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I am trying to understand the phenomena of super-conductivity from a broader level. What I understand for now is that for super-conductivity to be possible in a system, a necessary requirement is that of high energy difference between ground state of the system and the next excited state of the system (called energy gap). But what I do not understand is why (in BCS theory of superconductivity) do electrons form Cooper-pairs, and behave like bosons. So I have following question about the Hamiltonian that acts on superconducting system, to make things more clear.

We are given a Hilbert space $\mathcal{H}$ which is associated to the superconducting system. Let the Hamiltonian that acts on the superconducting system be $H$, which is a hermitian matrix on $\mathcal{H}$. Let $\Pi_{sym}$ be the projector onto the symmetric subspace of $\mathcal{H}$ (this corresponds to the subspace spanned by bosonic wave-functions) and $\Pi_{anti}$ be the projector onto the anti-symmetric subspace of $\mathcal{H}$ (this corresponds to the subspace spanned by fermionic wave-functions). Let $H_{sym}$ be a restriction of $H$ on symmetric subspace, that is $H_{sym} = \Pi_{sym}H\Pi_{sym}$ and similarly we have $H_{anti} = \Pi_{anti}H\Pi_{anti}$. Is it true that the energy gap of $H_{sym}$ is much larger than energy gap of $H_{anti}$ for the aforementioned hamiltonian $H$ that acts on a superconducting system? If yes then it makes sense that electrons pair-up to act like bosons.

Clearly, I would not expect every hermitian matrix $H$ to have above property, hence if above property is really true then it must be something quite specific to superconducting systems. But what are other hamitonians appearing in condensed matter physics that have this property?

I might also be completely misunderstanding superconductors, so any clarification shall be highly appreciated as well.

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  • $\begingroup$ This question is really puzzling indeed. There are many concepts behind the notion of superconductivity. It's difficult to know your background from this question, so let me sum up rather sketchily the problem of bosonic electromagnetism in superconductors. Superconductivity is the phase transition changing the ground state of an electron gas to a condensate of Cooper pairs. The associated electromagnetism is the same as the electromagnetism of charged bosons. In that sense one says that Cooper pairs have bosonic properties, or behave like bosons. $\endgroup$ – FraSchelle Apr 22 '16 at 10:56
  • $\begingroup$ If you know a bit more about ground-states and quasi-particles, I can be more precise and say that either for the normal metal (when they are called Landau's quasi-particles sometimes) or for the superconducting metal (when they are called Bogoliubov quasi-particle sometimes), the excitations are fermionic, i.e. their exchange statistics is the one of fermions, or i.e. the creation and annihilation operators anti-commute. $\endgroup$ – FraSchelle Apr 22 '16 at 11:01
  • $\begingroup$ Your projectors are clearly orthogonal to each other, so it is difficult to see whether one can pass from the symmetric to the antisymmetric part of the Hilbert space. I mean any $H=H_{anti}+H_{sym}$, so how would you allow the transition from one sub-space to its orthogonal one to occur ? This, as far as I understand your first paragraph, would be your vision of superconductivity, am I wrong ? Your sentence What I understand for now is that for super-conductivity to be possible in a system, a necessary requirement is that of [...] energy gap Is misleading, you confound cause and (...) $\endgroup$ – FraSchelle Apr 22 '16 at 11:03
  • $\begingroup$ (...) consequences. The gap may be (it is not) a consequence of the superconducting transition, but this is not required to generate superconductivity. What generates superconductivity is the instability of the Fermi sea toward the creation of Cooper pairs (or Cooper correlations). Consequences are Meißner effet, Abrikosov vortex, Josephson and proximity effects, Little -Parks effect, London momentum effect, ... from which Meißner and Josephson clearly require a gap, but for the other I'm not even sure you absolutely need a gap. $\endgroup$ – FraSchelle Apr 22 '16 at 11:09
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  1. It is not true that all superconductors are gapped. For example, d-wave superconductors in cuprates are gapless.
  2. The energy gap in the superconductor arises from the fact that breaking the Cooper pair requires finite energy. The low-lying quasi-particle excitations are all pair breaking excitations, so they are gapped from the ground state by the amount of the pairing energy.
  3. For a fermionic many-body system, the many-body Hamiltonian $H$ must be anti-symmetric. So under the symmetric projection, one obtains $H_\text{sym}=0$. The dynamics of both electrons and Cooper pairs are described by the antisymmetric Hamiltonian $H_\text{anti}$. Cooper pairs are just collective modes in $H_\text{anti}$.
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  • $\begingroup$ Point 2: Assume ground energy is zero. Existence of energy gap cannot be explained just from energy needed to break cooper pairs. One has also to prove that if all cooper pairs stay intact, much more energy will be needed to reach next excited state. Point 3: I am assuming cooper pairs are bosons (i hope its correct). Then the "effective hamiltonian" that acts on them can't be anti-symmetric. Otherwise, all symmetric wavefunctions would be its ground states and energy gap would be zero. Please note that my original question is about "effective hamiltonian" acting on cooper pairs. $\endgroup$ – anurag anshu Apr 22 '16 at 10:25
  • $\begingroup$ Point 1. Interesting! Then in this case, how does the system stay superconducting? According to my understanding, a system is superconducting if small perturbations cannot change ground state to next excited state (due to large energy gap). This counters the 'resistance' that can occur due to external influence. $\endgroup$ – anurag anshu Apr 22 '16 at 10:28

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