[molpro-user] error in qsdpath by ccsd(t)-f12 method

Tue Apr 11 15:15:04 CEST 2017

Dear 李永方,

I have had a look at your problem. The error message says in a way that 
your transition state structure is not ok. The transition state (TS) 
structure must correspond to a first-order saddle point on the potential 
energy surface. It is a good practice to check the vibrational 
frequencies of the TS structure you found. You should get exactly one 
imaginary frequency.

Note that a good starting guess must be given otherwise the TS cannot be 
located, or you may find another TS. Note also that convergence is 
usually much more difficult to achieve than in equilibrium geometry 
optimizations.

As a test, I was trying to locate an "easier" TS for your system. I have 
also changed the basis set from cc-pVTZ to cc-pVDZ, and the method from 
CCSD(T)-F12 to Hartree-Fock, as I would like to show you only that the 
calculation can be done without any error. After a transition state 
search I found the following TS structure:

---pathts.xyz---------------------------------------------------
      6

   Cl         0.2325794338       -0.0000000002       -1.3774990539
   C         -0.8817436356        0.0000000067        0.1672343344
   H         -1.5074750431        0.8810750019       -0.0132413941
   H         -1.5074748555       -0.8810751680       -0.0132412156
   H         -0.1843022995        0.0000000857        1.3439756228
   F          0.2931635044        0.0000000004        2.3949312193
----------------------------------------------------------------

In the calculation I tightened some thresholds. Furthermore, it is more 
reliable to compute the numerical Hessian by central gradient 
differences than forward gradient differences.

The calculation was finished without any trouble. Please find attached 
the modified input file.

I hope it helps.

Regards,

Werner.

On 20.03.2017 11:57, 李永方 wrote:
> *Dear Molpro users,*
> *    I got a problem when i was doing qsdpath calculation.*
> *    I have already done the saddle search and calculated the frequences
> ,and the piont has only one imaginary vibration frequence which means *
> *it is a transation state.*
> *   But when i run the qsd module,I got this error:*
>
>   Task list generated. Total number of displacements:    156
>    15 tasks completed, CPU=1h 10m 4s Elapsed=1h 17m 34s
>    30 tasks completed, CPU=2h 20m 8s Elapsed=2h 34m 52s
>    45 tasks completed, CPU=3h 30m 11s Elapsed=3h 51m 58s
>    60 tasks completed, CPU=4h 40m 14s Elapsed=5h 9m 25s
>    75 tasks completed, CPU=5h 50m 22s Elapsed=6h 26m 39s
>    90 tasks completed, CPU=7h 0m 25s Elapsed=7h 44m 25s
>   105 tasks completed, CPU=8h 10m 30s Elapsed=9h 1m 53s
>   120 tasks completed, CPU=9h 20m 36s Elapsed=10h 19m 30s
>   135 tasks completed, CPU=10h 30m 43s Elapsed=11h 36m 58s
>   150 tasks completed, CPU=11h 40m 44s Elapsed=12h 54m 29s
>  Tasks on processor    0 finished.  CPU=  43722.84 sec, Elapsed=
> 48318.93 sec. All tasks on all processors finished.
>
>  CCSD(T)-F12 hessian saved to record  5301.2
>
>  Combined Powell-Murtagh-Sargent Update of Hessian
>
>  Quadratic Steepest Descent -  Reaction Path Following using exact Hessian
>
>  Hessian eigenvalues:    -0.033385  0.000257  0.005602  0.015645
> 0.026065  0.068585  0.128212  0.208396  0.265296  0.288874
>                           0.305995  0.560727
>
>  IDIR= 1 requested but starting point is not a stationary point.
> Gradient norm=  0.40D-04  Step length=  0.18D-02
>  ? Error
>  ? Not a stationary point
>  ? The problem occurs in QSDPATH1
>
> *So why isn't it a stationary point??*
> *Here is my input:*
>
> ***,  Reaction Path
>  memory,250,m
>  charge=-1
>  basis=vtz
>  r1=1.95891 ang
>  r2=1.12844 ang
>  r3=1.13476 ang
>  r4=2.0873  ang
>  r5=0.92512 ang
>  a1=97.80701 degree
>  a2=98.16029 degree
>  a3=138.52460 degree
>  a4=173.67715 degree
>  a5=-167.40668 degree
>  a6=102.85072 degree
>  a7=41.71448 degree
>  geometry={
>            Cl
>            C,Cl,r1
>            H1,C,r2,Cl,a1
>            H2,C,r3,Cl,a2,H1,a6
>            H3,C,r4,Cl,a3,H1,a7
>            F,H3,r5,C,a4,Cl,a5}
>  int
>  rhf
>  ccsd(t)-f12
>  optg,root=2,saveact=geo_ts,rewind            ! Find and store the TS
>  {optg,method=qsdpath,dir=1,
> numhess=5,hesscentral,saveact=geo_path}         ! find IRC in positive
> direction
>  readvar,geo_ts.act                           ! Reset geometry to TS
>  {optg,method=qsdpath,dir=-1,numhess=5,hesscentral,saveact=geo_path,append}
> !find IRC in negative direction
>
>  readvar,geo_path.act
>
>  alpha=alpha*pi/180    !convert angle to radian
>
>  table,irc,r1,r2,r3,r4,r5,a1,a2,a3,a4,a5,e_opt   !tabulate results
>
>  {table,irc,e_opt                !plot energy profile as function of irc
>   plot,file='geo_eopt.plot'}
>
>  {table,irc,r1,r2,r3,r4,r5,a1,a2,a3,a4,a5        !plot distances and
> angle as function of irc
>   plot,file='geo_dist.plot'}
>
> Please give me a hint.
> Thank you.
>
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