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It is often said that Feynman diagrams for fermions do not have symmetry factors.

Consider I have a spinless fermionic quantum many-body system with Hamiltonian: $$H=\int_{r}\psi^{\dagger}(r)\frac{\nabla^{2}}{2m}\psi(r)+\frac{1}{2}\int_{r,r^{\prime}}\psi^{\dagger}(r)\psi^{\dagger}(r')V(r,r')\psi(r')\psi(r), $$ with $r$ being position and $V$ a symmetry potential. The detailed form is not important at all. But just to the give the question a clear ground.

But if the two self-energy diagrams below considered to be topologically identical:

enter image description here

Then this diagram would have a factor 2 (up to a minus sign due to the closed fermion loop.). But one can check that they will contribute to the self-energy exactly the same since they one can twist the top hat 180 degrees without touching the external lines.

Also, are the following 4 diagrams considered topologically the same or not?

enter image description here

It seems that there is some ambiguity here by what people mean by "topologically identical"?

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  • $\begingroup$ Should there be a directional arrow on the fermion loop? If so the two diagrams will have the arow in opposite directions, and therefore be distinct. $\endgroup$
    – mike stone
    Commented May 22 at 20:51
  • $\begingroup$ @mikestone Yes. In QED, there are two types of arrows, one for particle/anti-particle and the other for momentum. Here, I am considering a quantum many body systems (non-relativistic). And therefore I assume you mean the momenta arrows. Yes, the momentum arrows are the opposite on the Fermion loop. But the two diagrams gives identical contribution to the self-energy. I am just a bit confused if they should be classified as the same. How do you think of the bottom 4 diagrams? $\endgroup$
    – John
    Commented May 22 at 21:41
  • $\begingroup$ @mikestone Actually arrows are not important if you don’t have anti-particles, as long as each fermion loop has only one direction, i.e., no reverse arrows in one fermion loop or along a continuous fermion line. $\endgroup$
    – John
    Commented May 22 at 21:49

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I do not know whether this answers your question. Forget about Fermions for a moment, and thing this situation from purely graph theoretical context. You can see as long as the twisting does not have any interaction (formally singularity), they are isomorphic to each other (https://www2.math.upenn.edu/~mlazar/math170/notes05-2.pdf). In other words, the topologically equivalent graphs can be smoothly deformed to each other. You are right about the self energy. This is same as the situation of $e^-e^-\to e^- e^-$ scattering in $t$ and $u$ channel, where the amplitude is exactly the same but with difference of Mandelstam $t$ and $u$. Here $s=t=u$ since there is only one external line.

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  • $\begingroup$ But if I identify the two diagrams are topologically the same, then it would violate the common notations that "Feynman diagrams for fermions (except for diagrams with closed ends, i.e. directly computing partition function. In that case, those diagrams would have symmetry factors) do not have symmetry factors." $\endgroup$
    – John
    Commented May 23 at 11:32
  • $\begingroup$ I agree that in graph theory, they are isomorphic: en.wikipedia.org/wiki/Graph_isomorphism. But it seems that physicists would have different notions. Since the golden lesson: "Feynman diagrams for fermions do not have symmetry factors" are very well known and I do not see any book discussing this subtlety. $\endgroup$
    – John
    Commented May 23 at 11:35
  • $\begingroup$ The arrows run from the $\psi^\dagger$ to the $\psi$. As long as the fermions have a particle number charge (and yours do) they need an arrow to indicate the direction of the number current. $\endgroup$
    – mike stone
    Commented May 23 at 12:21
  • $\begingroup$ @mikestone, yes. But only for external lines. Internal lines can be assigned randomly as long as there is no continuous line that have opposite arrows. In the case, the ambiguity of “topologically identical” appears. $\endgroup$
    – John
    Commented May 23 at 18:30
  • $\begingroup$ no, for internal lines also do have notion of arrows $\endgroup$
    – Tanmoy Pati
    Commented May 27 at 4:50

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