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I've been reading a lot of condensed matter textbooks, which state both that the net momentum of a Cooper pair in a superconductor is zero, and that Cooper pairs have momentum when they carry current.

How can these two statements be consistent? If a Cooper pair has zero momentum, how can current flow in a superconductor?

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This is an excellent question. One of the very few textbooks, that I found actually talking about it is Ibach and Lüth's "Solid-State Physics: An Introduction to Theory and Experiment", chapter 10.6. Supercurrents and Critical Currents.

They show that a Galilean transformation applied to Cooper pairs only changes the phase of the superconducting order parameter. Since the BCS ground-state energy depends on the magnitude of the order parameter and not the phase, this doesn't cost any energy.

This also means, that in the Ginzburg-Landau approach, when the U(1) symmetry of the order parameter gets spontaneously broken, the Galilean invariance breaks too. So in effect, the Cooper pair condensate behaves like a superfluid and "chooses" the inertial reference frame in which to be stationary. This reference frame is not necessarily fixed to the laboratory reference frame.

To me, this is the essence of superconductivity: the fact that the ground-state itself can carry current. This current has to be dissipationless, because you can't take away any energy from a system, that is already in it's ground state.

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Recall that the Fermi surface of the free electron gas, the total momentum is zero. How can electron gas be conductive? Well, when you apply the electric field, it will be non zero. The story is same for superconductivity.

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  • $\begingroup$ My understanding is that in a conductor, electrons move because they are excited into a state of higher momentum. However, the next excited state for electrons in a Cooper pair breaks apart the pair. So, I still don't understand how the pair can be in a state of higher momentum while still being in the ground state. $\endgroup$ – rupertonline Nov 9 '14 at 11:50
  • $\begingroup$ @rupertonline The electric field need to be large enough to break the cooper pair. When less than that, the cooper pair will get a net momentum. $\endgroup$ – 喵喵是我的猫猫 Nov 9 '14 at 12:03
  • $\begingroup$ But if they have a net momentum, then they are in an excited state. The textbooks I've read seemed to imply (though they were a bit vague) that there was just one state the Cooper pair existed at, and the next excited state involved breaking the pair. Are you saying there are in fact multiple states the Cooper pair can inhabit before it breaks? $\endgroup$ – rupertonline Nov 9 '14 at 12:48
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    $\begingroup$ Maybe they mean, before you apply the electric field, the system is in its ground state. After the electric field is applied, the system has simply changed, the ground state thus is not the same as the previous one. If you are in 0T, it still in the ground state, but the cooper pair got a net momentum. $\endgroup$ – 喵喵是我的猫猫 Nov 9 '14 at 13:26
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    $\begingroup$ This is nonsense. A resitance-free conductor and a electric field cannot coexist. Imagine a cirular superconductor with a current as it is in any magnet for NMR apparatus. Where does the electric field "begin" or "end" in a circle? $\endgroup$ – Georg Jan 13 '15 at 11:51
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I know this is an old question, but I don't believe any of the above posts have provided the right answer (although that is for you to judge). The key to this is the difference between the canonical momentum and the physical momentum of the Cooper pair. The below image from a presentation I gave shows the canonical momentum (this was from London's theory, not BCS, so that's why the charge is e not 2e, but the basic principle remains the same). If we set the canonical momentum to zero, then we can relate the momentum of the pair to the vector potential, and then write an expression for the current density.

Current density arising from canonical momentum being set equal to zero

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This is a interesting question. I think the above two answers have already provided what you want to understand.

However, here I propose a new question: is it possible for the Cooper pair to carry a supercurrent when their total momentum is exactly zero? (Note that in the above discussion, the supercurrent occurs when the Cooper pair carry finite momentum.)

I believe it is possible. Why? Because the current is determined by the velocity not the momentum. This can be seen from that the current operator is always defined by the velocity operator. Therefore, we need only to search for the Cooper pair carry finite velocity but zero momentum, which can be sketched as the following picture.

enter image description here

The left panel is the usual case, where the band bottom lies at $k=0$, so that the current carrying state need the Cooper pair possess finite momentum, $k_{cp}>0$. Let's now investigate the right panel, where the band bottom has finite momentum, $k<0$ for example. Then when the Cooper pair momentum $k_{cp}=0$, they actually contribute finite supercurrent. The two picture are different in the Cooper pair momentum only, which are the same in the Cooper pair velocity! In this sense, the momentum seems unobservable, at least by the current.

The above argument bases on the toy model. Is there any real material can achieve the right panel case? I believe it is feasible. The crystal lattice can provide different band structure, which leads to different deformation of the band structure. When the ground state of the superconductor breaks inverse symmetry, supercurrent with zero Cooper pair momentum occurs.

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The Reason for ongoing flow is very simple: the cooper pairs do not have a choice. There is no resistance and the magnetic field of the current acts like mass in mechanics. The current in a circular super conductor just "goes on". The real question (which is not so easy) is, how do You start some current in the first place? I think marad has the appropriate answer. The fact that lumings wrong answer got most appreciation, shows the limitation of this democratic forum. Truth and facts cannot be found by ballot.

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