# [molpro-user] Expectation of second moment and kinetic energy

Yannis Petousis petousis at stanford.edu
Sat Mar 1 05:23:56 GMT 2014

```Dear Peter,

Thank you very much for the reply.

I still have one question about the second moment.
I am looking to calculate the terms: <ψ|x^2|ψ> and <ψ|x|ψ> however, I believe I mistakenly thought that SM would give me the value of the first of the 2 terms. I looked for a mathematical definition of the operators but could not find one.
If I am right, is it possible to calculate the above quantities in molpro either directly or indirectly by extracting the wavefunction and position operator?

Thank you very much again,
Yannis

----- Original Message -----
From: "Peter Knowles" <KnowlesPJ at cardiff.ac.uk>
To: "Yannis Petousis" <petousis at stanford.edu>
Cc: "Molpro user mailing list" <molpro-user at molpro.net>
Sent: Friday, February 28, 2014 12:00:54 AM
Subject: Re: [molpro-user] Expectation of second moment and kinetic energy

On 27 Feb 2014, at 20:04, Yannis Petousis <petousis at stanford.edu> wrote:

> Dear molpro users,
>
> In a simple rhf calculation of around 50 atoms, I am trying to find the expectation values of kinetic energy and second moment of position.
> My understanding is that this is done using the respective operators, EKIN and SM.
> However, the expectation of the kinetic energy I am getting, violates the virial theorem

Without seeing your whole job, it's difficult to comment. Hartree-Fock satisfies the virial theorem, but only in the limit of a complete basis set. You should not expect the kinetic energy to be equal to minus the total energy unless the forces are zero.

> and the expectation of the second moment components are negative!

The second moment of course depends on the origin, and you will in any case get contributions of opposite sign from nuclei and electrons, so it is difficult to comment.

Peter

>
> Here is my input:
> {hf;
> start,2100.2;
> wf,250,1,0;
> expec,pot,ekin,ov,sm,delta,tm;
> }
> show,epot,kin,ov,sm,delta,tm;
>
>
> And here is the output:
> !RHF expec            <1.1|XX|1.1>  -532.748689831615
> !RHF expec            <1.1|XX|1.1>  -532.748689831615
> !RHF expec            <1.1|ZZ|1.1>  -522.658296533770
> !RHF expec          <1.1|EKIN|1.1>   122.891738342572
>
>
> Note: I am using no symmetry so my only state is 1.1
>
> Can anyone help?
>
> Thank you,
> Yannis
>
> Ioannis Petousis
> PhD Student
> Nanoscale Prototyping Laboratory
> Mechanical Engineering dpt.
> Stanford University
> email: petousis at stanford.edu
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--
Prof. Peter J. Knowles
School of Chemistry, Cardiff University, Main Building, Park Place, Cardiff CF10 3AT, UK
Telephone +44 29 208 79182 Email KnowlesPJ at Cardiff.ac.uk