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This is the solution given in the textbook (my analysis below it):

Enter image description here

From my understanding, I believe this solution is wrong. My understanding is that the more accurately we know the position, the less accurately we know its velocity, or its momentum. I have two major arguments to contradict this which are from my understanding of the topic, which may be wrong →

  1. Conceptual Logic: The uncertainty has no relation with the real values. It is a degree of measuring how wrong we are rather than what wrong value we obtain. The Δv term should have nothing to do with the actual velocity and hence I believe this is wrong.
  2. Manipulative Logic: Nowhere do we use any concept/any information of the electron being present in the nucleus. Now, I split up the universe into small spheres, each of radius similar to nucleus. Applying Heisenberg's uncertainty principle to each such sphere, the electron can't exist at any of these. Hence, the electron doesn't exist.

Are my concepts wrong?

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    $\begingroup$ Welcome! Please see this guidance about screenshots (summary: avoid screenshots). $\endgroup$
    – rob
    Commented Nov 29, 2021 at 16:54
  • $\begingroup$ Just out of curiosity; what’s the name of the textbook? $\endgroup$ Commented Nov 30, 2021 at 4:46
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    $\begingroup$ The fact that it's not a formal publication is irrelevant. Surely it has a title and one or more authors. Any material that you post here which isn't your own work needs to be given proper attribution: state (at least) the title of the work, and its author(s), if possible. If the work is available online, it's good if you can also provide a link to it. Please see physics.stackexchange.com/help/referencing $\endgroup$
    – PM 2Ring
    Commented Nov 30, 2021 at 9:25
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    $\begingroup$ It would have been more accurate for the book to say the electron can't be confined to the nucleus rather then saying it can't exist there. $\endgroup$ Commented Dec 1, 2021 at 11:53
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    $\begingroup$ Ok @PM2Ring, I understand! $\endgroup$
    – Ravi Arora
    Commented Dec 3, 2021 at 14:33

7 Answers 7

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It’s a mistake to invoke relativity on the speed $v$ without also using the relativistic momentum $p$. But if this is a text for students who may be a little weak on relativity, it’s perhaps a pedagogically useful mistake. Most students come into a class knowing that the speed of light is a “speed limit,” even if they have only a vague idea of what happens as an object approaches that limit. Introducing all of the concepts you need to make this argument in a consistent way might be a lot to expect for the target audience of this text.

Physicists like to think in energy units. A correct version of this argument might be to replace

$$ \Delta x \Delta p \gtrsim \hbar $$

with

$$ \Delta x \cdot c\Delta p \gtrsim \hbar c \approx 200\rm\,MeV\,fm $$

(We won’t need to fuss about factors of two.) We know experimentally that an electron can be removed from an atom with an energy of a few eV. We suspect that the electron is held near the nucleus by electrical attraction, which is subject to the virial theorem, so its kinetic energy $T$ and its binding energy $U$ have the same magnitude, $T \approx -U$ (neglecting a factor of two). The uncertainty principle says that any particle confined to a nucleus, $\Delta x \lesssim 1\rm\,fm$, must have a momentum uncertainty $c\Delta p \gtrsim 200\rm\,MeV$. For an electron, mass $m_e c^2 \approx \frac12\rm\,MeV$, this momentum is hyperrelativistic, and the corresponding kinetic energy is $T\approx E\approx pc$. But if the electron were bound in a potential well with depth 200 MeV, you couldn’t ionize an atom with an eV-scale photon. Something’s got to give.

Note that if we make the same argument for a nucleon trapped in a nucleus, with mass $m_N c^2 \approx 1000\rm\,MeV$, we can mostly get away with using the nonrelativistic kinetic energy:

$$ T \approx \frac{p^2}{2m_N} = \frac{(cp)^2}{2m_N c^2} \approx 20\rm\,MeV $$

Actual nucleon-separation energies tend to be around 10 MeV, so if we assume $T\approx |U|$ this isn’t a bad guess at all. It’s reasonable to say that the size of the nucleus is directly related to the energies involved in the nuclear interaction by the uncertainty principle. And for that matter, the size of the atom is directly related to the scale of electron binding energies:

\begin{align} T_e &= \frac{p^2}{2m_e} \approx 10\rm\,eV \\ (pc)^2 &\approx 2 m_e c^2 T_e \approx 10^7\rm\,eV^2 \\ \Delta x &\approx \frac{\hbar c}{pc} \approx \frac{200\rm\,eV\,nm}{3\times10^3\rm\,eV} \approx \frac23 Å \end{align}

