# Does the real part of the inverse dielectric function have to be negative at some point for Cooper pairs to form?

Electrons naturally repel one another. However, in a superconductor, a phonon-mediated interaction causes the electrons to have a weak attractive interaction. Suppose that the interaction between two electrons is $V(\textbf{q})$ (the bare Coulomb interaction, which is $>0$) and that the effective interaction between two electrons is $V(\textbf{q})/\epsilon(\textbf q, \omega)$. My question is threefold:

(1) Is the only way to have an attractive interaction between electrons is for the $\Re[\frac{1}{\epsilon(\textbf{q},\omega)}]$ to become negative at a certain value of $\textbf{q}$ and $\omega$?

(2) (If yes to Question (1)): does $\Re[\frac{1}{\epsilon(\textbf{q},\omega)}]$ have to become negative at the values of $\textbf{q}$ and $\omega$ that correspond to a phonon?

(3) (If yes to Question (1)): Since high-temperature superconductors form Cooper pairs, does this mean that in the space of $\textbf{q}$ and $\omega$ such that $\Re[\frac{1}{\epsilon(\textbf{q},\omega)}] < 0$, there are clues lurking about high-temperature superconductivity yet unbeknownst to us?

• I think you ignoring several possibilities: non-local response, where the response is not translationally symmetric, but more fundamentally you are requiring the ground state to look free-electron like. Otherwise the interaction term is still not captured by $V(\mathbf{q})/\epsilon(\mathbf{q},\omega)$. The original BCS argument requires any attractive potential between time-reversed charge carrying states, which may or may not be captured by a screened Coulomb interaction like the one you wrote down. – KF Gauss Feb 26 '17 at 0:37