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Analytical expression for Hartree-Fock diagram for electron gas

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The picture above is taken from "introduction to many-body physics" by Coleman. When I try to write down the momentum space expression for the two diagrams, I got the same expression for the first one, but for the second one, the expression I got is (ignoring the constants, and the variable is in 4D spacetime):
\[ \int dp'V(p-p')G^0(p') =\int d\textbf{p'}dw'V(\textbf{p}-\textbf{p'},w-w')G^0(\textbf{p'},w')\\ =\int d\textbf{p'}dwV(\textbf{p}-\textbf{p'})\delta(w-w')G^0(\textbf{p'},w'), \]
where the second equality follows from the fact that the electron-electron interaction is time-independent, and thus we eventually get $\int d\textbf{p'}V(\textbf{p}-\textbf{p'})G^0(\textbf{p'},w)$.

First of all, for the factor $e^{iw0^+}$, I don't understand why this factor should appear in the second term as well. It appears in the first term because the first term contains a fermion loop, which in real space is $G(0,0^-)$ and when put into momentum space representation there will be such a factor. However, this does not happen for the second diagram, even if I start with the real space expression and do all the Fourier transforms.

Secondly, I used 4-vector representations to do the integrations and thus the there will be a $\delta(w-w')$ in the Fourier transform of the pair interaction because it's time independent as mentioned above. Consequently, in my result, the integration over frequency does not appear. 

There must be something wrong in my argument. Help is greatly appreciated!

asked Jul 20 in Theoretical Physics by M. Zeng (25 points) [ revision history ]
edited Jul 20 by M. Zeng
Most voted comments show all comments

What is the numerical value of this exponential?

@VladimirKalitvianski What do you mean by numerical value? If you are talking about $e^{iw0^+}$, it's just some convergence factor.

So it is unity at the both terms, isn't it?

even if I put it to be 1, my result is still different from what the book has.

Are the variables $\bf{p}$ and $\omega$ related with some dispersion law? If so, the make the variable change and then compare your result with the textbook's one.

Most recent comments show all comments

No, they are independent variables.

Formula (7.153) does not depend on omega. Then why does $\Sigma$ depend on omega? As well, the first integral in (7.153) is written as $\int_{\bf{p}'}$. Does it mean an integration over $\bf{p}'$?

1 Answer

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I figured out that the problem of my calculation actually lies in the incorrect use of 4-vector representation for the pair interaction, which was essentially due to the incorrect use of Feynman rules. We can clear the confusion by going back to look at the contractions of the time-ordered operator products. The natural form of the potential, in the case of potential scattering, or the pair interaction, in the case of electron-electron interaction, when put to interaction representation, does not explicitly contain time $t$. Therefore, it is at beginning more natural to directly work with usual 3-vectors. For the Hartree-Fock term, it is the lowest order correction coming from the following expression:

$$\langle \phi| T\int dtd\textbf{x}_1d\textbf{x}_2 \psi^{\dagger}(\textbf{x}_1,t)\psi^{\dagger}(\textbf{x}_2,t)V(\textbf{x}_1-\textbf{x}_2)\psi(\textbf{x}_2,t)\psi(\textbf{x}_1,t) |\phi\rangle,$$

which after contraction will be something like

$$\int dtd\textbf{x}_1d\textbf{x}_2V(\textbf{x}_1-\textbf{x}_2)(G^2(0,0^-)+G(\textbf{x}_1-\textbf{x}_2,0^-)G(\textbf{x}_2-\textbf{x}_1,0^-))$$

which after being put into momentum space would be

$$\int dt d\textbf{p}_1dw_1d\textbf{p}_2dw_2dqe^{iw_10^+}e^{iw_20^+}G(\textbf{p}_1,w_1)G(\textbf{p}_2,w_2)V(\textbf{q})[\delta^2(\textbf{q})+\delta^2(\textbf{p}_1-\textbf{p}_2-\textbf{q})]$$

$$=TV\int  d\textbf{p}_1dw_1d\textbf{p}_2dw_2e^{iw_10^+}e^{iw_20^+}G(\textbf{p}_1,w_1)G(\textbf{p}_2,w_2)[V(\textbf{q}=0)+V(\textbf{p}_1-\textbf{p}_2)]$$

From this expression we can clearly see why propagator part can be put compactly as 4-vectors, while the interaction part cannot. 

What I have been shown here are for the Hartree-Fock diagram without loose ends. But with the same correct idea, the expression for the diagrams in the original question can be readily written down. 

Painful lesson learnt: for people trying to get hands on Feynman diagrams, try not to become overly obsessed with the fancy diagrams before one has done sufficiently number of pages of calculations with the conventional "dumb" way of calculating correlation functions. The diagrams work very nicely only when we know exactly what they represent.

answered Jul 28 by M. Zeng (25 points) [ no revision ]

The diagrams work very nicely only when we know exactly what they represent.

That's why J. Schwinger tried to avoid the Feynman diagrams.

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