What are Quantum Numbers?
In the course of my PhD, I was often working with quantum numbers. These are quite mysterious entities and though we know a fair amount about them, its often a lot of folklore.
Nonetheless, they are a fascinating generalisation of “normal” numbers, so I’m going write a series about them. In this first installment, we’ll define them, and play with them, and then ask a question I honestly don’t know how to answer.
If you already know a little about quantum numbers, please do stick with it. I know this article starts slowly, but it will pick up pace, I promise.
“Quantum”? numbers
So the first problem we’ll come up against is that if you search “quantum number” and scroll past the ads, you’ll come across this Wikipedia page.
Now, this is a pure maths series, so I’m not actually talking about energy levels of quantum (physical) systems. However, if you know where to look (under the “$q$analogue” redirection) you can find what we are going to be talking about.
The problem is that the word “quantum” is too overloaded. We could call these “Gaussian numbers”, but that’s too close to Gaussian integers^{1}. They are technically “Chebyshev polynomials of the second kind” (sorta), but that’s a mouthful and not quite correct.
Perhaps “$q$numbers” is the best option, but it ties our mental model to one possible manifestation of quantum numbers (more on that later) as we can define them without using anything labelled $q$.
Plus, “quantum” comes with the verb “to quantise”, which means we’ll be able to say things like “lets quantise this equation” instead of the less sexy “lets take the $q$analogue of this equation”.
So I’m going to stick to “quantum number”. It’s confusing if you’re used to the physics term, but it could be worse.
Show me a quantum number already!
Ok, ok! Like… $\gdef\qs#1{[\![#1]\!]} \qs 7$? Or $\qs 3$? Or $\qs {12}$?
Oh hey! You’re me, but… with a blue tie?
I feel like we’ve met before…
Are you sure?
Alrighty then.
Well, to answer your question, yes. For now, a quantum number is a normal, whole number in funny square brackets. But before you turn away and think we already know everything about them, let me just tell you the following facts.
$$ \qs 1 + \qs 1 \neq \qs 2 \tag1 $$ and $$ \qs 2 \times \qs 7 \neq \qs {14}, \tag2 $$ but $$ \qs 3 \times \qs 4 = \qs 6 + \qs 4 + \qs 2. \tag3 $$
Yes. So sometimes arithmetic works and sometimes it doesn’t. Asking why this happens and when this happens is the point of this post.
Start with the normal numbers
But before we play these games and try any arithmetic, we should really get to the bottom of what a quantum number is.
And to do that, we need to understand what a normal number is.
No, although that would be a fascinating topic^{2}. I’m more interested in how we might rigorously define numbers. Why don’t we see how lean, a theorem prover does it?
inductive nat
 zero : nat
 succ (n : nat) : nat
Unpacking this, we could say that every natural number is either
 a special number called zero
 or the successor of a different natural number.
In this way, we would write 3
as succ succ succ 0
, which is the number after the number after the number after the number called zero.
succ a = succ b
then a = b
next?
No, that’s handled by what “=
” means to lean, and its not really the point here.
I’m more interested in the idea that we define a special number and then give a relation that recursively defines the rest.
See, if we know that $\texttt{succ}\,\bullet$ really means $\bullet\, + (1)$ and we’d decided on a naming for the natural numbers, lets say ${(0), (1), (2), \ldots}$ we could write this as $$(n+1) = (n) + (1).$$
No, no. Its only stupidly obvious if you think that you can get rid of the brackets, but we can’t yet. Remember $(n)$ is a name of some number  not necessarily the number you know of as $n$. The point of this is to tell you what $(n+1)$ is.
It might as well be. For now, assume it has its usual properties, like $a + b = b + a$ and $(0) + a = a$.
First though, lets correct an error in my above statement. I really should have told you what $(0)$ and $(1)$ were. You see, the equation can’t tell you what $(0)$ is (since $n+1 > 0$ for all $n$)^{3} and if we try substituting in $n=0$ we get $$(1) = (0) + (1) = (1),$$ which doesn’t help us any.
