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Pai-Wei,
attached is a file with the input tensors you need + all the metadata. This is for water molecule with cc-pVDZ basis.
Here's a short primer:
nirreps: number of symmetry blocks (of occupied and virtual indices)
nmo: total number of indices = O + V
nocc_act_alpha: O for alpha spin
nocc_act_beta: O for beta spin
nvir_act_alpha: V for alpha spin
nvir_act_beta: V for beta spin
obs_mosym_alpha: symmetry block of each alpha spin index; these values can be [0,nirreps-1];
occupied come first (e,g, the occupied orbitals have symmetries 0, 0, 3, 0, 2)
obs_mosym_beta: symmetry block of each beta spin index
next come nonzero elements of f tensor. In this case it is diagonal, but you cannot assume that it is
in general CC method. I did not split it up into blocks (oo, ov, vv, etc.), but that's straightforward: indices
of the given tensor are combined o+v indices, i.e. 0 is the first O index, 1 is the second O index, … 4 is the 5th
O index, 5 is the first V index, etc. These indices were described by obs_mosym_alpha and obs_mosym_beta.
Also, in this case f tensor is the same for alpha and beta spin, but you cannot use this fact and should treat them
separately (if you consider a different molecule alpha and beta spins will no longer be equivalent).
next are nonzero elements of v tensor. Again, it's given in combined o+v form.
**WARNING: v tensor is given in so-called chemists notation, whereas all equations you have use v tensor in physicist's notation. When you read v tensor just
swap second and third indices, that will convert to physicists notation. Henceforth in this document I use physicists notation.**
There is an additional complication regarding spin.
What is given is alpha-beta v tensor (v_ab), but you can compute v_aa from it (in this case v_aa and v_bb are the same, but you cannot
use this fact -- you must treat them separately):
v_aa [p1][p2][p3][p4] = v_ab[p1][p2][p3][p4] - v_ab[p1][p2][p4][p3]
Note that v_aa is therefore antisymmetric with respect to permutation of indices p3 and p4. It is also antisymmetric with
respect to permutation of indices p1 and p2 because v_ab has what you guys called "vertex" symmetry:
v_ab[p1][p2][p3][p4] = v_ab[p2][p1][p4][p3]
------------------
Lastly, about how to do CC iterations. It's very simple. Set initial t amplitudes to zero. Changes in t1 and t2 amplitudes are obtained
by scaling the residuals of the CC amplitude equations as follows:
delta_t1_a[o1][v1] = r1_a[o1][v1] / ( f_a_o[o1][o1] - f_a_v[v1][v1] )
t1_a += delta_t1_a;
same for beta spin
delta_t2_ab[o1][o2][v1][v2] = r2_ab[o1][o2][v1][v2] / ( f_a_o[o1][o1] + f_b_o[o2][o2] - f_a_v[v1][v1] - f_b_v[v2][v2] )
t2_ab += delta_t2_ab;
same for aa and ab
The interpretation is that residuals are like gradients of t amplitudes and the differences of diagonal f tensor elements are
like approximate hessian for t amplitudes. So this is an inexact Newton update.
Do this until norms of the residuals fall below a certain small threshold, like 10^-10. You should also compute the CC energy
after each iteration to monitor progress.
To help troubleshooting: the first iteration residual r1 should be zero and r2 should equal corresponding v tensor.
Hence after the first iteration t1 will still be zero but t2 will not be.
Good luck!
Ed.
P.S. Molecular geometry:
unit = bohr
H [ 0.000000000 -1.430129828 0.983062487 ]
H [ 0.000000000 1.430129828 0.983062487 ]
O [ 0.000000000 0.000000000 -0.123863145 ]
P.P.S. Reference energies:
E(nucl rep) = 9.1968871381
E(HF) = -76.0268078691
iter CCSD energy residual RMS Wall
======================================================
0 -0.2089129826632 0.0003843661760 0.23
1 -0.2128381951574 0.0000692168122 0.21
2 -0.2132011089634 0.0000414116502 0.19
3 -0.2132690605106 0.0000063919035 0.18
4 -0.2132691174800 0.0000053098418 0.18
5 -0.2132697629510 0.0000024172773 0.23
6 -0.2132698525217 0.0000009594535 0.18
7 -0.2132699188466 0.0000004035898 0.17
8 -0.2132699194842 0.0000002637598 0.20
9 -0.2132699498062 0.0000000828022 0.18
10 -0.2132699333151 0.0000000213339 0.84
11 -0.2132699352275 0.0000000022403 0.19
12 -0.2132699349395 0.0000000025833 0.31
13 -0.2132699349021 0.0000000017643 0.19
14 -0.2132699349371 0.0000000010290 0.16
15 -0.2132699349939 0.0000000000330 0.16
======================================================
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