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?


2 Answers 2


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.

  • $\begingroup$ 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

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.

  • $\begingroup$ "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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.