First of all ,what I understand about "condensation" is that: there exists macroscopic occupation in one or more than one states, i.e. there exists a state $|i\rangle$ with occupation number $N_i$ and: $$\lim_{N\rightarrow+\infty}N_i/N \neq0,$$ and its ground states can be written as a coherent state, e.g, for BEC: $$|\text{BEC}\rangle \propto e^{\sqrt{N_0}a_{k=0}^\dagger}\prod_{k\neq0}e^{-\phi_ka_k^\dagger a_{-k}^\dagger}|0\rangle$$ where the first term is the coherent state with the average occupation number $N_0$ in $k=0$ state, which is the feature of "condense", and the second term origins from the interaction with the condensate.

Then, what I understand about "BCS" is that:


We can construct a Cooper pairs creation operator $\Lambda^\dagger$: $$\Lambda^\dagger=\sum_k \phi_k c_{k\downarrow}^\dagger c_{-k\uparrow}^\dagger$$

and the ground state of BCS is: $$|\text{BCS}\rangle \propto \prod_k(1+\sum_k \phi_k c_{k\downarrow}^\dagger c_{-k\uparrow}^\dagger)|0\rangle= e^{\sum_k \phi_k c_{k\downarrow}^\dagger c_{-k\uparrow}^\dagger}|0\rangle=e^{\Lambda^\dagger}|0\rangle$$ compare with $|\text{BEC}\rangle$, we can find that the $N_0\rightarrow1 $,this means there only exists one Cooper pair, which is not consistent with the definition of condense above. Why can we call it the "condensation" of Cooper pairs?

I read many questions about concepts about Cooper pairs, but I still cannot understand the meaning of "condensation" here.


2 Answers 2


In the standard treatment of BCS theory, the connection to condensation is not very clear. Here is an slightly unconventional one that makes it more apparent. Some notation first: $\hat{a}_{k,\sigma}$ destroys a boson, $\hat{c}_{k,\sigma}$ destroys a fermion, and $\hat{d}_{K,q}=\hat{c}_{K-q,\downarrow}\hat{c}_{K+q,\uparrow} $ destroys a pair of fermions with relative momentum $q$, overall momentum $K$, and opposite spins.

In this notation, a BEC might look like: $$|\Psi_{BEC}\rangle=(\hat{a}_{0,\sigma}^\dagger)^{N_0} | 0 \rangle$$ corresponding to macroscopic occupation of the zero-momentum state. Note that I am using a particle-conserved formalism for the moment.

Now, here is the BCS ground state wavefunction. Instead of writing it from the vacuum as you did, I will write it relative to the normal-state Fermi sea $|\Psi_0\rangle=\prod_{|k|<k_F,\sigma}c^\dagger_{k,\sigma}|0\rangle$: $$ |\Psi_{BCS}\rangle=\left( \sum_{|q|>k_F}\phi_q d^\dagger_{0,q} \right)^{N_+} \left( \sum_{|q|<k_F}(\phi_q)^{-1} d_{0,q} \right)^{N_-}|\Psi_0\rangle$$ where $N_+=N_-=\sum_{|k|>k_F}\frac{|\phi_k|^2}{1+|\phi_k|^2}$.

So you can see that this is somewhat like you are creating a macroscopic occupancy of excited pairs with zero net momentum, relative to the normal Fermi sea. $N_+$ is proportional to, but not quite the same as, the usual "number of Cooper pairs."

Here's another view: for a BEC you have $$\langle a_0 \rangle=\sqrt{n_0},$$ which is the order parameter of the condensate (now taking for convenience the usual formalism where particle number conservation is relaxed). The corresponding statement in BCS is: $$\langle d_{0,q} \rangle=\frac{\phi_q}{1+|\phi_q|^2}=F_q,$$ $$ \sum_q |F_q|^2=n_0$$

In general, one can construct an ansatz that smoothly interpolates between the BEC and BCS limits as a function of inter particle attraction (BEC-BCS crossover). In the BEC limit, in which the "Cooper pairs" become tightly bound molecules, the internal degrees of freedom represented by $q$ are still there but are no longer relevant (see many previous questions about BECs of composite particles, e.g. (1), (2) ).

This entire treatment is right from the book Quantum Liquids by Leggett, which develops both theories in a unified way. I've glossed over many details that are treated carefully there. This development leads to a slightly nonstandard approach with the following appealing result: BEC is defined as a macroscopic eigenvalue of a single-particle density matrix, while BCS condensation is a macroscopic eigenvalue of the two-particle density matrix.


As usual, it is the draft version of the answer.

As you can know, the key point in derivation of Green functions for SC in BCS theory is the computation of 4-fermion average, $$\langle N|T\psi_{\alpha}\psi_{\beta}\psi_{\gamma}^{\dagger}\psi_{\delta}^{\dagger}|N\rangle.$$ This average is non-trivial because our theory has attractive interaction. However, we take into account interaction only in sense that it modifies vacuum state, $$\langle N|T\psi_{\alpha}\psi_{\beta}\psi_{\gamma}^{\dagger}\psi_{\delta}^{\dagger}|N\rangle\approx-\langle N|T\psi_{\alpha}\psi^{\dagger}_{\gamma}|N\rangle\langle N|T\psi_{\beta}\psi_{\delta}^{\dagger}|N\rangle+\text{permutated term}+\\+\langle N|T\psi_{\alpha}\psi_{\beta}|N+2\rangle\langle N+2|T\psi^{\dagger}_{\gamma}\psi_{\delta}^{\dagger}|N\rangle.$$ In these approximation we neglect all the scattering processes and now we focus on the last term. I hope that it is clear that this terms is nothing more then density of Cooper pairs (up to several factors).

You are right in you question, the situation is different to Bose-Einstein condensation.

In further derivation, we drop out two first terms. In Bose gas, at $T=0$ almost all particles are in condensate and exciatations above condensate are not so important. In our system, the condensate origins from weak interaction (indeed, previously we have neglected scattering processes, this corresponds that interaction is really weak). So, we neglect large terms.

However, these two terms do not change a spectrum and everything is ok.

In terms of operators, you introduce non-vanishing anomalous averages, $\Delta$ & $\bar{\Delta}$. For instance, $$\Delta\sim \langle\text{BCS}|cc|\text{BCS}\rangle,$$ where I omit momenta and spins. Then, you rewrite initial BCS hamiltonian, perfrom Bogoliubov transformation (=find new fermion operators) and obtain your last expression. Then, you should substitute $|\text{BCS}\rangle$ into anomalous averages. At this stage, you obtain the gap equation. As was mentioned, $\Delta$ desribes the Cooper pairs density (=condensate density) and you can find that $$\Delta\sim e^{-1/g}, $$ where $g$ is coupling in BCS Hamiltonian. Roughly speaking, $g$ is really small and so $\Delta$ is small to.

Finally, the condensation occurs in the sense that ground state of theory is modified by formation Cooper pairs condensate.

Hope this helps.

  • $\begingroup$ Thanks for your wonderful answer! But I am confused that: the density of Cooper pairs $\Delta$ is finite, which means the "total number" of Cooper pairs are extensive, i.e. $N_i/N\neq 0$; but if we focus on the ground state and write it as the coherent state(like in the question), it seems the average number of pairs $\Lambda^\dagger | 0 \rangle$ is intensive(only one), i.e. $N_i/N=1/N\rightarrow 0$. How to unify these two pictures? $\endgroup$ Apr 27, 2020 at 10:18

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