Of course, when you actually find a wavefunction, you don’t use the uncertainty principle: you use the wave mechanics of the Schrödinger equation. But the uncertainty principle is fundamentally a statement about how waves behave mathematically, so in factor-of-two land the result is fine. And if you leave factor-of-two land and compute carefully, you find that the uncertainties in position and momentum associated with an actual wavefunction are always larger than the minimum set by the uncertainty principle.

As for your specific complaints:

  1. The uncertainty has no relation with the real values.

This just isn’t so. Instead of a single atom, imagine an ensemble of 100 atoms, all fixed in place. If the electrons don’t leave their atoms, we know that the average vector momentum of the electrons must be zero. The uncertainty principle describes “one-sigma” uncertainties, so if you measured the electron momenta for your 100 atoms you expect about 32 of them to have momentum magnitude $|p|$ larger than $\Delta p$, and about 5 of them to have momentum magnitude $|p| > 2\Delta p$. (We talk more often about 68% and 95% “confidence intervals.”) If you picked an electron at random and guessed its momentum magnitude before measuring, $\Delta p$ is a better guess than zero.

  1. Now, I split up the universe into small spheres, each of radius similar to nucleus. Applying the Heisenberg's uncertainty principle to each such sphere, the electron can't exist at any of these.

This is actually a reasonable conclusion. The next step isn’t the one you take (“Hence, the electron doesn't exist”), but instead a counterintuitive statement about size: cold electrons are big, and can’t be confined. If you want an electron to be confined to a small volume, it has to be participating in some high-momentum interaction.

We often talk about an electron as a “point particle” with “zero size.” We say this because there doesn’t appear to be any other interaction that turns on when you probe an electron at short length scales. Such a short-range interaction would be a sign that the electron has some substructure. (For example, the nuclear interaction changes character at distances closer than about a femtometer, which is related indirectly to the composition of nucleons from quarks.) In the Copenhagen interpretation, and its bastard child the pilot-wave theory, we imagine there is a pointlike “real electron” that we can locate someplace. But as you have discovered, that’s inconsistent with the uncertainty principle. There are lots of situations where the mental model that “cold electrons are big” is helpful.

One physical consequence of “cold electrons are big” happens in white dwarf stars, which are held up by electron degeneracy. As you make the star hotter, the uncertainty on the momenta of the electrons $\Delta p$ gets bigger (because each electron is storing more energy on average). The extra uncertainty in $\Delta p$ allows the volume $(\Delta x)^3$ associated with each electron to shrink. White dwarfs get smaller as you heat them up, because cold electrons are big.

When I say that the mistake in your quoted textbook might be pedagogically useful, what I mean is that the textbook’s argument fits into five sentences and four lines of mathematics. My more-correct answer is substantially longer, and is shortened by plenty of mathematical “pay no attention to the man behind the curtain” which is easily justifiable but would make an intro student uncomfortable.

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  • $\begingroup$ I am really grateful for your answer, even though I understand only 10% of it haha. I am actually in grade 11, preparing for a college entrance exam, and about your point related to "for students weak in relativity", the answer is we have no idea what relativity is. This chapter is labelled atomic structure, and touches it just to the extent of the shapes, names and nodes of orbitals, and the formulae related to De Broglie Wavelength and Uncertainty Principle. So you are right about it being simplified for us I believe, but I just have a nature to go beyond in oversimplified things. $\endgroup$
    – Ravi Arora
    Commented Nov 30, 2021 at 6:52
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    $\begingroup$ Pilot wave theory is not inconsistent with HUP. HUP is about measurements, and in this regard PWT is totally compatible with it. The exact trajectories in PWT are hidden variables, which are irrelevant to HUP. $\endgroup$
    – Ruslan
    Commented Nov 30, 2021 at 11:46
  • $\begingroup$ If the mention of pilot waves is a distraction, I can edit it out. The problem of an unrealizable “real electron” is also present in the Copenhagen interpretation; it is a semiclassical crutch which doesn’t really survive into relativistic field theories. $\endgroup$
    – rob
    Commented Nov 30, 2021 at 16:21
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The solution given here is wrong. It uses the non-relativistic expression for momentum, and then compares the velocity with the speed of light, which is the velocity limit of relativistic mechanics. If we consistently use the relativistic expression for momentum,

$$p=\frac{mv}{\sqrt{1-v^2/c^2}},$$

the value of velocity will be much smaller, and less than $c$, leading to no contradiction with the assumption.

