# Resolvent (Green's function) of this random matrix

I have a matrix $$A$$ as follows: $$A=\begin{pmatrix} 0 & \boldsymbol{W} \\ \boldsymbol{W}^{\dagger} & \boldsymbol{H} \end{pmatrix}$$ where $$H$$ and $$W$$ are a random Hermitian $$N\times N$$ matrix and an $$N$$-component vector of independently distributed complex variables, respectively. The matrix elements have zero mean and variances $$\langle H_{kl}H^{*}_{mn}\rangle_H=\frac{\lambda^2}{N}\delta_{km}\delta_{ln},\ \langle W_kW^{*}_l\rangle_W=\frac{g\lambda^2}{N}\delta_{kl}.$$ The definition of resolvent is $$G(z)=\frac{1}{z-A}$$ after a ensemble average. My question is: Why the diagonal element $$G_{11}(z)=\left(1\ 0\ 0\ \dots\right)G(z)\left(1\ 0\ 0\ \dots\right)^T$$ is $$G_{11}(z)=\frac{1}{z-ig\lambda}$$ as written in the textbook.

## My solution

A generic way to calculate the GF is using projection operators method. Denoting the projector $$P=\left(1\ 0\ 0\ \dots\right)^T\left(1\ 0\ 0\ \dots\right),Q=I_{N+1,N+1}-P,$$ that satisfy $$P^2=P,Q^2=Q,QP=PQ=0$$. Rewritting $$A$$ as $$A=H_0+V$$ with $$H_0=\begin{pmatrix} 0 & 0 \\ 0 & H \end{pmatrix},$$ which satisfies $$QH_0P=PH_0Q=0,QVQ=PVP=0$$. After some algebras, the projected GF $$PG(z)P$$, which is a $$1\times1$$ matrix with element given by $$G_{11}(z)$$, reads as $$PG(z)P=\frac{P}{z-PH_0P-PR(z)P}.$$ Here the matrix $$R(z)$$ is $$R(z)=V+V\frac{Q}{z-H_0}V+V\frac{Q}{z-H_0}V\frac{Q}{z-H_0}V+\cdots.$$ Using the relation $$QVQ=PVP=0$$, the projected GF has this form $$PG(z)P=\frac{P}{z-PV\frac{Q}{z-H_0}VP}.$$ Now calculating the denominator $$PV\frac{Q}{z-H_0}VP=\sum_{i,j,m}W_iW^*_j\frac{c^m_i{c^m_j}^*}{z-E_m}=\sum_{i,j,m}W_iW^*_j\frac{1}{z-E_m}\frac{1}{N},$$ where $$E_m$$ is the $$m$$-th eigenvalue of $$H$$ and $$c^m_i$$ the $$i$$-th component of the normalized eigenvector associated with the jth eigenvalue of H. Taking a ensemble average, this gives $$G_{11}(z)=\frac{1}{z-g\lambda^2\left(z-\sqrt{z^2-4\lambda^2}\right)}.$$ This answer is obvious not consistent with $$1/\left(z-ig\lambda\right)$$ in the limit $$z=0$$ for $$PR(z)P$$. Where did I go wrong with the calculation?

• which text book? is this supposed to hold for any $N$? Oct 12, 2022 at 16:36
• @CarloBeenakker. It's in a paper, actually. And the result only hold for the large $N\to\infty$. See link Oct 13, 2022 at 0:52
• I worked out the derivation in the answer box, but do note that the formula you wrote down for $G_{11}(z)$ neglects higher order terms in $z$ in the denominator. Oct 14, 2022 at 10:48

This is a small varation on Pastur's derivation of the semicircle law.

We seek the average $$\langle G(z)\rangle$$ of the Green's function $$$$G(z)=(z-A)^{-1}=z^{-1}\textstyle{\sum_{p=0}^{\infty}}(A/z)^{p}.$$$$ Gaussian averages of $$A^{p}$$ consist of sums of all pairwise contractions. For $$N\gg 1$$ only non-intersecting contractions are kept, resulting in the nonlinear equation$$^\ast$$ $$$$\langle G(z)\rangle=z^{-1}+z^{-1}\langle A\langle G(z)\rangle A\rangle\langle G(z)\rangle.$$$$ This can be rearranged in the form $$$$\langle G(z)\rangle=\bigl[z-\Sigma(z)\bigr]^{-1},\;\;\Sigma(z)=\langle A\langle G(z)\rangle A\rangle.$$$$

