BCS state with well-defined particle number - interpretation?

Question

It's common knowledge (and has been discussed in other questions on this site) that the standard BCS ground state $ \left|\Psi_{BCS}\right\rangle = \prod_k \left( u_k + v_k c_{k\uparrow}^{\dagger} c_{-k\downarrow}^{\dagger}\right) \left|0\right\rangle$ does not have a well-defined particle number and that this doesn't matter in bulk superconductors because the standard deviation is $\Delta N \propto \sqrt{N}$ and hence irrelevant for $N\rightarrow \infty$.

But I also read that you can arrive at a BCS state with well-defined particle number by first defining

$$\left|\Psi_{BCS}(\phi)\right\rangle = \left( |u_k| + e^{i\phi} |v_k| c_{k\uparrow}^{\dagger} c_{-k\downarrow}^{\dagger}\right) \left|0\right\rangle $$

and then "integrating out" the phase according to

$$\left|\Psi_{BCS}(N)\right\rangle = \int_{0}^{2\pi} \mathrm{d}\phi\, e^{-iN\phi/2} \left|\Psi_{BCS}(\phi)\right\rangle \,\, ,$$ which gives you a BCS state with precisely N particles at the cost of having a completely ill-defined phase.

If that is true, then surely this is the actual "physical" state of a superconductor and the original BCS state is merely used for convenience.

But that, in turn, would make the well-defined phase of the superconducting state a mere mathematical artifact, when every other textbook highlights it as something very fundamental (and if I remember correctly, it's very important for things like the Josephson effect as well).

So, can anyone point to the error in the logic above?

Answer 1

The important point is that $\left[N,\phi\right]=i$, with $N$ the number of electron et $\phi$ the phase. So you can apply the Heisenberg uncertainty principle, of which you gave the two extremum situations of 1) either perfectly defined $N$ and completely unknown $\phi$ or 2) perfectly defined $\phi$ and completely unknown $N$. In general varying a bit the phase $\phi$ let a bit of electron $N$ flowing, which is the basic situation in mesoscopic systems (e.g. Josephson junctions).

Conceptually, isolate a part of a superconductor (or any macroscopic system behaving as quantum mechanics), if you can count the number of electrons precisely, there is no current and so no phase difference, so $\phi$ is not a good quantum number. If you can define a current (and so fix a phase difference) you can not count the electrons in your control volume of superconductor.

Note that it is not that obvious to demonstrate the commutation relation $\left[N,\phi\right]=i$ in the general case, and discussions using this relation are usually subtle. For a harmonic oscillator the relation is easy to prove, using the coherent states, see e.g.

• http://hitoshi.berkeley.edu/221a/coherentstate.pdf [PDF file directly]

or the book by Nagaosa on Quantum Field Theory in Condensed Matter (or a title like that).