Eigenvalue equation for interacting Green's functions


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4 months ago by
Studying the articles "Topological Hamiltonian as an exact tool for topological invariants" (<https://arxiv.org/abs/1207.7341>) and "Simplified Topological Invariants for Interacting Insulators" (<https://arxiv.org/abs/1201.6431>) I stumbled upon the following eigenvalue equation

\begin{align}
G^{-1}(k,i\omega) |\alpha(k,i\omega)> &= \mu_\alpha(k,i\omega) |\alpha(k,i\omega)>
\end{align}

where

\begin{align}
G^{-1}(k,i\omega) &= i\omega - H_0(k) - \Sigma(k,i\omega)
\end{align}

is the full Green's function of an interacting problem evaluated at the Matsubara frequencies \(\omega_n = (2n+1)\pi/\beta\) (Probably a similar formular also holds for real frequencies, then just replace \(i\omega \rightarrow \omega\)). \(H_0(k)\) is the Hamiltonian matrix of the noninteracting problem and \(\Sigma(k,i\omega)\) the \(\omega\) dependent self-energy that appears in Dyson's equation \(G = G_0 + G_0 \Sigma G\). On a mathematical level, \(G^{-1}(k,i\omega) \in\ \)GL(N,C) is an invertible complex \(N\times N\) matrix, and \(\{|\alpha(k,i\omega)>\}\) a set of orthogonal eigenvectors of that matrix with eigenvalues \(\mu_\alpha(k,i\omega)\).

The eigenvalue equation can also be written as

\begin{align}
\left(i\omega - H_0(k) - \Sigma(k,i\omega)\right) |\alpha(k,i\omega)> &= \mu_\alpha(k,i\omega) |\alpha(k,i\omega)>\\
\left(H_0(k) + \Sigma(k,i\omega)\right) |\alpha(k,i\omega)> &= -(\mu_\alpha(k,i\omega)+i\omega) |\alpha(k,i\omega)>
\end{align}

where the last line looks pretty similar to the time independent Schroedinger equation of a noninteracting system

\begin{align}
H_0(k) |\alpha(k)> &= \epsilon(k)|\alpha(k)>
\end{align}

with the big difference that the first equation depends on \(\omega\) and the second does not. I want to understand that equation and have the following questions

- What is the meaning of an \(\omega\)-dependent eigenstate?
- What is the meaning of an \(\omega\)-dependent (energy?) eigenvalue?
- Where can I find more about this topic? (Books, articles, ...)

2 Answers


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4 months ago by
I do not understand the meaning of an expression like
\( \Big( i\omega - H_0(k) - \Sigma(k,i\omega) \Big) | \alpha(k, i\omega) \rangle \)

AFAIU, \( H_0(k) \) and \( \Sigma(k,i\omega) \) are not operators, but scalars, unless you define precisely what you mean.

“On a mathematical level, \( G^{-1}(k,i\omega) \in \) GL(N,C) is an invertible complex \( N \times N \) matrix...” is quite vague for me. That's perhaps the reason you arrive at incomprehensible objects.
In these equations, \(H_0(k)\), \(\Sigma(k,i\omega)\) as well as \(G^{-1}(k,i\omega)\) are all matrices acting on the vectors \(|\alpha(k,i\omega)>\).
written 4 months ago by LidoDiCamaiore  
It's better to start from a very simple problem; like a non-interacting problem or a two-level mixing (or hybridization of two levels). Then try to represent or interpret the Green's function as a matrix. You might then get an idea.

For me, it is totally unclear what is meant by that matrix representation.
written 4 months ago by AlQuemist  
I agree with the Boss; the statement

\(\Big( i\omega - H_0(k) - \Sigma(k,i\omega) \Big) | \alpha(k, i\omega) \rangle\)

is quite ambiguous. If it is indeed, as you say, a multiplication of matrices with vectors, then this implies that the operators and states have been projected to a specific basis, and in doing this one sees that there will be integrals involved (due to inserting a complete set of states). These integrals are missing in the above equation.
written 4 months ago by Master Dragon  
In my case I have \(H=H_{MF}+H_{fl}\). The quadratic part can be written in matrix form as

\begin{align}
   H_{MF} &= \sum_{k}
      \begin{pmatrix} c_k^\dagger \\ f_k^\dagger \end{pmatrix}
      \underbrace{\begin{pmatrix} \xi_k^c\cdot 1 & b^* V_0\cdot\Phi(\vec{k}) \\ b V_0\cdot\Phi(\vec{k}) & \xi_k^f\cdot 1
      \end{pmatrix}}_{=H_0(k)}
      \begin{pmatrix} c_k \\ f_k \end{pmatrix}
\end{align}

The self energy \(\Sigma\), on the other hand, is obtained by perturbation theory and there are plenty of sums and integrals involved. But in the mentioned eigenvalue equation, I don't care how to calculate it in practice, I just act as if I knew it. It does not really matter.

I have the suspicion, however, that the equation does not have much physical meaning in the first place, but is used as a mathematical trick in the mentioned papers. Did you have a look at them? They are indeed interesting papers.
written 4 months ago by LidoDiCamaiore  
You seem to be mixing apples and coconuts ;)
The ingredients of the Hamiltonian are operators on the Fock space.
The self-energy is a ‘simple’ scalar-valued function.

At any case, you cannot represent the quartic (or higher-order parts) of a Hamiltonian with matrices. The reason is fundamental: Matrices represent linear transformations (operators) while such interactions are nonlinear operators.
written 4 months ago by AlQuemist  
First: you can gather the functions, that the self-energy is, in a matrix if you have different particle types.
Second: I do not represent the Hamiltonian by matrices, but the Green's function. And that comes straight from Dyson's equation
written 4 months ago by LidoDiCamaiore  
Just make clear what you want to achieve with such a “matrix-method”. Perhaps, then we can find a proper way to do that.
written 4 months ago by AlQuemist  
Aaaand... I think the problem is solved by now, you can read everything in my thesis later ;)
written 4 months ago by LidoDiCamaiore  
Well then; good luck!
But that remains still unclear.
written 4 months ago by AlQuemist  
0
4 months ago by
Actually, notice that taking this statement literally,

“On a mathematical level, \( G^{-1}(k,i\omega) \in \) GL(N,C) is an invertible complex \( N \times N \) matrix...”

implies that any function, \( f(\mathbf{x}, y) \in GL(N,C) \)! This is absurd.
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