week183

July 30, 2002

This Week's Finds in Mathematical Physics (Week 183)

John Baez

I'm now in England, visiting the category theorists in Cambridge. Before coming here I went to a wonderful conference in honor of Graeme Segal's 60th birthday. Most of the talks there described the marvelous different ways in which ideas from string theory are spreading throughout mathematics. I should really tell you about this stuff... it's very cool... but right now I'm in the mood for talking about something simpler: some ways in which ideas from quantum theory are spreading throughout mathematics.

Quantum theory is digging its way ever deeper into the mathematical psyche: for every branch of math, people seem to be developing a corresponding quantized version, from "quantum groups" to "quantum cohomology". Now there is even a textbook on "quantum calculus", suitable for undergraduates:

1) Victor Kac and Pokman Cheung, Quantum Calculus, Springer, Berlin, 2002.

Indeed, we'll soon see that quantum calculus is based on an even simpler subject that deserves to be called "quantum arithmetic"!

This book talks about two modified versions of calculus: the "h-calculus" and the "q-calculus". The letter h stands for Planck's constant, while the letter q stands for quantum. They are adjustable parameters related by the formula

q = exp(ih).

In particular, these modified versions of calculus reduce to Sir Isaac Newton's good old "classical calculus" in the limit where

h → 0

or alternatively,

q → 1.

One eerie thing about these modified versions of calculus is that people discovered them before quantum mechanics - and they even used the letters "h" and "q" in their formulas! In particular, the use of the letter "q" seems to go back all the way to Gauss, who wrote about a q-analogue of the binomial formula and other things.

So what's the idea? Like many great ideas, it's pathetically simple. To get the h-calculus, you just leave out the limit in the definition of the derivative, using this instead:

              f(x+h) - f(x)
              -------------
                    h

In the limit as h → 0, this reduces to the usual derivative.

There is a lot to say about this, but deeper and more mysterious mathematics arises from the q-calculus, where we use the "q-derivative":

              f(qx) - f(x)
              ------------
                 qx - x

This reduces to the usual derivative as q → 1. Note that the h-derivative says how f(x) changes when you add something to x, while the q-calculus says how it changes when you multiply x by something.

Some choices of q are more interesting than others. If q is a complex number with |q| = 1, we can take the q-derivative of a function that's defined only on the unit circle in the complex plane! Multiplying by a unit complex number rotates the unit circle a bit, just as adding a real number translates the real line. If you think about this for a while you'll see the relationship between the h-calculus and the q-calculus, and how it's especially nice when we set q = exp(ih).

Alternatively, if q is an integer, we can take the q-derivative of a function that's defined only on the integers! This is especially cool when q is a prime or a power of a prime; then there are nice connections to algebra.

Pretty much anything you can do with calculus, you can do with the q-calculus. There are q-integrals, q-trigonometric functions, q-exponentials, and so on. If you try books like this:

2) George E. Andrews, Richard Askey, Ranjan Roy, Special Functions, Cambridge U. Press, Cambridge, 1999.

you'll see there are even q-analogues of all the special functions you know and love - Bessel functions, hypergeometric functions and so on. And like I said, the really weird thing is that people invented them before their relation to quantum mechanics was understood.

I can't possibly explain all this stuff here, but a good way to get started is to look at the q-analogue of Taylor's formula. In ordinary calculus this formula says how to reconstruct any sufficiently nice function from its derivatives at zero:

f(x) = f(0) + f'(0) x + f''(0) x²/2! + ...

In q-calculus we can write down the exact same formula using q-derivatives and q-factorials! The nth q-derivative of a function is defined in the obvious way, by taking the q-derivative over and over. Let's do this to the function xⁿ. If we take its q-derivative once we get:

     (qx)ⁿ - xⁿ         qⁿ - 1
     -----------   =   ------- x^n-1 
       qx - x           q - 1

We can make this look almost like the usual derivative of xⁿ if we define the "q-integer" [n] by

            qⁿ - 1
  [n]  =   -------   
            q - 1

       = 1 + q + q² + ... + q^n-1

Then the q-derivative of xⁿ is just

[n] x^n-1

This implies that the nth q-derivative of xⁿ is the "q-factorial"

[n]! = [1] [2] ... [n]

This in turn means that the usual Taylor formula still works if we replace derivatives by q-derivatives and factorials by q-factorials.

