The conventional superconductors can be explained using the BCS theory. Usually, the BCS theory is introduced as follows:

We would like to consider the Hamiltonian that describes the system of electrons. This system is specific in a sense that electrons with opposite momenta and spins $(\mathbf{k},\uparrow)$, $(\mathbf{-k},\downarrow)$ have some kind of attractive interaction (no matter how weak it is): $$ H_{eff} = \sum_{\mathbf{k}\sigma} \xi_{\mathbf{k}} c^\dagger_{\mathbf k, \sigma} c_{\mathbf k, \sigma} + \frac{1}{N} \sum_{\mathbf k \mathbf k'} V_{\mathbf k \mathbf k'} c^\dagger_{\mathbf k \uparrow} c^\dagger_{-\mathbf k \downarrow} c_{-\mathbf k' \downarrow} c_{\mathbf k' \uparrow} $$ This unusual behavior can be explained due to the presence of electron-phonon interaction. This effective Hamiltonian is very complicated, but it can be simplified using the mean field theory. Then it can be diagonalized using the Bogoliubov transformations: $$ H = \sum_{\mathbf k, \sigma} E_{\mathbf k} \;\gamma^\dagger_{\mathbf k,\sigma} \gamma_{\mathbf k,\sigma} + const. $$ where $\gamma_{\mathbf k,\sigma}$ is the annihilation operator of newly obtained Bogoliubons (new particles) and $E_{\mathbf k} = \sqrt{\xi_{\mathbf k}^2 + |\Delta_{\mathbf k}|^2}$ is the corresponding dispersion relation. We can see that this new dispersion relation has a gap $|\Delta_{\mathbf k}|$ (I could write the analytical expression for this quantity, but it doesn't matter at this point).

Everything so far is clear to me. But how do we see that the system which is described with this formalism describes the superconductor? How to show that the resistivity is indeed vanishing?

Also, I've often seen that at finite temperature, the current has a contribution that correspond to 'carriers of superconductor' (Cooper pairs) and normal charge carriers (electrons). Even if these Cooper pairs have no resistivity (which I also do not understand, this is asked in first part of the question), why don't we get any dissipation due to the presence of normal electrons that are not bound in Cooper pairs?

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    $\begingroup$ Does this answer your question? What is the link between the BCS ground state and superconductivity? In a nutshell, your Hamiltonian is written in the center-of-mass reference frame Cooper pairs. $\endgroup$
    – Roger V.
    Dec 3, 2021 at 14:18
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    $\begingroup$ Well, no. The answers that you are referring to are only talking about the first part of my question. They do not say anything about normal and supperconducting terms in the current. One of the things I am interested in is: how is it possible to have a state with no dissipation if there is normal current present. $\endgroup$
    – RedGiant
    Dec 3, 2021 at 15:51
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    $\begingroup$ no dissipation means no excitations above the gap. It is like a scattering state: there's a flux, but the energy doesn't change. $\endgroup$
    – Roger V.
    Dec 3, 2021 at 16:47
  • $\begingroup$ Very interesting question. Related: physics.stackexchange.com/q/651336/226902 (this is the "experimental" complement to this "theoretical" question). More about the nature of BCS state: physics.stackexchange.com/q/345321/226902 $\endgroup$
    – Quillo
    Jun 2, 2022 at 18:01
  • $\begingroup$ The argument in the book of Mahan "Many-Particle Physics" (Plenum Press, 1981, page 793) goes like this: A current flow corresponds to a shift of the center of the Fermi sphere away from the origin of momentum space. Normal conductor: the electrons can relax "independently" via scattering (the center of the Fermi sphere will relax to the origin because of individual scattering processes). Superconductor: all Cooper pairs have to relax at once as they all share a common state (pair condensation): they are correlated to the displaced centre of the Fermi sphere. $\endgroup$
    – Quillo
    Jun 2, 2022 at 18:35

2 Answers 2


I think the first part of your question was previously disussed in this SE question "What is the link between the BCS ground state and superconductivity?".

However, I don't think the accepted answer is quite right (or, more precisely, I think the accepted answer is mostly hand waiving).

