Timeline for How to dress free Green functions with constant broadening?
Current License: CC BY-SA 4.0
27 events
| when toggle format | what | by | license | comment | |
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| Jan 11, 2021 at 12:52 | comment | added | xiaohuamao | OK, we are on the same page and I don't object to any of the math. So you think the guess is actually wrong for $G^<$ and we just don't have a simple enough real-time form? I presume adding broadening like this is computationally convenient common practice to avoid singular $\delta$-functions or so. This is a bit surprising to me. | |
| Jan 11, 2021 at 12:36 | comment | added | Roger V. | The physical intuition here is that when a level has a finite width, one has a finite probability that it is occupied or empty, i.e., one cannot say unambiguously that it is below/above the Fermi level. | |
| Jan 11, 2021 at 12:34 | comment | added | Roger V. | 5th and 6th comments indicate how one can restore the known result in case of zero broadening ($\Gamma \rightarrow 0^+$), which is your first set of equations. Your second set of equations is indeed different from mine (if by $n_F$ you mean the value of the distribution function at Fermi energy) - my understanding is that this was a guess, and as you see that guess is not quite correct for finite broadening. | |
| Jan 11, 2021 at 12:20 | comment | added | xiaohuamao | Just the 4th, 5th, and 6th comments here. | |
| Jan 11, 2021 at 11:20 | comment | added | Roger V. | Could you indicate more precisely which step you are talking about? Note also that $g_{dd}$ and $g_k$ are undressed/non-interacting Green's functions, so they have zero broadening (I use $\eta$ instead of writing $0^+$). | |
| Jan 11, 2021 at 10:40 | comment | added | xiaohuamao | $\lambda$ (or $\Gamma$) is a constant in the final result in the form $\frac{\lambda}{\omega^2+\lambda^2}$. But you already have used $\frac{1}{\pi}\frac{\lambda}{\omega^2+\lambda^2}\rightarrow \delta(\omega)$ in an intermediate step, which is not exact and is just an approximation when $\lambda\rightarrow0$. I mean this looks suspicious. Sorry if I didn't express my question clearly in the previous comments. | |
| Jan 11, 2021 at 10:23 | comment | added | Roger V. | I am still not sure what you mean: from the point where I assumed the constant Gamma, my derivations are mathematically exact. | |
| Jan 11, 2021 at 10:16 | comment | added | xiaohuamao | I know what you mean - this is indeed a representation of delta-function. Let me repeat my concern: all we want to do in this whole post is just like trying to replace $\delta(\omega)$ by $\frac{1}{\pi}\frac{\lambda}{\omega^2+\lambda^2}$ as shown in Eq.(1) and above in my original question. This intermediate step, however, uses the reverse, $\frac{1}{\pi}\frac{\lambda}{\omega^2+\lambda^2}\rightarrow \delta(\omega)$ in order to get a final result in the form $\frac{1}{\pi}\frac{\lambda}{\omega^2+\lambda^2}$. This looks quite suspicious. I was wondering if any physical justification of this. | |
| Jan 11, 2021 at 6:20 | comment | added | Roger V. | Sorry, I misunderstood your question. This is one of the representations of the delta-function - as a Lorentz function of infinitesimal width. | |
| Jan 11, 2021 at 6:19 | comment | added | Roger V. | Sorry, I misunderstood your question. This is one of the representations of the delta-function. | |
| Jan 11, 2021 at 2:28 | comment | added | xiaohuamao | Sorry, I don't get it. Why does a constant DOS justify the intermediate step $\frac{1}{\pi}\frac{\lambda}{\omega^2+\lambda^2}\rightarrow\delta(\omega)$? | |
| Jan 10, 2021 at 20:09 | comment | added | Roger V. | Typically variation of the density of states on the scale of applied bias and frequencies is small, so it can be considered constant. Moreover, for nanostructures fabricated on the basis of 2d electrin gas -the 2d density of states is constant. But in principle, applicability of every approximation depends on the problem at hand. There are no approximations that always work. | |
| Jan 9, 2021 at 12:06 | comment | added | xiaohuamao | Thanks. So in the context you mentioned like tunneling through quantum dots, why can we do that? | |
| Jan 9, 2021 at 11:45 | comment | added | Roger V. | Physical justification depends on the specific situation. Whay I described is used in the context of tunneling through quantum dots. Note that in case of only one Fermi sea one does not really need Keldysh - it is en equilibrium problem. But my solution is easily generalizable to several Fermi seas - as described in the paper that I cited. | |
| Jan 9, 2021 at 11:20 | history | bounty ended | xiaohuamao | ||
| Jan 9, 2021 at 11:19 | vote | accept | xiaohuamao | ||
| Jan 9, 2021 at 11:15 | comment | added | xiaohuamao | I know what you mean. But all we do is trying to replace $\delta(\omega)$ by $\frac{1}{\pi}\frac{\lambda}{\omega^2+\lambda^2}$. This very step, however, seemingly goes back and forth a little arbitrarily. So I was wondering if any physical justification. | |
| Jan 9, 2021 at 8:50 | comment | added | Roger V. | Also, $f(\omega)\delta(\omega-\epsilon_d)=f(\epsilon_d)\delta(\omega-\epsilon_d)$, it is a property of teh delta-function. | |
| Jan 9, 2021 at 8:49 | comment | added | Roger V. | You can use $\frac{1}{\pi}\frac{\lambda}{\omega^2+\lambda^2}\longrightarrow \delta(\omega)$ as $\lambda\rightarrow 0$ to recover the expressions for the infinitesimally small broadening. | |
| Jan 9, 2021 at 0:33 | comment | added | xiaohuamao | This is really a nice answer! Only one question left. Your last expression $\frac{\Sigma^{>,<}(\omega)}{(\omega-\epsilon_d)^2+\frac{\Gamma^2}{4}}$ has a complex form transformed to time. To get what I wrote in Eq.(1), one needs $\frac{\Sigma^{>,<}(\epsilon_d)}{(\omega-\epsilon_d)^2+\frac{\Gamma^2}{4}}$ instead. Any way to justify this? | |
| Jan 8, 2021 at 10:26 | comment | added | Roger V. | I have added the derivations. | |
| Jan 8, 2021 at 10:26 | history | edited | Roger V. | CC BY-SA 4.0 |
added derivations
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| Jan 6, 2021 at 14:43 | comment | added | Roger V. | Indeed, I have missed $\epsilon_k$. I will try to sketch the derivation when I have a bit more time. | |
| Jan 6, 2021 at 14:42 | history | edited | Roger V. | CC BY-SA 4.0 |
corrected equations
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| Jan 6, 2021 at 12:53 | comment | added | xiaohuamao | Thank you for your answer. Is there an $\epsilon_k$ missing in the second term of $H$? It seems to give the desired result. But could you please elaborate on how to get these self-energies from $H$? | |
| Jan 6, 2021 at 12:29 | history | edited | Roger V. | CC BY-SA 4.0 |
corrected equations
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| Jan 6, 2021 at 8:46 | history | answered | Roger V. | CC BY-SA 4.0 |