What do principal quantum numbers actually represent?

The modern structure represent that electrons do not revolve around the nucleus. It says there are no shells or orbits for electron rather there is just a probability of finding them in a certain region of space.

The principle quantum number represents the shell or orbit or energy level of electron - so it simply doesn't make sense.

My question is what do principal quantum numbers actually represent? If they actually represent the energy level of the electron, then in what sense do they do so? Since an electron in one orbit or having same principal quantum numbers does not have the same energy. The energy of orbitals of an orbit or shell differ in energy.

So, it is nonsense to say that principal quantum numbers represent the energy level of the electron, because electron with the same principal quantum number can differ in energy.

So, if we classify different electrons to have same quantum number, then what is the reason behind this?


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  • $\begingroup$ en.wikipedia.org/wiki/Azimuthal_quantum_number#/media/… $\endgroup$ – safesphere Nov 13 '18 at 17:34
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    $\begingroup$ I don't think this question deserves to be downvoted. I think the simple answer is $n$ is simply an index and doesn't correspond to any interesting quantity but maybe someone has an interpretation. $\endgroup$ – jacob1729 Nov 13 '18 at 17:43
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    $\begingroup$ @jacob1729 $n$ is a degree of a Legendre polynomial associated with the spherical harmonic: en.wikipedia.org/wiki/Spherical_harmonics $\endgroup$ – safesphere Nov 13 '18 at 18:16
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    $\begingroup$ @safesphere the associated Legendre functions are only the radial solution for hydrogen. In a general atom $n$ is just a label. $\endgroup$ – jacob1729 Nov 13 '18 at 18:56
  • $\begingroup$ @jacob1729 Orbitals of any atom are spherical harmonics (s, p, d, f orbitals exist in any atom). So $n$ is a degree that defines a polynomial order (the range of the azimuthal quantum number). $\endgroup$ – safesphere Nov 13 '18 at 19:09

For hydrogen atom, all orbitals with equal principal quantum number $n$ have equal energy. That can be taken as a significance of $n$.

Also, for any atom, the principal quantum number determines the range of azimuthal quantum number, so it determines the range of angular momentum the electron can have.

  • $\begingroup$ This is because the less energy the electron has, the less energy is available to go into rotational energy/angular momentum. The more energy, the greater the possible angular momenta. $\endgroup$ – Mr. HelloBye Nov 14 '18 at 2:05

The principal quantum number is the label we use for unique energy eigenvalues for a given Hamiltonian. Based on your question, I believe I need to step back and explain what I mean by this.

The orbitals that you learn about in chemistry are technically the orbitals for a single electron and a nucleus (a "Hydrogenic" atom). When you have multiple electrons, it gets far more complex, as the energy of one electron depends on where all of the other electrons are, and the nucleus can be screened by the other electrons. So it’s not accurate for multielectron systems, but it’s close enough to get the rough idea across, and solving the entire problem would be too complicated, so it is good enough.

Ok, fine, so what is the problem that we are solving with the hydrogen atom then? Well with any quantum system, solving the full time dependent Schrödinger Equation is pretty hard, and for all but a few systems (infinite well, harmonic oscillator, step function, linear, hydrogen atom) you can’t even solve the time independent equation at all by hand. I’m not sure how much you know, but the Schrödinger equation describes the motion of non-relativistic particles, that’s why I’m referring to it.

It turns out that if the Hamiltonian is time independent (which is the case for hydrogen), that the solutions are of the form $\Psi(\vec{r},t) = \sum_n c_n\psi_n(\vec{r}) e^{\frac{E_n}{i\hbar}t}$, i.e. you can separate the time dependence. This means that if you can find the $\psi_n(\vec{r})$ (the so-called "energy eigenfunctions"), then you can expand any initial state in terms of those functions. Then to see how the wave function changes in time, you just plug in the time into that exponential.

The idea is that we find states whose amplitude (where we interpret the square of the amplitude as the probability density) is time independent, i.e. constant in time, (however phase changes at a rate proportional to $E_n$, the energy eigenvalue). This is what solving the time independent Schrödinger equation is.

When you find the set of energy eigenfunctions, it is possible that multiple functions share the same eigenvalue. This is the case for hydrogenic atoms, where different $l$ and $m$, which are the eigenvalues that correspond to $\vec{L}^2$ and $L_z$, the magnitude of the angular momentum squared, and the z component of the angular momentum respectively. These operators commute with the Hamiltonian, which in this context means that their eigenstates are also time independent.

So in short, the principal quantum number $n$ is a label that tells you what the energy of the state is, and the others are related to orbital angular momentum. The reason that different $l$ values for the same orbital give you different $n$ is that some of the energy is rotational.

The solutions for the hydrogen atom can be separated into a radial and angular part, $\psi_{n,l,m} = R_{n,l}(r)Y_{l,m}(\theta, \phi)$, where $Y_{l,m}$ are the spherical harmonics and $R_{n,l}$ are the radial wavefunctions https://quantummechanics.ucsd.edu/ph130a/130_notes/node237.html#derive:Hradial

Note that the radial wavefunctions are determined by BOTH $n$ and $l$. And the higher $l$ is, the fewer antinodes there are in the radial component. However, a general trend in quantum is that the more antinodes there are, the higher the energy is. So where did the antinodes go? They went to the spherical harmonics. If you look at a table of spherical harmonics, you see that the higher $l$ is, the more antinodes there are, while $m$ just changes where the antinodes are. https://en.m.wikipedia.org/wiki/Spherical_harmonics#/media/File%3ARotating_spherical_harmonics.gif

Something to note, since the electron is moving rather fast around the nucleus, the electric field becomes Lorentz transformed a slight amount into a magnetic field which couples to the spin of the electron. This means that spin up has a slightly different energy from spin down. This breaks the symmetry of the Hamiltonian with respect to spin, and as a result “lifts the degeneracy” of the states. This is called hyperfine splitting, and causes spectral lines to split, which you can see if you look very closely at the spectrum.


It is correct that the quantum mechanical solution gives orbitals and not orbits.

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The solution represents the locus of the orbitals, which are not over the whole space but constrained by the wavefunction probability solution, as seen above for the hydrogen atom.

In the momentum enegy phase space you misunderstand the probabilistic nature of quantum mechanics:

So, it is nonsense to say that principal quantum numbers represent the energy level of the electron, because electron with the same principal quantum number can differ in energy.

There is a width to the energy levels that the electron can occupy, so the energy has an uncertainty, but the value is specific and thus one gets the light spectra from transitions between energy levels, which have a width. See the answers here to see how a width appears. For detailed calculation see here.

  • $\begingroup$ I don't think this answers the OPs question which to my reading is more "what is the physical significance of n". Since it by itself does not determine the mean distance or even the energy when discussing multi-electron atoms. $\endgroup$ – jacob1729 Nov 13 '18 at 17:41
  • $\begingroup$ @jacob1729 That is why I discussed the simple hydtrogen atom where the n corresponds to the energy levels as studied experimentally through the spectra and the series that fit them. For hydrogen it is the main quantum number for energy, the others just modify , adding extra levels around it. $\endgroup$ – anna v Nov 13 '18 at 18:11
  • $\begingroup$ continued "fine structure and hyperfine structure " hyperphysics.phy-astr.gsu.edu/hbase/hyde.html#c6 $\endgroup$ – anna v Nov 13 '18 at 18:39

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