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I once read that an atomic orbital can be conceptualised as a cloud of "electron-ness". That is, the electron literally is the cloud, and the probability density only relates to the probability of the electron interacting with another particle (say, a photon) at any given point in space.

This metaphor is fairly intuitive and got me through an undergrad course in physics and chemistry without issue. Yet I often read the caveat "the electron is not actually 'smeared out' across the probability distribution" with no evidence for why that is the case.

Is there an experiment/observation/result that disqualifies this interpretation in some circumstances?

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    $\begingroup$ Not an answer, of course, but I would be careful about "reifying our successful abstractions". An electron is not literally a cloud. Some might say an electron is literally an excitation of a quantum field, but I would call that reifying a successful abstraction, too. I'm not sure that we can ever say what an electron really is, because, well, it's whatever it is. Conceptual models of what things "are" are only useful to the extent that they aid our calculations and such. (I'm not totally a philosophical pragmatist, though.) $\endgroup$ – march Mar 19 at 17:33
  • $\begingroup$ Can you cite a source for the claim that the electron is not actually smeared out across the probability distribution? While I take march's point about taking mathematical models too seriously I would guess most physicists would take the view that the electron is smeared out. The electron has no precise position, well not unless it's in a position eigenstate and those are unphysical anyway. $\endgroup$ – John Rennie Mar 19 at 17:37
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    $\begingroup$ @JohnRennie It depends on what "smeared out" really means $\endgroup$ – Aaron Stevens Mar 19 at 17:43
  • $\begingroup$ @AaronStevens yes, and I agree with your interpretation that it means the electron does not have any location i.e. asking what the electron position is constitutes a meaningless question (at least I think that's what your answer says). But we need to say what the sources the OP refers to say in order to answer the OP's question definitively. $\endgroup$ – John Rennie Mar 19 at 17:46
  • $\begingroup$ @JohnRennie Yes I am saying asking what the electron's position is meaningless. I also agree that the source would be very useful. I would adjust my answer accordingly if needed. $\endgroup$ – Aaron Stevens Mar 19 at 17:49
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I think what we mean when we say that the electron is not smeared out is the following. "Smeared out" means that some of the electron is over here, and some of it is over there (like smearing out a ball of cream cheese on your bagel). Instead, according to some interpretations, the electron doesn't have any defined location (with uncertainty) until the interaction you mention at the beginning of your question.

So saying the electron is not smeared out is not going against the wavefunction/field picture. Rather, it is going against incorrect interpretations of the wavefunction/field picture by just saying the electron is not located "everywhere" in this "cloud".

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  • $\begingroup$ One could add that the electron is considered to be a "point" particle w/o underlying structure. $\endgroup$ – Paul Mar 19 at 21:29
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I'd like to stress that it's wrong to think only of "probability density". A wave function (an "orbital" in chemical jargon) carries more information than just its modulus squared. Generally speaking it's a complex function and its being complex has physical meaning.

As an instance consider $2p$ wavefunctions for hydrogen atom. There are three independent. Although chemists' use is to consider $2p_x$, $2p_y$, $2p_z$ in physics eigenfunctions of $L_z$ ($z$-component of angular momentum) are more important. Their expressions are $$\eqalign{ m=+1:\ & f(r)\,\sin\theta\,e^{i\phi} \cr m=0:\ &g(r)\,\cos\theta \cr m=-1:\ & f(r)\,\sin\theta\,e^{-i\phi} \cr}$$ You can see that as far as modulus is concerned the $m=+1$ and $m=-1$ eigenfunctions are identical. Yet they represent different states.

Another example: if you only looked at modulus square (the "electron cloud") you could know nothing about momentum. E.g. its expectation value is zero if $\psi$ is real, but may be not zero if $\psi$ is complex.

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  • $\begingroup$ Thank you. I only mentioned position because I thought it was the only property relevant to my example but I am aware all information about the particle in encoded in the wavefunction. Interesting that the m=0 expression is even while the others are odd (unless g(r) is also odd?). My understanding was that the "p-orbitals" used by chemists were derived from some linear combination of these expressions, resulting in 3 equivalent, non-complex shapes that together form a spherically symmetric set. As I type this out I realise how off base that might be though. $\endgroup$ – Cian O'Connor Mar 20 at 18:51

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