# What significance can we assign to 'breathing' of $s$-shell orbitals?

In my atomic lecture notes it says

We can visualise an $$l=0$$ $$s$$-state as a spherical cloud expanding and contracting - breathing, as the the electron moves in space.

Similarly in the video enter link description here

it has an animation of the spherical harmonic growing and contracting in time.

How much can we read into these description of the orbitals - can we therefore infer that there is only intrinsic 'radial' motion for the electron in s-orbitals and hence they have $$l=0$$? Or are these results of the oscillations of the spherical harmonics in time unphysical since they are only in $$\psi$$ and not $$|\psi|^2$$?

• What exactly do the spheres in that video represent? s-orbitals are stationary states of a hydrogen-like atom, and so their only time evolution is an overall phase, so I don't think it is reasonable to think of them as 'breathing in and out' – By Symmetry Jan 24 at 14:11
• That video is extremely misleading - pending a detailed look, I would say: stay well away from it. Atomic eigenstates are stationary, and they don't evolve in time at all. – Emilio Pisanty Jan 24 at 14:31
• That video addresses spherical harmonics in general, not spherical harmonics as applied to quantum mechanics. Spherical harmonics describe the angular geometry of a system and say nothing about the radial geometry. In general, the radial geometry of a system can change ("breath") and still be described by the same spherical harmonic. But in quantum mechanics the radial behavior is fixed. There is no breathing. – garyp Jan 24 at 15:48
• I so wish text books wouldn't come up with that kind of crap! – Gert Jan 24 at 15:48
• @garyp - I think what you're saying makes sense, but as a bonus question, for the electron states with orbital angular momentum, their wavefunction gains a phase which makes it rotate azimuthally over time (even if this isnt true for the wavefunction squared). Can we interpret this rotation as somehow physical? – Alex Gower Jan 24 at 16:03

That video was about spherical modes of vibration, so these modes are certainly present (approximately, as the earth is not a sphere) in the the Earth after a large earthquake, or in a spherical ballon that is driven into oscillation.

Before jumping to atomic orbitals, consider what the video did: it took waves on a string, and imagined them on a spherical surface.

So let's consider waves on a string and the modes of the particle in an infinite well. The fundamental mode of a string vibrates up and down, meanwhile, the ground state of a particle in a box looks like:

$$\psi_1(x) \propto \sin{\pi x/L}$$

Does that mean the electron is vibrating up and down? No. $$\psi(x)$$ doesn't even vibrate in the vertical direction...there is no vertical direction. The particle in a box has but one dimension, $$x$$.

The reason they look the same is because the functions are solving the same (or extremely similar) wave equations, but in atomic orbitals, $$Y_l^m(\theta, \phi)$$ is by no means describing a radial excitation, or a physical oscillation in any direction.

So if look at a state like $$|l=n, m=n\rangle$$, the angular and time dependence is:

$$\sin^n{\theta}e^{+ni\phi}e^{-i\omega t}= \sin^n{\theta}e^{i(n\phi-\omega t)}$$

which describes an equatorial wave going around in the "north" direction. Of course:

$$||\psi||^2 \propto \sin^{2n}{\theta}$$

is stationary, but the probability current is non-zero.

Note that all the "motion" is occurring in the complex phase, and not in the probability distribution. This is also true of plane waves.

I've noticed that most serious youtube videos devoted to $$\psi$$ do not show that the action is in the complex part of the wave function, rather, they show an bulge in $$x$$ moving around, or a real wave oscillating through zero. Wave-packets should be imagined more like a cork screw..the real and imaginary projections go though zero, but the magnitude of the wave function does not.

Appreciating this nuance is key to understanding why $$\hat {\bf p} = -i/\hbar{\bf \nabla}$$ and the energy dependence goes like $$\exp{(-iE/\hbar)}$$.

• I think what you're saying makes sense, but as a bonus question, for the electron states with orbital angular momentum, their wavefunction gains a phase which makes it rotate azimuthally over time (even if this isnt true for the wavefunction squared). Can we interpret this rotation as somehow physical? After all, we do get a non-zero probability current. – Alex Gower Jan 24 at 17:41
• @AlexGower think of the similar picture of eigenfunctions of the momentum operator: they are also constant everywhere in space, and their phase also moves along the direction of the momentum. But it would be wrong to think that the phase speed has any relation to the expectation value of absolute speed of the electron. E.g. it depends on the reference zero point of energy (which is arbitrary!). What you can actually do is try to introduce an envelope over this complex exponential to get a wave packet, and see that it moves on average with group speed. – Ruslan Jan 24 at 21:54
• Ah okay, is there any itnerpretation at all to be made about phase speed? – Alex Gower Jan 25 at 12:05

The video you give is talking about the appearance of the spherical harmonics in a different context from their appearance in atomic orbitals: it refers to their appearance as describing vibrations on a spherical, but two-dimensional, surface made of an idealized elastic material. The spherical harmonics that appear in atomic orbitals are mathematically the same, but the signification is quite different. There are analogies, but they are not a complete 1:1.

