Lorentz transformation of wavefunction Consider two  frames $S$ and $S'$ that are related by a Lorentz transformation.
The wavefunction in $S$ is $\psi(x)$ where $x$ is the spacetime coordinates. In $S'$, the transformed wavefunction is $\psi'(x')$.
If the wavefunction in $S$ has a value of $A$ at a spacetime point $x_1$ (i.e. $\psi(x_1)=A$), does it also mean that the transformed wavefunction in $S'$ has a value of $A$ at the spacetime point $x_1'$ (i.e. $\psi'(x_1')=A$)?
If this is true, why? If it is not true, why not?
 A: There is no relativistic wave-function of the sort you are thinking of. The Shroedinger equation is only covariant under Galilean boosts, under which
$$
\psi(x,t)\to e^{i\left(mV -\frac 12 mV^2t\right)/\hbar}\psi(x-Vt,t).
$$
Relativistic wave equations,  such as Dirac and Kein-Gordon, describe quantum fields, and do not have a well-defined single particle wavefunction associated with them.
A: First of all, keep in mind that relativistic quantum mechanics is not usually formulated in terms of wavefunctions. This is mainly due to the fact that relativity allows the creation and annihilation of particles, which requires at least a second-quantized formalism.
This being said, for simple non-interacting systems one can still somewhat use a description based on wavefunctions. In this case the transformation law you wrote down, namely
$$
\psi'(x')=\psi(x)\ ,
$$
where $x'$ and $x$ are related by a Lorentz transformation, does indeed hold for spin-0 particles. The reason for this is that the wavefunction of a spin-0 particle is a scalar function which, as such, transforms as above. If on the other hand the particle had a different spin, say spin $j$, then $\psi$ would be a $(2j+1)$-tuple also containing information on the particle's spin projection. Then, since the spin projection is not left invariant by a Lorentz-transformation, you would have something like
$$
\psi'(x')=U\psi(x)
$$
where $U$ - a matrix - is part of a representation of the Lorentz group onto the space of $(2j+1)$-tuples. Similar (and more rigorous) relations hold for the fields which define a Quantum Field Theory.
P.S.: The explanation I gave above is sketchy in that it does not discuss issues such as the actual interpretation (or interpretability) of a relativistic wavefunction, the unitarity (or, better, the lack thereof) of the finite-dimensional representations of the Lorentz group, etc. Again, keep in mind that a serious discussion of relativistic quantum mechanics should rest on second quantization and Quantum Field Theory. Nonetheless, your general intuition was headed in the right direction.
