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I ask something makes me confused.

In generating singlet oxygen molecule, I initially thought a light (photon) can excite the ground state oxygen molecule.

However, my colleague said that process is forbidden as fermion and boson hate each other.

I insisted these are all boson because ground state oxygen molecule has its intrinsic angular momentum of S=1.

I know there must be wrong reasoning in my thought. Otherwise liquid oxygen has to show Bose-Einstein condensation.

Cannot oxygen molecule really absorb a photon to be singlet oxygen molecule?

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    $\begingroup$ I see three different questions here, Please use three submissions. $\endgroup$
    – my2cts
    Commented Sep 28, 2019 at 20:14
  • $\begingroup$ Doesn't it simply and definitely depend on the specific case and if the molecules probability distribution respect Pauli exclusion principle or not? $\endgroup$
    – marshal craft
    Commented Jan 9, 2020 at 11:46

4 Answers 4

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Regardless of energy state, regardless of the choice of isotope, if we have two identical atoms in this molecule (i.e. same isotope of oxygen), this pair must have even numbers of all nucleons and electrons. This then precludes half-integer total spin, which leads to bosonic properties of the system.

I initially thought a light (photon) can excite the ground state oxygen molecule.

Yes, it can, and this has been done in real experiments, see [1] (PDF available).

However, my colleague said that process is forbidden as fermion and boson hate each other.

This is irrelevant, because simple chemical processes (without ionization) don't change number of fermions, so parity remains the same.

References:

[1]: Steffen Jockusch, Nicholas J. Turro, Elizabeth K. Thompson, Martin Gouterman, James B. Callis, Gamal E. Khalil. Singlet molecular oxygen by direct excitation. Photochem. Photobiol. Sci. 2008, 7 (2) , 235-239. DOI: 10.1039/B714286B.

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  • $\begingroup$ The assumption of identical nuclei is incorrect for natural oxygen. $\endgroup$
    – my2cts
    Commented Sep 28, 2019 at 20:20
  • $\begingroup$ @my2cts Given its atomic weight of 15.999 (with $^{16}\mathrm{O}$ abundance of 99.76%), this assumption is pretty reasonable. $\endgroup$
    – Ruslan
    Commented Sep 28, 2019 at 21:35
  • $\begingroup$ Whether it is reasonable depends on what is required. $\endgroup$
    – my2cts
    Commented Sep 28, 2019 at 21:54
  • $\begingroup$ @my2cts well, the OP has mutliple questions in it, so I've answered that part using one of the possibilities as an example. As I elaborated further, this boson/fermion distinction is actually irrelevant here. $\endgroup$
    – Ruslan
    Commented Sep 28, 2019 at 21:56
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The rule is simple:

A combination of particles is a fermion is it contains an odd number of fermions otherwise it is boson.

Since $\text{O}_2$ is two of the same particle (assuming both are the same isotope), it is necessarily bosonic.

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    $\begingroup$ The assumption that both are same isotope is incorrect. $\endgroup$
    – my2cts
    Commented Sep 28, 2019 at 20:25
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When the nuclei are disregarded $O_2$ is a boson, as there are an even number of fermions (electrons). On earth, 0.04% of the nuclei is $^{17}O$ which is a fermion with $S=5/2$ and the rest is bosonic $^{16}O$ and $^{18}O$. In conclusion, 0.08% of the natural $O_2$ is fermion and the rest is bosonic. However this is irrelevant for the triplet-singlet transition.

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QM requires that the total wavefunction be either symmetric or antisymmetric with respect to exchange of any identical nuclei.

The total wavefunction must be symmetric if nuclei are bosons, and antisymmetric if nuclei are fermions.

The O nuclei have spin I=0, so they are bosons.

enter image description here

Atomic oxygen is a boson.

Thus, the total wavefunction of O2 molecule must be symmetric or even.

https://physics.nist.gov/PhysRefData/Handbook/Tables/oxygentable1.htm

http://www.webmineral.com/help/FermionIndex.shtml#.XY-JiyhKhPa

https://ocw.mit.edu/courses/chemistry/5-62-physical-chemistry-ii-spring-2008/lecture-notes/13_562ln08.pdf

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    $\begingroup$ Oxygen nuclei can have non-zero spin. $\endgroup$
    – my2cts
    Commented Sep 28, 2019 at 20:21

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