What is the link between the BCS ground state and superconductivity?

The link between the BCS ground state $$\left|\Psi_\mathrm{BCS}\right\rangle = \prod_k \left( u_k - v_ke^{i \phi} c_{k\uparrow}^{\dagger} c_{-k\downarrow}^{\dagger}\right) \left|0\right\rangle$$ and the property of superconductivity (carrying current with no resistance) does not seem obvious to me, but I can't find a satisfying explanation anywhere. According to some authors, the gap explains superconductivity, but I am not convinced by this, since (1) insulators are gapped too, and (2) some superconductors have a vanishing gap for some values in $k$-space.

Isn't it possible to just calculate the linear response under DC field and compare with the normal state expectation value to see how this works? Does anyone know how to perform such a calculation?

• If I recall correctly, this calculation is done in the textbook "Quantum Liquids" be Leggett. You are very correct in saying that an energy gap does NOT imply zero resistance. This is a subtle point mentioned in the book by de Gennes. – KF Gauss Jun 17 '17 at 7:23
• @user157879 I think the BCS gap does imply zero resistance. However this is not a necessary condition. – Diracology Jun 29 '17 at 1:01
• @Diracology, I'm not sure what you mean by BCS gap. If you mean to say the BCS state implies zero resistance then that's obviously true. But the existence of a gap alone does not imply zero resistance. An insulator has a gap and yet does not have zero resistance. There are superconductors that are not gapped and have zero resistance. Take a look at the texts by de Gennes and Leggett. – KF Gauss Jun 29 '17 at 3:54
• Here are some nice clear-and-compact-yet-exhaustive lecture notes by Leggett on this: courses.physics.illinois.edu/phys598sc1/fa2015/Lectures/… – Ruben Verresen Aug 22 at 22:01

An electron-excitation-gap is necessary helpful for superconductivity. (Normally electrons scatter, but if there's a gap then they don't scatter, because there's no state to scatter into, as long as the temperature is low enough that they cannot jump the gap.**) But it is not sufficient. The filled states also have to be able to c

arry a current! The electrons in a filled semiconductor valence band have an electron-excitation-gap, like you say, but they do not carry a current. (If you think about it, in intrinsic GaAs near absolute zero, there are no electron scattering events!) On the other hand, the filled states in a superconductor CAN carry a current because...well I wasn't sure but I read the old BCS paper and they have a pretty basic explanation:

Our theory also accounts in a qualitative way for those aspects of superconductivity associated with infinite conductivity...the paired states $(k_{1\uparrow}, k_{2\downarrow})$ have a net momentum $k_1+k_2=q$, where $q$ is the same for all virtual pairs. For each value of $q$, there is a metastable state with a minimum in free energy and a unique current density. Scattering of individual electrons will not change the value of $q$ common to virtual pair states, and so can only produce fluctuations about the current determined by $q$." And these scattering events raise the free energy, unless all of the electrons scatter simultaneously in exactly the right way to make a new metastable state centered at a different q, which is extremely unlikely.

Well that makes sense to me.... In your question you wrote the BCS ground state with $q=0$, but that's just one of the family of BCS meta-stable (ground-ish) states with different $q$, with different $q$s corresponding to different current flows.

In other words, BCS theory explains how the electrons pair up and then there is an energy gap for single-particle-excitations. Slightly changing the $q$ requires little or no energy (or in some cases even lowers the energy), but will not happen spontaneously because it requires trillions of electrons to change their state simultaneously in a coordinated way. (An electric field can cause this kind of coordinated change, but it can't just happen spontaneously. Generally, only single-particle-excitations happen spontaneously, and these are gapped.) So it is metastable. And the fact that you can have a metastable state which carries current is just another way of saying that current can keep flowing and flowing even with no electric field pushing it.

**Update: OK, yes there is such a thing as "gapless superconductivity". My mistake was conflating "superconductor" with "superconductor with no dissipation whatsoever". The latter doesn't exist—even with a full proper superconducting energy gap, remember the superconducting transition is above absolute zero, so there is certain to be some small but nonzero electron scattering rate that is tolerable without destroying the superconducting order. So by that logic, it's not surprising that a partial or nonexistent gap is compatible with superconductivity at a very low transition temperature.

