Green's function and density of states

[Karthiksrao]

Dear all,

In his book chapter " Green’s Function Methods for Phonon Transport Through Nano-Contacts", Mingo arrives at the Green's function for the end atom of a one dimensional lattice chain (each atom modeled as a mass connected to neighbouring atoms through springs). He gives the green function as

$G = \frac{2}{\omega^2 + \sqrt{\omega^4 + 4 k \omega^2}}$

From this the intention is to find the spectral density of states. He directly gives it as

$\rho = \frac{1}{2 \pi} \frac{\sqrt{\omega^4 + 4 k \omega^2}}{2 k \omega^2}$.

I have not been able to show this.

Earlier in the chapter he uses the standard representation for the spectral density of states as
$G - G^* = 2 \pi \rho$.

Using this I attempted the following way. Let $\omega^2 = z$ (he had used such a representation earlier, hence I tried this), which gives:

$G(z) = \frac{2}{z + \sqrt{z^2 + 4 k z}}$

which can also be written as:

$G(z) = \frac{ z - \sqrt{z^2 + 4 k z}}{- 2 k z}$

Because of the $\sqrt{z}$ factor there is a branch cut along negative real axis. That's probably the only reason why $G - G^*$ would have a non-negative value - since there would be a non-zero difference in the values across the branch cut.

So what we need is :
$\lim_{\delta \rightarrow 0} G(z+ i \delta) - G(z - i \delta)$

This gives:

$\rho = \lim_{\delta \rightarrow 0} \frac{1}{2 \pi} \left( \frac{ \sqrt{1 + \frac{4 k}{z+ i \delta}} - 1}{ 2 k } - \frac{ \sqrt{1 + \frac{4 k }{z - i \delta}} - 1}{ 2 k } \right) = \frac{1}{2 \pi} \left( \frac{ \sqrt{1 + \frac{4 k }{z+ i \delta}} - \sqrt{1 + \frac{4 k }{z - i \delta}}}{ 2 k } \right)$

If you take the limit, this just goes to zero.

What am I doing wrong?

Thanks!

 

[DrDu]

I would have to take a closer look, but I think you have to write the square root as $f(z)^{1/2}$ and you have to continue it around the branch cut the square root whence it will pick up a factor -1. Summa summarum you will end up with the sum of the two square roots and not it's difference.