# How to calculate molecular volume from the wavefunction?

Very short question but maybe not a short answer ! As the probability of presence of electrons in a specific spatial position is defined as square of the wave function, I am wondering if there is a technique for molecular volume calculation based on the wavefunction. Is there such a technique?

Update: I could run the volume keyword in Gaussian. This is the output for methane , but I can not find the volume in it !

• Usually molecular volume is computed simply as the volume inside a contour of a particular electron density. The choice of the contour is basically up to you, Gaussian, for instance, uses the value of $0.001 \, \mathrm{electrons}/\mathrm{Bohr}^3$. – Wildcat Apr 9 '15 at 14:12
• @Wildcat thanks for the reply. I have not access to Gaussian ( I am in Colorado state ! ) . Can you give a suggestion for GAMESS or using third parties as Avogadro over GAMESS output ? – Aug Apr 9 '15 at 14:30
• MoCalc according to its description can handle Gamess output and calculate "molecular surface area and volume", so I suggest you give it a try. – Wildcat Apr 10 '15 at 14:11
• @Wildcat I installed MoCalc but can not find where I can calculate that. I just opened the GAMESS output but there is no option for volume calculation. Can you tell me how to find it please? – Aug Apr 11 '15 at 2:18
• Molecular volume can be defined different ways (one may say, rather arbitrary), so I think it is an important question what you would like to use the result / what kind of experimental data you want to compare to. Also, whatever definition you use it is sensitive to geometry, therefore you may need a thermal average... – Greg Apr 12 '15 at 8:39

The way to determine the volume of an atom in a molecule (Therefore, the molecular volume) is by doing a quantum calculation to obtain the wave function and, then, integrate it.

The way to obtain the wave function from GAMESS is with the AIMPAC keyword. Look at page 2-18 of the GAMESS manual. :D

You can obtain the volume of your desire molecule by following these steps:

1. Do a quantum calculation with the desire method and a very good wave function.
2. In the input file, you have to specify to the software of your preference to print out the wave function file (in gaussian, you can do it with output=wfn, more directions: Directions to print wave function from gaussian).
3. After the quantum calculation is done, you have to integrate the wave function with a specialized software (see at external links of QTAIM software).
4. After integrating the wave function, you will get the integrated values of several properties, one of them is the volume of each atom in the desire molecule. (See page 16 of The Quantum Theory of Atoms in Molecules).
5. Finally, sum up all the atomic volumes to get the molecular one.
• Hope this answer would be useful for your purposes – Another.Chemist Apr 15 '15 at 23:41
• And even for the title of your question :D – Another.Chemist Apr 16 '15 at 3:27
• I think that QTAIM is too much of an (computational) effort to obtain the molecular volume. You could calculate the van der Waals volume of a molecule from knowing the positions (and the vdW radii) of the involved elements alone. I also think that the van der Waals volume is not equal to the volume you would obtain from a cutoff value of the electron density. – Martin - マーチン Apr 16 '15 at 15:16
• Related to vdW is true that QTAIM does not obtaine it, I will edited my post to be purer. In QTAIM, an atom is based purely on the electronic charge density and uses what are called zero flux surfaces to divide atoms. On this surface, the charge density is a minimum perpendicular to the surface, which is a minimum between atoms and this is a natural place to separate atoms from each other. Therefore, Based on this partition, it defines several electronic properties: volume per atom. This point of view can be the correct way to obtain the molecular volume from the wave function by summing up – Another.Chemist Apr 16 '15 at 15:29
• QTAIM is meant to divide the volume among atoms in a molecule. For division of the atom with the external space, it uses the total density criteria--meaning that the total volume really isn't an AIM property; rather, its the volume in a fixed density contour, so my nature is exactly the same as what Gaussian does by default. But this certainly gives you a number. – Xiaolei Zhu Apr 18 '15 at 21:57

the volume seems to be already in the output:

$\text{Molar volume}=305.994~\mathrm{bohr^3/mol}\ (27.304~\mathrm{cm^3/mol})$

But of course, this is defined with the volume inside a density contour of $0.001~\mathrm{e/bohr^3}$. That may be different from what you measure, because volume is not a quantity with well defined meaning, and what you find in an experiment depends on how you measure it.

Mathematically any atom or molecule have a volume of infinity since the electron spreads to the entire space, but this is clearly not the physically meaning of volume as we usually know it.

• Yes but I really can't understand it. As I ran it for hydrogen atom it gives 111 bohr^3/mol. As I know bohr is radius of hydrogen and its volume can not be 111 bohr^3/mol. On the other hand it is divided by mol. Should I divide it by Avogadro number ? – Aug Apr 12 '15 at 17:38
• As I said, the definition of volume itself is somewhat ambiguous. Bohr is the expectation value of the radius. The radius of hydrogen atom is 1 born only in the Bohr model. In the real quantum mechanical treatment of hydrogen atom, the wave function stretches to infinity and takes the entire space, so volume depends on where you define the probability density to be "too small" to be considered almost zero. In Gaussian this value is set to be 0.001, which is quite small so i can imagine that would cause the radius to be much larger than the Bohr radius. – Xiaolei Zhu Apr 12 '15 at 18:21
• In fact, even the expectation value of distance between electron and the nuclei is larger than the Bohr radius in a hydrogen atom. On the ground state this expectation value is 1.5 times Bohr radius, and that is the distance you expect to find the electron on average. Gaussian defines radius differently, namely the distance where you can confidently say the electron will almost never go beyond. – Xiaolei Zhu Apr 12 '15 at 18:23
• The "distance to expect to find the electron on average" can be really promising. Is the a technique to convert that 0.001 e/bohr into this average? Such a conversion can be the definite answer to the question . – Aug Apr 12 '15 at 19:21
• For hydrogen atom, you basically just do the integral and calculate the expectation value through <ϕ0|r|ϕ0>. For molecules that becomes a vague definition because there is no clear center, and a molecule isn't exactly a ball like hydrogen atom. I do not think this is available in QC packages. – Xiaolei Zhu Apr 12 '15 at 19:24