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At $T=0K$, do all the electrons form Cooper pairs, or just the electrons near the Fermi Surface do?

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    $\begingroup$ Do you mean all electrons in the conduction band? $\endgroup$ – John Rennie Jul 16 '14 at 8:25
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    $\begingroup$ If you mean by $T=0\text{ K}$ absolute zero, then it's impossible to reach that temperature $\endgroup$ – Gigi Butbaia Jul 16 '14 at 8:47
  • $\begingroup$ Sorry, maybe I didn't ask clearly. I am not emphasizing T=0K, but the latter part: we know for a metal, at T=0K, in the momentum space, the electrons occupy the energy states below the Fermi energy, forming a Fermi sphere. My question is in superconductor, do all the electrons form pairs (e.g. including the electrons far below the Fermi sphere) or only the electrons near the Fermi sphere surface form pairs? $\endgroup$ – user43713 Jul 16 '14 at 19:55
  • $\begingroup$ Yes, I mean all the electrons in the conduction band. $\endgroup$ – user43713 Jul 16 '14 at 20:32
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    $\begingroup$ Short answer: no ! But all conduction electrons are in principle all forming Cooper pairs at zero-temperature. Of course this is valid only in the thermodynamics limit. If you have only few conduction electrons, and an odd number of them, it is clear that non all of them will form Cooper pairs, since one will always remain at the end. $\endgroup$ – FraSchelle Jul 18 '14 at 14:53
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Introduction (BCS Theory):

Cooper pairs in BCS theory are explained like this: the energy that can break cooper pairs in material is for example $10^{-4}\text{ }eV$ and thermal energy of that material is $E=kT$ where $k$ is Boltzman's Constant, so if thermal energy is lower than the energy that can break cooper pairs (in this example $kT<10^{-4}\text{ }eV$) then electrons form pairs called cooper pairs and can flow freely (forever) without any resistance, resistance decreases as temperature goes down until it reaches critical point ($T_c$) where resistance is exactly zero (this phenomenon is called Superconductivity), here's graph of that:

enter image description here

so if you decreased kinetic energy of molecules (temperature) one by one below critical temperature $$T_c=\frac{E_{min}}{k}$$ then all electrons that are flowing come in pairs (cooper pairs)

Conclusion:

So answer is that all electrons that are flowing form cooper pairs (and you don't have to cool it down to $T= 0 \text{ Kelvin}$ (absolute zero which is impossible) to get cooper pairs, decreasing all molecules kinetic energy (decresing temperature) below critical temperature ($\frac{mv^2}{2} = kT\Rightarrow v = \sqrt{\frac{2kT}{m}}$ where $v$ is molecule's velocity) is enough to form cooper pairs), heres how it looks like:

enter image description here

Flow of electrons is called current but in superconductor current can flow freely without any resistance and current that flows is called supercurrent, In practice you can measure current of material (which is supercooled below it's critical temperature $T_c$) and calculate number of electrons from this equation: $$ I = \frac{\partial q}{\partial t} = \dot q $$ where $q=ne$, (where $n$ is number of particles and $e$ is elementary charge) $$ne=\int I\text{ }dt\Rightarrow n=\frac{1}{e}\int I\text{ }dt$$

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    $\begingroup$ Are you sure if $kT<10eV$, the electrons form pairs? because $10eV$ corresponds to a very high energy scale (keep in mind the room temperature T=300K is only 0.026eV. $\endgroup$ – user43713 Jul 16 '14 at 20:01
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    $\begingroup$ thanks for mentioning that @user43713 , it actually should be $\approx 10^{-4}\text{ eV}$ $\endgroup$ – Gigi Butbaia Jul 16 '14 at 20:08
  • $\begingroup$ No, not all electrons form Cooper pairs. This can be seen from the density of states (DOS) of the superconducting state, where only the electrons in the gap (less than a bandgap away from the Fermi energy, $\varepsilon_F-\Delta<\varepsilon<\varepsilon_F+\Delta$) form pairs where the constituent electrons have energies at the peaks (at $\varepsilon_F\pm\Delta$). At these peaks, the electrons have opposite momenta relative to the $\bf{p}_F$ (which can point in any direction). $\endgroup$ – Betohaku Feb 12 '18 at 16:05
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The answer is no. I'm going to quote this passage of "Introduction to Superconductivity" by Rose-Innes that explains really good the reason:

One might at first think that there is no limit to the number of electrons which may be raised from $p < p_F$ to form Cooper pairs with a resultant lowering of the total energy, so that we should end up with all the electrons having $p > p_F$. This, however, is clearly absurd, and the reason why is not hard to find. In order that a pair of electrons may be scattered from $(\mathbf{p_i} \uparrow, -\mathbf{p_i} \downarrow)$ to $(\mathbf{p_j} \uparrow, —\mathbf{p_j} \downarrow)$ the states $(\mathbf{p_i} \uparrow, —\mathbf{p_i} \downarrow)$ must first be occupied and the states $(\mathbf{p_j} \uparrow, —\mathbf{p_j} \downarrow)$ must be empty. As more and more electrons form Cooper pairs with $p > p_F$ the chance of finding the states $(\mathbf{p_j} \uparrow, —\mathbf{p_j} \downarrow)$ empty becomes progressively smaller and smaller, so the number of scattering processes which may take place is reduced, with a consequent decrease in the magnitude of the negative potential energy. Eventually a condition is reached in which the lowering of the potential energy is insufficient to outweigh the increase in the kinetic energy, and it is no longer possible to lower the total energy of the electrons by forming Cooper pairs. There will be an optimum arrangement which gives the lowest overall energy, and this arrangement can be described by specifying the probability $v_\mathbf{k}$ of the pair state $(\mathbf{p_i} \uparrow, —\mathbf{p_i} \downarrow)$ being occupied in the wavefunction $\psi_G$.

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