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My question concerns which electrons in a superconductor form Cooper pairs in the BCS ground state, i.e., all of them or only part of them. I am currently reading about superconductivity from Chap. 10, G. Mahan, Many-Particle Physics, 3rd Ed.

Mahan writes on page 627: "The basic idea of BCS theory is that the electrons in the metal form bound pairs. Not all electrons do this, but only those within a Debye energy of the Fermi surface." Good, this is how I have always understood BCS theory. The BCS ground state is given by \begin{equation} |BCS\rangle = \prod_{\bf k} (u_{\bf k}+v_{\bf k}\hat{c}^{\dagger}_{\bf k,\uparrow}\hat{c}^{\dagger}_{-\bf k,\downarrow})|vac\rangle. \end{equation} This is just the standard Fermi sea, if $u_{\bf k}=0$ and $v_{\bf k}=1$, for $|{\bf k}|<|{\bf k}_F|$, and $u_{\bf k}=1$ and $v_{\bf k}=0$, for $|{\bf k}|>|{\bf k}_F|$. Now put the BCS interaction in. I have always understood Cooper pairs as those electrons for which \begin{equation} u_{\bf k}^2=\frac{1}{2}\left(1 + \frac{\xi_{\bf k}}{\sqrt{\xi_{\bf k}^2+\Delta^2}} \right), \quad v_{\bf k}^2=\frac{1}{2}\left(1 - \frac{\xi_{\bf k}}{\sqrt{\xi_{\bf k}^2+\Delta^2}} \right), \end{equation} where $\xi_{\bf k}=\hbar^2|{\bf k}|^2/2m-\mu$. The gap $\Delta=\Delta_{\bf k}=-\sum_{\bf k'}V_{\bf kk'}u_{\bf k'} v_{\bf k'}$ is non-zero only for states within the Debye energy from $\mu=E_F$ (at $T=0$), since BCS assumes an interaction of the form \begin{equation} V_{\bf kk'}=\left\{ \begin{array}{ll} -V, & \textrm{for } -\hbar \omega_{\rm Debye} < \xi_{\bf k}, \xi_{\bf k'} < \hbar \omega_{\rm Debye} \\ 0, & \textrm{otherwise} \end{array} \right.. \end{equation}

So, I understand the superconducting BCS ground state as describing a state where electrons within the Debye energy of $E_F$ form Cooper pairs, and the rest of the electrons remain unpaired, i.e. do not form a bound Cooper pair. However, the whole system is superconducting as no excitations can be created unless their energy is above the gap. The unpaired particles are not quasiparticles, since the BCS state is a vacuum state for the quasiparticle annihilation operator.

Now, I am not sure if my understanding is actually correct, especially since excitations of the BCS state must be created or destroyed in pairs. That's why the excitation gap is $2\Delta$.

In addition, Mahan writes on page 647: "At zero temperature, all the electrons in the superconductor are in the pair states at the chemical potential." But this clearly contradicts with what he wrote above. Also Ashcroft and Mermin, Solid State Physics writes that in the BCS ground state all of the (conduction) electrons are paired. If all of the electrons are paired, where does the attractive interaction required for forming a Cooper pair emerge for those electrons whose energies are not within $\hbar \omega_{\rm Debye}$ of $E_F$? It is assumed to be zero in the BCS approach.

tl;dr Do all of the electrons in the BCS ground state form Cooper pairs? If yes, where does the attractive interaction come for electrons with $|\xi_{\bf k}|>\hbar \omega_{\rm Debye}$? If not, why do excitations need to created or destroyed in pairs if there are non-Cooper-paired electrons? I am aware that the BCS state is a coherent state, etc., but I would like to have a more physical picture of the ground state.

