If K(x,y) is such that
- ∫dx K(x,x)=α
- ∫dy K(x,y)K(y,z)=K(x,z)
For simplicity I have written ∫ instead of ∫R. This formula is used when calculating n-point functions in the Gaussian Unitary Ensemble (GUE). Tao gives a short proof of (1) based on induction and the Laplace expansion of determinants. In this post, I give a proof using integration over Grassmann variables. The reason I am interested in this alternative proof is that I want to compress the calculation of n-point functions in the GUE as much as possible.
The proof of (1) proceeds as follows. I express the determinant as a Gaussian integral over Grassmann variables deti,j≤n+1(K(xi,xj))=∫(n+1∏i=1dθidˉθi)exp(n+1∑i,j=1ˉθiK(xi,xj)θj)
The LHS of (1) is
∫dxn+1∫(n+1∏i=1dθidˉθi)exp(n∑i,j=1ˉθiK(xi,xj)θj+X)
with
X=n∑i=1ˉθiK(xi,xn+1)θn+1+n∑j=1ˉθn+1K(xn+1,xj)θj+ˉθn+1K(xn+1,xn+1)θn+1
I now perform the integration over θn+1,ˉθn+1. I therefore expand the exponential
eX=(1+n∑i=1ˉθiK(xi,xn+1)θn+1)(1+n∑j=1ˉθn+1K(xn+1,xj)θj)(1+ˉθn+1K(xn+1,xn+1)θn+1)
Integrating over θn+1,ˉθn+1 picks the coefficient of ˉθn+1θn+1, this gives
K(xn+1,xn+1)−n∑i,j=1ˉθiK(xi,xn+1)K(xn+1,xj)θj
At this point I have
LHS=∫dxn+1∫(n∏i=1dθidˉθi)exp(n∑i,j=1ˉθiK(xi,xj)θj)(K(xn+1,xn+1)−n∑i,j=1ˉθiK(xi,xn+1)K(xn+1,xj)θj)
Integrating over xn+1 then gives
LHS=∫(n∏i=1dθidˉθi)exp(n∑i,j=1ˉθiK(xi,xj)θj)(α−n∑i,j=1ˉθiK(xi,xj)θj)
I elaborate a bit on the second term:
∫(n∏i=1dθidˉθi)exp(n∑i,j=1ˉθiK(xi,xj)θj)(n∑i,j=1ˉθiK(xi,xj)θj)=∫(n∏i=1dθidˉθi)1(n−1)!(n∑i,j=1ˉθiK(xi,xj)θj)n−1(n∑i,j=1ˉθiK(xi,xj)θj)=n∫(n∏i=1dθidˉθi)1n!(n∑i,j=1ˉθiK(xi,xj)θj)n=n∫(n∏i=1dθidˉθi)exp(n∑i,j=1ˉθiK(xi,xj)θj)
It follows that
LHS=∫(n∏i=1dθidˉθi)exp(n∑i,j=1ˉθiK(xi,xj)θj)(α−n)=(α−n)deti,j≤n(K(xi,xj))
Further reading
- An introduction to integrating over Grassmann variables can be found in Path Integrals in Quantum Mechanics by Jean Zinn-Justin
- Details and many formulas about integrating over Grassmann variables can also be found in appendix A in this paper.
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