Hartree-Fock: Feynman diagrams vs perturbation theory

  • #1
pines-demon
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TL;DR Summary
What is the relation between perturbation theory and many-body diagrams?
I am trying to understand Green's functions in many-body theory for condensed matter. After much struggle, I managed to calculate my first diagrammatic expansion. However I am perplexed by getting more of the usual results.

The Hartree–Fock energy result I know from second quantization can be derived either by doing some mean-field approximation or by just calculating the first order perturbation energy of an electron gas. This is straightforward using ordinary quantum mechanics.

However now that I can write it using diagrams, I was expecting that doing a sum over a infinite number of diagrams would be equivalent to going over many orders of perturbation theory (even if some diagrams are not included). However the final result is the same as the one from the mean-field/1st order perturbation theory (hence it being called again Hartree–Fock).

What is going on here? Am I wrong at thinking that diagrammatic expansion is different from first order perturbation theory (when excluding some diagrams)? Or is there some sort of cancelation going on? Or are the results subtlely different?
 
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  • #2
Feynman diagrams are firmly based on perturbation theory. But they definitely go beyond first order. An infinite number of forward scattering events, e.g. for photons, result only in a change of the refractive index.
 
  • #3
WernerQH said:
Feynman diagrams are firmly based on perturbation theory. But they definitely go beyond first order. An infinite number of forward scattering events, e.g. for photons, result only in a change of the refractive index.
Sure but what does it mean to carry out an infinite order of diagrams (even if not all) and still get the same result as first order perturbation theory?
 
  • #4
pines-demon said:
Sure but what does it mean to carry out an infinite order of diagrams (even if not all) and still get the same result as first order perturbation theory?
Is it the exact same result? Or is it the same result with a re-interpreted (renormalized) parameter? I don 't really know what you have calculated.
 
  • #5
WernerQH said:
Is it the exact same result? Or is it the same result with a re-interpreted (renormalized) parameter? I don 't really know what you have calculated.
Maybe the latter but I am not sure. You have a Hamiltonian ##H=H_0+H_I## (##H_0=\epsilon_kc_k^\dagger c_K##)

$$H_I=\sum_{klmn}c_l^\dagger c_k^\dagger c_m c_n$$

Let us stick to just Hartree interactions (I will not bother to put the Coulomb interaction here), you have only ##V_{klkl}##(##k\neq l##) the rest are zero. It is easy to see that the new first order perturbation energy is just
$$E_1=\sum_k \epsilon_k +\sum_{kl} V_{lklk} \langle c_l^\dagger c_k^\dagger c_k c_l\rangle= \sum_k( \epsilon_k+\sum_l V_{klkl} ) $$
where both sums in the rhs are carried up to the Fermi surface.

Now, adults prefer to use some diagrammatic calculation
Screenshot 2024-04-16 at 11.54.27.png

(from Mattuck A Guide to Feynman Diagrams in the Many-Body Problem) which leads to the propagator
##i G (k,\omega)= \displaystyle \frac{1}{[iG_0(k,\omega)]^{-1}-(-1)(-i)(-1)\sum_l V_{klkl}}##
or simply
##G = (k,\omega)= \displaystyle \frac{1}{\omega-\epsilon_k+i\delta -\sum_l V_{klkl}}##
where you can read directly from the denominator that the energy is going to be the same as the one from perturbation theory. That's why both methods are called Hartree. Same happens with Hartree–Fock term.

But what did diagrams achieve here?
 
  • #6
pines-demon said:
But what did diagrams achieve here?
Everything depends on the potential that you assume acting between the electrons. And that is supposed to be chosen self-consistently. So there you have that can of worms of renormalization. John Ziman in his excellent book Elements of Advanced Quantum Theory (Cambridge University Press, 1969) describes both approaches:
Even in the case of the electron gas, attempts to calculate the excitation spectrum by treating the residual particle--particle interactions as perturbations on the Hartree--Fock ground state also lead to nonsense. It is well known, for example, that the 'exchange correction' to the density of states at the Fermi level is divergent in this approximation.

To understand the techniques that have been evolved to avoid these errors, it is helpful to interpret the Hartree--Fock method in the language of diagrams, so that we may get some idea of what has been left out.
(section 5.5, Diagrammatic interpretation of Hartree--Fock theory)

Perhaps you can find a copy in the library.
 
  • #7
WernerQH said:
Everything depends on the potential that you assume acting between the electrons. And that is supposed to be chosen self-consistently. So there you have that can of worms of renormalization. John Ziman in his excellent book Elements of Advanced Quantum Theory (Cambridge University Press, 1969) describes both approaches:

(section 5.5, Diagrammatic interpretation of Hartree--Fock theory)

Perhaps you can find a copy in the library.
Yeah, maybe it is that. It seems that this is a pathological case.

I know that carrying second order perturbation on a Hartree-Fock-like calculation is going to blow up, so it is kind of magic that there is even a Hartree-Fock diagram series.

I am going take a look at it thanks!
 
Last edited:
  • #8
Basically, Feynman diagrams are related to a perturbation expansion of the resolvent (H-lambda)^{-1} and not of the hamiltonian itself. So it is not astonishing that a first order correction to energy yields a geometric series for the resolvent and vice versa to get exactly coincident expressions.
 
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  • #9
DrDu said:
Basically, Feynman diagrams are related to a perturbation expansion of the resolvent (H-lambda)^{-1} and not of the hamiltonian itself. So it is not astonishing that a first order correction to energy yields a geometric series for the resolvent and vice versa to get exactly coincident expressions.
Can you expand on this or give a mathematical example?
 
  • #10
Well, if ##H=H_0+\epsilon V## and ##R=(H-\lambda)^{-1}## and similarly ##R_0##, then the one particle Greensfunction is something like ##G=\langle 1 | R| 1\rangle> ## where ##| 1 \rangle ## is some one particle state. Expanding R in powers of ##\epsilon## yields the Dyson equation for ##G##. If you only insert 1 particle states, this gives a geometric series. Summing the series yields the same result as Brillouin-Wigner perturbation theory for the Hamiltonian (which by the way, in deed infinite orders as the denominator does not contain the unperturbed E_0 but the perturbed E.
 

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