Resolvent (Green's function) of this random matrix

Question

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?

Answer 1

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 \begin{equation} G(z)=(z-A)^{-1}=z^{-1}\textstyle{\sum_{p=0}^{\infty}}(A/z)^{p}. \end{equation} 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$^\ast$ \begin{equation} \langle G(z)\rangle=z^{-1}+z^{-1}\langle A\langle G(z)\rangle A\rangle\langle G(z)\rangle. \end{equation} This can be rearranged in the form \begin{equation} \langle G(z)\rangle=\bigl[z-\Sigma(z)\bigr]^{-1},\;\;\Sigma(z)=\langle A\langle G(z)\rangle A\rangle. \end{equation}

As a tentative solution we substitute a block-diagonal $\Sigma$, \begin{equation} \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}, \end{equation} 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 \begin{equation} 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).\end{equation} For $N\gg 1$ I may neglect the last term, resulting in \begin{equation} a(z)=\tfrac{1}{2}g\left(z-\sqrt{z^2-4\lambda^2}\right),\;\; b(z)=g^{-1}a(z). \end{equation} 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, \begin{equation} \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}. \end{equation} 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.

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$^\ast$ 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 $$\langle G(z)\rangle = z^{-1} + z^{-1}\langle A_1\langle G(z)\rangle A_2\rangle\langle G(z)\rangle + \text{intersecting contractions}.$$