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Example in C
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@ -8,6 +8,9 @@ in a [[https://github.com/TREX-CoE/trexio][TREXIO]] file.
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* Python
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** Check numerically that MOs are orthonormal
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:PROPERTIES:
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:header-args: :tangle mo_ortho.py
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:END:
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In this example, we will compute numerically the overlap
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between the molecular orbitals:
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@ -127,7 +130,7 @@ before = time.time()
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mo_value = qmckl.get_mo_basis_mo_value(context, point_num*mo_num)
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after = time.time()
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mo_value = np.reshape( mo_value, (point_num, mo_num) )
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mo_value = np.reshape( mo_value, (point_num, mo_num) ).T # Transpose to get mo_num x point_num
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print("Number of MOs: ", mo_num)
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print("Number of grid points: ", point_num)
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@ -138,14 +141,14 @@ print("Execution time : ", (after - before), "seconds")
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#+begin_example
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Number of MOs: 201
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Number of grid points: 1728000
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Execution time : 3.511528968811035 seconds
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Execution time : 5.577778577804565 seconds
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#+end_example
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and finally we compute the overlap between all the MOs as
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$M^\dagger M$.
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$M.M^\dagger$.
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#+begin_src python :exports code
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overlap = mo_value.T @ mo_value * dr
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overlap = mo_value @ mo_value.T * dr
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print (overlap)
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#+end_src
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@ -167,6 +170,9 @@ print (overlap)
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* C
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** Check numerically that MOs are orthonormal, with cusp fitting
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:PROPERTIES:
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:header-args: :tangle mo_ortho.c
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:END:
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In this example, we will compute numerically the overlap
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between the molecular orbitals:
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@ -188,6 +194,9 @@ print (overlap)
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#+begin_src c :exports code
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#include <qmckl.h>
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#include <stdio.h>
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#include <string.h>
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//#include <time.h>
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#include <sys/time.h>
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int main(int argc, char** argv)
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{
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@ -201,10 +210,6 @@ int main(int argc, char** argv)
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#+begin_src c :exports code
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qmckl_context context = qmckl_context_create();
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if (context == NULL) {
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fprintf(stderr, "Error creating context\n");
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exit(1);
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}
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rc = qmckl_trexio_read(context, trexio_filename, strlen(trexio_filename));
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@ -221,7 +226,7 @@ int main(int argc, char** argv)
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To identify which atom is an oxygen and which are hydrogens, we
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fetch the nuclear charges from the context.
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#+begin_src python :exports code
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#+begin_src c :exports code
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int64_t nucl_num;
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rc = qmckl_get_nucleus_num(context, &nucl_num);
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@ -234,7 +239,7 @@ int main(int argc, char** argv)
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double nucl_charge[nucl_num];
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rc = qmckl_get_nucleus_charge(context, &(nucl_charge[0]));
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rc = qmckl_get_nucleus_charge(context, &(nucl_charge[0]), nucl_num);
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if (rc != QMCKL_SUCCESS) {
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fprintf(stderr, "Error getting nucl_charge:\n%s\n", qmckl_string_of_error(rc));
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@ -242,28 +247,43 @@ int main(int argc, char** argv)
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}
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double r_c[nucl_num];
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double r_cusp[nucl_num];
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for (size_t i=0 ; i<nucl_num ; ++i) {
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switch ((int) nucl_charge[i]) {
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case 1:
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nucl_charge[i] = 0.5;
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break;
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case 8:
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nucl_charge[i] = 0.1;
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break;
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}
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switch ((int) nucl_charge[i]) {
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case 1:
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r_cusp[i] = 0.5;
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break;
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case 8:
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r_cusp[i] = 0.1;
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break;
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}
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}
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rc = qmckl_set_mo_basis_r_cusp(context, &(r_cusp[0]), nucl_num);
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if (rc != QMCKL_SUCCESS) {
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fprintf(stderr, "Error setting r_cusp:\n%s\n", qmckl_string_of_error(rc));
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exit(1);
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}
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#+end_src
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#+begin_src python :exports code
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We now define the grid points $\mathbf{r}_k$ as a regular grid around the
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molecule.
