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Consider the equilibrium time ordered and retarded Green's function for two operators $A,B$, defined as follows $$ G^T(t,t') \equiv -i\langle T A(t)B(t') \rangle=-i\theta(t-t')\langle A(t)B(t')\rangle -i\theta(t'-t)\langle B(t')A(t)\rangle\\ G^R(t,t') \equiv -i\theta(t-t') [A(t),B(t')] \rangle $$ At equilibrium, the geens functions only depend on $t-t'$ so WLOG can set $t'=0$. The two greens function are related by $$ G^T(t,t'=0) = -i\theta(t)\langle A(t)B(0)\rangle -i\theta(-t)\langle B(0)A(t)\rangle = -i\theta(t)\langle [A(t),B(0)] \rangle-i\langle B(0)A(t) \rangle = G^R(t)-i\langle B(0)A(t)\rangle $$

I have heard statements that the two Green's functions are equal at zero temperature and for positive energies, i.e. $G^T(\omega) = G^R(\omega)$ for $\omega>0$. This basically says that the fourier transform of the second term above vanishes for $\omega>0$. Why is it so?

I have assumed bosons, but I believe the same is true for fermions.

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  • $\begingroup$ Can you give some reference about where you have encountered this statement? for example in any book or paper or article etc. $\endgroup$ – abhijit975 Sep 3 at 1:37
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I don't see why your statement would be true in general for all Green's functions, in fact I have a hunch this not the case for two-particle Green's functions such as current-current correlation functions, but let's simply work with one-particle Green's functions to show why this is true.

Consider a many-electron system. It could be an insulator with the chemical potential in the band gap, or it could be a conductor. Now consider the following one-particle Green's functions, where $c_k$ creates an electron with momentum $k$ in a state above the Fermi energy:

$$G^>(k,t)=-i\langle c_k(t)c_k^\dagger(0)\rangle$$ $$G^<(k,t)=i\langle c_k^\dagger(0) c_k(t)\rangle$$ $$G^T(k,t)=\Theta(t)G^>(k,t)+\Theta(-t)G^<(k,t)$$ $$G^R(k,t)=G^T(k,t)-G^<(k,t)$$

At zero temperature, the conduction band state created $c_k^\dagger$ is empty, such that

$$G^<(k,t)=0$$

Therefore, at zero temperature

$$G^R(k,t)=G^T(k,t)$$

One can also show that the definition of $G^R(k,t)$ given here is equivalent to your definition above. So we have determined that the retarded and time-ordered Green's functions are equal for one-particle excitations above the Fermi level, i.e. $\omega>0$ excitations, as long as we're working at zero temperature where the states above the Fermi energy are empty at equilibrium. One could probably be even more formal in proving that the creation of an electron necessarily raises the energy of the system, but this seems to be intuitively plausible.

Now why is this result only true for $\omega>0$? This is because if we instead create a hole in a state below the Fermi energy, we find that instead

$$G^>(k,t)=0$$

such that $G^R(k,t)$ is still nonzero only at positive times, while $G^T(k,t)$ is now only nonzero at negative times, and these functions are now clearly different. The analog of your claim for negative energy excitations is that the advanced Green's function and time-ordered Green's function are identical for negative energy excitations.

Now you may be using a different definition of operators in your Green's functions, i.e. $$c_k \rightarrow \psi(r)$$ however you should find that the statements made here generalize when you write spatial Green's functions in terms of the momentum Green's functions discussed here.

In this way you'll find that a spatial Green's function can be written in terms of the electron and hole Green's functions described here. The poles of the time-ordered Green's function will match those of the retarded Green's function at positive energies and those of the advanced Green's function at negative energies.

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  • $\begingroup$ Thanks, in your argument that $G^<(k,t)=0$ are you saying that $c_k(0)|GS\rangle =0$? Does it hold for interacting systems as well? what about bosons? $\endgroup$ – user1830663 Sep 9 at 14:01
  • $\begingroup$ This is exactly what I'm saying and everything that I argue above is generally true for an interacting Fermi gas, possibly coupled to bosonic modes, as long as all N+1 particle states are higher in energy than all all N particle states. For the bosonic case I'm not 100% sure, but it seems that we should simply replace the Fermi level with the BE condensate energy, and then the same should hold. $\endgroup$ – Ian Sep 10 at 15:10
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EDIT: I changed the first part of my answer to bring it together with your notation.