• I like this, but it raises the natural question of why those "ground-ish" (actually highly excited in general) metastable states are metastable at all, or equivalently why there is an extensive free energy barrier for scattering out of them. – Rococo Jun 30 '17 at 4:04
• To me, this seems like hand-waving, and it fails to answer the question how I get a current in the limit of zero field. – Thomas Jun 30 '17 at 4:59
• @Rococo and Thomas , I added another paragraph with more details – Steve Byrnes Jun 30 '17 at 13:31
• Pretty interesting. Can you provide (or give a link to) a more detailed calculation showing the metastability of those states? – Undead Jul 2 '17 at 4:09
• You are right. In fact, what I really wanted to know is in which way those $q \neq 0$ states are "ground-ish" states. – Undead Jul 2 '17 at 20:52

As you rightly mention, the presence of a gap explains nothing of the superconducting phenomenology, except its DC (and quasi-DC) behaviour. This is quite natural : same causes, same consequences. So a superconducting behaves as a semi-conductor because it has a gap. This gap being quite small, conventional superconductors are not really interesting semi-conductors.

So, what are the crucial aspects of superconductivity hidden in the BCS Ansatz you wrote ? Well, many many, for instance

• this is an example of a coherent fermionic state
• it comes from the microscopic picture of the Cooper pairing (as explained in the first paragraph of Joshuah Heath answer (the rest of his answer is non-sense)
• it gives a phase (in the sense of $\varphi$ in the writing $e^{i\varphi}$, not in the sense of phase of matter) to the condensate, hence you got a macroscopic wave function associated to the condensate
• it explains the temperature dependency of the gap
• it makes the condensate a perfect diamagnetic material
• ...

Perfect diamagnetism goes hand to hand with no resistance. The demonstration that there is no current / perfect diamagnetism / no resistance associated to the BCS Ansatz is explained in great details in the historical account presenting the microscopic theory, namely

Bardeen, J., Cooper, L. N., & Schrieffer, J. R. (1957). Theory of Superconductivity. Physical Review, 108, 1175–1204.

Calculation details can also be found in

Tinkham, M. (1996). Introduction to superconductivity (second edition). Dover Publications, Inc.

As far as I remember, Leggett also provide many different calculations of this effect, as

Leggett, A. J. (1975). A theoretical description of the new phases of liquid He3. Reviews of Modern Physics, 47, 331–414.

The calculation is nevertheless a bit cumbersome, so I'm reluctant to try it on this platform. Feel free to ask about unclear details in the linked references.

The BCS wave function you write down depends on a parameter $\phi$, but the ground state energy is independent of it. This implies that $\phi$ is the (would be) Goldstone mode that governs the low energy dynamics of the system. The gradient of $\phi$ is the conserved $U(1)$ current $\vec\jmath \sim \vec\nabla\phi$, and the ordinary charged current is $\vec\jmath_s =n_s e \vec\nabla\phi /m$, where $n_s$ is the superfluid density of electrons.

Because $\phi$ is a Goldstone mode the effective low energy action can only depend on gradients of $\phi$. By gauge invariance the effective action is of the form $S[A_\mu-e\nabla_\mu\phi]$. The explicit form of $S$ can be computed from the BCS wave function, or more easily determined using diagrammatic methods. For our purposes the only important point is that $S$ has at least a local minimum if the field vanishes. This means that solutions of the classical equation of motion are of the form $A_\mu=e\nabla_\mu \phi$ (This is the London equation). Let's consider an applied electric field $\vec{E}=-\vec\nabla A_0$. I find $$\vec{E}=-e\vec\nabla \dot\phi=\frac{m}{n_s}\frac{d\vec\jmath}{dt}$$ which shows that a static current corresponds to zero field, and the resisitivity is zero.

The effective action also governs other properties of the system, such as the Meissner effect, the critical current, and fluctuations of the current in a thermal ensemble.

Postscript: A commenter argues that I really need to show that $S$ has a minimum $$S \sim \gamma (A-e\nabla\phi)^2 + \ldots$$ First, note that $\gamma$ determines the Meissner mass, so even without a calculation I have shown that the Meissner effect implies superconductivity. Beyond that I do indeed have to do a calculation of $\gamma$ based on the BCS wave function (I could appeal to the Landau-Ginzburg functional, but this only shifts the question to the gradient term in the LG functional). Fortunately, the calculation is straightforward, and can be found in many text books. For people with more of a particle physics interest there is a beautiful explanation in Vol II of Weinberg's QFT book. There is Anderson's famous paper on gauge invariance and the Higgs effect. I provided a version of the calculation in Sect. 3.4 of these lecture notes https://arxiv.org/abs/nucl-th/0609075

Post-Postscript: How is this different from a weakly interacting electron gas? In the electron gas I have a low energy description in terms of electrons and phonons (and other degrees of freedom). For simplicity consider the high temperature limit, where a classical description applies (as explained by Landau Fermi liquid theory, this generalizes to low T). The equation of motion for a single electron is just $m\dot v=e E$, which superficially looks like the London equation. However, this is not a macroscopic current. When I pass from microscopic to macroscopic equations there is no symmetry that forbids the appearance of dissipative terms, so the conductivity is non-zero. There is indeed a subtlety in the coupling of electrons and phonons, because without the umklapp process momentum conservation would force the conductivity to vanish.

