Why the generator of $U(1)$ is the particle number operator $N$? When $U(1)$ symmetry is broken, the particle number is not conserved any more. What is the relation between u1 and particle number. Why the generator of $U(1)$ is the particle number operator $N$?
 A: The lagrangian for a complex scalar field $\phi$ exhibits a global $U(1)$ symmetry which gives rise to a conserved current and charge, by virtue of Noether's theorem. The theorem tells us that $$Q_{U(1)} = \int \text{d}^3 x\,j^0 = \int \text{d}^3 x \,\phi^{\dagger} \overset{\leftrightarrow}{\partial_0} \phi$$ is conserved. If we now rewrite $\phi$ in terms of its mode expansion and substitute this into the right hand side of the above, after some manipulation and normal ordering one arrives at $$Q_{U(1)} = \int \frac{\text{d}^3p}{(2\pi)^3} (a_p^{\dagger}a_p - b_p^{\dagger} b_p) = N_a - N_b,$$ which is identifiable with the number operator for the quanta created by $a_p^{\dagger}$ (dubbed particle) minus that for $b_p^{\dagger}$ (dubbed antiparticle). This tells us that the two species have similar quantum numbers but an opposite $Q_{U(1)}$ charge. To check that indeed $N_a$ is a number operator we can look at its action on an $n$ particle state, for example, and see that $$N_a |\mathbf p_1 \dots \mathbf p_n \rangle = n | \mathbf p_1 \dots \mathbf p_n \rangle$$ so that the eigenvalue of the operator $N_a$ on a state in the Fock space of states is just the number of particles.
Note that it is the $U(1)$ charge that is equal to the number of particles minus the number of antiparticles (and not the $U(1)$ generator which is simply the identity operator if the global transformation is written like $\phi \rightarrow e^{i \theta T} \phi$, where $T \equiv 1$).
All of this generalises straightforwardly in the case of considering a Dirac spinor $\Psi$, where the conserved current is associated with quark number conservation as the zero component of $j^{\mu} = \bar \Psi \gamma^{\mu} \Psi$ makes clear.
A: For a single-particle state, a $U(1)$ rotation operation is simply multiplying a phase factor of the form $e^{-i\theta}$. The same operation on an $N$-particle state multiplies $e^{-i\theta}$ for each particle and hence $e^{-iN\theta}$ on the quantum state in question.
It is easy to see that the operator acting on the Fock space with the above property is $e^{-i\hat{N}\theta}$, where $\hat{N}$ is the number operator. Thus, the generator of $U(1)$ rotation is the number operator.
