# 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.

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.
• Probably we need to use another Lagrangian since your expression doesn't give the complex conjugate Schrödinger equation $-i\hbar\frac{\partial\psi^*}{\partial t}=V\psi^*$. It's important because $\psi$ and $\psi^*$ are not independent, and if the Lagrangian gives different equations for them then we have more restrictions for a solution $\psi$. I suggest to use this form: $\mathcal{L}\left(\psi,\frac{\partial \psi}{\partial t},\psi^*,\frac{\partial \psi^*}{\partial t}\right)=\left(V-\frac{\hbar^2}{2m}\Delta \right )|\psi|^2+i\hbar\psi\frac{\partial\psi^*}{\partial t}$ – Olexot Nov 12 '19 at 20:38