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We know that the total energy of the hydrogen atom is proportional to the inverse of the square of the principal quantum number $n$:

$$E_n \propto -\frac{1}{n^2}$$

So at high quantum numbers the energy spectrum tends towards a continuum.

Shown below, a representation of one of the seven $\text{6f}$ orbitals (courtesy of The Orbitron):

f orbital

However, due to the Correspondence Principle, high quantum number hydrogen atoms should show wave functions that tend toward Classical orbit-like (instead of orbital) shapes:

electron orbit

This is so, at least according to a video I watched yesterday (unfortunately I don't have the web address)

If this is true, high equantum number orbitals would to become Bohrian in nature and thus emit electromagnetic radiation.

Is this true?

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  • $\begingroup$ I think you are confusing the binding energies of a neutral hydrogen atom with the possible kinetic energy of an atom of hydrogen, different variables. High energy orbitals (at most ~13ev) can be occupied by the electron if energy is absorbed in some form, and will be given up by cascading to the ground level. A neutral hydrogen atom can gain very high energy in a thermal environment ( as the sun for example). $\endgroup$
    – anna v
    Commented Nov 23, 2021 at 18:21
  • $\begingroup$ @annav I don't think the OP was referring to the kinetic energy with 'high energy; but to the excitation energy. $\endgroup$
    – Thomas
    Commented Nov 23, 2021 at 18:47
  • $\begingroup$ @Thomas does 13 ev sound high to you? $\endgroup$
    – anna v
    Commented Nov 23, 2021 at 18:51
  • $\begingroup$ @annav High energy orbitals are those approaching 0 eV from below. The ground state of hydrogen has an energy -13.6 eV. $\endgroup$
    – Thomas
    Commented Nov 23, 2021 at 18:56
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    $\begingroup$ "high quantum number hydrogen atoms should show wave functions that represent Classical orbit-like (instead of orbital) shapes" — no, high-$n$ eigenfunctions are still the same Laguerre-times-exponential functions multiplied by spherical harmonics. To get a classical-like orbit you need a superposition of eigenfunctions, as I constructed here. $\endgroup$
    – Ruslan
    Commented Nov 24, 2021 at 11:43

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A "Bohr orbit" is related to a classical orbit via the correspondence principle, but not all sets of quantum numbers for hydrogen-like atoms correspond to Bohr orbits. Quantum mechanics is richer than Bohr initially imagined.

Let's consider a hydrogen-like atom with quantum numbers $(n,\ell,m)$. The radial part of the wavefunction is

$$ R_{n\ell}(r) = \sqrt{\text{stuff}}\times e^{-u/2} u^\ell L_{n-\ell-1}^{2\ell+1}(u) \quad\quad \text{where } u = \frac{2Zr}{n a} $$

where $r$ is the radial coordinate, $Z$ the nuclear charge, and $a$ the Bohr radius. The $L_\alpha^\beta$ are the associated Laguerre polynomials, which are polynomials of order $\alpha$. So for an orbital with maximal angular momentum $l=n-1$, the Laguerre stuff is just a constant, and the radial wavefunction $R\sim e^{-r} r^\ell$ has a zero at the origin (for nonzero $\ell$) and a single maximum at some finite $r$. For large $n,\ell$ this single peak is narrow and it makes sense to think of the electron as being "radially localized," like a particle in a circular orbit.

Similarly, the spherical harmonics are given by an associated Legendre polynomial in the variable $\cos\theta$, multiplied by an azimuthal phase $e^{im\phi}$. The extremal Legendre polynomials all have the form

\begin{align} P_\ell^\ell (x) &= (\text{constant}) (1-x^2)^{\ell/2} \\ P_\ell^\ell (\cos\theta) &= (\text{constant}) \sin^{\ell}\theta \end{align}

So for a hydrogen electron where the angular momentum projection is maximized, $|m|=\ell$, the angular probability distribution is strongly peaked at the equator; that peak gets narrower for larger $\ell$. The complex phase increases linearly as you go around in $\phi$.

Taken together, an electron in a hydrogen-like orbital with maximum angular momentum and maximum angular momentum projection onto the $z$-axis, with quantum numbers $(n,\ell,m) = (\ell+1,\ell,\pm\ell)$, has its probability concentrated in a narrow ring around the equator of the coordinate system, with the phase changing around the ring. If you include the time-dependent part of the wavefunction, multiplying by $e^{-i\omega t}$, you have a phase change with time which you can use to find a "probability current." This is the Schrodinger version of a Bohr orbit.

If $\ell$ is large enough that the Bohr orbit is a good description, then the atom can in fact radiate —— not continuously, but by emitting photons and going to orbits with smaller $n$, and eventually to the ground state.

Beware of visualizations like the orbitron, linked in your post. For the case of the $p$-orbitals ($\ell=1$), chemists like to refer to the spatially-oriented $p_x$ and $p_y$, which are rotated versions of $p_z$. However, $p_z$ has definite $m=0$; the $m=\pm1$ states are linear combinations of $p_x \pm i p_y$. For the higher $\ell$, there are more choices to make. I think the chemists' approach is to combine wavefunctions with equal-magnitude $m$, so that the combined wavefunctions are real.

