In symmetry-adapted perturbation theory, the interaction energy of a non-covalently bound pair of molecules (it could be done to arbitrary order but is usually just pairs) is built up from a perturbative expansion of the interaction potential between two monomers.

The first-order perturbation drops out a term which we call induction. This has components associated with the induction on monomer B by monomer A and vice versa. The second-order perturbation drops out a term which we call dispersion (this is what people usually refer to when they say van der Waal's force (although that's not technically correct...)).

My question is, can that second-order perturbation term--dispersion-- have a non-zero contribution to the total energy while the first-order induction terms do not contribute?

When I imagine this situation physically, it doesn't make much sense because it means that the presence of monomer A near monomer B does not perturb the eigenstates of monomer B (beyond electrostatics, exchange, etc.) and vice versa, but somehow there is a mutual dispersion effect.

So basically what I'm asking is if it is mathematically possible to have a non-zero dispersion term while simultaneously having zero contributions from the first-order induction terms in the interaction potential expansion?

Also, if it is not possible, it seems as if we ought to be able to write the dispersion energy in terms of the induction energies, but clearly we can't because dispersion is a second-order perturbation. If the effects are related, however, can we approximate dispersion using induction? I imagine this to be similar to how in Moller-Plesset perturbation theory you can approximate the fourth-order perturbation using the second-order perturbation.

I don't know too much about the details of what I'm asking about, so if I've said something quite stupid please let me know.

  • 1
    $\begingroup$ This is an excellent question but I don't have time to think about it properly right now. In case you didn't know about it already, here is a review containing all the terms you mention. $\endgroup$ Commented Oct 18, 2016 at 2:47
  • $\begingroup$ Thanks for the link! I was actually just reading that before you linked it. It's very interesting stuff. $\endgroup$
    – jheindel
    Commented Oct 18, 2016 at 7:11

1 Answer 1


The first-order correction in the interaction operator is an integral over the frozen reference densities, so in a product basis $|i_A,j_B\rangle$ that would read $$ E^{(1)} = \langle 00|V|00 \rangle $$ and as such is the electrostatic interaction. This is the same as the Coulomb interaction of the two non-interacting reference densities.

Induction is defined as the polarization of one monomer due to the reference density of the other monomer, so must involve integrals over the virtual orbitals of the monomer being polarized in order to capture its polarization. So induction is a second-order correction and has the form $$ E^{(2)}_{ind} = \sum_{j\neq 0} \frac{|\langle 00|V|0j \rangle|^2}{E_{00}^{(0)}-E_{0j}^{(0)}} + \sum_{i\neq 0}\frac{|\langle 00|V|i0 \rangle|^2}{E_{00}^{(0)}-E_{i0}^{(0)}}$$ where the first term represents the polarization of monomer $B$ due to the reference density of monomer $A$ and the second term is the reverse process.

What sets the dispersion energy apart from the induction energy is that it is the simultaneous polarization of both monomers, and so the integrals run over the virtual orbitals of both monomers as in $$ E^{(2)}_{disp} = \sum_{i\neq 0} \sum_{j\neq 0}\frac{|\langle 00|V|ij \rangle|^2}{E_{00}^{(0)}-E_{ij}^{(0)}} $$ Mathematically, the dispersion and induction terms are both second-order correction terms in the energy expansion, but they have been grouped in a physically-intuitive way to give two different types of interactions. There are systems where dispersion is the dominant interaction, such as rare gas dimers.


An example of a dispersion-dominated bound system is the neon dimer. Below is the SAPT decomposition of the binding curve at the HF-SAPT(2)/aug-cc-pVDZ level.

enter image description here

At the equilibrium distance, dispersion dominates the binding interaction but there is a non-zero induction contribution. I am inclined to say that although the ratio of dispersion and induction may become large, you can never truly have zero induction. Some naive mathematical reasoning follows.

We need $ \langle 00|\frac{1}{r_{12}}|0i\rangle, i\neq 0$ to be zero in a Gaussian product basis. At a long-enough distance we can expand the Coulomb operator in a multipole expansion to give something like $$ \frac{1}{r_{12}} = \alpha x_1x_2+\beta(x_1x_2^2-x_1^2x_2)+... $$ in a one-dimensional model picture. I've also folded all the coefficients into the constants $\alpha,\beta,...,etc.$ The important thing here is the symmetry of the integrals. So, lets take $i=1$ and see what happens... $$ \langle 00|\alpha x_1x_2+\beta(x_1x_2^2-x_1^2x_2)|01\rangle = \langle 00|\alpha x_1x_2|01\rangle + \langle 00|\beta x_1x_2^2|01\rangle-\langle 00|\beta x_1^2x_2|01\rangle+ ... $$ The first two integrals are zero because $\langle 0|x_1|0\rangle=0$. The third integral is non-zero and survives in the sum. This can happen any time the operator in monomer $A$ is an even function. As $i$ gets larger, higher-order terms in the expansion contribute to the induction energy. So mathematically, there is a tiny but non-zero induction contribution.

  • $\begingroup$ Mathematically that helps me to understand the distinction between dispersion and induction better so thanks for that, but I don't think you have said yet if it is possible for the induction integrals to evaluate to zero but have the dispersion integral be non-zero. It feels to me like there should be a way of proving it one or the other, but I don't know if I can quite work it out myself. $\endgroup$
    – jheindel
    Commented Oct 19, 2016 at 4:14
  • $\begingroup$ @jheindel, I've expanded my answer to include why induction may be small but not zero. $\endgroup$
    – qntmnmd
    Commented Oct 19, 2016 at 16:19
  • $\begingroup$ Thanks! That's a very helpful contribution. Thanks for the answer. $\endgroup$
    – jheindel
    Commented Oct 19, 2016 at 17:13

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