RE: RE: AMBER: about QMMM output

From: Ross Walker <>
Date: Fri, 4 Jul 2008 12:55:10 -0700

Hi Linfu,

> Thank you very much for profs. Ross and Gustavo's answer.
> >Note that E(QM/MM)VDW is
> >calculated classically and so this gets included in the general VDWaals
> >energy term.
> However, from tutorial
> the result below shows
> when NSTEP =0, in my opinion, sander hasn't started doing MD and just
> calculated energy(?)
> at this time VDWAALS should be the same for Classical and QM/MM method,
> However,
> they differ much in VDWAALS term (VDWAALS=930.8446 in classical and
> VDWAALS=881.6951 in QM/MM)

Ah, but you are forgetting that classically VDW includes the
MMligand-MMligand interaction but when you run a QM/MM calculation there is
no QM-QM VDW term since that is implicit in the calculation of ESCF (and
unfortunately can't really be decomposed from that). Hence what you see for
VDW when you run the QM/MM calculation is the sum of MM-MM and QM-MM VDW
terms. QM-QM terms are zero. From your example above you can conclude that
the difference 49.1495 Kcal/mol is from what would have been QM-QM VDW

All the best

|\oss Walker

| Assistant Research Professor |
| San Diego Supercomputer Center |
| Tel: +1 858 822 0854 | EMail:- |
| | PGP Key available on request |

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Received on Sun Jul 06 2008 - 06:07:51 PDT
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