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  • $\begingroup$ I think you are absolutely right, personally, I think this is a weakness of BCS theory. (When I was talking with a famous professor, I was told that "if you cheat, you won the Nobel prize, if you want to be accurate, you get nothing...") $\endgroup$ – Chuan Chen Jul 25 '17 at 13:54
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First of all, at zero temperature not all of the electrons in the BCS ground sate (GS) form Cooper pairs. One way to think about this is that deep inside the Fermi surface, there is no available states for scattering events, including the phonon-mediated ones, and thus we wouldn't have the effective attraction to form the Cooper pairs.

This can also be seen from the coefficients $u_k$ and $v_k$ in the postulated BCS GS given in the original post. When $k$ is way below the Fermi surface, then $|\xi_k|>>\Delta$, with $\xi_k<0$. In this case, $|u_k|\to 0$ and $v_k\to 1$, i.e. the states far below the Fermi surface are almost fully occupied by electrons created by $c^{\dagger}_{k\uparrow}c^{\dagger}_{-k\downarrow}$. These electrons are created in pairs, but they are NOT Cooper pairs!!! The pair creation operator does not tell us anything regarding whether the pair is a Cooper or just a normal pair. Only when it is close to the Fermi surface, and thus possible to have phonon-mediated scattering, is the Cooper pair creation possible.

In summary, the BCS GS looks like the following: deep down inside the Fermi surface, it's just normal electrons. Close to the Fermi surface, it's mostly a superposition $\left(u_k+v_kc^{\dagger}_{k\uparrow}c^{\dagger}_{-k\downarrow}\right)|0\rangle$ of empty state, with amplitude $u_k$, and occupied Cooper pairs, with amplitude $v_k$.

EDIT:

After I submitted the above answer, I realized that I missed out the other part of your question, which is about the excitations of the BCS GS and here it goes:

You mentioned that "the excitations of the BCS state must be created or destroyed in pairs". This is not correct. The excitations of the BCS GS are fundamentally different from Cooper pairs or the normal electrons. Instead, it is the quasi-particle created by $$\gamma^{\dagger}_{p\uparrow}=u^*_pc^{\dagger}_{p\uparrow}-v^*_pc_{-p\downarrow}$$ Not surprisingly, this is just the Bogolyubov transformation used to diagonalize the BCS mean field hamiltonian and this why it is the legitimate excitations of the GS. The corresponding eigenvalue is exactly $$E_k=\sqrt{\xi_k^2+\Delta^2}$$ This is where the gap $\Delta$ to the BCS excitation comes from. This kind of excitation is slightly less intuitive compared to objects like single electrons or Cooper pairs. It has spin-1/2, but it doesn't carry well-defined charge, whereas a Cooper pair has spin 0 and charge $2e$.

Also, You can easily check that $$\gamma_{p\uparrow}|BCS\rangle=\gamma_{-p\downarrow}|BCS\rangle=0$$ which makes perfect sense since the annihilation operator for the excitation annihilates the GS.

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Answering your first question.

Where does the attractive interaction required for forming a Cooper pair emerge for those electrons whose energies are not within $\hbar\omega_\mathrm{Debye}$ of EF?

They just don't feel an attractive interaction. The attractive interaction comes from electron-phonon interaction. If their energy is higher than $\omega_\mathrm{Debye}$ they feel a repulsive interaction that prevents them to forming Cooper pairs, and hence they would be in an exited state.

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So, I understand the superconducting BCS ground state as describing a state where electrons within the Debye energy of EF form Cooper pairs, and the rest of the electrons remain unpaired, i.e. do not form a bound Cooper pair. However, the whole system is superconducting as no excitations can be created unless their energy is above the gap. The unpaired particles are not quasiparticles, since the BCS state is a vacuum state for the quasiparticle annihilation operator.

I will write what I learned from Bose-Einstein condensate. While it is a chargeless superfluid, the phenomenon of condensation shares some similarity with what is happening in superconductor. For a condensate, not all particles need to be in the "bare" ground state (state with net momentum zero). In the presence of interaction, the condensate acquires a small admixture of Bogoliubov quasiparticles. This is called quantum depletion, and is most notable in superfluid helium (a paradigmatic example of strongly interacting superfluid).

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