In a hydrogen atom an electron is orbiting around a proton, similarly to a moon around a planet. The orbit of a moon around a planet is flat (2D) whereas the orbit of an electron around a proton is spherical (3D). Why is this?

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    $\begingroup$ This is why students should be weened off the Bohr atom at the very earliest opportunity, and one should always clearly distinguish between "orbit" and "orbital". $\endgroup$ Nov 22, 2015 at 18:48
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    $\begingroup$ "In a hydrogen atom an electron is orbiting around a proton, similarly to a moon around a planet"...no, it isn't, see e.g. What is wrong with the Bohr model?, Trouble understanding the Bohr model of the atom $\endgroup$
    – ACuriousMind
    Nov 22, 2015 at 18:48
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    $\begingroup$ @qftishard on the contrary, the answer is that it isn't in an orbit at all. $\endgroup$ Nov 22, 2015 at 18:55
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    $\begingroup$ @dmckee: there's an argument to be made for teaching modern QM of the H atom first and later discussing Bohr as a matter of history. Education systems are full of baggage due to inertia. $\endgroup$
    – Gert
    Nov 22, 2015 at 18:58
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    $\begingroup$ @Laff70: "It just doesn't make any sense from a logical perspective". No, it does make perfect logical sense. QM seems at first counter-intuitive but one gets used to it very quickly. To see why your model is bound to fail, just study the paradoxes Bohr's model leads to. $\endgroup$
    – Gert
    Nov 22, 2015 at 19:07

2 Answers 2


One has to distinguish between, on one hand, an orbit and an orbital motion, which are classical notions; and on the other hand, an orbital, which is a quantum mechanical notion, cf. above comment by dmckee.

If the question is really Why quantum mechanics?, then have a look at e.g. this Phys.SE post and links therein.

Here we will assume that OP accepts quantum mechanics as is, but is genuinely puzzled why the electron is not confined to a 2D plane?

By rotation of the coordinate system, we may assume without loss of generality, that the hypothetical 2D plane corresponds to $z=0$.

Quantum mechanically, this question actually can be realized. It corresponds to a wave function $\psi$ that has support at $z=0$, say, because of a wave function collapse after measuring that $z=0$.

However, if one calculates the average energy $\langle \psi |\hat{H}| \psi \rangle$, it would be positive, because of the non-constancy of the wavefunction in the $z$-direction packs a lot of positive kinetic energy.

So the electron would no longer be bound to the nucleus. The precise measurement $z=0$ intuitively transfered so much energy to the electron that it is no longer bounded.

Now the observable $\hat{z}$ does not commute with the Hamiltonian $\hat{H}$. If we want the electron to be bounded to the nucleus, and hence that the average energy $\langle \psi |\hat{H}| \psi \rangle$ is negative, we cannot know the $z$-position very well, cf. Heisenberg's uncertainty principle (HUP).

In other words, the wave function must be have support in a 3D bulk rather than in a 2D plane. Phrased differently, a 3D spherical symmetric wave function is energetically favored, in order to have as little energy as possible.


Electron in a ground state hydrogen atom has zero angular momentum $L^2$, l=0.

Moon has a huge angular momentum. Therefore it is a poor comparison.

If moon would have zero angular momentum, in classical physics, it would fall down and hit earth.

Electron in an hydrogen atom, in l=0 state gets constantly pulled to the center, but this is countered by the quantum mechanical kinetic energy making the orbital finite.

If there would be an electron with the mass, angular momentum and position uncerainity of the moon, this would be a linear combination of very high (l=1853728172728993937272292662182829 and so) angular momentum states. In other words, It is possible to create angular wave functions which are planar. They exist already in molecular level.


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