Dyson’s Brownian Motion

Posted on April 4, 2010 by


See also: Tao’s blog on Dyson’s Brownian motion

This is an outline of Dyson’s theorem. It isn’t meant to be completely rigorous, and please feel free to expand it. Let B_{ij}(t) be i.i.d. Brownian motions (not starting at 0), and form an N\times N matrix A with these entries. A second possibility would be to have B_{ij}(t) be complex Brownian motions, but we will stick with the real case. Note that the distribution of A is invariant under rotation of coordinates, namely O^TAO and A have the same distribution whenever O is an orthogonal matrix. The same is true for A+A^T, since it is true for each of these matrices individually, and since (O^TAO)^T=O^TA^TO. This leads to diagonal entries with a coefficient of 2, and off-diagonal entries with a coefficient of \sqrt 2. For convenience, we divide the whole matrix by \sqrt 2, and call the resulting matrix:

M(t) = \frac{A(t) + A^T(t)}{\sqrt{2}}

Let \lambda_i(t) denote the eigenvalues of M(t). We wish to find the SDE for the \lambda_i(t). In fact,

d\lambda_i = \sqrt{2}dB_i + \sum_{j\ne i} \frac{1}{\lambda_i-\lambda_j}dt.

In other words, the eigenvalues perform independent Brownian motions with a repelling drift.

By the Markov property, it suffices to do this computation at t=0. Suppose we are initially in a basis for which M is diagonal, with diagonal entries \lambda_i(0). By Ito’s lemma, it suffices to find the differential arising from each B_{ij}, and add the results together. For B_{ii}, we just get d\lambda_i=\sqrt{2}dB_{ii}, which is a term we want to have. Now consider off-diagonal entries. It is enough to consider B_{12}. Note that the corresponding matrix will have top left 2×2 block

\left( \begin{array}{cc} \lambda_1(0) & B_{12}(t)  \\ B_{12}(t) & \lambda_2(0) \end{array} \right)

Let a(t)=\lambda_1(t)+\lambda_2(t) and b(t)=\lambda_1(t)-\lambda_2(t). The trace of the displayed matrix is constant, so da=0. The determinant of the displayed matrix is \lambda_1(0)\lambda_2(0)-B^2, where we have dropped the subscripts on B for convenience. Note that this determinant equals (a^2-b^2)/4. Taking differentials, we find that


since we are assuming that B(0)=0 and working at time t=0. Thus we have an ODE for b^2, namely bb'/2=1 and so


This is the drift for db=d(\lambda_1-\lambda_2).  Each individual eigenvalue would get just half of this drift (since \lambda_1+\lambda_2 is constant in this scenario), so for d\lambda_1 we get the drift term


which repels the eigenvalues.

We can also answer the question of whether the eigenvalues meet.  Denote X(t)=\lambda_1(t)-\lambda_2(t).  Then, combining Brownian motions, we get


Using Brownian scaling, we get that 2dB(t)=dW(4t) for another Brownian motion W.  Then setting s=4t and Y(s)=X(s/4)=X(t), we get


which is the critical Bessel process, known not to hit 0.  Therefore, pairs of eigenvalues don’t meet.

Questions and Comments

I don’t see why when you consider a change of basis such that M is diagonal it turns out that the differential of \lambda_i is dB_i. If you change the basis you get a linear combination of the original brownian motions ( or are you using that sum of two normals is normal and then this is a different brownian motion? ).

  • Yes, an orthogonal transformation of an i.i.d. Gaussian vector is again an i.i.d. Gaussian vector.

My second question is when you consider M to be 2x2. I don’t quite understand why does the \lambda appear in the diagonal, nor why these are constants in the calculations latter. I have the impression that the reason is because of Itô’s formula but I don’t see it.

  • We are computing the effect of dB_{ij} individually, so the effect of dB_{ij} on the eigenvalues would be determined by the 4 entries whose indices are either i or j.

This is a comment, it seems that the argument handles 2x2 cases but for bigger matrices it would be necessary to conisder more functions than trace (a) and b ( trace with sign ? ). What are these functions? or from this argument we can obtain the result for bigger matrices?.

  • See the above comment, the argument is meant to handle matrices of all sizes.

This is another comment, it seems that there is a typo when you calculate the a^2 - b^2 since a^2 - b^2 = 4\lambda_1 \lambda_2 so when you calculate the differential I don’t quite understand why there is the differential of B_{12}^2 there, maybe I’m missing something?. ( maybe the functions a and b have a 1/2 ? )

  • This is a quick post, I wrote “modulo constants”. Please correct the constants if you’re so inclined.

When you calculate the differential of b^2 you obtain a term that is 2BdB but this term is 0 since you are working at time 0, but in the first sentence of the article you mention that the Brownian Motions do not start at 0.

  • We switched Brownian motions, so that instead of B_t with B_0=x we have x+B_t.
Posted in: Uncategorized