# What do atomic orbitals represent in quantum mechanics?

I am learning the basics of quantum mechanics and am familiar with the Schrödinger equation and its solution, but I was confused about what the familiar atomic orbital shapes represent?

Do they represent nothing physical and are just plots of the wavefunction in 3D polar co-ordinates? Or do they represent the region where probability of finding an electron is $$90\%$$? Or something else?

Levine 7th ed. states that

An atomic orbital is just the wavefunction of the electron

In atomic theory and quantum mechanics, an atomic orbital is a mathematical function describing the location and wave-like behavior of an electron in an atom. This function can be used to calculate the probability of finding any electron of an atom in any specific region around the atom's nucleus. The term atomic orbital may also refer to the physical region or space where the electron can be calculated to be present, as predicted by the particular mathematical form of the orbital

• If you read carefully the answer is in the boxes. As for if you know X then you know a function of it. As a chemist I would mostly call orbital the region in which there is a certain probability of finding the electron. But it is fine or at least clear that the wavefunction can be named orbital. Jan 14, 2021 at 10:58
• The last sentence in the quote from Wikipedia is a widespread nonsense. If one states that the orbital is a function, think how weird it would sound the following statement: "the function $e^{-x^2}$ is the subset of the real numbers where it is not zero". I have just applied the quoted definition of orbital to a gaussian orbital in 1D! Even other rephrasings mentioning a probability threshold wouldn't be much better. Jan 14, 2021 at 13:34
• @GiorgioP the sentence you define as non sense it is basically what I wrote in my comment above. I am not sure of the use of words among quantum specialists, but surely the term orbital can refer to both the wavefunction as well as to the region in which, according to the same wavefunction, one calculate the probability of finding the electron satisfies a threshold. The complex nature of the wavefunction is considered by signs. I even don't see the reason for the question, especially considering that OP seems to know things already. Jan 15, 2021 at 7:57

(Disclaimer: I am only a highschool student and have learned the following mostly on my own. If there are any mistakes, please feel free to correct me!)

An atomic orbital represents the probability distribution* of the location of an electron around the nucleus and is mathematically described by a wave function.

Now what does this mean? Let's start with what an atomic orbital isn't:

• An orbital is not a fixed spatial region or a "container" in which an electron can move around - In Quantum mechanics, an electron does not have a specific location.

So what is an atomic orbital?

• As mentioned before, the electrons don't have a fixed position (and momentum, but this seems less relevant to me at this point), so we cannot determine its position to a single point - this only happens when we measure the position.

• When we measure the position, we find it to be more likely to be present at some points than at other points. This is what is meant by the probability distribution - it simply describes the probability of "finding" an electron when measuring its position for every point in space. So theoretically, there is a probability that at any point in time, some electron is 100km away from the atom it belongs to, but this probability is extremely small. (see What is the probability for an electron of an atom on Earth to lie outside the galaxy?)

• Now assume that we measure the position of the electrons for 1000 times and plot the measured positions to some 3-dimensional model of our atom. We will find that in 90% of the cases the electron is in a certain area of space and this is usually depicted by the familiar atomic orbital shapes:

(Source)

So the shapes of the orbitals as they are most often depicted is usually chosen in such a way that the probability of finding the electron inside this shape (when measuring its position) is at least 90%. However, note that the electron is not constrained to this shape and there is a probability that it is measured outside.

There are some other things to mention about orbitals apart from their "shape". One of these is that every orbital has a certain energy level associated with it. This means that when an electron is in an orbital $$A$$ it has the exact energy associated with $$A$$.

If there is another orbital $$B$$ with higher energy level than $$A$$, the electron in $$A$$ can "jump" to $$B$$ if it absorbs the exact amount energy which is the difference between the energy levels of $$A$$ and $$B$$. The most common example is an electron absorbing a photon which has the wavelength that corresponds to the energy differents of the orbitals. Likewise, electrons can jump to an orbital with lower energy by emitting a photon with the wavelength corresponding to the difference in energy between the orbitals.

Here is a graph showing the relative energy levels of some atomic orbitals:

(Source)

I hope this somewhat clears up the confusion.

*As mentioned in the comments, the wavefunction $$\psi$$ describing an atomic orbital does not directly give the probability density, but the probability amplitude. The probability density can be obtained by $$|\psi |^2$$ for complex orbitals or $$\psi ^2$$ for real orbitals.

• That’s a good answer for someone in high school! Actually, it’s a good answer full stop. You have my upvote! Jan 14, 2021 at 8:43
• Great! There's no rush, since Is there any significance of atomic orbitals? linked above by A.P. explains the connection between orbitals & energy. Jan 14, 2021 at 9:24
• Good answer for a high school student. However, at least a key point should be stated explicitly if you don't want to misguide people. The orbital does not represent the probability distribution directly. It is the ingredient to get the probability distribution, after evaluating the square modulus, if it is a complex orbital or the square if it is a real orbital. It is true that the usual graphic representation often uses squared modulus, but it is also true that orbitals are used with the algebraic sign to discuss their linear combination. That is the basis to understand chemical bonds. Jan 14, 2021 at 13:26
• @Jonas GiorgioP's comment is correct and extremely important. There is a very important difference between the probability density (which is compatible with classical probability) and the probability amplitude (which carries a complex phase (or, in simpler cases, a sign) that forms a crucial ingredient of the physics). Orbitals are probability amplitudes. Jan 14, 2021 at 16:40
• @Jonas great answer indeed but unfortunately this doesn't explain my fundamental question to the degree I would like. Jan 14, 2021 at 16:51

Let me split up your sources into Levine

An atomic orbital is just the wavefunction of the electron

as well as Wikipedia part 1

In atomic theory and quantum mechanics, an atomic orbital is a mathematical function describing the location and wave-like behavior of an electron in an atom. This function can be used to calculate the probability of finding any electron of an atom in any specific region around the atom's nucleus.

and Wikipedia part 2.

