Particle number conservation equals $U(1)$-symmetry? If have by now frequently read the above but never really understood it. It is said that the particle number conservations is related to the phase of the wave function, but how?
 A: The concept of number operator is most natural in the context of second quantization. I assume you are familiar with it, and denote by $a(x)$ and $a^*(x)$ the creation and annihilation operators(-valued distributions). They satisfy the commutation relation $[a(x),a^*(y)]=\delta(x-y)$.
The number operator $N$ is the second quantization of the identity, and it is often denoted by $$N=\int a^*(x)a(x)dx\;. $$
A system preserves the number of particles if its Hamiltonian $H$ commutes with $N$. Now the commutation rules of $N$ with $a^\#(x)$ are very simple ($a^\#$ stands for either $a$ or $a^*$):
$$N a^\#(x)= a^\# (x)  (N \mp 1)\; .$$
It is then obvious that any polynomial with an equal number of operators $a$ and $a^*$ will yield (I choose normal ordering since this is the usual choice for second quantization Hamiltonians, but it is not important):
$$\Bigl(a^*(x_1)a^*(x_2)\dotsm a^*(x_n)a(y_1)\dotsm a(y_n)\Bigr) N=N \Bigl(a^*(x_1)a^*(x_2)\dotsm a^*(x_n)a(y_1)\dotsm a(y_n)\Bigr)\; .$$
The classical correspondent of $a(x)$ is a function, denote it as $\alpha(x)$, and to $a^*(x)$ corresponds the complex conjugate $\bar{\alpha}(x)$. It is clear that a global $U(1)$ transformation of $\alpha(x)$ and $\bar{\alpha}(x)$ will be
$$\alpha(x)\to e^{i\theta}\alpha(x)\; ,\; \bar{\alpha}(x)\to e^{-i\theta}\bar{\alpha}(x)\; ,$$
for some $\theta\in\mathbb{R}$. Thus any polynomial with the same number of $\alpha$ and $\bar{\alpha}$ would be $U(1)$ invariant.
Now suppose to have a classical Hamiltonian function $h(\alpha,\bar{\alpha})$ that is $U(1)$ invariant. Then every quantization of $h(\alpha,\bar{\alpha})$ (i.e. any operator where we replace $\alpha$ with $a$ and $\bar{\alpha}$ with $a^*$, following a prescribed ordering) $H(a,a^*)$ will contain polynomials with the same number of $a$ and $a^*$, and thus will commute with the number operator $N$.
A simple example is the Hamiltonian of many bosons in an external potential $V_{ext}$, interacting via a symmetric two body potential $V_{pair}$, that in the language of second quantization is written:
$$H=\int a^*(x)\Bigl(-\frac{\Delta_x}{2M}+V_{ext}(x)\Bigr)a(x)dx+\frac{1}{2N}\int V_{pair}(x-y) a^*(x)a^*(y)a(x)a(y)dxdy\; ;$$
that commutes with $N$, and obviously its classical counterpart
$$h(\alpha,\bar{\alpha})=\int \Bigl(\lvert\nabla \alpha(x)\rvert^2+V_{ext}(x)\lvert\alpha(x)\rvert^2\Bigr)dx +\frac{1}{2N}\int V_{pair}(x-y)\lvert\alpha(x)\rvert^2\lvert\alpha(y)\rvert^2dxdy\; ,$$
is $U(1)$ invariant.
Notation remark: In QFT, the interaction part of the Hamiltonian is a polynomial of the self-adjoint field $\frac{1}{\sqrt{2}}(a(x)+a^*(x))$, that is not $U(1)$ invariant in the sense above (and in fact usually QFTs does not preserve the number of particles).
Also, again in QFTs, it is usual to investigate symmetries of the classical field action because it leads to quantities conserved in time and to gauge theories. In that context, the $U(1)$ invariance (of the action) is often associated (if I recall correctly) with conservation of charge, and thus your question may seem misleading for the ones used to that language. 
A: In single particle picture (with wave functions) it is clear that probability distribution $|\phi(x)|^2$ won't change under the $U(1)$ transformation $\phi(x) \rightarrow \phi(x) e^{i\phi},\quad \phi^* (x) \rightarrow \phi^* (x) e^{-i\phi}$ 
In many-particle picture, using field operators, simplest $U(1)$ symmetric Hamiltonian can be rewritten as:
$$ \hat{H} = \sum_{x} \epsilon(x) \ \hat{\phi}^\dagger (x) \ \hat{\phi} (x) $$.
As you can easily show, number of particles is conserved (commutator $\left [ \hat{H}, \hat{\phi}^\dagger (x) \ \hat{\phi} (x) \right]$ vanishes). This is a direct consequence of $U(1)$ symmetry (here: $\ \hat{\phi} (x) \rightarrow \ \hat{\phi} (x) e^{i\phi},\quad \hat{\phi}^\dagger (x) \rightarrow \hat{\phi}^\dagger (x) e^{-i\phi}$ ) - Hamiltonian won't change under this transformation.
If you would add a term e.g. $ \lambda \hat{\phi}(x) $ Hamiltonian will not longer have $U(1)$ symmetry ($ \lambda \hat{\phi}(x) $ would change to $ \lambda \hat{\phi}(x)e^{i\phi} $) and as can be shown independently Hamiltonian would not conserve number of particles (commutator $\left [ \hat{H}, \lambda \hat{\phi}(x)\right]$ would not vanish).
