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Fri Apr 4 12:07:02 BST 2014
r^5 Y_{50} =3D constant * ( 63z^5 - 70z^3*r^2 + 15 z*r^4 )

All these components, z^5, z^3x^2 etc does not appear in the output, thus i=
zero. (It could be checked by slightly distorted from Td symmetry)

Sorry for reply so late, hopefully this helps............
Best wishes
Cong Wang

Ph. D. Student
Department of Chemistry
Laboratory for Instruction in Swedish
University of Helsinki
A.I. Virtanens plats 1
P.O. Box 55
FI-00014 University of Helsinki

2010/2/25 Yulia Kalugina <amora at>

> Hello,
> I have calculated the multipole moments of the ch4 molecule at hf and
> mp2,but in the output hf values are equal to the mp2 ones.
> Else: for the CH4 molecule the multipole moment of the 5th order is equal
> to zero. here it is not zero. The hf values of octupole and hexadecapole
> moments are not correct at all (~5 times for hexa and ~3 times for oct mo=
> Please, help me, what I did wrong?
> Best regards, Yulia
> _______________________________________________
> Molpro-user mailing list
> Molpro-user at

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Dear Yulia, <br><br>=A0 Hello,<br><br>>Else: for the CH4 molecule the mu=
ltipole moment of the 5th order is=20
equal to zero. here it is not zero. The hf values of >octupole and=20
hexadecapole moments are not correct at all (~5 times for hexa and ~3=20
times for oct mom). <br><br>=A0 I am not sure which data were compared for =
the octupole  and=A0
hexadecapole  moments. Seems experimental groups assigned quite different v=
alues of octupole moments for methane. e.g.<br><br><a href=3D"http://zon8.p=">
5-6 * 10^{-34} esu, cm^3<br><br><a href=3D"
2.12 * 10^{-34} esu, cm^3<br><br>This could produce a factor of 2-3 differe=
nce. (I didn't check the experimental detail, temperature, whether adop=
ted the Buckingham convention for prefactor etc....)<br><br><br>If we stick=
 to computational vaules, e.g. G. Maroulis Chem. Phys. Lett. 226, 420-426 (=
\Omega 2.4601<br>\Phi -7.984 (in. a.u.)<br><br>and your output is (Molpro s=
eems print out the non-zero expectation values of the Cartesian product, ex=
cept "qm")<br>!RHF expec=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0 <1.1|XY=
Z|1.1>=A0=A0=A0=A0=A0 .989606160302<br>
<br>The agreement is quite good . Since the CPL article uses the Buckingham=
 convention, \Omega_{ijk} =3D \sum_a e_a 5/2 a_i a_j a_k - 1/2 a^2 (a_i \de=
lta_{jk} + a_j \delta_{ik} + a_k \delta_{ij} ). For  \Omega_{xyz} there is =
a prefactor of 5/2. By 5/2*=A0=A0 .989606160302 =3D2.474 it is quite close =
to the CPL value.<br>
<br>For the hexadecapole moment of Td symmetry<br>\Phi_{xxxx} =3D=A0 \Phi_{=
yyyy} =3D\Phi_{zzzz} =3D z^4 - 3(x^2 +y^2) z^2 + 3/8 (x^4 + 2 x^2y^2 + y^4)=
=A0 from Stone's the theory of intermolecular forces, P228<br>=3D7/4 z^=
4 - 21/4 x^2 y^2<br>
<br>with your data<br>=A0!RHF expec=A0=A0=A0=A0=A0=A0=A0=A0=A0 <1.1|XXXX=
|1.1>=A0=A0 -44.391072212534<br>=A0!RHF expec=A0=A0=A0=A0=A0=A0=A0=A0=A0=
 <1.1|XXYY|1.1>=A0=A0 -13.290016794368<br>=A0!RHF expec=A0=A0=A0=A0=
=A0=A0=A0=A0=A0 <1.1|XXZZ|1.1>=A0=A0 -13.290016817165<br>=A0!RHF expe=
c=A0=A0=A0=A0=A0=A0=A0=A0=A0 <1.1|YYYY|1.1>=A0=A0 -44.391072212538<br=
=A0!RHF expec=A0=A0=A0=A0=A0=A0=A0=A0=A0 <1.1|YYZZ|1.1>=A0=A0 -13.290=
016817164<br>=A0!RHF expec=A0=A0=A0=A0=A0=A0=A0=A0=A0 <1.1|ZZZZ|1.1>=
=A0=A0 -44.391072185833<br><br>=3D7/4 *( -44.391072185833) - 21/4 * (-13.29=
0016794368)<br>=3D-7.91179<br><br>also agrees with the CPL value. <br>
<br>The 32th- pole of Td point group symmetry is zero only for the spherica=
l harmonics components, as expanded by the irreducible representations, doe=
s not contain A_1; but not for any Cartesian product. e.g., for Td, <xx&=
gt; !=3D0, <3xx - r^2>=3D0<br>

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