I bumped into a book, where Resolvent $R^{\pm}(E)$ is defined as

$e^{\mp iHt/\hbar}=\pm\frac{i}{2\pi}\int_{-\infty}^{\infty}dER^{\pm}(E)e^{\mp iEt/\hbar}$ and $R^{\pm}(E)=\frac{1}{\pm i\hbar}\int_0^{\infty}dte^{\mp iHt/\hbar}e^{\pm iEt/\hbar}e^{-\eta t/\hbar}$. It is easy to show that $R^{\pm}(E)=\frac{1}{E-H\pm i\eta}$. Here H is the full Hamiltonian. So can anyone tell me the difference between it and Green Function?

  • $\begingroup$ More on Greens functions etc: physics.stackexchange.com/q/20797/2451 $\endgroup$
    – Qmechanic
    Nov 25, 2012 at 16:08
  • 1
    $\begingroup$ What is the name the book? I have some trouble finding the resolvent of the Liouville superoperator $\mathcal{L}=\frac{1}{i\hbar}[H,\;]$. I could not find it in any book. Any comments and suggestions are welcomed. $\endgroup$
    – user16864
    Dec 13, 2012 at 11:30
  • $\begingroup$ Statistical Mechanics of Nonequilibrium Process by Dimitri Zubarev $\endgroup$
    – Brioschi
    Dec 13, 2012 at 12:10
  • $\begingroup$ See above @harken . $\endgroup$
    – Brioschi
    Dec 13, 2012 at 12:11

1 Answer 1


They are closely related: the resolvent of the eigenvalue equation of a self-adjoint operator $\hat{A}$ is the operator valued function defined as

$$G_\lambda=(\hat{A}-\lambda \hat{1})^{-1}$$

We call Green's function the kernel of the resolvent [the kernel of an integral transform] which is the solution of the homogeneous differential equation

$$(\hat{A} -\lambda)G_\lambda (\mathbf{x},\mathbf{y})=\delta^{(3)}(\mathbf{x}-\mathbf{y})$$

for suitable boundary conditions. Thus

$$(\hat{A}-\lambda)\int_{\mathbb{R}^3}d\mathbf{y} \,G_\lambda(\mathbf{x},\mathbf{y}) \psi(\mathbf{y})=\psi(\mathbf{x})$$

for any continuous $\psi(\mathbf{x})$ in $L_2(\mathbb{R}^3)$ in the case $\hat{A}$ is a differential operator.


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