# Motivation for truncating the expansions

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• #4145 Score: 0
WoutStoffels
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It has been explained in the classes that both the multipole expansion and the perturbation series can be truncated to only the leading terms.

Could someone explain mathematically why the conditions of ‘small interaction’ and ‘no overlap’ allow for these approximations?

#4157 Score: 0
MarieDeseyn
Participant

I think that the no overlap approximation is in both cases not really a condition to truncate, but more a condition to make our calculations simpler and to be able to do the laplace expansion in a nice way.

The small interaction for perturbation is due to the fact that for perturbation we treat a system with H = H_0 + epsilon H_1, for which epsilon is small. In our case we have H_0 + H_1, the epsilon is thus put inside the H_1 therefore we need H_1 to be small, which is the same as saying that the interactions de to the complicated shape of the nucleus need to be small.

For the multipole expansion if we for example go back to the relativistic case, we see that the n’th term in the expansion goes with r_<^n/r_>^{n+1}, in our case this becomes r_n^n/r_e^{n+1}, so we integrate over the domain of the nucleus and over the domain of the electron cloud, if the nucleus is spatially less spread than the electron cloud this series will rapidly converge (if you have the no overlap condition this is also fullfilled, however the no overlap condition is no ‘nessecairy condition’ but a ‘sufficient condition’, i.e. the condition that I state here can also have for example some r_e’s that are smaller than an r_n since integration is robust)

This has also an impact on the perturbation series, if we look at slide 11 of A03-02 we have rn^n/r_e^n+1, so the no overlap condition helps to make the the multipole expansion small which then leads to a smaller H_1, which makes the higher order perturbation terms negligable, which leads to a rapid convergence of the perturbation series. (again this is a sufficient but not a nessecairy condition, a nucleus that is less spread out than an electron cloud gives also fast convergence)

[I hope this is written in an understandable way, I am also not sure about this answer and I have also used the internet a little bit]

#4158 Score: 0
MarieDeseyn
Participant

[I have edited my answer a little bit by thinking a little bit deeper, the previous answer can be neglected, since there were some weird things in there]

I think that the no overlap approximation is in both cases not really a condition to truncate, but more a condition to make our calculations simpler and to be able to do the laplace expansion in a nice way.

The small interaction for perturbation is due to the fact that for perturbation we treat a system with H = H_0 + epsilon H_1, for which epsilon is small. In our case we have H_0 + H_1, the epsilon is thus put inside the H_1 therefore we need H_1 to be small, which is the same as saying that the interactions de to the complicated shape of the nucleus need to be small.

For the multipole expansion if we for example go back to the relativistic case, we see that the n’th term in the expansion goes with r_<^n/r_>^{n+1}, in our case this becomes r_n^n/r_e^{n+1}, so we integrate over the domain of the nucleus and over the domain of the electron cloud, if the nucleus is spatially less spread than the electron cloud this series will rapidly converge (if you have the no overlap condition this is also fullfilled, however the no overlap condition is no ‘nessecairy condition’ but a ‘sufficient condition’, i.e. the condition that I state here can also have for example some r_e’s that are smaller than an r_n since the infinite summation is robust (I am not completely sure about this, in my head it is logical that every r_n doesn’t need to be for every r_e smaller, but I cannot formulate why))

This has also an impact on the perturbation series, if we look at slide 11 of A03-02 we have rn^n/r_e^n+1, so the no overlap condition helps to make the the multipole expansion small which then leads to a smaller H_1, which makes the higher order perturbation terms negligable, which leads to a rapid convergence of the perturbation series. (again this is a sufficient but not a nessecairy condition, a nucleus that is less spread out than an electron cloud gives also fast convergence)

[I hope this is written in an understandable way, I am also not sure about this answer and I have also used the internet a little bit]

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