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The answer by @DinosaurEgg addresses the problem of how a neutral hydrogen atom can emit electromagnetic radiation. It does not address the misconceptions in the question.

Would a high energy Hydrogen atom start emanating electromagnetic radiation?

We know that the total energy of the hydrogen atom is proportional to the inverse of the square of the principal quantum number n:

$$E_n \propto -\frac{1}{n^2}$$

This is not the energy of the hydrogen atom, but the difference in energy from the ionization level. Interpreted correctly it means that for high $n$ levels very little energy is needed to ionize the atom.

Nothing to do with high energy.

high energy orbitals would be Bohrian in nature

The italics should be corrected to "high n orbitals"

the high n orbitals are closer to the semiclassical Bohr model

The following is a non sequitur,

and thus emit electromagnetic radiation.,

It does not follow, as explained in the answer by @DinosaurEgg. Once an electron is raised to a high $n$ level by some incoming radiation or field, i.e. energy is absorbed by the neutral hydrogen, a soft cascade of photons from the great multiplicity of levels at high n is possible. But it is also possible that it will go directly to a lower level or the ground state, depending on the quantum mechanical probabilities.

Maybe this link will help in studying the hydrogen atom .

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  • $\begingroup$ Thanks. +1. I've amended the question. But I don't really need a basic course in hydrogen QM, thank you! ;-) Question aside: the $\Psi_{n,l,m}$ are functions of $r$ only. How then, at very high $n$, would these 'flatten' to semi-classical orbits? $\endgroup$
    – Gert
    Commented Nov 24, 2021 at 11:41
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    $\begingroup$ see what orbitals look like commons.wikimedia.org/wiki/File:Hydrogen_Density_Plots.png . For very large n the bright, high probability region, may become thin, enough to approximate a bohr orbit. $\endgroup$
    – anna v
    Commented Nov 24, 2021 at 13:14
  • $\begingroup$ Certainly the maximum (stemming from the Laguerre polynomials) of $|\Psi|^2$ of a high $n$ $s$ orbital would become very high, compared to the maxima left and right of that highest maximum, so that region is where the electron would predominantly 'be'. $\endgroup$
    – Gert
    Commented Nov 24, 2021 at 15:01
  • $\begingroup$ @annav By convention, the energy of a bound system is negative, of an unbound system positive. It is zero at the ionization energy. So the energy of bound hydrogen state is $E_n=-1/n^2$ $\endgroup$
    – Thomas
    Commented Nov 24, 2021 at 19:43
  • $\begingroup$ @Thomas you can call it the potential energy, since it is negative, and it is avery very small negative number for high n $\endgroup$
    – anna v
    Commented Nov 25, 2021 at 5:08
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No, you did not break quantum mechanics just because the high energy orbitals are more classical in nature. Think about it like so: suppose an electron is barely bound to the hydrogen atom. Suppose there is an EM field at its ground state with no photons present (we need the field to induce transitions in the hydrogen atom). The electron can now decay to a lower energy state by emitting photons. Because the electron is in such a high quantum level of the hydrogen atom, it can decay by emitting photons of very small energy. To the observer, the emission radiation of a highly excited hydrogen atom gas would approximate a continuum at high energies, and therefore these electrons seem to behave classically at temperatures of about $\sim10-13 ~\text{eV}$, emitting continuous spectrum radiation and supposedly violating quantum mechanics.

However, this classical radiation pattern cannot be continued for arbitrarily low energy. Suppose the electron has been emitting small classical looking packets of energy for a while, becoming more and more strongly bound to the hydrogen atom. At some point, given by the resolution of the experiment and the relevant decay times, the experimentalist is going to start seeing radiation only at certain, clearly discrete wavelengths. The electron has become strongly bound again and has lost its classical character because it came closer to the nucleus, where the transition energies cannot be approximated by a continuum, so they can be deemed classical.

Lesson of the day: Just because something can be approximated as behaving classically in certain regimes of the parameters of the system (high temperature etc.) does not mean it's actually classical. To invoke a theory that doesn't hold in the regime it is to be used to infer results constitutes a logical fallacy. The correct microscopic theory is decidedly non-classical and it is the one that results like stability of nuclei need to be based on.

