While trying to understand this paper, I read the following method for calculating the polarizability of different Xe isotopes:

Quantum chemical calculation of the exact polarizability of each xenon isotope was performed by using Gaussian 09 software (version D.01, Gaussian Inc., USA). The exact polarizability of each xenon isotope was optimized by b31yp/3-21G and the density functional theory method.

My understanding is that DFT can be used to assess polarizability of larger systems, but I'm a little befuddled by how it's applied here.

I couldn't identify what is meant by "b31yp" but it was fairly easy to read a description of the "3-21G" basis online, and it's apparent that the polarization of Xe atoms won't be included in that basis:

the 3-21G* basis set has polarization functions on second row atoms only.

So...can someone decipher what is meant here by "b31yp"? EDIT: I think it must be a typo for the "B3LYP" DFT method, described here:

uses the non-local correlation provided by the LYP expression, and VWN functional III for local correlation (not functional V). Note that since LYP includes both local and non-local terms, the correlation functional used is actually:


In other words, VWN is used to provide the excess local correlation required, since LYP contains a local term essentially equivalent to VWN.

Does it seem plausible that the researchers correctly calculated that four different isotopes of Xe have the same polarizabilities to three significant figures?

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    $\begingroup$ I don't think it is of topic here, but it might be more on topic at Matter Modeling. $\endgroup$ Dec 27, 2022 at 1:45
  • $\begingroup$ thanks! I didn't know about Matter Modeling! I think you're probably right, but I'm very happy by the quick and cogent responses below $\endgroup$
    – Ryan
    Dec 27, 2022 at 3:56

3 Answers 3


In addition to the other answers, to do a proper calculation of the polarizability of Xe or any other "heavy" element, one needs to perform proper relativistic calculations in a larger basis set - ideally at the basis set limit since it's only a single atom. (The basis set used is entirely too small, as mentioned in other answers.)

So the number in the paper ($3.60 Å^3$) is simply wrong. Atomic polarizabilities, particularly of noble gases have been measured accurately and used for benchmark calculations:

(Incidentally, the NIST reference cites $4.005 Å^3$ as the atoms polarizability of Xe.)

My guess is that one of the authors had used Gaussian, knew it handled isotopes (e.g., for vibrations), saw that it could calculate polarizabilities, and just ran the calculations without thinking too deeply.


As the other answer already says: it's entirely plausible and not surprising that these calculations came out this way. This maybe answered your question, but I'd like to offer some more context.

For all intents and purposes, B3LYP/3-21G is a terrible choice of computational method. By the way, you have (without me reading this paper) correctly identified this. I'm my personal opinion, using a phone book will produce more consistent results.
The basis set is way too small for any productive work, and in a computational setup of today's hardware, there's hardly any difference by using a better balanced one.
(If you'd really needed to cut down computations, go with a minimal basis, i.e. STO-3G. 6-31G is already a bad, but popular, standard, and only slightly more demanding than 3-21G. Most other split valence basis sets perform computationally better. With a dual core computer, a gigabyte RAM, for small molecules and on a human time scale, is not noticeable.)

Gaussian 09 is old and about 6 years past is prime. Revision D.01 isn't even the last one. However, with this level of theory, nothing much has changed, so it likely doesn't matter at all. However, if there were any bugs in the program, they would be in the final numbers.

Density Functional Theory is nothing without calibration.
The holy grail is still hunted for. And until we know about it with certainty, some functionals will be optimised for one scenario and others for something completely different. That's why there are quite a few benchmarks. Unless there's one that covers your field of study, you (them/they) should do those yourself (themselves/their selves).

N.B.: Most other quantum chemical programs use VWN5 for B3LYP.

Anyway, answering the titular question:

Can one calculate the polarizability of Xe isotopes using Gaussian 09 and the 3-21G basis?

You can calculate a number, but it's not reliable; so: no.

Answering a question you have not asked:

Can the exact polarizability of each xenon isotope be optimized with B31YP/3-21G?

Probably (not very likely) not. With DFA (density functional approximations) nothing is exact (yet, due to the limitations of the method.)


Different isotopes (of Xe but also any other element) only differ in the mass of the nucleus; the charge of the nucleus and the electron shells are exactly the same. In the Born Oppenheimer approximation the electrons have their own wave function which includes point potentials from the nuclei (neglecting electron-nuclei cross terms). Because of that, the mass of the nuclei (or kinetic energy of the nuclei) has no effect on the electrons at all.

From that perspective, it is not surprising that different isotopes have the same polarizability (or indeed any other properties) when calculated with DFT using the BO approximation. What is maybe surprising is that the authors of the paper would think that is not the case. I haven’t read the paper but this result would absolutely be expected in any formulation which doesn’t involve a wave function for the nuclei.

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    $\begingroup$ The BO approximation doesn't seem particularly relevant since they're just calculating the polarizability of individual atoms. $\endgroup$ Dec 27, 2022 at 3:50
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    $\begingroup$ I was wondering this myself... whether BOA was applied here, because then yes, it's impossible for nuclear dynamics to impact electronic dynamics. Am I right that DFT as applied here implicitly uses the BOA? $\endgroup$
    – Ryan
    Dec 27, 2022 at 4:07
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    $\begingroup$ Yes, the calculation would imply the BO approximation.. but it's irrelevant for an atomic calculation. $\endgroup$ Dec 27, 2022 at 4:15
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    $\begingroup$ @GeoffHutchison Not using BO is the only way nuclear mass can possibly affect the results here, it seems $\endgroup$
    – Alex I
    Dec 27, 2022 at 17:00

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