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In BCS theory, it's said that electrons that form the pairs only lie within $\omega_D$ (Debye frequency) of the Fermi energy. Why is that?

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In Quantum field theory for the gifted amateur, Chapter 44 - Superconductors, In page-no 404, it says,

Since the electrons that form the pairs only lie within $\omega_D$ of the Fermi energy, we can restrict our sum to $|\epsilon_\mathbf{p}-\mu|<\omega_D$.

in units $\hbar=1$.

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    $\begingroup$ What is $\omega_D$? (Yes, people familiar with BCS can guess, but it is always better to be explicit) What do you mean by "it's said"? Surely, whatever told you that this is true made some sort of argument for that? Can you be a bit more specific about where you expect an answer to start here? (E.g. rehashing the entire derivation of BCS superconductivity might answer this question, but would be very much wasted if you are only missing one specific step) $\endgroup$
    – ACuriousMind ♦
    Commented May 7, 2022 at 14:02
  • $\begingroup$ Related: physics.stackexchange.com/q/126742/226902 physics.stackexchange.com/q/483318/226902 physics.stackexchange.com/q/597484/226902 physics.stackexchange.com/q/126484/226902 physics.stackexchange.com/q/483318/226902 . Regarding the Debye frequency $\omega_D$, physics.stackexchange.com/q/65352/226902 $\endgroup$
    – Quillo
    Commented May 7, 2022 at 14:02
  • $\begingroup$ $\epsilon_p - \mu$ is an energy. I presume $\omega_D$ is an (angular) Debye frequency, not a wavelength and should be multiplied by $\hbar$ ? $\endgroup$
    – ProfRob
    Commented May 7, 2022 at 17:42
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    $\begingroup$ Just to be clear: Your edit does nothing to address my comment, since you still haven't supplied a definition of $\omega_D$, nor explained why the text you're quoting claims that only electrons within that distance of the Fermi surface participate in pair formation. (The quoted sentence starting with "since" means that there should be such an argument in the preceding text, otherwise it's just badly written) $\endgroup$
    – ACuriousMind ♦
    Commented May 9, 2022 at 23:50
  • $\begingroup$ @ProfRob You are right. It's the frequency (pardon). The units are chosen so that $\hbar=1$. $\endgroup$
    – Young Kindaichi
    Commented May 10, 2022 at 7:53

2 Answers 2

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The maximum energy of the phonon interactions is $\hbar \omega_D$. Since the electrons are almost completely degenerate, only electrons within $\hbar \omega_D$ of the Fermi energy can then participate in the pairing interactions and lower their energies.

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  • $\begingroup$ What do you mean by "the electrons are almost completely degenerate"? $\endgroup$
    – Lucas Baldo
    Commented May 7, 2022 at 19:26
  • $\begingroup$ I mean the Fermi Energy is $\gg k_B T$ @LucasBaldo $\endgroup$
    – ProfRob
    Commented May 7, 2022 at 20:55
  • $\begingroup$ To elaborate on ProfRobs answer: most of the conduction electrons are deep in the "Fermi Sea", filling the energy states of the normal metal according to Pauli exclusion principle. The term degenerate can be slightly confusing because these states do have different energies. But the point is, those electrons don't have anywhere to "go" and cannot participate in the pairing, as Rob pointed out. $\endgroup$
    – pmal
    Commented May 13, 2022 at 6:48
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    $\begingroup$ Interestingly enough, the presence of the Fermi sea is actually essential for the formation of superconductivity. In three dimensions, quantum mechanical bound states are not guaranteed and require a minimum interaction strength. But the presence of the Fermi surface renders the phase space of the electrons participating in pairing effectively two-dimensional, and in two dimensions a bound state always exists for arbitrarily small negative potential. The simplest BCS formulation predicts that superconductivity will always occur for an arbitrarily small effective attractive interaction. $\endgroup$
    – pmal
    Commented May 13, 2022 at 6:53
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In order for two pairs of electrons to be scattered from $| \vec k_1 \rangle, |-\vec k_1 \rangle$ to $| \vec k_1 + \vec q \rangle, |-\vec k_1 - \vec q \rangle$, the states $| \vec k_1 \rangle, |-\vec k_1 \rangle$ must be occupied whilst the states $| \vec k_1 + \vec q \rangle, |-\vec k_1 - \vec q \rangle$ must be unoccupied. Deep below the Fermi level, all four of these states are occupied so the electrons have no where to go. Far above the Fermi level, all of them are unoccupied. In both cases, no scattering can take place, so we only consider scattering between electrons that are within a thin belt $\pm \hbar \omega_D$ of $\epsilon_F$ ($\omega_D$ is the Debeye frequency and $\epsilon_F$ is the Fermi level).

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