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In a lecture, the tutor mentioned that

"when the discrete energy spectrum becomes continuous and the poles of the resolvent shrink into a continuous line. Therefore it becomes a branch cut".

This is not clear to me. I understand that the poles of the resolvent are the energy eigenvalues, but how come the singularity become a branch point in the continuous case? It would be great if someone could make this clear to me.

Qmechanic
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  • By resolvent, you mean for a Hamiltonian $H$ and one of its eigenvalue $E$, the operator $(H-E\mathbb{1})^{-1}$, don't you? –  Jun 13 '17 at 13:33
  • Yes, the resolvent is $(H−E)^−1$ – ABCD Jun 13 '17 at 16:15

1 Answers1

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As nicely explained by Ron Maimon in this answer, one can think of a branch cut as a continuous line of poles each with an infinitesimal residue. For example, quoting him, $$\int_a^b \frac{1}{z-u} \mathrm du= \log \left( \frac{z-a}{z-b} \right)$$ with the latter indeed having a branch cut between $z =a $ and $z=b$.

For example if $b= - a$, the multi-valued function on the right-hand side looks like

enter image description here

(image courtesy of MIT OpenCourseWare)

Ruben Verresen
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  • Thank you very much. What is the physical meaning of this? Also, does this mean that any complex function with a branch cut can be approximated using poles? For example, if I have a square root function which has a branch cut singularity, can this be thought of as "continuous line of poles each with an infinitesimal residue". Please make clear. –  Jun 13 '17 at 13:11
  • (1) Well, what do you mean by 'physical meaning'? You can still think of having a continuous line of poles, but due to there being an uncountable number of states one has to change the normalization accordingly. So whatever 'physical meaning' you assign to a single pole is still relevant here. (2) That is a good question. In fact, if you look at Ron Maimon's answer, he explains how one can rewrite the square root function in this way :) I don't have a proof one can do this for any function with a branch cut, but it seems plausible. – Ruben Verresen Jun 13 '17 at 13:17