# Green's function for Anderson impurity model?

In this article "PHYSICAL REVIEW B 90, 155136 (2014) " (or here):

Title: "Machine learning for many-body physics: The case of the Anderson impurity model"

the Exact diagonalization (ED) method has been used to solve the Anderson impurity model (AIM). According to the ED method, the bath part is truncated as follows:

$$H = \sum\limits_\sigma {{\varepsilon _d}d_\sigma ^\dagger {d_\sigma }} + Ud_ \uparrow ^\dagger {d_ \uparrow }d_ \downarrow ^\dagger {d_ \downarrow } + \sum\limits_{l = 1,\sigma }^{{N_b}} {{\varepsilon _l}c_{l\sigma }^\dagger {c_{l\sigma }}} + \sum\limits_{l = 1,\sigma }^{{N_b}} {{V_l}\left( {d_\sigma ^\dagger {c_{l\sigma }} + c_{l\sigma }^\dagger {d_\sigma }} \right)} ,$$ where $$N_b$$ is the number oa bath sites and $$\varepsilon _d$$ is the bath on-site interaction.
Most of the figures contained in the article have been plotted at different values of $$U$$, $$V$$, and $$n_d$$, where $$n_d$$ is the impurity occupation. I have two inquiries about $$V$$ and $$n_d$$:

1. The authors replaced the on-site energy $$\varepsilon _d$$ by the occupation $$n_d$$! how to do this during the calculation? i.e. what is the value of $$\varepsilon _d$$ I have to use if, for example, $$n_d = 1$$.

2. The Hamiltonian doesn't contain $$V$$! Only $$V_l$$ and$$\varepsilon _l$$. So what $$V$$ refers to?

• Everything is explained after equation (14)
– LPZ
Commented Jun 2, 2023 at 7:26
• @LPZ Yes, but the calculation will be made based on the truncated Hamiltonian, which dos't contain $V$. The problem that I faced is that most of the ED codes need to inter the values of $U$, $\varepsilon_d$, $V_l$, and $\varepsilon_l$, so, how to set $\varepsilon_d$ and $V$? Commented Jun 2, 2023 at 7:36
• The link you provide is behind a paywall. Please provide a link to e.g. Arxiv if possible or at least the paper's title and authors. Commented Jun 2, 2023 at 8:26
• @StephenG-HelpUkraine Edited. Commented Jun 2, 2023 at 8:34
• As the authors say, they use $V_{\ell} = V$ $\forall \ell$, so just use the same value for all the $V_{\ell}$ coefficients. The problem with $\varepsilon_d$ is more subtle, and in fact you don't know in advance what is the corresponding value of $n_d$, so I would just do many calculations at different $\varepsilon_d$ and compute the associated densities Commented Jun 2, 2023 at 8:45

If you are only interested in the impurity’s Green’s function, the precise value of the individual $$V_l,\epsilon_l$$ is irrelevant. The Hamiltonian is more a crutch, that can be useful for some methods like exact diagonalisation for example.

However, you can define the dynamics of the impurity (in particular the interacting Green’s function) without using the Hamiltonian. The only thing that matters physically is the hybridization function: $$\Delta = \sum\frac{|V_l|^2}{i\omega-\epsilon_l}$$ or more physically the non interacting Green’s function: $$G_0 =\frac{1}{i\omega-\epsilon_d+\mu-\Delta}$$ One way to express this mathematically is to use the path integral: $$S=-\int \int(\bar d_\uparrow (\tau’)G_0^{-1} d_\uparrow (\tau)+\bar d_\downarrow (\tau’)G_0^{-1} d_\downarrow(\tau) )d\tau’+U\bar d_\uparrow d_\uparrow \bar d_\downarrow d_\downarrow)d\tau$$ This formalism is an easier entry point for developing CTQMC methods for example.

This is why specific choices of $$V_l,\epsilon_l$$ does not matter much. What matters is how you parametrize the hybridisation function.

As they describe it, it is simply: $$\Delta=\frac{2|V|^2}{\pi}\int_{-1}^1\frac{\sqrt{1-\epsilon^2}d\epsilon}{i\omega-\epsilon}$$

The choice of the value of $$\epsilon_d$$ is equivalent to modifying the chemical potential. This is why it is more natural to think in terms of particle number.

Due to particle hole symmetry, you need $$\epsilon_d=0$$ for half filling, i.e. $$n_d=1$$. In general, you’ll need to solve for the interacting Green’s function which is precisely what you want to simulate. There is therefore no known formula.

Hope this helps.