Wrong explanation for why "electron can't exist in the nucleus"?

Question

When are asked about why an electron cannot fall into the nucleus we put together the argument that:

as Heisenberg's uncertainty principle says

$\Delta x \Delta p$ $\geq \frac{h}{4 \pi}$

$\Delta x \Delta v$ $\geq \frac{h}{4 \pi m}$

now we say that "as the size of the nucleus is 2fm(diameter)"

$\Delta v$ $\geq \frac{6.626 \cdot 10^{-34}}{2 \cdot 10^{-15} 9.1 \cdot 10^{-31} 4 \pi}$

$\Delta v$ $\geq 2.89871 \cdot 10^{10} ms^{-1}$

and we say that as "for electron to be in the nucleus it must have velocity greater than the speed of light; therefore, it can't exist inside the nucleus."

But I feel that there is a problem with that because $\Delta v$ represents the uncertainty in the velocity of electron, but my point is the same argument could be put together for a sphere present in the electron's probabilistic cloud, for example for the $1s$ orbital if we take a sphere of the same diameter as that of the nucleus at any point in the $1s$ orbital other than the nucleus itself we will be able to say that the electron should also not be able to exist in that sphere, and hence we can show by taking pieces of the $1s$ orbital that the electron can't exist anywhere in that orbital.

Answer 1

Electrons do "spend time" inside the nucleus, just as they "spend time" in other spherical regions of a similar size. You can see that from the fact that the $1s$ orbital wave function is nonzero in those regions.

Arguments from the uncertainty principle are meant to show that the electron can't permanently inhabit the nucleus, i.e. its wave function can't square-integrate to roughly 1 inside the nucleus and be roughly zero everywhere else over a significant span of time. The reason is that the momentum uncertainty of that state would be high enough that the electron would immediately leave the region. That argument does apply to any other region of similar size.

Answer 2

When are asked about why an electron cannot fall into the nucleus...

This is a common "problem" when talking about the issues with the classical picture of an electron orbiting the nucleus, since the electron would radiate energy away and move closer to the nucleus, eventually colliding with it.

When we move to the quantum world, the electrons are not following classical trajectories, and so it's odd to talk about the electron "falling" into the nucleus.

Still, we can reframe the question in the way that the analysis suggests: can the wavefunction of the electron be sufficiently peaked at the location of the nucleus?

From here, we need to be careful about what we are talking about. If we are talking about electrons in orbitals that are eigenstates of the Coulomb potential, then we don't really need to revert to using the uncertainty principle; the wavefunctions for such states are already known, and they are not sharply peaked at the nucleus.

If you want to just talk about some arbitrary state that is sharply peaked at the nucleus, as the argument you have outlined does, that is fine, but you are no longer talking about states with definite energy anymore. These are not the orbitals we usually discuss in the context of hydrogen or hydrogen-like atoms.

the same argument could be put together for a sphere present in the electron's probabilistic cloud, for example for the $1s$ orbital if we take a sphere of the same diameter as that of the nucleus at any point in the $1s$ orbital other than the nucleus itself we will be able to say that the electron should also not be able to exist in that sphere, and hence we can show by taking pieces of the $1s$ orbital that the electron can't exist anywhere in that orbital.

The uncertainty principle applies to entire states. Each quantum state has an associated $\Delta x$ and $\Delta p$, and the uncertainty principle gives a lower limit on the product of these two values. You don't just pick some region in space and then say "well this is $\Delta x$ here."

The argument you outline is considering a state that is localized at the nucleus and has $\Delta x$ roughly equal to the nucleus diameter. The $\Delta x$ is set by the wavefunction, not by the region of space being looked at. However, if you set up a wavefunction with this property, then you can outline the arguments you have made.

Basically... you definitely could make the same argument for a point outside of the nucleus, but not using the wavefunction of the $1s$ orbital. You would just have a different wavefunction that is sharply peaked somewhere else. If you are looking at the $1s$ wavefunction, $\Delta x$ is already determined by the wavefunction itself.

In general, I try to take hand-wavy arguments like these using the uncertainty principle with a grain of salt. They are good "back of the envelope" calculations, but usually the simplifications leave out information or cause confusion of what the principle is really talking about.

Answer 3

Everything @BioPhysicist is correct, but in addition:

Your argument is semi-classical, quantum mechanics is non-relativistic (there is no $c$ in the Schrödinger Eq.), and the problem involves relativistic quantum field theory, so in leu of formalism, we have to be a bit hand-wavy.