So I need to specify $(0)$ and $(1)$.
zero
above, remember?
Yes, but that was before I changed the recursive definition to involve the symbol $(1)$.
Yes, I lied. It seemed like a good idea at the time. Even lean thinks they are different and has to explicitly prove a lemma that they are the same.
lemma add_one (n : â„•) : n + 1 = succ n :=
rfl
Oh that’s cool. Looks like lean accepts unicode identifiers like â„•
.
That’s an alias for nat
that we used above.
Unicode in source is great  it lets you do things like have subscripts (sometimes) or variables with interesting names.
I just hope it never gets confusing and nobody uses multiple very similar characters.
But now I’m confused again. What does it mean exactly to “specify” $(0)$ and $(1)$? Can you give an example?
I thought you’d never ask.
Example (Clock arithmetic): Consider the usual twentyfour hour day, measured in hours since midnight. Let’s define $(0)$ to mean midnight and $(1)$ to mean one hour. In other words,^{4} $$(1) = (0) + (1) = \text{midnight} + \text{one hour} = 1\text{ o’clock}.$$ Then, naturally, $(2)$ is 2 o’clock, $(3)$ is 3 o’clock and so on. Indeed $$ (3) = (2) + (1) = \text{2 o’clock} + \text{1 hour} = \text{3 o’clock}.$$ But what is $(24)$? Well its one hour after 11 PM, which is midnight. That is to say $(24) = (0)$ which means $(25) = (1)$, $(26) = (2)$ and so on.
Mathematicians call this structure $\gdef\Z{\mathbb{Z}} \Z_{24}$, pronounced “the integers modulo 24”.^{5}
Maybe, but what about this?
Example (Clock arithmetic with the minute hand): Lets look at time again, but now measured in minutes (so we allow times like “half past three in the afternoon”). Again we’ll let $(0)$ be midnight and $(1)$ be one hour. Again we recover that $(n)$ is just $n$ o’clock.
But now, the mathematical structure we are looking at is $\Z_{1440}$, since there are 1440 minutes in the day. Since we are counting minutes, we might say $(1) = 60 \text{ minutes}$.
Now we’ve done something strange. Ignoring the mismatch of units (truthfully they were only there to distinguish when we are working in $\Z_{1440}$), $$ \begin{aligned} (1) \times (3) &= 60 \text{ minutes} \times 180 \text{ minutes}\\&= 10\,800 \text{ minutes}\\&= 180 \text{ hours}\\&= (180)\\&= (12) \end{aligned} $$
Yes.
Sort of, but we are not done here yet.
Lets just make one more observation/definition. Just so you know, when $p$ is a prime number, $\Z_p$ is most often written $\gdef\F{\mathbb{F}} \F_p$. This is because its a field which just means that all the numbers $x$ have multiplicative inverses $\frac 1x$.
For example, in $\F_7$, we have that $\frac 12 = 4$. Indeed, $2 \times 4 = 8 = 1$ modulo $7$.
But we can go further. If $i$ is the imaginary unit (that is, $i^2 = 1$ like in the complex numbers), then we could consider something like $\F_3[i]$. This just means the numbers $\{0, 1, 2, i, 1 + i, 2 + i, 2i, 1 + 2i, 2 + 2i\}$ with the wraparound addition.
But now if I tell you that $(0)$ is $0$ and $(1)$ is $1$, then $(2) = 1 + 1 = 2$ and $3 = 2 + 1 = 0$. So the set $\{(0), (1), (2), (3), \ldots \}$ is actually just $\{0,1,2\} = \F_3\subset F_3[i]$. This is called the prime subfield.
Definition Let us take some ring (a mathematical structure with addition and multiplication that are nicely behaved) and define $(0)$ to be the zero and $(1)$ to be the multiplicative identity (that is, $(1) \times x = x$).
Then the smallest $p$ such that $(p) = 0$ is called the (normal) characteristic of the ring. If this never happens we’ll say that $p = \infty$.
The reason we choose $p$ is that if the ring is sufficiently nice (that is, if $x\times y = 0$ means either $x = 0$ or $y = 0$) then $p$ is always a prime number!
Yes! The smallest $\ell$ such that $\qs \ell = 0$ is called the quantum characteristic^{6}! We’ll get back to that in a later post.
But will you define quantum numbers already?
That’s enough about natural numbers I think.
Henceforth, instead of writing $(n)$ I’ll mostly just write $n$. We’ll be justified in this because $(0)$ will always be such that $(0) + x = x$ and $(1)$ will always be such that $(1) \times x = x$.^{7}.
Alright, now lets define quantum numbers $\qs n$.
