Ether before relativity in quantum theory also

[jostpuur]

I think this all makes sense to me, but I've never heard of this in lectures or in books, so I'll check if I'm getting this right.

In one dimension Galilean coordinate transformations are $$x'=x-ut$$ and $$t'=t$$. Momentum transforms as $$p'=p-mu$$, and energy is $$E=p^2/(2m)$$. With a quick calculation I get $$E't'-p'x'=Et-px-\frac{1}{2}mu^2t-mux$$. So this means, that a wave plane solution $$\psi(t,x)=\exp(i(Et-px))$$ of the Shrodinger equation is not a solution in other inertial frames? In relativistic theory it goes better, as $$p_\mu x^\mu$$ is Lorentz invariant, and a solution of Klein-Gordon equation is always a solution also in other inertial frames.

So if I assume, that a wave function is real in the sense, that there truly is some complex number associated with each space time point, in non-relativistic theory I must assume an ether coordinate set? I think it somehow remarkable, that a Shrodinger equation is not Galilean invariant like the Klein-Gordon or Dirac equations are Lorentz invariant.

 

[azntoon]

Yes, you are correct. In non-relativistic quantum mechanics, the wave function is not invariant under Galilean coordinate transformations. This means that in different frames of reference, the same wave function will appear differently. For example, a wave packet traveling in one direction will appear to travel in the opposite direction when observed from a different frame of reference. Additionally, the energy of the system is not preserved under Galilean coordinate transformations, and so solutions of the Schrödinger equation must be redefined in order to remain valid in other frames of reference. In contrast, solutions of the Klein-Gordon and Dirac equations are Lorentz invariant and remain valid in different frames of reference without requiring redefinition.