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It is stated in many texts that the number operator $N$ which counts the number of Cooper pairs and the phase operator $\phi$ which counts the superconducting order parameter's phase $\text{Arg}(\Delta)=\phi$ are canonically conjugate and thus satisfy \begin{equation} N = -i\frac{\partial}{\partial \phi} \iff \phi = i\frac{\partial}{\partial N} \end{equation} This makes sense when one looks at the BCS ground state \begin{equation} |BCS\rangle = \prod_k (u_k+v_kc^\dagger_{k\uparrow}c^\dagger_{-k\downarrow})|0\rangle \end{equation} and since $u^*_kv_k=|u_k||v_k|e^{i\phi}$ this can be written as \begin{align} |BCS\rangle &= \prod_k (|u_k|+|v_k|e^{i\phi}c^\dagger_{k\uparrow}c^\dagger_{-k\downarrow})|0\rangle \end{align} up to some irrelevant global phase.

We see that every Cooper pair comes with a phase $e^{i\phi}$ so differentiating with respect to $\phi$ gives the same result as counting the number of Cooper pairs.

Due to this well-defined phase $\phi$, $|BCS \rangle$ is a coherent superposition of states with different particle number, and there does not have a well defined $N$. But now let's consider what happens when we project the BCS state on a the subspace with fixed particle number $N$ as explained in https://canvas.harvard.edu/courses/79258/files/folder/Problem%20Sets?preview=12301649. We find that \begin{equation} |BCS_N\rangle\equiv \int_0^{2\pi} \frac{d\phi}{2\pi} e^{-iN\phi} |BCS\rangle = C \bigg(\sum_k \frac{|v_k|}{|u_k|} c^\dagger_{k\uparrow}c^\dagger_{-k\downarrow}\bigg)^N |0\rangle \end{equation} which looks a lot like a BEC state formed by $N$ Cooper pairs. This state clearly has $N$ cooper pairs. However when I use the definition $N = -i\frac{\partial}{\partial \phi}$, since there is no $\phi$ dependence in this state (it got integrated out) I am expected to get $N=0$?!

So what's wrong with this reasoning?

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The identification of $N$ with $-i\partial_\phi$ is mathematicaly inconsistent as $N$ cannot take negative values. As a consequence it is not surprising that there are some paradoxes.

For example, from $[\phi, \hat N]=i$ we can derive an uncertainly relation $\Delta N \Delta \phi\ge 1/2$, but when $\Delta N$ is small (zero in your fixed number state) this implies $\Delta \phi>2\pi$, which is not possible as $\phi$ is an angle with bounded range.

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  • $\begingroup$ Your statement " ... Δϕ>2π ...is not possible as ϕ is an angle with bounded range" is closely related to univalence superselection, which is not uncontroversial. $\endgroup$ Commented Apr 25, 2022 at 12:47
  • $\begingroup$ @Andrea Alciato What is"univalence superselection"? $\endgroup$
    – mike stone
    Commented Apr 25, 2022 at 13:13
  • $\begingroup$ See en.wikipedia.org/wiki/Superselection#Examples for something directly related to your statement and pirsa.org/14030114 for the full story on univalence superselection. $\endgroup$ Commented Apr 25, 2022 at 15:18
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It is useful to first review the physics of a quantum planar rotor. Consider the Hilbert space spanned by $|N\rangle$. They are eigenstates of the number operator $\hat{N}|{N}\rangle=N|{N}\rangle$. The phase eigenstates are defined as $|\phi\rangle=\sum_N e^{iN\phi}|N\rangle$. The $|\phi\rangle$'s also form an orthonormal basis. The meaning of $N=-i\partial_\phi$ is that for a given state written in the $|\phi\rangle$ basis:

$$ |\psi\rangle=\int_0^{2\pi}d\phi\, \psi(\phi)|\phi\rangle $$

We have

$$ \hat{N}|\psi\rangle=\int_0^{2\pi}d\phi\, \Big(-i\partial_\phi\psi(\phi)\Big)|\phi\rangle $$

Now, the BCS case is not really like a rotor, since $N$ can not be negative. In this case, it is better to think of $N$ as measuring the deviation from the average number of Cooper pairs, which is supposed to be huge, and approximately consider the range $N$ to be all integers.

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  • $\begingroup$ "The $|\phi\rangle$'s form an orthonormal basis". Is this really true? As far as I can see, these states are essentially coherent states, which wikipedia states are not orthogonal. In fact, as far as I know, the coherent states form an over-complete basis (see e.g.). $\endgroup$ Commented Dec 18, 2023 at 22:49
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    $\begingroup$ They are NOT the coherent states as described in wikipedia. $\endgroup$
    – Meng Cheng
    Commented Dec 19, 2023 at 16:00
  • $\begingroup$ You're right, I'm sorry. $\endgroup$ Commented Dec 19, 2023 at 17:00

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