If $(n+1) = (n) + (1)$ tells us something about the additive structure of numbers but sometimes messes up multiplication, perhaps we can try a different recurrence relationship that does use multiplication.
We still need it to tell us about $\qs {n+1}$ in terms of smaller things though.
Wow! Its almost as if you already know about quantum numbers.
I’ll keep that in mind.
Ok, ok. I’m going to stop you right there. You know you’re jumping the gun here. Lets stick to the first option:
$$ \qs 2 \qs n = \qs {n+1} + \qs {n1}. \tag4 $$Similarly to before, we’ll also have to specify $\qs 0$ and $\qs 1$. Because we’d like to mix quantum and nonquantum numbers, lets insist that $\qs 0 = 0$ and $\qs 1 = (1) = 1$. Now, though we also need to know what $\qs 2$ is. You can check that equation 4 doesn’t define $\qs 2$ if we know $\qs 1$ and $\qs 0$. Indeed, all it says is $$\qs 2 = \qs 2 \qs 1  \qs 0 = \qs 2  0 = \qs 2,$$ which gives us nothing.
Well, that’s the beauty of it. Looking at an inductive argument on equation 4 we see that $\qs n = n$ in that case! Lets actually do the proof, to get an idea of how to work with quantum numbers.
Lemma: If $\qs 2 = 2$ then $\qs n = n$ for all integers $n$.
Proof: Note that this is true for $n = 0$ and $n=1$. Thus, for induction, assume it holds for all $0 \le n < N$. The base case of $N=1$ has just been shown. Now consider $\qs N$. We know that $$\qs 2 \qs {N1} = \qs N + \qs {N2}$$ By the inductive hypothesis, $\qs N = N$ and $\qs {N1} = N1$. By the assumption, $\qs 2 = 2$^{8} so $$2 (N1) = \qs N + N2.$$ This has solution $\qs N = N$, and so the statement is shown for all positive integers.
Now note that $0 = \qs 2 \qs 0 = \qs 1 + \qs {1}$ so $\qs{1} = 1$. Assume, for an inductive proof, that $\qs {n} = \qs n$ for all $0\le n < N$. Again we have a base case of $N=1$. Now $$\qs 2 \qs{N} = \qs {(N+1)} + \qs {(N1)},$$ so $\qs 2\qs N = \qs{(N+1)} \qs{N  1}$ by the inductive hypothesis. But we know $\qs 2\qs N = \qs{N+1}  \qs {N1}$ Hence $\qs {n} = \qs n$ for all $n$ and the result is proven for all integers $\qs n$.
Indeed, and in fact $\qs{n} = \qs n$ no matter what $\qs 2$ is… it could be 2, it could be 1 hour, it could be “sheep”.
So lets leave behind specifying $\qs 2$ and just leave it as is — a symbol and see what happens. For example, it follows from the definition that $$ \qs 3 = \qs 2^2  1. $$ and $$ \qs 4 = \qs 2^3  2\qs 2. $$ In fact, we can build a whole table
$$ \begin{aligned} \qs 0 &= 0\\\qs 1 &= 1\\\qs 2 &= \qs 2\\\qs 3 &= \qs 2^2 1\\ \qs 4 &=\qs 2 ^3  2\qs 2\\\qs 5 &= \qs 2 ^ 4  3 \qs 2 ^ 2 + 1\\\qs 6 &= \qs 2 ^ 5  4 \qs 2 ^ 3 + 3\qs 2 \end{aligned} $$
Now we can go check equation 1 , equation 2 and equation 3 . In the first case, $\qs 1 + \qs 1 = 2 \neq \qs 2$. In the second, $\qs 2\qs 7 = \qs 6 + \qs 8$. Noticing a pattern, we expect $\qs{14}$ to have a $\qs 2^{13}$ term in it, but neither $\qs 6$ nor $\qs 8$ do. On the other hand $$\qs 3\qs 4 = (\qs 2^2 1)(\qs 2^3  2\qs 2) = \qs 2^5 3 \qs 2^3 + 2 \qs 2 = \qs 6 + \qs 4 + \qs 2$$ which is why equation 3 works.
We can formalise that pattern and start bringing some higher level maths into this with the following lemma. Remember, the set (ring) of polynomials with integer coefficients is denoted $\gdef\Z{\mathbb{Z}}\Z[X]$ and contains all expressions of the form $a_0 + a_1 X + a_2 X^2 + \cdots + a_n X^n$. The polynomial is called monic if $a_n = 1$ and we say it has degree $n$. We can, of course replace $X$ with whatever we like.
Lemma: The quantum number $\qs n$ can be written as a degree $n1$ monic polynomial in $\qs 2$ with integer coefficients.
Actually, we know the exact form of the polynomial too.
Lemma: $$ \qs n = \sum_{i = 0}^{\lfloor \frac n2\rfloor}(1)^i \binom{ni1}{i} \qs 2^{n2i1} \tag5 $$
Here $\binom{n}{r}$ is the (normal) binomial coefficient defined as $$\binom {n}{r} = \frac{n!(nr)!}{r!}.$$ It is always an integer.
Sort of. All in good time…
Actually, speaking of time, I think we should wrap up this post. We’ve set the stage for a delve into quantum numbers and quantum arithmetic. We’ve defined what $\qs n$ is, and are comfortable for why $\qs 3\qs 4 = \qs 6 + \qs 4 + \qs 2$ but $\qs 2 + \qs 3 \neq \qs 5$, for example.
Next time, we’ll bring $q$ into the equation and show why these numbers get the term “quantum”!