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  • $\begingroup$ What gives you the idea that highly excited atoms would emit continuum radiation? They emit a discrete line spectrum like the lower states as well (see en.wikipedia.org/wiki/Hydrogen_spectral_series ). These lines are merely broadened either by collisions with other atoms or by the finite lifetime of the level as for lower levels As the lifetime of a level increases strongly with n, lines originating from highly excited states tend actually to be sharper than for lower states. $\endgroup$
    – Thomas
    Commented Nov 24, 2021 at 19:36
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    $\begingroup$ You clearly do not understand what the gist of my answer is. I am not talking about broadening effects coming from many-particle considerations. What I am talking about is a hypothetical situation where you would be able to measure soft photons coming from transitions like $n\to n\pm1$ for large $n$. These peaks are so close together in energy that they effectively would look like a continuum around $E=0$ to a machine with a certain energy resolution. All I'm saying is that the classical radiation law should emerge for VERY low transition energies. $\endgroup$ Commented Nov 24, 2021 at 19:49
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    $\begingroup$ Yes there will be decays $n\to 1$, sure, but I believe multiphoton decays with small energy denominators should exist and be detectable with a sufficiently low energy resolution. I also have a hard time believing that the $n-th$ energy level is more stable in the presence of an EM field than the ground state. If anything I would assume the opposite. Can you cite your sources or explain why this is the case? $\endgroup$ Commented Nov 24, 2021 at 19:56
  • $\begingroup$ The classical radiation law would not produce any lines. It would make the electron to continuously lose energy and spiral into the nucleus. This is what the OP was referring to, but this is obviously not observed. $n\rightarrow n-1$ lines for $n$ into the hundreds are indeed observed in astronomy. Their line width decreases like $1/n^5$ (see my website plasmaphysics.org.uk/#atdecay (Atomic Decay Probability)), which is faster than the distance between lines decreases ($1/n^3$). Only collisional broadening and the instrumental resolution sets a limit to this. $\endgroup$
    – Thomas
    Commented Nov 24, 2021 at 22:16
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    $\begingroup$ I don't think we disagree so I fail to see what this conversation is leading us to. My point is that at $n$ large the peaks are getting denser. That's all. After all it's the OP that shows a misunderstanding of the concepts and I'm defending QM. Thank you for the reference. $\endgroup$ Commented Nov 25, 2021 at 1:15
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High energy orbitals can in some respects be represented by classical Bohr orbits but have strictly speaking still to be described my quantum mechanical wave functions, especially when it comes to calculating the transition probabilities to other states. High energy orbitals of all neutral atoms become increasingly hydrogen-like though for increasing n, which makes the transition probabilities easier to calculate. In any case, radiation is only emitted when the electron makes a transition to a lower level. This is a result of excited atomic states being quantum mechanically unstable (whether n is large or not), not because of the electron radiating as a classical particle. A classically radiating electron would continuously lose energy (producing a continuous spectrum over a wide frequency range in the process) and eventually spiral into the nucleus. This is obviously not observed. One only observes the discrete lines resulting from quantum mechanical transitions from level n to lower states.

The plots below (produced at https://keisan.casio.com/exec/system/1224054805 ) show that, as mentioned above, the wave functions become relatively more sharply peaked with increasing n, but still are wave functions with a continuous spread and do not represent classical orbits.


                                       n=3 (l=2)

probability density for n=3 (l=2)


                                       n=10 (l=9)

probability density for n=10 (l=9)


                                       n=100 (l=99)

probability density for n=100 (l=99)

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  • $\begingroup$ In any case, radiation is only emitted when the electron makes a transition to a lower level. Sure, but I'm talking about EM that results from a charged particles running in circles. It's what 'killed' the Bohm model of atoms. $\endgroup$
    – Gert
    Commented Nov 23, 2021 at 19:03
  • $\begingroup$ @Gert As I said, the concept of QM wave functions applies to high orbitals as well. It is just that the maximum of these wave functions is relatively more concentrated at a radius where you would expect the electron in the Bohr model as well. It does not mean you are suddenly dealing with classical particles. $\endgroup$
    – Thomas
    Commented Nov 23, 2021 at 19:31
  • $\begingroup$ I don't think you're understanding me. Nowhere did I intimate that 'you are suddenly dealing with classical particles' Your answer doesn't address the question. $\endgroup$
    – Gert
    Commented Nov 23, 2021 at 19:33
  • $\begingroup$ @Gert Yes, you were suggesting high energy orbits are Bohrian in nature whereas low energy orbitals are not. This is completely unfounded and wrong. The nature of the orbits is in either case only described by the wave functions, conceptually and quantitatively. $\endgroup$
    – Thomas
    Commented Nov 23, 2021 at 19:42
  • $\begingroup$ I like the graphs. They reflect what I commented on another answer. Thanks. +1. $\endgroup$
    – Gert
    Commented Nov 25, 2021 at 1:09
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The physics others said in their answers are bang on. But we can also get to the error by saying that the Correspondence principle doesn't say what you say it does. That is because there is no single Correspondence principle.

Bohr coined the principle. Philosophers, historians, and physicists who study these things can't extract one principle from Bohr's writings. Scholars find he had three different versions of the principle throughout his career.

In addition, Bohr was notoriously vague. He even lamented that he couldn't articulate himself to his colleagues. I would go further and say his thinking and writing were incomplete and wishy-washy. Even if he had a single Correspondence principle, I would not trust it as a tool to make broad conclusions.

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