# How does the creation of Cooper pairs and their condensation explains superconductivity?

Notes I follow provide an introduction to the BCS theory where one considers a Hamiltonian consisting of a kinetic term and an attractive potential between electrons. Next by the mean-field approximation and the Boguliubov transformation, one gets the diagonalised Hamiltonian with the Boguliubov dispersion given by $$\begin{equation} E_{\vec{k}}=\sqrt{\left(\varepsilon_{\vec{k}}-\mu\right)^2+\left|\Delta\right|^2} \end{equation}$$ where $$\varepsilon_{\vec{k}}$$ and $$\mu$$ are respectively kinetic energy and chemical potential from the original Hamiltonian whereas $$\Delta$$ is gap energy that "built up" during the diagonalisation. At that point, the notes consider the model as solved.

My question is: How does that dispersion relation explain the condensation of pairs? And more importantly, how does that explain superconductivity? How one could see that with such an excitation spectrum one could have resistance less flow of currents?

It is the sketch of answer, can be changed further

When one diagonalized the Hamiltonian of BCS theory, the resulting form is $$H\sim \sum_{k,\sigma}\sqrt{\Delta^2+\xi_k^2}\alpha^{\dagger}_{k,\sigma}\alpha_{k,\sigma}+...,$$ where I have omitted some terms (they are not interesting for the answer), $$\sigma$$ denotes spin, $$\xi_k\equiv\epsilon_k-\mu$$ and $$\alpha_{k,\sigma}$$ is Bogoliuobv quasi-particle.

From this structure it is visible that the minimum energy is simply $$\Delta$$. This Hamiltonian also tells us that it is difficult to excite the quasi-particle at low temperature: there is the gap $$\Delta$$ between filled and empty quasi-particle states. It means that ground state (with minimum energy) is rigid.

Next, we should remind the Landau criterion for superfluidity (for details, check vol. 9, chapter 3, par. 23 of Landau Course in Theoretical Physics), which tells us if $$\min(E_k/k)>0$$, then the superfluid behavior is possible. The Cooper pairs appear when one introduce new particle creation operators for Hamiltonian diagonalization. So, we have: 1) effective bosons appeared, 2) resulting spectrum obeys Landau criterion for superfluidity. Nice.

Next step is to demonstrate somehow that $$\Delta$$ appears due to the 2nd order phase transition. There are at leas two possibilities: 1) if you are familiar with field-theoretical formalism and Hubbard transformation, you can immediately perform four fermion interaction in Cooper channel and obtain famous Ginzburg-Landau (GL) effective action, 2) you can also verify (from Bogoliubov transformation) that $$\Delta$$ depends on temperature by power-law. It tells us that there is a continuous phase transition w.r.t to temperature and we can write down expansion of free energy near $$T_c$$ in terms of $$|\Delta|^2$$. It is the mentioned GL-expansion.

Next step is to add external electromagnetic field. To do it, you should couple the this field to $$\Delta$$ as usual. By minimization of free energy, you can derive the GL-equations. From GL-equations, one can immediately obtain so-called Londons' equation, which demonstrates that magnetic field does not penetrate into the superconductor except the thing layer. Nice.

I have seen a lot of books where Londons' equation is written down only for magnetic field, but to illustrate superconductivity one should use electric field. This derivation gives $$\frac{\partial\bf{j}}{\partial t}=\frac{1}{\delta^2}{\bf E},$$ where relation $${\bf E}=-\partial_t{\bf A}$$ is taken into account and $$\delta$$ is the penetration length. Then, you shold rewrite this expression in $$(\omega,{\bf q})$$-space, $${\bf j}(\omega,{\bf q}=0)=\frac{i}{\omega\delta^2}{\bf E}(\omega,{\bf q}=0).$$ Therefore, the conductivity has pole at $$\omega=0$$. Nice, let us keep it in mind.

Let me remind that similar expression for normal metal the simplest framework to derive conductivity is Drude model. This model results the following expression for AC-conductivity, $${\bf j}(t)=\frac{ne^2\tau}{m}\frac{1}{1-i\omega\tau}{\bf E}(t)$$ with $$E(t)=Ee^{-i\omega t}$$ and $$j(t)=je^{-i\omega t}$$. The conductivity looks like $$\text{Re}\,\sigma(\omega)=\frac{ne^2\tau}{m}\frac{1+\omega^2\tau^2},$$ $$\text{Im}\,\sigma(\omega)=\frac{ne^2\tau}{m}\frac{\omega\tau}{1+\omega^2\tau^2}.$$ In case of an ideal conductor, we can just take limit $$\tau\rightarrow\infty$$, which results $$\text{Re}\,\sigma(\omega)\xrightarrow{\tau\rightarrow \infty}\frac{ne^2}{m}\delta(\omega),$$ $$\text{Im}\,\sigma(\omega)\xrightarrow{\tau\rightarrow\infty}=\frac{ne^2}{m\omega}.$$ So, we coclude that imaginary part of AC-conductivity of an ideal conductor has the pole at $$\omega=0$$. Moreover, real part of AC-conductiviy diverging at $$\omega=0$$. The DC-conductivity can be obtained from AC-conductivity by simply setting $$\omega\rightarrow 0$$.