As a tentative solution we substitute a block-diagonal $$\Sigma$$, $$$$\Sigma(z)=\begin{pmatrix} a(z)&0\\ 0&b(z)I \end{pmatrix}\Rightarrow \langle G(z)\rangle=\begin{pmatrix} [z-a(z)]^{-1}&0\\ 0&[z-b(z)]^{-1}I \end{pmatrix},$$$$ with $$I$$ the $$N\times N$$ identity matrix. We then compute \begin{align} \langle A\langle G(z)\rangle A\rangle={}&\left\langle\begin{pmatrix} 0&W\\ W^\dagger&H \end{pmatrix} \begin{pmatrix} [z-a(z)]^{-1}&0\\ 0&[z-b(z)]^{-1}I \end{pmatrix}\begin{pmatrix} 0&W\\ W^\dagger&H \end{pmatrix}\right\rangle\nonumber\\ ={}&\begin{pmatrix} [z-b(z)]^{-1}\langle WW^\dagger\rangle&0\\ 0&[z-b(z)]^{-1}\langle H^2\rangle+[z-a(z)]^{-1}\langle W^\dagger W\rangle \end{pmatrix}\nonumber\\ ={}&\begin{pmatrix} [z-b(z)]^{-1}g\lambda^2&0\\ 0&[z-b(z)]^{-1}\lambda^2 I+[z-a(z)]^{-1}(g\lambda^2/N)I \end{pmatrix}. \end{align} This must be equal to $$\Sigma(z)$$, hence we have the two equations $$$$a(z)=[z-b(z)]^{-1}g\lambda^2,\;\;b(z)=[z-b(z)]^{-1}\lambda^2 +[z-a(z)]^{-1}(g\lambda^2/N).$$$$ For $$N\gg 1$$ I may neglect the last term, resulting in $$$$a(z)=\tfrac{1}{2}g\left(z-\sqrt{z^2-4\lambda^2}\right),\;\; b(z)=g^{-1}a(z).$$$$ The sign of the square root is fixed by the requirement that $$\Sigma\rightarrow 0$$ for $$z\rightarrow\infty$$.

Collecting results I thus obtain the $$1,1$$ element of the average Green's function, $$$$\langle G(z)\rangle)_{1,1}=[z-a(z)]^{-1}=\left[z-\tfrac{1}{2}g\left(z-\sqrt{z^2-4\lambda^2}\right)\right]^{-1}.$$$$ The imaginary part of $$a(z)$$ gives the semicircle density of states $$\tfrac{1}{2}g\sqrt{4\lambda^2-z^2}$$, for $$|z|<2\lambda$$. Note that the formula in the OP only gives the $$z=0$$ limit.

$$^\ast$$ The OP asks for some insight into the derivation of the nonlinear equation for the average Green function. This is known as the Dyson equation in quantum physics, I guess most quantum theory text books will have a derivation, let me summarize the key steps.

The first step is to note that a Gaussian average is a sum over all pairwise averages, or contractions. A contraction of two $$A$$'s gives a factor $$1/N$$ and a Kronecker delta. The summation over indices can contribute a factor of $$N$$, so that this contraction becomes of order unity, but only if the contraction of the two $$A$$'s does not intersect with another contraction. Otherwise the Kronecker delta's restrict the summation and prevent the appearance of a factor $$N$$ to cancel the $$1/N$$. So what we learn from this first step is that to leading order in $$N$$ only non-intersecting contractions contribute.

The second step is to look at the Taylor series of the Green function $$G(z)$$ in powers of $$A/z$$. Take the first $$A$$, let me call it $$A_1$$ and contract it with another $$A$$, say $$A_2$$. In between $$A_1$$ and $$A_2$$ there appear other contractions, which give you back $$\langle G(z)\rangle$$. Beyond $$A_2$$ there also appear other contractions, which also give you $$\langle G(z)\rangle$$. So you find the desired equation $$\langle G(z)\rangle = z^{-1} + z^{-1}\langle A_1\langle G(z)\rangle A_2\rangle\langle G(z)\rangle + \text{intersecting contractions}.$$

• the thing you are averaging is in the denominator of the Green function; you need the average of the Green function, not the average of the denominator. Oct 16, 2022 at 11:22
• I have added the steps of the derivation. Oct 21, 2022 at 10:30
• certainly, in your case $\langle H\rangle=0$, otherwise you will have extract the non-fluctuating part, and the initial $z^{-1}$ will contain that contribution. Oct 24, 2022 at 9:34
• in the more general case, decompose $H=A+B$ where $B$ is nonfluctuating and $A$ has a normal distribution; then define $G(z)=(z-H)^{-1}$ and $G_0=(z-B)^{-1}$, and the equation you wish to solve is $$\langle G(z)\rangle=G_0(z)+G_0(z)\langle A\langle G(z)\rangle A\rangle\langle G(z)\rangle.$$ Oct 27, 2022 at 9:31
• this is not how I understand the method; you can write down some formal expression which is correct to all orders in $1/N$, but then to evaluate this you will need to make a saddle point approximation, which is equivalent to the leading order in the $1/N$ expansion. Nov 10, 2022 at 7:47