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We fetch the nuclear coordinates from the context,
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#+begin_src c :exports code
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double nucl_coord[nucl_num][3];
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rc = qmckl_get_nucleus_coord(context, 'N', &(nucl_coord[0][0]), nucl_num*3)
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rc = qmckl_get_nucleus_coord(context, 'N', &(nucl_coord[0][0]), nucl_num*3);
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if (rc != QMCKL_SUCCESS) {
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fprintf(stderr, "Error getting nucl_coord:\n%s\n", qmckl_string_of_error(rc));
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@ -271,10 +291,12 @@ int main(int argc, char** argv)
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}
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for (size_t i=0 ; i<nucl_num ; ++i) {
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printf("%d %+f %+f %+f\n", (int32_t) nucl_charge[i],
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nucl_coord[i][0],
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nucl_coord[i][1],
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nucl_coord[i][2]) );
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printf("%d %+f %+f %+f\n",
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(int32_t) nucl_charge[i],
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nucl_coord[i][0],
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nucl_coord[i][1],
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nucl_coord[i][2]);
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}
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#+end_src
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#+begin_example
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@ -283,103 +305,157 @@ int main(int argc, char** argv)
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1 +1.430429 +0.000000 -1.107157
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#+end_example
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We now define the grid points $\mathbf{r}_k$ as a regular grid around the
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molecule.
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We fetch the nuclear coordinates from the context,
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and compute the coordinates of the grid points:
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#+begin_src python :exports code
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nx = ( 120, 120, 120 )
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shift = np.array([5.,5.,5.])
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point_num = nx[0] * nx[1] * nx[2]
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#+begin_src c :exports code
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size_t nx[3] = { 120, 120, 120 };
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double shift[3] = {5.,5.,5.};
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int64_t point_num = nx[0] * nx[1] * nx[2];
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rmin = np.array( list([ np.min(nucl_coord[:,a]) for a in range(3) ]) )
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rmax = np.array( list([ np.max(nucl_coord[:,a]) for a in range(3) ]) )
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double rmin[3] = { nucl_coord[0][0], nucl_coord[0][1], nucl_coord[0][2] } ;
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double rmax[3] = { nucl_coord[0][0], nucl_coord[0][1], nucl_coord[0][2] } ;
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for (size_t i=0 ; i<nucl_num ; ++i) {
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for (int j=0 ; j<3 ; ++j) {
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rmin[j] = nucl_coord[i][j] < rmin[j] ? nucl_coord[i][j] : rmin[j];
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rmax[j] = nucl_coord[i][j] > rmax[j] ? nucl_coord[i][j] : rmax[j];
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}
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}
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linspace = [ None for i in range(3) ]
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step = [ None for i in range(3) ]
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for a in range(3):
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linspace[a], step[a] = np.linspace(rmin[a]-shift[a],
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rmax[a]+shift[a],
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num=nx[a],
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retstep=True)
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rmin[0] -= shift[0]; rmin[1] -= shift[1]; rmin[2] -= shift[2];
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rmax[0] += shift[0]; rmax[1] += shift[1]; rmax[2] += shift[2];
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dr = step[0] * step[1] * step[2]
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double step[3];
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double* linspace[3];
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for (int i=0 ; i<3 ; ++i) {
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linspace[i] = (double*) calloc( sizeof(double), nx[i] );
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if (linspace[i] == NULL) {
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fprintf(stderr, "Allocation failed (linspace)\n");
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exit(1);
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}
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step[i] = (rmax[i] - rmin[i]) / ((double) (nx[i]-1));
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for (size_t j=0 ; j<nx[i] ; ++j) {
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linspace[i][j] = rmin[i] + j*step[i];
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}
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}
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double dr = step[0] * step[1] * step[2];
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#+end_src
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#+RESULTS:
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Now the grid is ready, we can create the list of grid points
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$\mathbf{r}_k$ on which the MOs $\phi_i$ will be evaluated, and
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transfer them to the QMCkl context:
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#+begin_src python :exports code
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point = []
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for x in linspace[0]:
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for y in linspace[1]:
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for z in linspace[2]:
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point += [ [x, y, z] ]
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#+begin_src c :exports code
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double* point = (double*) calloc(sizeof(double), 3*point_num);
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point = np.array(point)
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point_num = len(point)
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qmckl.set_point(context, 'N', point_num, np.reshape(point, (point_num*3)))
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if (point == NULL) {
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fprintf(stderr, "Allocation failed (point)\n");
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exit(1);
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}
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size_t m = 0;
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for (size_t i=0 ; i<nx[0] ; ++i) {
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for (size_t j=0 ; j<nx[1] ; ++j) {
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for (size_t k=0 ; k<nx[2] ; ++k) {
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point[m] = linspace[0][i];
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m++;
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point[m] = linspace[1][j];
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m++;
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point[m] = linspace[2][k];
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m++;
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}
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}
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}
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rc = qmckl_set_point(context, 'N', point_num, point, (point_num*3));
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if (rc != QMCKL_SUCCESS) {
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fprintf(stderr, "Error setting points:\n%s\n", qmckl_string_of_error(rc));
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exit(1);
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}
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#+end_src
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#+RESULTS:
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: None
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Then, we evaluate all the MOs at the grid points (and time the execution),
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and thus obtain the matrix $M_{ki} = \langle \mathbf{r}_k | \phi_i \rangle =
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\phi_i(\mathbf{r}_k)$.