Therefore, in order to see diverging conductivity, you should:

1. Convince yourself somehow that anomalous average $$\Delta$$ appears in the Fermi system with attractive interaction
2. Verify that there is continuous phase transition and $$\Delta$$ is the order parameter
3. Add external electromagnetic field to consideration and obtain corresponding equations of motion for EM field (GL-equations, Londons' equation)
4. Perform comparison between Londons' equation with electric field and Drude model of normal metal
5. Consider the case of an ideal conductor and observe that conductivity is divergent.

Last point: it is worth mentioning that case of ideal conductor differs from superconductor. One can demonstrate that in case of ideal conductor we have $$\partial{\bf B}/\partial t=0$$, whereas in case of superconductor we have $${\bf B}=0$$, where $${\bf B}$$ is the magnetic field.

• Thank you for the answer. Unfortunately, I can't see how particular parts of your answer relate to each other. Firstly I can't see how that correction to the current from quasiparticles proves superconductivity. I could imagine that e.g. the divergence of that correction could imply it. But is it divergent? Also what you wrote about that potential gap $\Delta$ contradicts my intuition. I thought that a superconductive state is achieved especially in low temperatures and lowering the temperature promotes the creation of quasiparticles. Sep 10, 2022 at 10:25
• So do I understand correctly that that gapped excitation spectrum has nothing to do with superconductivity? That spectrum is just a property of a condensate. What matters is the fact that we have a ground state with a large (because of condensation) number of quasiparticles that contribute to the current. Current is proportional to the average value of a number of pairs operator, which is large due to condensation. And gaped potential has nothing to do with that. Is it an explanation for my hesitation from the comment above? Sep 10, 2022 at 10:34
• @PawełKorzeb I understand your statements, will update the answer and provide derivations Sep 10, 2022 at 10:36
• @PawełKorzeb , after several months, I have updated the answer and I hope that it becomes clearer. Dec 8, 2022 at 14:15

How does that dispersion relation explain the condensation of pairs?

It does not. It is a result of pair condensation, since: $$\Delta \propto \sum_k \langle c_{k\uparrow}c_{-k\downarrow} \rangle$$

Using the resulting dispersion, we can further predict things like the behavior of specific heat or the tunneling current in a superconductor-metal junction.

The existence of an energy gap solely does not tell us anything. A band insulator for example, has an energy gap and has zero conductivity.

The relation between Bogoliubov quasi-particle operators and the original electronic operators is what causes superconductivity. As you can see in Artem Alexandrov's answer, the $$\Delta$$ term will produce an extra term in the formula for current that says $$J \propto A$$ which is exactly the London's formula for superconducting current that historically had been proposed phenomenologically before the BCS theory. If you put this formula for current in Maxwell's equations, you will get many of the interesting properties of superconductors, including the Meissner effect and also the fact that since $$J \propto v$$ then $$v\propto A$$ and so $$\dot{v} \propto E$$ which means that charges move without any friction in a superconductor.

• So I think that's the point I don't understand.$\Delta \propto \sum_k \langle c_{k\uparrow}c_{-k\downarrow} \rangle$ Indeed this is the definition of $\Delta$ I encountered, but how one could see that it follows from condensation? For example in BEC condensation is clear as one assumes the ground state to be a coherent and average value of the number operator on that state given the number of particles condensed in the ground state. Is $\langle c_{k\uparrow}c_{-k\downarrow} \rangle$ an analogue of a number operator for Cooper pairs? If yes, then why? Sep 10, 2022 at 19:33
• The operator $b_q=\sum_k f(k) c_{k+q,\uparrow}c_{-k,\downarrow}$ is like annihilation operator for a pair of fermions with the momentum of the center of mass equal to $q$ and so it is the annihilation operator for a boson(=a cooper pair). So $\langle c_{k\uparrow}c_{-k\downarrow}\rangle\neq 0$ means that the wave function is a coherent state for composite (cooper pair) bosons. Sep 10, 2022 at 21:57