4
$\begingroup$

While I'm reading the introduction of matrix models in Chapter 8 in Mariño's book(https://doi.org/10.1017/CBO9781107705968), I notice this description of matrix model:

We will begin by a drastic simplification: we will go from four to zero dimensions, and consider theories made out of constant fields in the adjoint representation of $U(N)$, i.e. theories based on Hermitian matrices, also known as matrix models. In these models, the path integral reduces to a conventional multidimensional integral.

Also, in this web (https://www.maths.dur.ac.uk/users/stefano.cremonesi/projects1819/MatrixModels.html), it says:

Physical observables in quantum gauge theories in more spacetime dimensions are also given by integrals, but these are hard-to-compute and mathematically ill-defined integrals over infinite-dimensional spaces of field configurations. In contrast, the integrals of interest in matrix models are well-defined ordinary finite-dimensional integrals of the type that you are familar with from high school, with the only complication that there are multiple integration variables (the entries of the matrix).

I know little about how the path integral in quantum field theory works, but I think it's also a multi-dimensional integral. And in matrix model the parameter $N$ is always infinite, so it's also a infinete dimensional integral.

So how do the path integral simplified in the matrix model?

$\endgroup$

2 Answers 2

5
$\begingroup$

I know little about how the path integral in quantum field theory works

This is not good news, but I will try give an answer anyhow. Let us consider gauge theory, with gauge group $G$, in four dimensions and in Euclidean time. Here the gauge field is a $G$-connection one-form $A$, which is a fancy way to say the following

  • $A_\mu$ transforms as a covector under spacetime coordinate transformations (in particular Lorentz transformations). $A_\mu$ are the components of the one-form $A = A_\mu \mathrm{d}x^\mu$. This is the meaning of the one-form part of the italic text above
  • $A_\mu$ has values in the Lie algebra of $G$ (i.e. $A_\mu = A_\mu^a t_a$ for $t_a$ a basis of the Lie algebra) and under a gauge transformation $g(x)$ transforms as $A^\prime_\mu = gA_\mu g^{-1} - g^{-1}\partial_\mu g$. This is the meaning of $G$-connection

Then you pick an action $S[A]$, which is some functional of $A$ which is gauge invariant and you construct the partition function for the quantum theory \begin{equation} Z = \int \mathcal{D}A \exp(-S[A]) \end{equation} This is not really the correct expression for the partition function as here we are integrating over values of $A$ which are gauge equivalent and some form of gauge fixing is required via, for example, BRST.

However what really matters here is what the integration really is. The path integral measure can be tought informally as \begin{equation} \mathcal{D}A = \prod_{x\in \mathbb{R^4}} \prod_{a=1}^{\dim \mathfrak{g}}\mathrm{d}A_1^a(x)\mathrm{d}A_2^a(x)\mathrm{d}A_3^a(x)\mathrm{d}A_4^a(x) \end{equation} You see it is a product over a continuous index, the position $x$ and a product over the Lie algebra index. It is also easy to see how the integration measure changes with the dimension of space, we could write \begin{equation} \mathcal{D}A = \prod_{x\in \mathbb{R^d}} \prod_{a=1}^{\dim \mathfrak{g}}\prod_{\mu=1}^d\mathrm{d}A_\mu^a(x) \end{equation}

Now what happens when we go to dimension $d = 0$. Well $\mathbb{R}^0$ is just a single point i.e. $\mathbb{R}^0 = \{0\}$, thus product over the continuos index disappears, also $A$ does not depend on the space coordinates $x$ anymore since space is just one point and it is just an element of the Lie algebra $\mathfrak{g}$ (a "matrix") and the measure is \begin{equation} \mathcal{D}A = \prod_{a=1}^{\dim \mathfrak{g}}\mathrm{d}A^a \end{equation} which is just the measure of a $(\dim \mathfrak{g})$-dimensional integral. I guess you confusion may come from the fact that when $G = U(N)$ we have $\dim \mathfrak{u}(N) = N^2$, then in when $N \to \infty$ we get another infinite dimensional integral. However I think the catch here is that first for the large $N$ limit you consider a $U(N)$ theory with arbitrary $N$, do your computations and then look at what happens when $N$ is large. The path integral is an $N^2$-dimensional integral with finite $N$ when you compute it. Second the matrix theory path integral does not have the "product of differential over a continuous index", which does not make sense and needs to be made sense of somehow (like via a discretization of space). For matrix models there are no functional integrals anymore and the path integral is just a regular multidimensional integral

$\endgroup$
2
  • $\begingroup$ Thank you very much! And that's how matrix model avoid the renormalization because they don't have the crossing terms $d A^{a}_{\mu} d A^{a}_{\nu}$. Am I correct? $\endgroup$
    – Errorbar
    Commented Dec 14, 2023 at 13:47
  • 1
    $\begingroup$ I would say that the main reason you don't need renormalization is that space is a single point. Renormalization is all about which length scale you're testing, but when there is a single point there is no renormalization to do (no high frequency modes to integrate away) $\endgroup$
    – lucabtz
    Commented Dec 14, 2023 at 14:51
3
$\begingroup$
  1. Technically speaking, for the (random) matrix model in 0-dimensional spacetime, the matrix size $N$ is finite, and hence there are only finitely many integrations over (orthogonal, hermitian, unitary, $\ldots$) matrix elements; not infinitely many.

  2. Physicists indeed often study the $1/N$ expansion and the large $N$ limit, but this is in principle different from facing the problem of actually integrating infinite-dimensional matrices, whatever that means (unless one from the very beginning simply define this path integral to be just the large $N$ limit).

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.