Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
More on Greens functions etc: physics.stackexchange.com/q/20797/2451 –  Qmechanic Nov 25 '12 at 16:08
1  
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. –  user16864 Dec 13 '12 at 11:30
    
Statistical Mechanics of Nonequilibrium Process by Dimitri Zubarev –  a0087946gy Dec 13 '12 at 12:10
    
See above @harken . –  a0087946gy Dec 13 '12 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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