# Orbital wave functions and probability density - interpretation issue

The chronology in which concepts are taught in university physics classes I feel sets students up to have to "unlearn" some earlier concepts in order to grasp new ones, and I think it's why I'm stuck on a conceptual issue with the atomic orbital wavefunction, which is hanging me up on wavefunction interpretation in general.

The assumptions I'm starting with are:

• The orbital wavefunction is a non-physical calculational tool, the absolute square of which is proportional to the probability of interacting with an electron in a given region

• The atomic orbitals are constrained both circumferentially and radially by the de Broglie wavelength of the electron

• Losing or gaining a whole number of wavelengths (excluding other factors like angular momentum, etc) neatly explains the electron transition between orbitals

• The transition from a higher energy orbital to a ground state is accompanied by the creation of a real photon of energy proportional to this transition.

Here's my issue:

If the wavefunction is non-physical, and the square of the wavefunction describes probability only, it's difficult for me to conceptualize why gaining or losing a whole number of wavelengths of probability should result in the creation of a very physically real and measureable photon of discrete energy.

It seems like we're having our cake and eating it at the same time. If the wavefunction describes a physically distributed electron wave that is gaining and losing whole-number wavelengths as it transitions orbitals, that these transitions would be associated with photon absorption/creation makes intuitive sense.

Once we say "The orbitals are probability densities only with no physical existence, the electron remains a point charge at all times" then the concept of gaining or losing a wavelength in the probability distribution seems like we're connecting an abstract mathematical construct to the creation of physically real particles, and then converting back again.

Add in that we're later taught that the Born rule is really a non-relativistic approximation that doesn't apply once you upmode to the Dirac equation, and I'm left wondering whether "probability waves" are really the most accurate way to think about atomic orbitals, and the wavefunction in general.

'Shut up and calculate' works too, but I don't think I'm alone in wanting to hold at least the most up-to-date conceptual framework in my head while I trudge through the maths.

John Rennie's answer here I think is close to what I'm looking for: If orbital shells are just probability functions, why are quantum numbers only ever integers?

Though if I'm interpreting his answer correctly we should stop referring to the square of the wavefunction as a probability density for the position of the electron because there simply IS no position for the electron while it's in an energy eigenstate, in which case it seems to me thinking of it as truly "smeared out" in space would be more accurate than picturing it as a point particle with a probability of being in a specific (x,y,z). But I've been repeatedly informed that the "smeared out" picture isn't right either.

• Add in that we're later taught that the Born rule is really a non-relativistic approximation that doesn't apply once you upmode to the Dirac equation I don't think this is quite right. If you take the Born rule to mean that the norm of a state vector is a probability measure, then this is the same in QFT as in nonrelativistic QM. What is different in QFT is that we don't have states that are simultaneous states of good position and good particle number. (Heuristically, a state of good position has infinite spread in energy, and therefore has enough energy to create any number of particles.)
– user4552
Feb 13, 2019 at 22:42
• Losing or gaining a whole number of wavelengths (excluding other factors like angular momentum, etc) neatly explains the electron transition between orbitals This sounds like a misunderstanding to me. Orbitals are what they are simply because they're solutions of the Schrodinger equation that have definite energy. This has nothing to do with photons or transitions. If you measure energy, you see states of definite energy.
– user4552
Feb 13, 2019 at 22:49
• Why do you say that something that tells you how likely the electron is to be found in various places is “non-physical”? Feb 13, 2019 at 22:55
• I think your emphasis on the de Broglie wavelength is misplaced. This notion is well-defined for a free electron but not for a bound one. All that matters is the Schrodnger equation. The wave function contains all the information about the state of the electron, so I think of it as very much “physical”. Feb 13, 2019 at 23:08
• Regarding my contention that the concept of the de Broglie wavelength is ill-defined for a bound electron... The ground-state wave function of hydrogen is $Ae^{-r/a_0}$ where $a_0$ is the Bohr radius and $A$ is a normalization constant. This doesn't look anything like a periodic spatial wave with a "wavelength", either in the radial direction or in the tangential direction. Feb 14, 2019 at 3:20

We're always converting between "abstract mathematical constructs" and "physically real entities" when we do physics. That's what having a mathematical model that predicts what happens in (an idealization of) the world means. This is not at all unique to quantum mechanics, it is inherent in the idea that mathematics can tell us anything about the real world.

For a purely classical instance of this, see e.g. the question " How can energy be useful when it is 'abstract'?".

For a famous essay pondering the larger philosophy behind the use of mathematics in physics, see Wigner's "The Unreasonable Effectiveness of Mathematics in the Natural Sciences".

Furthermore, the "most accurate way" is not always the most useful way to think about a problem. Yes, sure, our most "accurate" understanding of the quantum world is not non-relativistic quantum mechanics as wave mechanics, but relativistic quantum field theory, just like our most "accurate" understanding of gravity is not Newtonian gravity, but general relativity.

But we don't use the most "accurate" understanding for predicting everything. No one does general relativistic calculations to figure out how long a thrown rock will take to hit the ground, and likewise, no one uses full quantum field theory to get a picture of what orbitals are.

Finally, you should be careful to distinguish between the formal predictions of quantum mechanics - "The square of the wavefunction is the probability density to detect the electron at a particular place and time" - and its interpretations - e.g. "The electron really is smeared out and has no position", "The electron has a definite position and is guided by the pilot wave", "There is a world for every possible position of the electron", etc. It is the former which are objective and can be experimentally tested, while the latter cannot.

You are right. One has to learn stuff and then unlearn it again.

The picture of a de Broglie electron wave neatly fitting a whole number of wavelengths into the circumference of a Bohr orbit - which lies behind your assumptions 2 and 3 - and is nicely explained, for example, in https://physics.stackexchange.com/a/318638/194034 is, at a higher level, just plain wrong.

Electrons do not orbit the nucleus like planets going round the sun. They have a wave-function which has to be a solution of the Schrodinger equation (or the Dirac equation - relativity is not the issue here).

But we will go on teaching the Bohr model, as it is a useful step up the ladder of fuller understanding.