arXiv:hep-th/9304011 v1 Apr 5 1993

2
λ
Fig. 17: The Wigner semicircle distribution.
Exercise. Derivation of the Wigner Distribution
As an example of an eigenvalue distribution, we consider the Gaussian matrix
model. The leading term in the large N asymptotics of the eigenvalue distribution is
the famous Wigner distribution
K 1 (λ, λ) = 1 √
2 − λ
4π
2 θ(2 − λ 2 ) , (9.12)
shown in fig. 17.
Derive the Wigner distribution from the “Schwinger–Dyson” equations of the matrix
model using the above procedure. First show that for the Gaussian potential we
have
〈Ŵ(ζ) 〉
= 1 ( √
h=0 ζ − ζ2 − 2 ) , (9.13)
2
and from this obtain the Wigner distribution (9.12).
The finiteness of the support of the kernels has important implications for the nonanalyticity
in ζ. Consider for example the one-point function
〈Ŵ 〉 1
〈
1
〉
(ζ) ≡ tr = 1 ∫
K(λ, λ)
dλ
N ζ − φ N ζ − λ
∼ ∑ ∫
N χ dλ K χ(λ, λ)
.
χ I ζ − λ
(9.14)
The nonanalytic dependence on ζ we are looking for comes from the contributions in
the λ integrals from the integrals near the edge of the support I of the eigenvalue distribution.
In the last expression we may take ζ real and ζ > ζ c . We encounter nonanalytic
behavior as ζ hits the edge of the eigenvalue distribution.
Example: Let us verify the statements about analytic dependence on ζ in the example
of a Gaussian potential. Expanding ζ = ζ c + δζ = √ 2 + δζ, or equivalently, expanding λ
around the edge of the eigenvalue distribution, we obtain a nonanalytic function of (δζ) 1/2
corresponding (formally) to the one-loop amplitude 〈 W (l) 〉 = l −3/2 .
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