The time-ordered GF for 2 operators at different times $t$ and $t'$ is: $$ G^T(t,t')=-i\left< T A(t) B(t') \right>. $$

The time-ordered GF can then be decomposed by applying the time-ordering operator: \begin{align} G^T(t,t')&= \theta(t-t') G^>(t, t') -\theta(t'-t) G^<(t, t'). \end{align} The first one is called the greater GF and the second one the lesser GF: \begin{align} G^>(t, t') &= -i\left< A(t) B(t')\right>,\\ G^<(t, t') &= +i\left< B(t') A(t)\right>. \end{align} Now we look at the retarded GF: \begin{align} G^R(t, t') &= -i\theta(t-t')\left< [A(t), B(t')] \right>,\\ &=\theta(t-t')\left(G^>(t, t') + G^<(t, t')\right) \end{align} The difference to the time-ordered GF is given by the appearence of the theta-functions. We can use the last identity to replace the greater GF by the retarded one for the time-ordered GF:

\begin{align} G^T(t,t')&= G^R(t, t') + G^<(t, t'). \end{align}

Now we can set $t'= 0$ ($\rightarrow t>t'$) to get: $$ G^T(t) = G^R(t) + G^<(t). $$

If you Fourier transform this, none of them vanishes generally: \begin{align} G^T(\omega) = G^R(\omega) + G^<(\omega). \end{align}


EDIT: after your corrections I realized that my old ("hand-wavy") answer does not answer your question and I only have an idea left. I think you can show it using this version of the fluctuation dissipation theorem: $$ G^<(\omega) = i f(\omega) A(\omega), $$ where: $$ f(\omega) = \frac{1}{e^{\beta \omega} - 1}, $$ is the statistical distribution function for Bosons ($\beta$ is the inverse temperature) and $A(\omega)$ is the spectral density.

You may be able to show that this expression goes to zero in general or only for some special cases. It is clear that $f(\omega)\rightarrow0$ for $T\rightarrow0$ ($\beta\rightarrow\infty$) and $\omega>0$, but the whole limit including $A(\omega)$ may not.

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  • $\begingroup$ I have indeed made several typos. I updated my question. Sorry for the confusion. $\endgroup$ – user1830663 Sep 3 at 22:00
  • $\begingroup$ You still need to include the commutator after the first "=" sign. $\endgroup$ – P. U. Sep 4 at 5:57
  • $\begingroup$ This will will give you only the retarded GF in the end because the term with theta(negative time) will vanish. You can see this procedure in my answer above. $\endgroup$ – P. U. Sep 4 at 6:02
  • $\begingroup$ Every text or notes that I have dont include commutator in the first "=" sign for the time ordered GF. I am confused by your definitions. $\endgroup$ – user1830663 Sep 4 at 17:34
  • $\begingroup$ See new edit. I tried to make my notation even more precise... $\endgroup$ – user1830663 Sep 4 at 23:16
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First of all, the statement "the two Green's functions are equal at zero temperature and for positive energies, i.e. $𝐺^𝑇(\omega)=𝐺^𝑅(\omega)$ for $\omega>0$" is totally wrong.

In the case of zero temperature, one defines the time-ordered Green's function to formulate the many-body perturbation expansion, in which the Gell-mann and Low's theorem can be utilized. However, the retarded Green's fucntion which related physical quantities cannot be used to formulate the many-body perturbation expansion. But once the expansion has been formulated, one can build the connection between the retarded Green's function and the time-ordered Green's function with analytical contiuation technique.

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  • $\begingroup$ This is incorrect. It is known that retarded Green's functions have poles at positive energy particle excitations corresponding to causal dynamics while advanced Green's functions have negative energy hole excitations corresponding to anticausal dynamics. The so-called Feynman propagator or time-ordered Green's function contains all these poles in one object. $\endgroup$ – Ian Sep 10 at 22:38

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