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and thus obtain the matrix $M_{ki} = \langle \mathbf{r}_k | \phi_i
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\rangle = \phi_i(\mathbf{r}_k)$.
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#+begin_src python :exports code
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import time
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#+begin_src c :exports code
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mo_num = qmckl.get_mo_basis_mo_num(context)
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int64_t mo_num;
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rc = qmckl_get_mo_basis_mo_num(context, &mo_num);
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before = time.time()
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mo_value = qmckl.get_mo_basis_mo_value(context, point_num*mo_num)
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after = time.time()
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long before, after;
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struct timeval timecheck;
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mo_value = np.reshape( mo_value, (point_num, mo_num) )
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double* mo_value = (double*) calloc(sizeof(double), point_num*mo_num);
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if (mo_value == NULL) {
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fprintf(stderr, "Allocation failed (mo_value)\n");
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exit(1);
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}
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gettimeofday(&timecheck, NULL);
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before = (long)timecheck.tv_sec * 1000 + (long)timecheck.tv_usec / 1000;
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print("Number of MOs: ", mo_num)
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print("Number of grid points: ", point_num)
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print("Execution time : ", (after - before), "seconds")
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rc = qmckl_get_mo_basis_mo_value(context, mo_value, point_num*mo_num);
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if (rc != QMCKL_SUCCESS) {
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fprintf(stderr, "Error getting mo_value)\n");
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exit(1);
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}
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gettimeofday(&timecheck, NULL);
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after = (long)timecheck.tv_sec * 1000 + (long)timecheck.tv_usec / 1000;
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printf("Number of MOs: %ld\n", mo_num);
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printf("Number of grid points: %ld\n", point_num);
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printf("Execution time : %f seconds\n", (after - before)*1.e-3);
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#+end_src
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#+begin_example
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Number of MOs: 201
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Number of grid points: 1728000
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Execution time : 3.511528968811035 seconds
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Execution time : 5.608000 seconds
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#+end_example
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and finally we compute the overlap between all the MOs as
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$M^\dagger M$.
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$M.M^\dagger$.
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#+begin_src python :exports code
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overlap = mo_value.T @ mo_value * dr
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print (overlap)
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#+begin_src c :exports code
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double* overlap = (double*) malloc (mo_num*mo_num*sizeof(double));
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rc = qmckl_dgemm(context, 'N', 'T', mo_num, mo_num, point_num, dr,
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mo_value, mo_num, mo_value, mo_num, 0.0,
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overlap, mo_num);
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for (size_t i=0 ; i<mo_num ; ++i) {
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printf("%4ld", i);
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for (size_t j=0 ; j<mo_num ; ++j) {
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printf(" %f", overlap[i*mo_num+j]);
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}
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printf("\n");
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}
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}
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#+end_src
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#+begin_example
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[[ 9.88693941e-01 2.34719693e-03 -1.50518232e-08 ... 3.12084178e-09
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-5.81064929e-10 3.70130091e-02]
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[ 2.34719693e-03 9.99509628e-01 3.18930040e-09 ... -2.46888958e-10
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-1.06064273e-09 -7.65567973e-03]
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[-1.50518232e-08 3.18930040e-09 9.99995073e-01 ... -5.84882580e-06
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-1.21598117e-06 4.59036468e-08]
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...
