# How can I get the projected density of states within the Green's function method?

I have this Green's function matrix $$G^r= (E \mathbb{I} - \hat{H})^{-1}$$ in the $$|\textrm{site}\rangle\otimes|\textrm{orbital}\rangle\otimes|\textrm{spin}\rangle$$ Hilbert space, and I am curious if there is a way to pull the density of states of the projection onto different quantum operators.

It would be easy to get the density of states for an operator within the same basis, say $$S_z$$, by simply summing up all of the $$m = 1/2$$ states and $$m = -1/2$$ states separately, instead of just taking $$\frac{-1}{\pi}\textrm{Tr}(\textrm{Im}(G^r))$$.

But I am curious if one could do more with the green's function to get the density of states of $$S_x$$, for example.

Any insight or resources would be appreciated!

• +1 and welcome to our new community! Thanks very much for contributing your question here and we hope to see much more of you! How did you find us? By the way I've edited your MathJaX formatting so that it's nicer to read for the word "site" in the $| \textrm{ket} \rangle$, can you do the same for the words "orbital" and "spin"? Sep 22, 2022 at 18:14
• Got it. Thanks for the edit haha. I have been working with DFT codes and exploring electron transport for a few years now, but the lack of documentation frustrated me a lot, so I found the physics stack exchange on google while trying to debug Quantum Espresso. Saw there was talk of spinning off a matter modelling forum, so it's been on my mind for a while. :) I've been writing my own NEGF electron transport codes for about a year now, so I hope I can give back a little bit. Sep 22, 2022 at 23:32
• Can you define (or give a reference) what you mean with 'density of states of an operator'? Sep 23, 2022 at 13:32
• That is my bad, I think my wording was unclear. Normally, one might define a Projected Density of States (PDOS) by diagonalizing an operator and getting the eigenstates $|\textrm{O}_i\rangle$, then diagonalizing a hamiltonian and projecting $|\psi_j\rangle$ as $PDOS_i(E) = \sum_j\langle\psi_j|\textrm{O}_i\rangle \langle\textrm{O}_i|\psi_j\rangle \delta(\textrm{E-E}_j)$. This give a DOS for each eigenvalue of O. But I am not sure how/if this could be recovered from the Green's function matrix, defined in my OP, which is also widely used in nanoelectronics. Sep 23, 2022 at 14:38

I believe that you can get what you want using the spectral density,

$$\rho({\bf r},{\bf r'},E) = -\frac{1}{\pi}{\rm Im}[G({\bf r},{\bf r'},E)]\tag{1}$$

Then the total density of states would be an integral over all space r and r', while a projected density of states would be

$$\textrm{PDOS}_i(E) = -\frac{1}{\pi}{\rm Im} \langle O_i |G | O_i \rangle\tag{2}$$

where $$O_i$$ denotes whichever states you like, and can certainly be the eigenstates of an operator $$\bf O$$.

If you already have the Green's function in a basis $$|\alpha\rangle$$, then you would have,

$$\textrm{PDOS}_i(E) = -\frac{1}{\pi}{\rm Im} \langle O_i|G|O_i \rangle = -\frac{1}{\pi}{\rm Im}\sum_{\alpha,\alpha'}{\langle O_i |\alpha\rangle G_{\alpha\alpha'}\langle \alpha'|O_i\rangle}\tag{3}$$

• Oh great, thanks! Would you happen to have a reference for that? No problem if not. Would you know how to pull expectation values out of the green's function also? From my screwing around, it looks like $\langle O\rangle = -\frac{1}{\pi}{\rm Im (Tr}(\hat{O} G^r(E))$ but I haven't proved it or found a reference on it. Sep 28, 2022 at 22:06

I'm not sure I fully understand the question here. The local density of states is as you defined (see equation 10 of https://arxiv.org/pdf/1907.11302.pdf too)

Could you define what you mean by density of states of the projection onto different quantum operators. As you are measuring the inverse of a complex shifted Hamiltonian.

The only way I could think of getting G for certain operators is by projecting the quantum state into a desired subspace. For example one could project the quantum state into a subspace of singlet states or a subspace of triplet states (which fixes the spin angular momentum operator [S]).

I apologise if I've misunderstood your question, but I hope you understand what I mean.