The BCS Hamiltonian reads:

$$H_{BCS}=\sum_{k\sigma}\epsilon_k c_{k\sigma}^\dagger c_{k\sigma}-\Delta^*\sum_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger+h.c.$$

The particle number operator reads:

$$ \hat{N}=\sum_{k\sigma}c_{k\sigma}^\dagger c_{k\sigma} $$

The Heisenberg equations of motion reads:

$$ \frac{d \hat{N}}{dt}=i[H,N]=2i(\Delta^*\sum_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger-h.c) $$

Taking the average respect to the wavefuction, the right hand side is perfectly zero.

So why saying the particle number is not conserved?

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    $\begingroup$ The fact that $[H, N]\neq 0$ already says the particle number is not conserved. Nothing else is needed to make the statement. Also, if you compute $\langle \sum_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger\rangle$ on a BCS wavefunction $(u_k + v_kc_{k\uparrow}^\dagger c_{-k,\downarrow}^\dagger)|0\rangle$, the answer is not zero. $\endgroup$ – Meng Cheng Apr 22 '15 at 2:34
  • $\begingroup$ @MengCheng Thanks, I noticed the definition of a conserved quantity in quantum mechanics. So what my question really claim is that the average particle number is not change with time, but it may flutuates around this mean value. $\endgroup$ – an offer can't refuse Apr 22 '15 at 2:41
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    $\begingroup$ OK, but that does not say anything: the expectation value of any physical quantity, as long as the state is stationary (i.e. time-independent), does not change with time. $\endgroup$ – Meng Cheng Apr 22 '15 at 3:08
  • $\begingroup$ @MengCheng Thanks again for this further clarification. $\endgroup$ – an offer can't refuse Apr 22 '15 at 3:23

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