One and the same electron in a p orbital and taking part in a common π (pi) bond has two lobes visualized as connecting through the nucleus. There is however zero probability of finding an electron at the plane through the nucleus at right angles to the lobes. Where has the electron gone? Is the electron tunneling through the nucleus as a virtual photon?

Ed./ Without altering the meaning of the original question it is more clearly stated as: How does the electron cross the nucleus.

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    $\begingroup$ Note that the visualization with a pointlike presence in the plane assumes a pointlike charge distribution for the nucleus, which is a good approximation, but not true. Nuclear length scales are on order of $10^{-15}\text{ m}$, which is small but non-zero on atomic length scales. $\endgroup$ Jan 19, 2011 at 17:40
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    $\begingroup$ I'm more concerned about the nonzero probability of the electron being in the nucleus of a S-orbital :) $\endgroup$
    – Georg
    Jan 19, 2011 at 18:21
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    $\begingroup$ @Georg, when the energy balance works out right that gives you the halflife for decay by "electron capture", the rest of the time it's no big deal. $\endgroup$ Jan 19, 2011 at 19:49
  • $\begingroup$ Hello dmckee, I meant that mostly ironic, the electron never will be in the nucleus (close to, but not in) fermions refuse to share space to each other. What I intended to show by this, the orbits from Schrödinger equation are simply wrong in the nucleus. But compared to other errors made in MOs, this is really neglectable. $\endgroup$
    – Georg
    Jan 20, 2011 at 11:06
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    $\begingroup$ @Georg: Er...no. Electrons will be in the nucleus (or in more rigorous language have non-trivial probability density inside the nucleus). Exclusion only applies to identical fermions; thus not between nucleons and electrons. This really is the origin of the nuclear decay mode known as "electron capture", and goes on even in nuclei that won't decay that was due to energy considerations. $\endgroup$ Jan 21, 2011 at 20:26

2 Answers 2


the wave function describing an electron in a p-state ($L=1$) with $m=0$ indeed vanishes in the $z=0$ plane because the spherical harmonic $Y_{10}$ is proportional to $\cos(\theta)$ which vanishes in that plane - even at the very origin.

This wave function's squared absolute value describes the probability density that the electron is at a given point. This is true at any moment and it completely answers the question "where the electron has gone". Because the state is stationary, it has gone to the very place where it has always been - in the p-state.

You could think that the electron is jumping in between the positive $z$ and negative $z$ half-clouds. And indeed, Feynman's path integral approach to quantum mechanics tells you that you must sum over all possible trajectories to get the transition amplitudes. They will always include trajectories that go in between the half-clouds - or anywhere else in the Universe. It's just true that most of them will nearly cancel. And it's true that the $l=1$, $m=0$ wave function vanishes at the $z=0$ plane. There is no contradiction here.

In the (wrong) Bohmian theory, the (fictitiously real point-like) electron is influenced by the "quantum potential" that repels it from the places where $\psi=0$. So in the p-state, the Bohmian electron would be repelled by the $z=0$ plane, too. If it began in the upper half-cloud, it would stay there, and the same thing is true for the lower half-cloud.

But the Bohmian picture of physics is physically incorrect - I am just mentioning the point for the sake of completeness because you seem to be thinking about a "real position of the electron" at every moment - a fundamental misconception. According to quantum mechanics, the electron simply doesn't have any particular sharp position at every point.

The electron doesn't have to tunnel anywhere because you haven't showed that the sign of its $z$ has ever changed. But even if it changed, one wouldn't need any tunneling because there is no potential barrier that it would have to tunnel through. In particular, points where $\psi$ happens to vanish are not a "wall" in any sense. They're just zeros of the wave function. In the flawed Bohmian mechanics, points $x$ with $\psi(x)=0$ are singular (usually loci of bizarre solenoids), but in the real quantum mechanics which is totally linear in $\psi$, there is nothing special about points with $\psi(x)=0$.

But even if the electron had to tunnel somewhere - which is never the case for a single atom - it could never become a virtual photon because such a strange transformation would violate the conservation of electric charge and spin, among many other things.

Apologies for debunking so many things - but your question wasn't a real question. It was closer to a sequence of approximately 37 independent fundamental misconceptions about physics.

Best wishes Lubos

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    $\begingroup$ Well, thanks very much for exellently answering not-real aka dumb questions, questions one may at least learn from! It is usually not explained very well in elementary chemistry textbooks or it is said "almost zero probability". I did not however had the alleged misconception of a strict "real position", and even you mentioned trajectories between the lobes, but like to start from "common sense world" visualizations or visual models of which most, if not all of them, will eventually be "debunked". $\endgroup$
    – curious
    Jan 19, 2011 at 18:52

The nucleus is not at a single point in space. It's a quantum object and so does not have a determinate position. Consequently, in a real physical atom there is no single point that the electron has to go through.

The apparent problem arises only from our approximations. To get a problem where the nucleus didn't move we'd have to have an infinite mass nucleus and this would lead to a gravitational singularity. Even requiring the charge of the nucleus originate from a single point would require singularities. These are not physical problems, they are math problems.

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    $\begingroup$ This isn't relevant to the OP's question. In, for example, a hydrogen atom, the proton's position is perfectly correlated with the electron's because the center of mass doesn't move, so $m_pr_p+m_er_e=0$. This constraint has solutions both for $r_e=r_p=0$ and for $r_e\ne 0$, $r_p\ne 0$. $\endgroup$
    – user4552
    Jul 23, 2011 at 18:55
  • $\begingroup$ @Ben Crowell; You've written: "In, for example, a hydrogen atom, the proton's position is perfectly correlated with the electron's because the center of mass doesn't move ..." How do you reconcile your belief with the (widely held) belief that a hydrogen atom can be represented as a quantum object and therefore its position must be represented by a wave function? $\endgroup$ Jul 24, 2011 at 19:25

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