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  • $\begingroup$ so could you just take a moment to explain what exactly ΔV is? I'm a bit unclear on that. Is it like the range of possible velocities we might measure? I'll mark this answer as correct after that! $\endgroup$
    – Ravi Arora
    Commented Nov 29, 2021 at 17:18
  • $\begingroup$ See my answer here physics.stackexchange.com/questions/571412/… $\endgroup$ Commented Nov 29, 2021 at 19:18
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    $\begingroup$ If we use relativity, we still reach the conclusion that the trapped electron would be ultrarelativistic, with $\gamma\approx 10^3$, and that the electron cannot be trapped. See my answer on this question. $\endgroup$
    – rob
    Commented Nov 30, 2021 at 11:24
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The uncertainty principle is about x and p being Fourier pairs. That is, the standard deviation (loosely called the uncertainty) of the position of the wavefunction times the standard deviation of the momentum of the wavefunction must be greater than or equal to hbar/2. If the entirety of the wavefunction is compressed to the size of the nucleus, the corresponding minimum standard deviation in the momentum is so large that the attractive force of the nucleus cannot hold the electron anymore and the electron would very quickly escape into the vacuum (essentially, there are so many momentum states included in the electron wavefunction that are of greater energy then the attractive force that the electron just flies away). The bound state of the electron that is of the lowest energy (i.e. the 1s state) is essentially the point where the attractive force of the nucleus and the tendency for the electron to escape because of its inherent momentum are balanced out.

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This answer is complementary to the others: I feel the phrase "cannot exist inside the nucleus" is badly chosen, gives an impression inconsistent with the calculation it's paired with, and this may be an important source of the questioner's confusion.

The calculation shown (after corrections as described in the other answers) demonstrates that an electron cannot be confined inside a nucleus-sized sphere. That is, it cannot be in a state where its position is never outside that sphere.

But the plain-English meaning of "cannot exist inside" is different: "can never be inside". The calculation shown demonstrates no such thing, and indeed it's not true. If we think in terms of the "real pointlike electrons" that rob's answer objects to: atomic electrons are sometimes inside their associated nucleus. If we instead prefer the "cold electrons are big" mental model, atomic electrons occupy volumes of space that overlap the nucleus. (In terms of the actual math, atomic electrons' wave functions have amplitude $> \epsilon$ for space within as well as outside the nucleus. If I remember correctly, for the lowest-energy $s$ orbitals, the amplitude per unit volume is greatest inside the nucleus, even though the greatest overall probability is at a distance comparable to the Bohr radius.)

Atomic electrons being inside their associated nuclei "sometimes", has real observable physical consequences: electron capture could not occur otherwise.

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  • $\begingroup$ "the amplitude is greatest inside the nucleus" Amplitude isn't defined except with respect to some measure. If we look at amplitude per volume (which I can see why you would consider the natural interpretation), I think you're right, but it's not true for amplitude per $\Delta r$. $\endgroup$ Commented Nov 30, 2021 at 19:44
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    $\begingroup$ @Acccumulation I learned all this stuff from a theoretical chemistry perspective, so yeah, amplitude per volume is all we ever talked about. I don't even know what context would suggest the use of amplitude per $\Delta r$. $\endgroup$
    – zwol
    Commented Nov 30, 2021 at 23:29
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    $\begingroup$ There is a context to use amplitude per $\Delta r$: it's when you're calculating radial probability density. I.e. change to spherical coordinates, integrate out the angular degrees of freedom, and what's left to integrate over $r$ is this radial probability density. And, due to the $r^2$ term in the Jacobian of the Cartesian-to-spherical transformation, this density will be zero in the center, unlike the per-volume probability density. $\endgroup$
    – Ruslan
    Commented Dec 1, 2021 at 7:53
  • $\begingroup$ @Ruslan I missed your response until now -- I can see why, in these coordinates, the density will be zero at the origin, but nuclei have nonzero extent. What about density in shells spaced ~1fm apart, from the origin out to 11fm (uranium nucleus) versus density in shells close to the ~50pm Bohr radius? $\endgroup$
    – zwol
    Commented Dec 3, 2021 at 21:57
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    $\begingroup$ I'm not sure what your "what about" question is asking, but for a hydrogen-like atom/ion radial probability density is $\propto r^2e^{-2r/n}$, with the (single) maximum at the Bohr radius. So the probability of finding the electron inside the nucleus is tiny compared to finding it near the Bohr radius (assuming the same $\Delta r$ in both cases)—despite the volumetric probability density having a maximum inside the nucleus. $\endgroup$
    – Ruslan
    Commented Dec 3, 2021 at 22:14
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It seems a strange way to do it.