I think the correct answer is here. A quick summary: The BCS wave function depends on a phase $\varphi$, but the ground state energy is independent of $\varphi$. Fluctuations in this phase therefore correspond to Goldstone modes, and $\nabla\varphi$ is proportional to a current, the supercurrent $\jmath = n_s e\nabla\varphi/m$. The effective action for $\jmath$ that can be derived from the BCS wave function. It leads to the London equation $A_\mu=e\nabla_\mu\varphi$, which implies zero resistance (a non-zero electric current in the limit of zero electric field).

The second question was also previously discussed, for example in this post "The two-current model of superconductivity." The basic point is that at non-zero temperature there is indeed a supercurrent and a dissipative current. However, the dissipative current decays quickly, and the pure supercurrent state is metastable.

  • $\begingroup$ Thank you! the discussion in physics.stackexchange.com/q/320587/226902 is very interesting, even though I see no "definitive" answer there on the validity of the 2-fluid model. However, your arguments are convincing. The reference to Bardeen's ``Two fluid model of superconductivity'' (1958) is worth reading! $\endgroup$
    – Quillo
    Aug 29, 2022 at 11:26

The question is very insightful. Indeed, BCS theory (and its derivatives) cannot explain the zero resistance. The problem is that BCS considers superconducting (SC) electrons and normal electrons as indistinguishable (interchangeable) particles belonging to one shared momentum space, where every electron can take every SC or normal state with a non-zero probability. It is easy to show this makes an eternal supercurrent impossible. Indeed, at non-zero temperature, there are always normal and SC electrons in every superconductor. If SC-electrons were interchangeable with normal electrons, then the SC - electrons would become normal (dissipative) for a while, whereas the newly created SC - electrons did not experience any electromotive force. So the total current momentum (or angular momentum) would disappear and the supercurrent would inevitably stop (otherwise the momentum conservation law is violated; the atom lattice took the momentum Px of the “broken” initial SC-electrons, hence Px of the newly created SC-electrons must be zero). Thus, the SC – electrons cannot hop into normal states, i.e. the SC - electrons occupy the SC - states permanently in time. If so, then the SC - electrons must be distinguishable (separated) from the normal electrons in the momentum space. The same case is valid for superfluid helium: electrons of helium atoms are not interchangeable with any free single electrons below the helium ionization temperature, because the helium electrons cannot leave their 1s momentum states, and the free electrons cannot take the permanently occupied 1s states in helium. So the free band and 1s band are well separated by a gap and we can assert that the free electrons and helium electrons are distinguishable in the momentum space, although their wave functions can overlap in the real space. The ambiguity of (in)distinguishable electrons provides many theoretical problems in the matter.

  • $\begingroup$ are there unbounded electrons in superfluid helium??? Not sure I understand the part about electrons in Helium. Moreover: which is the definition of "superconducting electron"? $\endgroup$
    – Quillo
    Aug 26, 2022 at 14:59
  • $\begingroup$ In every real space (even filled by helium) can be distributed some free electrons emitted from an extra source, so - there are additional electrons not bound in helium atoms, although their density may be very low. "Superconducting electron" in this context is electron, participating in a long-lived supercurrent. Here is not defined paired or unpaired electron, relevant is that the electron doesn't hop into normal states. Therefore, SC electrons and helium electrons are in the same situation; both are permanently in their "super-states". $\endgroup$ Aug 26, 2022 at 17:11
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    $\begingroup$ So the Nobel committee was mistaken? $\endgroup$
    – Thomas
    Aug 29, 2022 at 0:59
  • $\begingroup$ It is not the first (and not the last) mistake of science institutions. Not surprising; people have made mistakes since the beginning of mankind. $\endgroup$ Aug 29, 2022 at 15:20
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    $\begingroup$ I understand all the doubts. Indeed, electrons, distinguishable in the momentum space, are difficult to imagine. I introduced the helium only to show, that the distinguishable electrons are not any exotic. So in solids we may also consider this case. Anyway, the current situation in the SC theory seems to be worse than my attempts to find the root cause of the problem. Maybe you find interesting a brief note “Limitations of Second Quantization Notation for Description of Superconductivity” hal.archives-ouvertes.fr/hal-03427535v2/document Feel free to comment. $\endgroup$ Sep 2, 2022 at 7:04

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