How can Noether's Theorem be used to prove that the probability density satisfies a continuity equation? How can I use Noether's Theorem to show that the probability density $\rho (x)=|\psi(x)|^2$ for a wave  function $\psi(x)$ satisfies the continuity equation $\frac{\partial \rho}{\partial t}+\nabla \cdot\vec{j}=0$, where $\vec{j}$ is the probability current defined in quantum mechanics?
I have solved this problem before by other means but I don't think I understand Noether's Theorem well enough to apply it in this case. Any help would be greatly appreciated.
 A: First note that Schrödinger's equation can be understood to come from an action. The Lagrangian is
$$L = \int~\mathrm d^3x \,\,\psi^†(x) \left(i \frac{\partial}{\partial t} - \frac{\nabla^2}{2m}\right)\psi(x) - \psi^†(x)\psi(x)V(x)$$
The Euler-Lagrange equation for $\psi^†(x)$ is exactly the Schrödinger equation. Since the dynamics of $\psi(x)$ are determined by Lagrangian mechanics in this way, Noether's theorem applies without any caveats.^^
In particular, this Schrödinger Lagrangian has a $U(1)$ symmetry corresponding to $\psi(x) \mapsto e^{i\alpha}\psi(x)$. The corresponding conserved charge current density is
$$\rho = j^0 = \frac{\partial L}{\partial \dot{\psi}}\delta \psi  = \psi^†\psi(x)$$
$$\vec{j}^i = \frac{\partial L}{\partial_i\psi}\delta \psi+\frac{\partial L}{\partial_i\psi^†}\delta \psi^†=\frac{i}{2m}\left((\partial^i\psi^†)\psi-\psi^†\partial^i\psi\right),$$
which is the well-known probability current density.
^^ In non-relativistic quantum mechanics the wavefunction $\psi(x)$ is a "classical" variable in that it is simply a function from space and time to $\mathbb{C}$. Noether's theorem works exactly the same for it as in classical mechanics. In quantum field theory the relevant objects $\psi(x)$ become quantum operators and the usual arguments have to be modified somewhat.
