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The Pauli exclusion principle applies to all fermions, right? And protons are fermions. So if you consider a water molecule, and swap the protons in the two hydrogens, shouldn't the wavefunction of the entire molecule acquire a negative sign?

And if so, what are the implications of this? Does this have a noticeable impact on the chemistry of water, or are the hydrogens too far apart for it to make any significant difference?

(In quantum chemistry, there's a big issue called electron-electron correlation that complicates calculations. Would proton-proton correlation have a similar effect?)

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    $\begingroup$ Did you look at ortho- and para-water? It's not trivial to separate the two species: arxiv.org/abs/1407.2056 (Separating Para and Ortho Water, Daniel A. Horke, Yuan-Pin Chang, Karol Długołęcki, Jochen Küpper) $\endgroup$ – CuriousOne Dec 16 '14 at 4:06
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    $\begingroup$ Why do you think that a negative sign of the wavefunction is relevant? Since $\lvert \psi \rvert^2$ is the physically relevant quantity, global phases don't matter. $\endgroup$ – ACuriousMind Dec 16 '14 at 11:27
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    $\begingroup$ I'm puzzled by the votes to close as unclear as it seems perfectly clear what Nick is asking. The question is certainly based on a misapprehension, but that doesn't make it unclear. $\endgroup$ – John Rennie Dec 16 '14 at 17:46
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You are quite correct. If you have a wavefunction $\Psi$ for the water molecule then swapping the two protons will multiply the wavefunction by -1.

However the wavefunction is not observable, by which I mean that there is no experiment we could do that will measure the value of the wavefunction. Typically to measure some property $Q$ there will be an associated Hermitian operator, $\hat{Q}$, and the expectation value of $Q$ that we actually measure will be given by:

$$ Q = \langle\Psi|\hat{Q}|\Psi\rangle = \int \Psi^* \hat{Q}\Psi $$

So multiplying the wavefunction by -1, or indeed by any complex number with a magnitude of unity, will make no difference to the end result.

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