10E-04
SPAC = .600
WARNING *** WRSMAT = .TRUE. BUT JTOTD= 5 .NE. 1
NOPRIN = .TRUE., SO IPRINT SET TO -1
** J = 20 JLPAR = 1 FINISHED; CPU: 00:00:00.01 WALL: 00:00:00.03 DATE: 18-Mar-96 22:14:12
SPAC = .480
WARNING *** WRSMAT = .TRUE. BUT JTOTD= 5 .NE. 1
NOPRIN = .TRUE., SO IPRINT SET TO -1
** J = 20 JLPAR = 1 FINISHED; CPU: 00:00:00.01 WALL: 00:00:00.03 DATE: 18-Mar-96 22:14:12
average difference between old and new S modulus = 43.21%
Largest difference between old and new S modulus = 157.14%
in (i = 9 j = 2) element of S modulus
SPAC = .384
WARNING *** WRSMAT = .TRUE. BUT JTOTD= 5 .NE. 1
NOPRIN = .TRUE., SO IPRINT SET TO -1
** J = 20 JLPAR = 1 FINISHED; CPU: 00:00:00.00 WALL: 00:00:00.03 DATE: 18-Mar-96 22:14:12
average difference between old and new S modulus = 34.86%
Largest difference between old and new S modulus = 133.88%
in (i = 9 j = 2) element of S modulus
SPAC = .307
WARNING *** WRSMAT = .TRUE. BUT JTOTD= 5 .NE. 1
NOPRIN = .TRUE., SO IPRINT SET TO -1
** J = 20 JLPAR = 1 FINISHED; CPU: 00:00:00.02 WALL: 00:00:00.04 DATE: 18-Mar-96 22:14:12
average difference between old and new S modulus = 29.78%
Largest difference between old and new S modulus = 99.97%
in (i = 9 j = 9) element of S modulus
SPAC = .246
WARNING *** WRSMAT = .TRUE. BUT JTOTD= 5 .NE. 1
NOPRIN = .TRUE., SO IPRINT SET TO -1
** J = 20 JLPAR = 1 FINISHED; CPU: 00:00:00.00 WALL: 00:00:00.03 DATE: 18-Mar-96 22:14:12
average difference between old and new S modulus = 19.88%
Largest difference between old and new S modulus = 56.73%
in (i = 8 j = 8) element of S modulus
SPAC = .197
WARNING *** WRSMAT = .TRUE. BUT JTOTD= 5 .NE. 1
NOPRIN = .TRUE., SO IPRINT SET TO -1
** J = 20 JLPAR = 1 FINISHED; CPU: 00:00:00.03 WALL: 00:00:00.04 DATE: 18-Mar-96 22:14:13
average difference between old and new S modulus = 4.58%
Largest difference between old and new S modulus = 28.40%
in (i = 6 j = 6) element of S modulus
SPAC = .157
WARNING *** WRSMAT = .TRUE. BUT JTOTD= 5 .NE. 1
NOPRIN = .TRUE., SO IPRINT SET TO -1
** J = 20 JLPAR = 1 FINISHED; CPU: 00:00:00.03 WALL: 00:00:00.04 DATE: 18-Mar-96 22:14:13
average difference between old and new S modulus = .36%
Largest difference between old and new S modulus = 2.23%
in (i = 6 j = 6) element of S modulus
optimized value for SPAC = .1573
average difference in old and new S modulus is .36%
Largest difference in old and new S modulus is 2.23%
in (i = 6 j = 6) element of S modulus
thus we see that 0.16 is a good value for SPAC
Now let us try to optimize the endpoint of the integration, RENDAI. We will
do so by starting at 10 and stepping outward 2 bohrs at a time
Hibridon> opt,rendai,10,30,1,2
Optimization of RENDAI for Jtot = 20
Start: 10.000 End: 30.000 Factor: 1.00 Increment: 2.000
Average error limit: 1.0%
Maximum error limit: 5.0%
Threshold for S-matrix elements: .10E-04
RENDAI = 10.000
WARNING *** WRSMAT = .TRUE. BUT JTOTD= 5 .NE. 1
NOPRIN = .TRUE., SO IPRINT SET TO -1
** J = 20 JLPAR = 1 FINISHED; CPU: 00:00:00.02 WALL: 00:00:00.03 DATE: 18-Mar-96 22:27:29
RENDAI = 12.000
WARNING *** WRSMAT = .TRUE. BUT JTOTD= 5 .NE. 1
NOPRIN = .TRUE., SO IPRINT SET TO -1
** J = 20 JLPAR = 1 FINISHED; CPU: 00:00:00.01 WALL: 00:00:00.04 DATE: 18-Mar-96 22:27:29
average difference between old and new S modulus = 1.33%
Largest difference between old and new S modulus = 11.70%
in (i = 6 j = 6) element of S modulus
RENDAI = 14.000
WARNING *** WRSMAT = .TRUE. BUT JTOTD= 5 .NE. 1
NOPRIN = .TRUE., SO IPRINT SET TO -1
** J = 20 JLPAR = 1 FINISHED; CPU: 00:00:00.01 WALL: 00:00:00.04 DATE: 18-Mar-96 22:27:30
average difference between old and new S modulus = 1.15%
Largest difference between old and new S modulus = 3.87%
in (i = 2 j = 2) element of S modulus
RENDAI = 16.000
WARNING *** WRSMAT = .TRUE. BUT JTOTD= 5 .NE. 1
NOPRIN = .TRUE., SO IPRINT SET TO -1
** J = 20 JLPAR = 1 FINISHED; CPU: 00:00:00.02 WALL: 00:00:00.03 DATE: 18-Mar-96 22:27:30
average difference between old and new S modulus = .35%
Largest difference between old and new S modulus = 1.23%
in (i = 8 j = 8) element of S modulus
optimized value for RENDAI = 16.00
average difference in old and new S modulus is .35%
Largest difference in old and new S modulus is 1.23%
in (i = 8 j = 8) element of S modulus
we conclude that RENDAI = 15 is probably sufficient
Hibridon> quit
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