In quantum mechanics, if you had a (pure) wave function that was spherically symmetric ($L=0$) and confined to the volume of the nucleus, it would not be an energy eigenstate, and you would write it as a sum over $S$ eigenstates (with $n, l, m$ quantum numbers):

$$ \psi_{N}(r) = \sum_{k=1}^{\infty} c_k\psi_{(n=k, l=0, m=0)}(r) $$

Then a computation of the energy expectation would proceed as:

$$ \bar E = \langle\psi^*_{N}|\hat H|\psi_N\rangle = \sum_{k=1}^{\infty}c^*_kc_k E_k $$

with the result:

$$ \bar E \gg 2m_ec^2 $$

That's straight quantum mechanics, where the extreme energy doesn't matter--but relativistic considerations make it problematic.

The problems are captured in the concept of reduced Compton wavelength (https://en.wikipedia.org/wiki/Compton_wavelength), which is the minimum radius to which an electron can be confined before the momentum uncertainty leads to $e^+e^-$ creation.

Fun Fact: The Bohr radius and Reduced Compton Wavelength are related by the strength of electromagnetism via the fine structure constant:

$$ a_0 = \frac 1 {\alpha} \frac{\lambda_C}{2\pi} \approx 137 \times \frac{\lambda_C}{2\pi} $$

To quote the wikipedia link:

"The inverse reduced Compton wavelength is a natural representation for mass on the quantum scale."

(So for an atom with $Z \ge 137$, the 1S shell should be...interesting).

One final caveat: The nucleus is not a stationary lump of charge. We solve the Hydrogen Atom (https://en.wikipedia.org/wiki/Hydrogen_atom) in terms of reduced mass coordinates, so irl, the proton is also not-localized, it is in a complementary $|nlm\rangle$ wave function, which for the ground state, has a radius:

$$ R_0 \approx \frac{m_e}{m_p}a_0 \approx 33 \times R_p $$

where $R_p = 0.88\,$ fm is the proton radius. That means you can confine it to nuclear size in reduced mass coordinates, but in standard one particle coordinates, it's in a much larger volume--I don't know how that would impact the analysis, but it needs to be considered.

Answer 4

This is too long for a comment as I want to add to some arguments made in a comment.

First using $\Delta x\Delta p_x$ is very dangerous and second is getting $\Delta v>c$ (while nonsensical) doesn’t really tell you what you think it does because you cannot substitute $\Delta v$ for $\Delta v_x$ in general.

First, you should really be using $\Delta r\Delta p_r$ as you are working in spherical coordinates. Now, $\Delta r$ is well defined since $\langle r\rangle$ and $\langle r^2\rangle$ are well-defined, but actually getting a good definition of $\Delta p_r$ in quantum mechanics is not clear; in particular, $p_r$ is not self-adjoint if you insist on $[r,p_r]=i\hbar$ (see this post or this post) so how to compute $\Delta p_r$ is not clear either.

Next, going from radial to 1d, i.e. replacing $\Delta r$ by $\Delta x$ and $\Delta p_r$ by $\Delta p_x$ is fraught with additional difficulties because (for instance), $v_x$ can take positive and negative values whereas $v_r$ cannot.

As an example, consider the two velocity vectors $\vec v_1=\frac{c}{2} \hat x$ and $\vec v_2=-\vec v_1$. Clearly here the average $\langle v_x\rangle=0$, and $\langle v_x^2\rangle=v_x^2$ so $\Delta v_x =\frac{c}{2}$.

Now the point is that, if you convert the previous $\vec v_1,\vec v_2$ to spherical coordinates, you have, for the radial component, $v_{1r}=+\frac12 c$ and $v_{2r}=+\frac12 c$, both of which are positive, and $\Delta v_r=0$ since there is only one outcome in your distribution of $v_r$’s. Thus, $\Delta v_x>\Delta v_r$.

So in general you cannot simply replace radial variables by the 1d version and expect to get the same results. Granted your $\Delta v_x>c$ doesn’t make sense, but this doesn’t necessarily imply that $\Delta v_r>c$.

The calculation you give is entirely unreliable to make solid arguments. It just happens to give you a reasonable answer here but that’s mostly because of luck, not because the argument is fundamentally correct. The correct way to approach this is to use Fourier analysis, wherein one can show that, to localize a signal in space, one needs wavevectors taken from an increasing range, and the wavectors of some of these plane waves will correspond to energies and velocities that are nonsense physically. The Fourier analysis argument does not depend on averages and variances, which is at the core of the HUR.