# I U(1) Invarience Lagrangians

1. Oct 31, 2016

### Staff: Mentor

Hi

On page 176 of Physics from Symmetry it says (note 9)

If we assume Ψ describes our particle directly in some way what would U(1) the transformed solution Ψ' = e^iθ Ψ be which is equally allowed describe.

He is speaking of allowed solutions of Lagrangian's. Its true for all Lagrangian's I know but why is it true generally? Or is he only speaking of the usual Lagrangian's?

Thanks
Bill

2. Nov 1, 2016

Staff Emeritus
Because the only thing that's observable is the probabiliy density, Ψ†Ψ. Ψ'† = e^-iθ Ψ†, i.e. puts a minus sign in front of the phase. The two phases multiply to 1.

3. Nov 1, 2016

### Staff: Mentor

That's true - but he hadn't reached that point (ie interpreting Ψ) so maybe he was thinking of something else - but what

Thanks
Bill

4. Nov 1, 2016

### Jilang

Would it correspond to a change of the basis?

5. Nov 1, 2016

### Staff: Mentor

I am now thinking ts simply because its a generator.

Thanks
Bill

6. Nov 2, 2016

### vanhees71

For the non-relativistic case (Schrödinger, Pauli equations) it's the existence of a cintinuity equation, ensuring the conservation of the norm of the wave function in the time evolution. This enables the probability interpretation via Born's rule, because $|\psi|^2$ is the density in the continuity equation.

7. Nov 4, 2016

### strangerep

For the benefit of other readers who don't have the book, the context is:

I think the way he argues this is little better than hand-waving gobbledegook. The question has no chance of being answered unless one first converts the words "in some way" to a precise meaning.

He does not mention "interferometer", nor "Aharonov-Bohm", nor even "diffraction" or "double slit" anywhere in the book, hence does not have to confront the misleading superficiality of his exposition.

Imho, he's just referring to the trivial result that a symmetry of the Lagrangian will necessarily be mirrored somehow (possibly trivially) in the space of solutions of the equations of motion (since the eom are unchanged by those symmetry transformations).