2
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ which text book? is this supposed to hold for any $N$? $\endgroup$ Oct 12, 2022 at 16:36
  • $\begingroup$ @CarloBeenakker. It's in a paper, actually. And the result only hold for the large $N\to\infty$. See link $\endgroup$
    – Guoqing
    Oct 13, 2022 at 0:52
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 14, 2022 at 10:48

1 Answer 1

3
$\begingroup$

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.


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

$\endgroup$
19
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 16, 2022 at 11:22
  • 1
    $\begingroup$ I have added the steps of the derivation. $\endgroup$ Oct 21, 2022 at 10:30
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 24, 2022 at 9:34
  • 1
    $\begingroup$ 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 $$\langle G(z)\rangle=G_0(z)+G_0(z)\langle A\langle G(z)\rangle A\rangle\langle G(z)\rangle.$$ $\endgroup$ Oct 27, 2022 at 9:31
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 10, 2022 at 7:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.