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.

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  • $\begingroup$ +1 you might mean, with caveats in your post. No big deal on this question, just a yes/no/ don't know reply is fine but with fields as operators, would the Ward identities make any kInd of sense in my comment above, if you know QFT? $\endgroup$ – user108787 Sep 12 '16 at 1:31
  • $\begingroup$ One minor nitpick: Noether's theorem cannot be used “just” like in classical mechanics, since you can formulate quantum mechanics in the Hamiltonian as well as Lagrangian formalism (see e. g. the Marsden and Ratiu's book on Classical Mechanics), it simply is Noether's theorem. $\endgroup$ – Max Lein Sep 12 '16 at 2:04
  • $\begingroup$ 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: $\mathcal{L}\left(\psi,\frac{\partial \psi}{\partial t},\psi^*,\frac{\partial \psi^*}{\partial t}\right)=\left(V-\frac{\hbar^2}{2m}\Delta \right )|\psi|^2+i\hbar\psi\frac{\partial\psi^*}{\partial t}$ $\endgroup$ – Olexot Nov 12 '19 at 20:38
  • $\begingroup$ Unfortunately, I didn't find any literature about a Lagrangian for the Schrödinger equation, so I also could make a mistake $\endgroup$ – Olexot Nov 12 '19 at 20:38

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