Epilogue: In which it gets very mathsy
Here I’m going to pose the questions we’ve been mulling over in the above post (and some new ones) in a more mathematical way. I am genuinely interested to know the answers to these and if you make any progress, please let me know.
Expressions
Consider the surjection of rings $\Phi: \Z[X_1, X_2, \ldots] \to \Z\big[\qs 2\big]$ given by $X_n \mapsto \qs n$. An expression $E$ is just an element of $\Z[X_1, X_2, \ldots]$ which we can think of as the preimage of something in $\Z\big[\qs 2\big]$.
For example, the element $\qs 3\qs 4  \qs 6  \qs 4  \qs 2$ corresponds to the expression $X_3X_4  X_6  X_4  X_2$.
Let $d$ be the degree of $E$, $n$ be the largest $N$ such that $X_N$ appears in $E$, and $t$ be the number of terms in $E$.
Now, the kernel of $\Phi$ is simply all expressions that hold over quantum numbers. It is generated (as an ideal) by the set $S = \big\{\qs 2 \qs n  \qs {n+1}  \qs {n1}\big\}_{n = 1}^\infty$.
Question: Is there an algorithm that determines whether or not an element $E$ belongs to $\ker \Phi$ in $\mathcal O(f(d, t))$ time and memory for any function $f$.
Here’s a naive solution that doesn’t obtain the time or space bound. Given an expression (read and stored in $\mathcal O(dt)$ time and memory), replace each $\qs n$ with its expanded form in terms of $\qs 2$. We have equation 5 so this takes $\mathcal O(dtn^2)$ (each binomial expression takes $\mathcal O(n)$ time to evaluate and then we have $\mathcal O(n)$ of them to calculate $\qs n$). Multiplying out take $\mathcal O(n^d)$ and then we have $\mathcal O(tn^d)$ terms to simplify. Thus this algorithm takes $\mathcal O(t n^d)$ which is not a function of $t$ and $d$ alone.
On the other hand, here is a very quick algorithm that identifies when an expression is not in $\ker \Phi$. Simply replace each of the $\mathcal O(dt)$ terms $\qs n$ by $n$ and check if the resulting equation in the naturals is zero. This runs in $\mathcal O(dt)$ space and time. However, equation 1 or equation 2 shows that this is not enough.
Or another: check that the term $X_{n_1}X_{n_2}\cdots X_{n_r}$ that maximises $(n_11)(n_21)\cdots (n_r1)$ is not unique^{9}.
Here the expression $X_5  X_3{}^2 + 4$ satisfies both of these algorithms, but doesn’t lie in $\ker \Phi$.
Question (weaker): Does there exist an algorithm that is polynomial in $n$ (but possibly not in $d$ or $t$)?
My guess is that in general, this is equivalent to some word problem, so the answer is “no”. However, there are some assumptions that make it more specific than a general word problem (upper unitriangularity being one), so there is some hope.
Two extensions
Repeat the above, but now with $\Phi_R : R[X_1, X_2, \ldots] \to R\big[\qs 2\big]$ for some arbitrary ring $R$. Does it only depend on the characteristic of $R$?
What about $\Phi_{(R, \delta)} : \R[X_1, X_2, \ldots] \to R$ where $\delta = \qs 2 \in R$?
Some combinatorics
Question: Is there a combinatorial motivation for equation 5 ? If we ignore gradings (defined in a later post), is there a counting argument for why $$n = \sum_{i = 0}^{\lfloor \frac n2\rfloor}(1)^i \binom{ni1}{i} 2^{n2i1}$$
Other formulations of numbers
Let $f(1, X_2, \ldots X_s; Y_1, \ldots, Y_s)$ be an integer polynomial in its arguments such that $$ f(1, 2,\ldots s; n1, \ldots, ns) = n $$ for all natural numbers $n$.
Then define $\qs n_f$ as $f(1, X_1, X_2, \ldots, X_s ; \qs{n1}_f, \ldots, \qs{ns}_f)$.
For example, we’ve been using $f(1, X_2; Y_1, Y_2) = X_2 Y_1  Y_2$ to define $\qs n$.
These form other kinds of quantum numbers which specialise to normal numbers upon $X_n \mapsto n$. How do our answers to the previous questions change?
Representation theory
Quantum numbers of this kind are closely tied to the representation theory of $U_q(\mathfrak{sl}_n)$, and perhaps more surprisingly to that of the cyclic groups, $C_n$^{10}. Does picking different $f$ give results in the representation theory of other objects?
Formal proofs
The kinds of proofs covered in this post seem very ripe for formalisation into a theorem checker (lean perhaps?).
They should be easy to prove on their own^{11}, but also tie into other structures like polynomial rings.
It would be good to see these included in mathlib
or something.