Now, starting with q-factorials we can define q-binomial coefficients:

            [n]! 
        -----------
        [m]! [n-m]!

and then cook up a q-Pascal's triangle, prove a q-binomial theorem, and so on. It's not just a matter of recapitulating ordinary calculus, either: eventually we run into lots of cool identities that have no classical analogues, like the "Jacobi triple product formula":

                 sum_n  q^n(n+1)/2 xⁿ  =  

        product_i  (1 + xqⁱ) (1 + x^-1q^i-1) (1 - qⁱ)

where the sum is over all integers n, while the product is over positive integers i. Now, personally I'm not a big fan of identities just for the sake of identities. However, I like taking identities and trying to find their "secret inner meaning" - mainly by seeing how they come from isomorphisms between interesting mathematical structures. The mysterious identities of q-mathematics provide an ample playground for this game, especially since they're all related in intricate ways.

If you ever get stuck on a desert island you can have lots of fun reinventing quantum calculus, and if you don't, you can read Kac and Cheung's book. So either way, there's no point in me describing its contents further; instead, I want to say more about how q-mathematics is related to physics.

For starters, let's see how the canonical commutation relations change when we use a q-derivative to define the momentum operator, instead of an ordinary derivative. Remember what Schroedinger said: a particle on a line is described by a "wavefunction", which is a complex function on the line, say ψ. The position operator Q multiplies a wavefunction by x:

(Q ψ)(x) = x ψ(x)

while the momentum operator P basically takes their derivative:

(P ψ)(x) = -i ψ'(x)

The canonical commutation relations say that

PQ - QP = -i

Now, how does this change if we define the momentum operator using the q-derivative instead? I could do this calculation for you, but you'll be a much better person if you do it yourself - it's incredibly easy, so please do it. The answer is

PQ - qQP = -i

In other words, we must replace the commutator PQ - QP by a "q-commutator". This is the tip of a big iceberg: the whole theory of Lie algebras has a "q-deformed" version where q-commutators of various sorts take the place of commutators - and just as Lie algebras go along with Lie groups, these q-deformed Lie algebras go along with "quantum groups".

Now let's check to see if you're paying attention. The alert reader should have already noticed an incredible glaring contradiction in what I've said! I put it there on purpose, to make an important point.

No? It's simple. I said that making q different from 1 is like making Planck's constant different from 0 - going from classical to quantum. People working on quantum groups often say this. But look what we just did! We took the canonical commutation relations, which are already quantum-mechanical, and modified them by making q different from 1. This is blatantly obvious if we put Planck's constant where it belongs in the above formulas, instead of hiding it by setting it equal to 1. The momentum operator is really

(P ψ)(x) = -i ħ ψ'(x)

so the canonical commutation relations are

PQ - QP = -i ħ

and when we use a q-derivative in the momentum operator they become

PQ - qQP = -i ħ.

So there really are two adjustable parameters floating around: Planck's constant and this mysterious new "q"!

In fact, I've been complaining about this for years: it's only in certain special contexts that you can think of the "q" or "h" in quantum calculus as related to Planck's constant; here's one in which they're obviously distinct. So what's the physical meaning of q-deformation?

One person to take a stab at this is Shahn Majid:

3) Shahn Majid, Foundations of Quantum Group Theory, Cambridge U. Press, Cambridge, 2000.

In this book he says q is related to Newton's gravitational constant. This would be cool, because then you could take your theory of quantum gravity, full of formulas like

PQ - qQP = -i ħ,

and make the quantum effects small by letting ħ → 0, or make the gravitational effects go away by setting q → 1. The problem is, I've never seen a theory of quantum gravity like this! Neither loop quantum gravity nor string theory work this way.

In fact, both loop quantum gravity people and string theorists agree on how to quantize gravity without matter in 3 spacetime dimensions. This is about the only thing they agree on. Quantum gravity in 3 dimensions is full of q-mathematics, and in this theory q is the exponential of something involving the cosmological constant. When q = 1 you get the quantum theory of flat 3d spacetime, since then Einstein's equations say that spacetime is flat - this is a peculiarity of 3 dimensions. But when q is different from 1, you get the quantum theory of a spacetime having constant curvature: a nonzero cosmological constant means the vacuum has energy density, which curves spacetime!

For some interesting new insights into this, see:

4) John Barrett, Geometrical measurements in three-dimensional quantum gravity, available as gr-qc/0203018.