It's conceptually easier if we make $\Delta x$ the subject and argue, if it's greater than $10^{-15}m$ that the electron can't be contained in the nucleus.

Since we know nothing about the speed of the electron if it were in a nucleus, $\Delta v$ can be put to the maximum value of $3\times 10^8$ giving $\Delta x = 2 \times 10^{-13}m$, so it can't be contained in the nucleus.

If $\Delta v$ were lower, $\Delta x$ becomes higher.

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    $\begingroup$ Heisenberg's principle actually uses relativistic momenta, so that $p$ and $\Delta p$ can be arbitrarily large. I do not think there is way to prove the statement with similar arguments. $\endgroup$ Commented Nov 29, 2021 at 17:00
  • $\begingroup$ @Valter Moretti Good point, the other answer on your link is interesting $\endgroup$ Commented Nov 29, 2021 at 20:23
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The uncertainty has no relation with the real values. It is a degree of measuring how wrong we are rather than what wrong value we obtain.

Well, I suppose this depends a bit on what quantum interpretation you're using (there may be some truth to this in, say, the pilot wave interpretation), but in the Copenhagen interpretation, this is definitely not true; a measurement of the velocity is a random variable, and the uncertainty refers to the standard deviation of that random variable.

The Δv term should have nothing to do with the actual velocity and hence I believe this is wrong.

In the Copenhagen interpretation, there is no fixed velocity. The velocity is spread out over a range of values, and $\Delta v$ is a measurement of how spread out those values are.

Applying the Heisenberg's uncertainty principle to each such sphere, the electron can't exist at any of these. Hence, the electron doesn't exist.

The electron's wavefunction can't be entirely contained within any of these spheres. That doesn't mean it doesn't exist. The wording in the book is misleading, in addition to its math being wrong, but the basic idea is true: for most of the probability mass of the electron to be contained with a sphere of radius $10^{-15}$m , its velocity would have to be spread of an interval that includes the escape velocity from the nucleus, and so this is not a stable state.

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In the full calculation, the size of the electron cloud around the nucleus is determined by solving Schrödinger's equation, the solution of which is simply much larger than the nucleus. Classically you'd expect that it'd be possible for the electron to simply "stick" to the nucleus. The given text example is an attempt to give some intuition to the quantum result by approaching it from a different angle, showing that the attraction of the Coulomb force must in some sense fight against the kinetic energy of a confined electron implicit in the Heisenberg uncertainty principle.

I think that the text makes a slight pedagogical mistake in focusing too much on the speed of light, which is not really the point, and is leading to some accusations that the text is just straight wrong. The main point of this common calculation is that in quantum mechanics, because of the uncertainty principle, a particle which is especially positionally confined must have a high kinetic energy.

The problem could have gone one step further and calculated the energy corresponding to this velocity, but for those with more experience, as soon as you do this calculation and see $v\gg c$ you realize that such an electron must be very relativistic and have a huge energy which is way, way larger than the attractive potential energy, and thus is immediately implausible. I suspect this the author's thought process.

To respond directly, however,

(1) This is basically just incorrect. I would encourage you to look at how $\Delta x$ and $\Delta p$ are calculated given the real wave function, in terms of expected values. The principle very much tells you about possible real values of $x$ and $p$.

(2) This is correct until the last step. By this argument, a non-relativistic electron couldn't exist in any single nucleus-sized sphere in the universe, but it can exist spread out between many, which gives it a much larger $\Delta x$

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