Why are the protons of ethene magnetically equivalent?

According to [this website](https://organicchemistrydata.org/hansreich/resources/nmr/?page=05-hmr-08-symmetry%2F),



> 
> For two protons to be magnetically equivalent they not only have to have the same chemical shift, but *they must also each have the same J coupling* to other magnetic nuclei in the molecule.
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> 


(Emphasis in orginal text)


One example of this is 1,1-difluoroethylene, where the two protons couple to the fluorines differently (each one is *cis*- to one fluorine and *trans*- to another, so JHF is different than JH'F and JHF' is different than JH'F'). Why does this same logic not apply to ethene? I would think that each proton couples differently to the protons on the opposite side of the double bond, leading to magnetic inequivalence. [This answer](https://chemistry.stackexchange.com/questions/39932/how-many-peaks-would-there-be-in-h1-nmr-of-ethene) says that the protons are equivalent through symmetry, but if the criterion for magnetic equivalence is having the same coupling constant with each magnetic nucleus, I think symmetric molecules could still have magnetically inequivalent protons. Are there cases where symmetry leads to magnetic equivalence, despite differences in coupling constants?


Thank you.




Magnetic equivalence is more rigorously defined in Levitt's *Spin Dynamics* (2nd ed., secction 17.5):



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> Magnetic equivalence is a strong form of chemical equivalence. A set of spins is magnetically equivalent IF
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> * Condition 1: the spins have the same chemical shifts
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> AND
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> * (Condition 2a: the spins have identical couplings to all other spins in the molecule
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> OR
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> * Condition 2b: there are no other spins in the molecule.)
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> 


Condition 1 is the so-called 'chemical equivalence', and condition 2 is what differs between the cases of ethene and 1,1-difluoroethene.


In 1,1-difluoroethene, condition 2a is not fulfilled (*cis* and *trans* H–F couplings are different), and condition 2b is not fulfilled (when considering the equivalence of the two protons, there are obviously other spins, namely the fluorines).


In ethene, condition 2b is fulfilled: we can consider the four protons as a single group because they are all chemically equivalent, and there are no other spins present (ignoring the low-natural abundance 13C, etc.)


A more mathematical explanation of this is shown in Appendix A.9 of Levitt. Long story short, the H–H J-coupling interactions in ethene (of the form $\sum\_{ij} 2\pi J\_{ij} (\mathbf{I}\_i \cdot \mathbf{I}\_j)$) commute fully with the chemical shift interactions ($\sum\_i \omega I\_{iz}$) as well as the detection operator ($\sum\_i I\_{ix}$), so can be dropped from the Hamiltonian when calculating expectation values. nb: in the copy I have, eqn. (A.37) has a typo. It should read


$$\hat{B} = 2\pi J'(\hat{\mathbf{I}}\_1\cdot\hat{\mathbf{I}}\_2
 + \hat{\mathbf{I}}\_2\cdot\hat{\mathbf{I}}\_3
 + \hat{\mathbf{I}}\_1\cdot\hat{\mathbf{I}}\_3).$$


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