When we make the cosmological constant nonzero in 3d quantum gravity we must replace the group SU(2) by the quantum group SU_q(2). Based on this, one can argue that quantum groups are misnamed - they should really be called "cosmological groups". Another way to put it is this: ordinary groups are already perfectly sufficient for most of quantum theory; quantum groups show up only in certain special contexts.

This goes to show that the deep inner meaning of the "q" in quantum groups is still up for debate. Mathematically it has a lot to do with replacing groups by non-cocommutative Hopf algebras, whose representations form a braided rather than symmetric monoidal category. Here Majid and I agree completely: Planck's constant is about deviations from commutativity while this "q" stuff is about deviations from co-commutativity, or the failure of braidings to be symmetric. Still, I think one should try to understand this more deeply. The amazing things that happen when q is a power of a prime number have got to be an important clue! I'll talk about this more next week.

Addendum: Toby Bartels brought up an important point in a reply on the newsgroup sci.physics.research:

John Baez wrote in small part:

>In fact, I've been complaining about this for years: it's only in
>certain special contexts that you can think of the "q" or "h" in
>quantum calculus as related to Planck's constant; here's one in
>which they're obviously distinct.  So what's the physical meaning
>of q-deformation?

If q = exp h, then h couldn't possibly be Planck's constant,
because Planck's constant is not dimensionless.
(Or when you make it dimensionless, you generally fix its value,
and then it makes no sense to speak of varying q.)
To get a dimensionless constant for h, use (ħ G Λ/c^3),
where ħ = Planck's constant, G = Newton's constant,
Λ = cosmological constant, and c = speed of light.

If you're coming from the POV where you only had 3 of these before,
with the 4th equal to 0 (or infinite in the case of c),
then you're going to view changing from q = 1 to some other q
as varying the value of the 4th constant.
Thus John (a quantum gravity theorist that often sets
ħ, G, and c to fixed values) thinks that it's Λ,
while Majid (who studied quantum field theory, which fixes ħ and c
and thinks of Λ as a fixed QFT effect)[*] thinks that it's G.
But it is the dimensionless ratio that matters to everybody.

[*]I'm being presumptuous here.

-- Toby Bartels

I replied:

Toby Bartels wrote:

>John Baez wrote in small part:

>>In fact, I've been complaining about this for years: it's only in
>>certain special contexts that you can think of the "q" or "h" in
>>quantum calculus as related to Planck's constant; here's one in
>>which they're obviously distinct.  So what's the physical meaning
>>of q-deformation?

>If q = exp h, then h couldn't possibly be Planck's constant,
>because Planck's constant is not dimensionless.
>(Or when you make it dimensionless, you generally fix its value,
>and then it makes no sense to speak of varying q.)

It might make sense to treat Planck's constant as dimensionless 
and still talk of varying its value.  

However, you're certainly right about this: in applications of 
q-mathematics to quantum gravity, we make Planck's constant 
dimensionless by combining it with Newton's gravitational 
constant, the speed of light, and the cosmological constant in this way:

>To get a dimensionless constant for h, use (ħ G Λ/c^3),
>where ħ = Planck's constant, G = Newton's constant,
>Λ = cosmological constant, and c = speed of light.

... or something like that.  I think the formula depends on 
the dimension of spacetime, and so far it's in (2+1)d spacetime
that all the really solid applications of q-mathematics to
quantum gravity arise.  But the basic idea is robust, and it
doesn't depend on the dimension of spacetime:

We get a dimensionless constant by measuring the density of
the vacuum in Planck masses per Planck volume!

In other words: using ħ G and c we can construct units of
length, time, mass and so on - and then we can talk about the
energy density of the vacuum, measured in those units, and
get something dimensionless.

This explains why q-mathematics only shows up when we do 
quantum gravity with a nonzero cosmological constant (or perhaps
matter).

>If you're coming from the POV where you only had 3 of these before,
>with the 4th equal to 0 (or infinite in the case of c),
>then you're going to view changing from q = 1 to some other q
>as varying the value of the 4th constant.
>Thus John (a quantum gravity theorist that often sets
>ħ, G, and c to fixed values) thinks that it's Λ [...]

Right.  Actually, the real reason I like to claim it's Λ
is that this is the most surprising of the four alternatives.

When we try to pick out anything by itself, we find it hitched to everything else in the Universe. - John Muir