And do we really want to name yet another thing after Gauss? ↩︎

One I’ve been itching to read up on ever since reading Anathem by Neal Stephenson. If you have any good references on Platonic realism vs nominalism or the syntax vs semantics themes, please hit me up. ↩︎

This is where the notation gets a little tricky. Since we have arabic numerals and are very familiar with natural numbers, when I write $(n+1)$ what I mean is “the symbol which is an open bracket, followed by the arabic numerals of the number represented by $n$ plus one, followed by a close bracket.” When I write $n+1 > 0$ I’m asking you to do the arithmetic $n+1$ in arabic numerals and note that it isn’t $0$.
It may help to imagine yourself a benevolent god, blessed with the understanding of Numbers attempting to teach a very stupid child. You might say “Oh, my child. This equation can’t define $(0)$.” You don’t have to explain to the child why. ↩︎ 
I am not getting into the difference between one hour and one o’clock right now. Especially if you start muttering the words “daylight saving” or “leap seconds”. Most programming (and human) languages distinguish time and duration, but we need a simple example here. ↩︎

The word “modulo” has latin roots with “modular”, “modulate”, and “module” and likely comes from the branch of definitions meaning “small measure” or “interval”. ↩︎

Note how that wasn’t the case in the “minutes” example above. ↩︎

This doesn’t follow from the inductive hypothesis too. ↩︎

And that the coefficients of such terms add to zero. ↩︎

Scroll through that paper to find a beautiful fractal picture relating to quantum numbers. ↩︎

Actually I know they are, because I’ve done it, but not very well. ↩︎