For one, while this surface is only two-dimensional, atomic orbitals occupy a whole spatial volume - they are actually three-dimensional. For another, there are composite orbital states that do not have any analogue to the vibrations on the surface, owing to that extra degree of freedom provided by the third dimension.

And those differences carry over into motion. While the elastic membrane does indeed flex, atomic orbitals actually do not change their shape at all.

That said, in their case, there is what one can think of as a sort of "subtler" motion suggested by the quantum equation. This subtle motion takes the form of the "beating" or throbbing, not of the shape, but of the phase, that is generated by the action of the Hamiltonian operator: for the orbital $$|n,l,m\rangle$$,

$$|\psi\rangle(t) = |n,l,m\rangle\ e^{iE_nt/\hbar}$$

where $$E_n$$ is the energy. The angular velocity of the subtle vibration is $$E_n/\hbar$$, and the frequency that corresponds to this is $$|E_n|/h$$.

This vibration is unobservable directly for an individual orbital, but it absolutely matters if you consider cases in which two or more orbitals can interfere with each other, which suggests it has some sort of physical realism, but of a kind not directly accessible to us. Hence why I call it, subtle motion.

Moreover, to further underline the differences here with the elastic case, because the total energies $$E_n$$ are negative, but increase (become less negative) with increasing principal quantum number $$n$$, this creates the interesting result that high-energy orbitals throb slower than low-energy ones, which is exactly the opposite of the perceptible vibration of the elastic surface, which beats more vigorously at higher-energy modes. For example, the $$n = 1$$ ground state of hydrogen throbs at about 3290 THz, while both states in the $$n = 2$$ excited family throb at only 1/4 this frequency, or about 823 THz, and the $$n = 3$$ family, at 366 THz.

The most interesting part about this motion is that its frequency is actually exactly twice that of the classical Kepler orbit with the given energy and angular momentum under the electric force. This suggests a possible interpretation, under the "quantum theory as a theory of systems with limited information content" paradigm[1], that this is the actual underlying movement of the electron, but there physically is no information giving what the true anomaly is, so the probability distribution of position doesn't track it and thus is static, as well as there being no information as to whether the orbit is even going clockwise or counterclockwise, which secondarily accounts for the factor of two in the frequency, because effectively, oppositional points of the orbit have become identified.

[1] for some preceding work on this direction, see https://www.perimeterinstitute.ca/videos/quantum-mechanics-theory-systems-limited-information-content

Yes, this is a nice interpretation. It may be easier to understand what's going on in the simpler example of a particle in a box, in one dimension. Consider the lowest-energy standing wave of this system, which is a wave where half a wavelength fits in the box. If you apply the de Broglie relation $$p=h/\lambda$$, you get a momentum. This seems paradoxical, since momentum suggests movement, but this is a standing wave, so there is no probability flux to the left or right. But actually the de Broglie relation only tells us the magnitude of the momentum, not its sign. Recall from your study of waves that a standing wave can be considered as a superposition of two traveling waves going in opposite directions.

The ground state of hydrogen is the same story, just with a little more complication because it's in three dimensions. Conceptually, it's the same idea. There is no probability flux because it's a standing wave. However, there is "motion" in the sense that this state is not a state with a definite value of momentum p=0.

How much can we read into these description of the orbitals - can we therefore infer that there is only intrinsic 'radial' motion for the electron in s-orbitals and hence they have 𝑙=0?

Yes, there is only radial motion in the sense that the momentum operator in quantum mechanics is the gradient operator, and this wavefunction only has a gradient in the radial direction.

Or are these results of the oscillations of the spherical harmonics in time unphysical since they are only in 𝜓 and not |ψ|2?

Actually you can make the oscillations go away completely just by redefining the zero level of potential energy in such a way that the state you're considering has a total energy of zero.

You just have to accept the idea that in quantum mechanics, there can be "motion without motion" -- that is, motion when no physical observable is changing over time. A simpler example is a plane wave.

• But actually the de Broglie relation only tells us the magnitude of the momentum, not its sign. So how come the momentum operator returns an imaginary number? – Gert Jan 24 at 16:21

A useful picture for the electronic $$s$$-orbits ($$L=0$$ states) in an atom is that the electron is oscillating through the center of the nucleus. However, we are not considering a single oscillation in one preferred direction, but we average the oscillations of all possible directions in 3D. This results in a spherical probability distribution with zero angular momentum, $$L=0$$. However, there is no "breathing" of the orbit involved -- at least not on a time scale on the order of $$1ns$$.

• I think there will be some "breathing" if you take a suitable superposition of $s$-states with different $n$. Namely, in a radial probability plot this will look like a wave packet going back and forth. – Ruslan Jan 24 at 21:46
• @Ruslan: I believe the OP considers the "simple" atomic state, such as $1s$ and not a superposition of states with different $n$. – Semoi Jan 24 at 22:20