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[ 3.12084178e-09 -2.46888958e-10 -5.84882580e-06 ... 1.00019107e+00
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-2.03342837e-04 -1.36954855e-08]
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[-5.81064929e-10 -1.06064273e-09 -1.21598117e-06 ... -2.03342837e-04
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9.99262427e-01 1.18264754e-09]
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[ 3.70130091e-02 -7.65567973e-03 4.59036468e-08 ... -1.36954855e-08
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1.18264754e-09 8.97215950e-01]]
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0 0.988765 0.002336 0.000000 -0.000734 0.000000 0.000530 0.000000 0.000446 0.000000 -0.000000 -0.000511 -0.000000 -0.000267 0.000000 0.000000 0.001007 0.000000 0.000168 -0.000000 -0.000000 -0.000670 -0.000000 0.000000 -0.000251 -0.000261 -0.000000 -0.000000 -0.000000 -0.000397 -0.000000 -0.000810 0.000000 0.000231 -0.000000 -0.000000 0.000000 -0.000000
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...
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200 0.039017 -0.013066 -0.000000 -0.001935 -0.000000 -0.003829 -0.000000 0.000996 -0.000000 0.000000 -0.003733 0.000000 0.000065 -0.000000 -0.000000 -0.002220 -0.000000 -0.001961 0.000000 0.000000 -0.004182 0.000000 -0.000000 -0.000165 -0.002445 0.000000 -0.000000 0.000000 0.001985 0.000000 0.001685 -0.000000 -0.002899 0.000000 0.000000 0.000000 -0.000000 0.002591 0.000000 -0.000000 0.000000 0.002385 0.000000 -0.011086 0.000000 -0.003885 0.000000 -0.000000 0.003602 -0.000000 0.000000 -0.003241 0.000000 0.000000 0.002613 -0.007344 -0.000000 -0.000000 0.000000 0.000017 0.000000 0.000000 0.000000 -0.008719 0.000000 -0.001358 -0.003233 0.000000 -0.000000 -0.000000 -0.000000 0.000000 0.003778 0.000000 0.000000 -0.000000 0.000000 0.000000 -0.001190 0.000000 0.000000 -0.000000 0.005578 -0.000000 -0.001502 0.003899 -0.000000 -0.000000 0.000000 -0.003291 -0.001775 -0.000000 -0.002374 0.000000 -0.000000 -0.000000 -0.000000 -0.001290 -0.000000 0.002178 0.000000 0.000000 0.000000 -0.001252 0.000000 -0.000000 -0.000926 0.000000 -0.000000 -0.013130 -0.000000 0.012124 0.000000 -0.000000 -0.000000 -0.000000 0.000000 0.025194 0.000343 -0.000000 0.000000 -0.000000 -0.004421 0.000000 0.000000 -0.000599 -0.000000 0.005289 0.000000 -0.000000 0.012826 -0.000000 0.000000 0.006190 0.000000 0.000000 -0.000000 0.000000 -0.000321 0.000000 -0.000000 -0.000000 0.000000 -0.000000 0.001499 -0.006629 0.000000 0.000000 0.000000 -0.000000 0.008737 -0.000000 0.006880 0.000000 -0.001851 -0.000000 -0.000000 0.000000 -0.007464 0.000000 0.010298 -0.000000 -0.000000 -0.000000 -0.000000 -0.000000 0.000000 0.000540 0.000000 -0.006616 -0.000000 0.000000 -0.002927 -0.000000 0.000000 0.010352 0.000000 -0.003103 -0.000000 -0.007640 -0.000000 -0.000000 0.005302 0.000000 0.000000 -0.000000 -0.000000 -0.010181 0.000000 -0.001108 0.000000 0.000000 -0.000000 0.000000 0.000000 -0.000998 -0.009754 0.013562 0.000000 -0.000000 0.887510
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#+end_example
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* Fortran
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