Season 9 Episode 7

We've seen waves in music on OOMC, but now in this episode we're going to the quantum realm to find out how studying waves in higher dimensions can help us to understand the shape of the periodic table!

OOMC Season 9 Episode 7

Watch on YouTube

Further Reading

Starter question

For which values of \(k\) is \(\sin(kx)\) the same function as \(\sin(k(x+360^\circ)\)?

I made a Desmos for this, which I forgot to show on stream.

The answer is “precisely when \(k\) is an integer”. Quantization!

 

Quantum Mechanics: Wavefunction

A free resource for this material is the lecture notes for A11 Quantum Theory at Oxford (warning: lots of content, much harder than A-level or equivalent).

Section 1 of the lecture notes gives a really good overview of why waves come into this at all, and why probability is involved.

The wavefunction describes the probability that a particle is found any some state, but (because of the use of complex numbers) it’s more than that; as well as the magnitude \(\left|\psi\right|\) that’s something to do with the probability, there’s also a phase \(\arg(\psi)\) for the wavefunction at each point in space.

The Schrödinger equation describes how the wavefunction changes over time. Because of the use of complex numbers, it’s possible for \(\psi\) to change without the magnitude of \(\psi\) changing anywhere; such solutions to the Schrödinger equation are called stationary solutions.

I also said that I’d link to some detail for my claim that Schrödinger not a great guy. Trigger warning: sexual abuse. There are some efforts to find something else to call the equation.

 

Spherical coordinates

We’ve had geographic coordinate systems for thousands of years; this is the sort of thing that ancient mathematicians worked on.

As a sort of geography tangent; it’s relatively easy to work out your latitude because the stars that you can see at night are different at different latitudes. But longitude is difficult, because as the Earth spins everyone on a particular line of latitude get to see the same stars, regardless of longitude. Solving the longitude problem required accurate clocks, and there’s a rich history of science here. That link points to the Royal Museums Greenwich, who guard the \(0^\circ\) longitude line. Or do they? This video by (previous OOMC guest) Matt Parker suggests otherwise! The Greenwich Meridian is in the wrong place | YouTube

Together with radial distance from the origin, this forms a coordinate system called spherical coordinates. They’re a good choice for describing problems with no preferred direction.

Another geography tangent at this point might talk about map projections like the Mercator projection and might mention that because of the non-trivial Gaussian curvature of a sphere, it's impossible to get a good map, but we need to get back to quantum mechanics via differential equations.

 

Differential Equations

I mentioned on the stream that you get two integers in the solution (quantization) from the behaviour of the solution near the poles, and under rotation around the polar axis.

I didn’t have time near the end to explain the third integer, for which I need a technique for analysing differential equations, even if the differential equation is so horrible that you can’t solve it. Let’s imagine that you’ve got something awful like $$x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+(1-x)\frac{\mathrm{d}y}{\mathrm{d}x}-cy=0$$ for some real constant \(c\). I don’t know how to find \(y(x)\) from that! But perhaps I could try a Taylor series for \(y\)? That would be a series like $$y(x)=a_0 + a_1 x + a_2 x^2 +a_3 x^3 +\dots.$$

The idea then is to plug that into the differential equation and compare coefficients. It’s a bit messy, but perhaps it turns out that you get a relationship between the coefficients like $$a_{k+1}=\frac{k-c}{(k+1)^2}a_k.$$

This is progress, because if I set \(a_0=1\) say, then I can calculate coefficients using that rule, and build a Taylor series for \(y\) from there, working up one value of \(k\) at a time. Even better, I can talk about the behaviour of those coefficients for large \(k\), and those are the terms that are important for large \(x\). For large \(k\), the number \(k\) is going to be much larger than \(c\), so we’ll have something like

$$a_{k+1}\approx \frac{1}{k+1}a_k$$

for large \(k\). And therefore \(a_k \approx B/k!\) for some constant \(B\). In terms of \(x\), that matches the Taylor series for \(B \exp(x)\).

Maybe that’s not OK – maybe I don’t want my solution to grow exponentially. If so, then I have a get-out-of-jail card; if, for some value of \(k\geq 0\), it’s the case that \(k-c=0\), then \(a_{k+1}\) will be zero. And then all further coefficients will be zero too. This gives a polynomial solution, which I might be OK with. And it happens when \(c\) is a positive integer. Quantization!

 

Putting it all together

Something like that happens in the problem for the (deep breath) radial part of the separable solution for the stationary state solution to the Schrödinger equation for an electron in the potential due to a proton. We get an integer \(n\geq 0\).

In total then, there are three integers to describe the solution; \(m\) which has something to do with periodicity around the direction of increasing longitude, \(l\) which has something to do with the dependence on latitude near the poles, and \(n\) which has something to do with wanting a solution that’s bounded for large distances from the origin.

We also have some relationships between the integers, and it's natural to try to count the solutions.

Given a positive integer \(N\geq 1\), find all solutions for integers \(n\), and \(l\) such that \(N=n+l+1\) and \(n\geq 0\) and \(l\geq 0\) and \(-l\leq m \leq l\).

These correspond to the different atomic orbitals that you might learn about in Chemistry A-level or equivalent.

 

Periodic table

There is one last ingredient, and that’s the Pauli exclusion principle, which states that you can fit up to two electrons in each orbital. They fill up the lowest energy states first.

Your solutions above (with a factor of two) for each \(N\) correspond to the number of elements in each row of the periodic table. I think that’s amazing, and totally worth all the maths and physics we needed to get to that point.

 

Physics

If this is your sort of thing, then you might like Physics at Oxford, although note that this sort of calculation is also (an optional) part of the Mathematics degree at Oxford.

The Comprehensive Oxford Mathematics and Physics Online School (COMPOS) is an innovative large-scale long-term outreach project within the Department of Physics, University of Oxford. Its mission is to help enthusiastic school students to systematically learn mathematics and physics at a deep level by offering them free tuition and mentoring.

There’s a fourth-year joint Maths and Theoretical Physics masters course at Oxford. That’s the course that Anjali is studying! You can’t apply for it on UCAS though; it’s something that people do after three years of either Maths or Physics at university. If you’re choosing between Maths and Physics at university, then it might be reassuring to know that there are opportunities like this later on, for people who have done either. Indeed, the Oxford Mathematics department has researchers like Prof Jon Keating FRS who did Physics as a degree before working in the Maths department (Jon's research includes Quantum Chaos, which is incredibly cool).

 

If you want to get in touch with us about any of the mathematics in the video or the further reading, feel free to email us on oomc [at] maths.ox.ac.uk.

Last updated on 1 Mar 2025, 10:24am. Please contact us with feedback and comments about this page.