The term atomic orbital may also refer to the physical region or space where the electron can be calculated to be present, as predicted by the particular mathematical form of the orbital.

With this in place:

• Levine and Wikipedia part 1 are in complete agreement. Wikipedia is a more detailed (but less precise and more talkative) description of the same concept.
• Wikipedia part 2 presents notation which (i) is indeed used in introductory textbooks, but which (ii) is not used in any professional capacity in research or engineering in quantum mechanics.

What orbitals really are is wavefunctions $$-$$ this is what the term is understood to mean in the full theory of quantum mechanics. And, as wavefunctions, orbitals are also associated with probability distributions (though it's important to remember that the wavefunction carries more information than just the probability distribution), and those probability distributions are similarly associated with the spatial regions where they are supported.

In introductory texts it is sometimes useful, for didactic purposes, to identify the orbital with this spatial region, and you can sometimes get relatively far on this notion, but it is important to keep in mind that this is a 'lie to children' and that in the full theory 'orbital' implies a wavefunction.

• This is the kind of precise answer I wanted.(not 100% precise though, that violates Heisenberg's uncertainty principle in the way that you know precisely what I want to know thus altering the original state of what I wanted to know) XD Jan 14, 2021 at 17:01
• Especially loved the "lie to children" inclusion! Jan 14, 2021 at 17:02

If you take any linear solution $$\Psi(r,\theta,\phi)$$ to Schrödinger's Equation in 3 dimensions (spherical coordinates $$(r,\theta,\varphi)$$) and a probability $$P = \vert \Psi \vert^2$$, representing the wave function of your atomic orbital, you can "split it" in both radial and angular functions:

$$\Psi(r,\theta,\varphi) = R(r)Y(\theta,\varphi)$$

(note that $$R$$ and $$Y$$ implicitly depend on atomic numbers, so are different for different atomic orbitals).

Then the representation we have of atomic orbitals is a 3-D plot of both radial probability density $$D_r = r^2\cdot R^2(r)=\frac{\mathrm{d}P(r)}{\mathrm{d}r}$$ and angular probability density $$D_a = Y^2(\theta,\phi) = \frac{\mathrm{d}^2P(\theta,\varphi)}{\sin\theta \mathrm{d}\theta\mathrm{d}\varphi}$$

evaluated and plotted in spherical coordinates around your atom.

It's important to note the atomic orbitals are approximations. In the context of the basic hydrogen atom Schrödinger Equation, they are exact eigenstates of energy, total angular momentum squared, and $$L_z$$, where $$z$$ points in any direction you want it to.

As energy eigenstates, they are stationary states, and their time evolution involves a global phase rotating with frequency $$E/\hbar$$. As such, they can never change, which obviously contradicts experiment. Call this "problem 1".

Also: in quantum mechanics, the electron is a point particle. This leads to problematic interpretations that have their uses, but are not fundamental. One of these interpretations is that electron moves randomly in a fashion that has it inside an orbital boundary 90% of the time. Call this "problem 2".

Both these problems are addressed in quantum field theory, in which the electron is no longer a point particle, but the minimum excitation of the electron field, a spinor field that fills all space. With that, an orbital describes how the electron field excitation of a single electron is spread out over space in an approximate energy eigenstate, and how it propagates in time.

The wave function then represents the complex quantum amplitude, whose modulus squared is the probability density of the electron's location. There really is no intuitive (or classical) way to understand coherent complex amplitudes of fermion fields, other than it's kind of like how we treat light...but with conserved quantum numbers, antiparticles, and Fermi-Dirac statistics.

The quantum field treatment also applies to the electromagnetic field, which then adds an interaction term to the hamiltonian, and allows transitions between states. It also adds virtual electron positron pairs to the binding, and that's only at the 1st order. The actual complexity of the state is beyond calculation.

With that, I would say the wave function is a mathematical approximation to something physical. I do believe this conundrum is the origins of Feynman's two famous quotes on quantum mechanics:

The disheartening,

" I think I can safely say that nobody understands quantum mechanics."

and the practical,

"Shut-up and calculate"

• this sure is enlightening! Jan 14, 2021 at 16:57
• I did not know QM considered the electron as being a point particle. Jan 15, 2021 at 8:38
• As to "problem 1" it is helpful to consider a 440 Hz tone produced by an tuning device. Unless this sound wave was around since eternity with constant amplitude and frequency, it is not a pure monochromatic wave but has a spectral distribution. So problem 1 is a universal statement about waves and not a specific QM problem. Jan 15, 2021 at 13:23