In a superconductor the gradient of the Goldstone boson automatically describes a macroscopic current ($\gamma$ is proportional to the density of electrons). S is a quantum effective action, and dissipative terms are automatically forbidden. At finite temperature things do get a little more complicated because the total current is in general the sum of a non-dissipative supercurrent, governed by $S$, and a dissipative normal current. However, below $T_c$ part of the response is carried by a supercurrent.

• I don't think this is really sufficient, unfortunately. For a vanishing external potential, the equilibrium current, as shown by the London equation, is zero. Nonetheless, a non-equilibrium state of an initial current in the absence of a potential is extremely metastable, and it is this dissipationless metastable current that leads to zero resistivity. The distinction is made very explicitly in these notes: courses.physics.illinois.edu/phys598sc1/fa2015/Lectures/… – Rococo Jun 30 '17 at 3:23
• I disagree. The question was how I can explain zero resistivity, that is the existence of a current in the limit of zero field. If there is a simpler (but still correct) explanation I'd be happy to hear it. Of course there are many other things that can be studied, such as critical currents and current fluctuations (at finite T). The important point is that these phenomena are encoded in the same effective action. – Thomas Jun 30 '17 at 4:43
• This derivation is very strange, you use essentially nothing from the actual BCS Hamiltonian or wavefunction. First of all, you can make the same argument for a free fermion system. In the free fermion case, you also have the same $U(1)$ conserved current, in which case you obtain the classical result that charged electrons obey Newton's law of motion $F=ma$ and have zero resistance since you ignore phonons! I recommend you not skip steps and repeat your argument rigorously, which is what op wants anyways. – KF Gauss Jun 30 '17 at 11:40
• @user157879 That's the whole point. The BCS wave function is just a variational state, and not a very good one for many superconductors. The explanation for superconductivity should be robust, and only depend on symmetries of the wave function. I am using one important property of the BCS wave function, which is that the $A=e\nabla\phi$ is at least a local minimum. In a free fermi gas the curvature (sometimes called the phase stiffness) is zero, – Thomas Jun 30 '17 at 12:47
• @Thomas, I appreciate the edit. However, there are still a few confusions in your explanation. The idea that the Meissner effect implies superconductivity is incorrect. The Meissner effect is an equilibrium phenomena, while supercurrent flow is a non equilibrium effect. Take a look at Quantum Liquids by Leggett for an elaboration. – KF Gauss Jun 30 '17 at 15:02

I am not sure about the second part of your question, but I think I can give you an answer to the first portion. The BCS trial wave function proposes a linear combination of a filled Fermi sea state (with probability $|u_k|^2$) and a state with a Cooper pair (with probability $|v_k|^2$). If one calculates the expectation value of the pairing Hamiltonian and minimizes the energy, one will find that the system favors being in the Cooper paired state. What this tells us is that, as long as we have some attractive potential (no matter how small), the system will prefer a Cooper paired state. In some lattice structure, phonon-electron interactions provide this attractive potential.

It is this inevitable formation of Cooper pairs that gives us zero resistivity. Small scale physics (like scattering) gets absorbed into the macroscopic quantum mechanics described by this system, as described in this answer. Hence, scattering processes won't effect the current. As suggested in your question and in the discussion in the link above, zero resistance is therefore not dependent upon the existence of a gap.

A good discussion of what causes superconductivity can be found here. For an excellent discussion of all things superconducting, see Tinkham's book.

• This answer's entire premise is incorrect. Having an energy gap does NOT imply zero resistance to current. Take a look at de Gennes ch7-8. You are making a very common misconception here. – KF Gauss Jun 17 '17 at 7:21
• What you need show is the response of the superconducting system to an electric field. What you are saying right now is simply that: "If there is current flowing already, the existence of a gap means the current should not decay in time" – KF Gauss Jun 17 '17 at 8:10
• @user157879 Okay--I edited my answer to make it clearer what I was saying. The gap doesn't imply zero resistance--the Cooper pair ground state does. It seems that my answer seemed to imply the opposite. – Joshuah Heath Jun 17 '17 at 14:20
• Your edit still dodges the main question, an explanation of the response function to an electric field. I don't mean to be aggressive, but the usual answers to this question always fall along the lines you've described. I've yet to see a simple and correct way to go from the BCS ground state to the conductivity without muddling into a mess of math that obscures the interpretation. This has been a suprise to me since zero resistance is the most famous property, the Meissner effect is secondary. – KF Gauss Jun 17 '17 at 21:35