# reperiendi

## Renormalization and Computation 2

Posted in General physics, Math, Quantum by Mike Stay on 2009 October 10

This is the second in a series of posts covering Yuri Manin’s ideas involving Hopf algebra renormalization of the Halting problem. Last time I showed how perturbing a quantum harmonic oscillator gave a sum over integrals involving $n$ interactions with the perturbation; we can keep track of the integrals using Feynman diagrams, though in the case of a single QHO they weren’t very interesting.

One point about the QHO needs emphasis at this point. Given a wavefunction $\psi = \sum_{n=0}^{\infty} \psi_n |n\rangle$ describing the state of the QHO, it must be the case that we get some value when we measure the energy; so if we sum up the norms of the probability amplitudes, we should get unity:

$\displaystyle \langle \psi|\psi \rangle = \sum_{n=0}^{\infty} \langle n | \psi_n^* \psi_n | n \rangle = \sum_{n=0}^{\infty} |\psi_n|^2 = 1.$

This is called the normalization condition.

When we perturb the QHO, the states $|n\rangle$ are no longer the energy eigenvectors of the new Hamiltonian. We can express the new eigenvectors $|m\rangle$ in terms of the old ones:

$\displaystyle |m\rangle = \sum_{n=0}^{\infty}\lambda^n m_n|n\rangle,$

where $\lambda$ is the strength of the perturbation, and we reexpress our wavefunction in this new basis:

$\displaystyle \psi = \sum_{m=0}^{\infty} \psi'_m |m\rangle$

Since we’re working with a new set of coefficients, we have to make sure they sum up to unity, too:

$\displaystyle \sum_{m=0}^{\infty} |\psi'_m|^2 = 1.$

This is the renormalization condition. So renormalization is about making sure things sum up right once you perturb the system.

I want to talk about renormalization in quantum field theory; the trouble is, I don’t actually know quantum field theory, so I’ll just be writing up what little I’ve gathered from reading various things and conversations with Dr. Baez. I’ve likely got some things wrong, so please let me know and I’ll fix them.

A field is a function defined on spacetime. Scalar fields are functions with a single output, whereas vector fields are functions with several outputs. The electromagnetic field assigns a single number, called the electric field, and a vector, called the magnetic field, to every point in spacetime. When you have two electrons and move one of them, it feels a reaction force and loses momentum; the other electron doesn’t move until the influence, traveling at the speed of light, reaches it. Conservation of momentum says that the momentum has to be somewhere; it’s useful to consider it to be in the electromagnetic field.

When you take the Fourier transform of the field, you get a function that assigns values to harmonics of the field; in the case of electromagnetism, the transformed field $\phi$ assigns a value to each color $k$ of light. Quantizing this transformed field amounts to making $\phi(k)$ into a creation operator, just like $z$ in the QHO example from last time. So we have a continuum of QHOs, each indexed by a color $k.$ (By the way—the zero-dimensional Fourier transform is the identity function, so the QHO example from last time can be thought of both as the field at the unique point in spacetime and the field at the unique frequency.)

When we move to positive-dimensional fields, we get more interesting pictures, like these from quantum electrodynamics:

Here, our coupling constant is the fine structure constant $\alpha = e^2/\hbar c,$ where $e$ is the charge of the electron. For each vertex, we write down our coupling constant times $-i$ times a delta function saying that the incoming momentum minus the outgoing momentum equals zero. For each internal line, we write down a propagator—a function representing the transfer of momentum from one point to another; it’s a function of the four-momentum $k$—and multiply all this stuff together. Then we integrate over all four-momenta and get something that looks like

$\displaystyle \int_0^\infty f(k) d^4 k.$

The trouble is, this integral usually gives infinity for an answer. We try to work around this in two steps: first, we regularize the integral by introducing a length scale $\Lambda.$ This represents the point at which gravity starts being important and we need to move to a more fundamental theory. In the quantum field theory of magnetic domains in iron crystals, the length scale is the inter-atom distance in the lattice. Regularization makes the integral finite for $\Lambda$ away from some singularity.

There are a few different ways of regularizing; one is to use $\Lambda$ as a momentum cutoff:

$\displaystyle \int_0^\Lambda f(k) d^4 k.$

This obviously converges, and solutions to this are always a sum of three parts:

• The first part diverges as $\Lambda \to \infty,$ either logarithmically or as a power of $\Lambda.$
• The second part is finite and independent of $\Lambda.$
• The third part vanishes as $\Lambda \to \infty.$

Renormalization in this case amounts to getting rid of the first part.

These three parts represent three different length scales: at lengths larger than $\Lambda,$ all quantum or statistical fluctuations are negligible, and we can use the mean field approximation and do classical physics. At lengths between $\Lambda$ and $1/\Lambda,$ we use QFT to calculate what’s going on. Finally, at lengths smaller than $1/\Lambda,$ we need a new theory to describe what’s going on. In the case of QED, the new theory is quantum gravity; string theory and loop quantum gravity are the serious contenders for the correct theory.

The problem with this regularization scheme is that it doesn’t preserve gauge invariance, so usually physicists use another regularization scheme, called “dimensional regularization”. Here, we compute

$\displaystyle \int_0^\infty f(k) d^\Lambda k$

which gives us an expression involving gamma functions of $\Lambda$, where gamma is the continuous factorial function, and then we set $\Lambda = 4 - \epsilon.$ The solutions to this are also a sum of three terms—a divergent part, a finite part, and a vanishing part—and then renormalization gets rid of the divergent part.

Assume we have some theory with a single free parameter $g$. We’d like to calculate a function $F(x)$ perturbatively in terms of $g$, where $F$ represents some physical quantity, and we know $F(x_0) = g'$. We assume $F$ takes the form

$\displaystyle F(x) = g + g^2 F_2(x) + g^3 F_3(x) + \cdots$

and assume that this definition gives us divergent integrals for the $F_n.$ The first step is regularization: instead of $F$ we have a new function

$\displaystyle F_\Lambda(x) = g + g^2 F_{2,\Lambda}(x) + g^3 F_{3,\Lambda}(x) + \cdots$

Now we get to the business of renormalization! We solve this problem at each order; if the theory is renormalizable, knowing the solution at the previous order will give us a constraint for the next order, and we can subtract off all the divergent terms in a consistent way:

1. Order $g.$

Here, $F_\Lambda(x) = g + O(g^2).$ Since it’s a constant, it has to match $F(x_0) = g',$ so $g = g' + O(g'^2).$ In this approximation, the coupling constant takes the classical value.

2. Order $g^2.$

Let $g = g' + G_2(g') + G_3(g') + \cdots,$ where $G_n(g') \sim O(g'^n).$ Plugging this into the definition of $F_\Lambda,$ we get

$\displaystyle F_\Lambda(x) = g' + G_2(g') + g'^2 F_{2,\Lambda}(x) + O(g'^3).$

Using $F(x_0) = g',$ we get $G_2(g') = -g'^2F_{2,\Lambda}(x_0),$ which diverges as $\Lambda \to \infty.$ In the case of QED, this says that the charge on the electron is infinite. While the preferred interpretation these days is that quantum gravity is a more fundamental theory that takes precedence on very small scales (a Planck length is to a proton as a proton is to a meter), when the theory was first introduced, there was no reason to think that we’d need another theory. So the interpretation was that with an infinite charge, an electron would be able to extract an infinite amount of energy from the electromagnetic field. Then the uncertainty principle would create virtual particles of all energies, which would exist for a time inversely proportional to the energy. The particles can be charged, so they line up with the field and dampen the strength just like dielectrics. In this interpretation, the charge on the electron depends on the energy of the particles you’re probing it with.

So to second order,

$\displaystyle F_\Lambda(x) = g' + g'^2\left(F_{2,\Lambda}(x) - F_{2,\Lambda}(x_0)\right) + O(g'^3).$

A theory is therefore only renormalizable if the divergent part of $F_{2,\Lambda}(x)$ is independent of $x.$ In QED it is. We can now define $F(x)$ as the limit

$\displaystyle F(x) = \lim_{\Lambda \to \infty} F_\Lambda(x).$

3. Higher orders.

In a renormalizable theory, the process continues, with the counterterms entirely specified by knowing $F(x_0).$