Change of wavefunction due to relativistic speed Imagine a spacecraft which is moving at a speed comparable to the speed of light relative to a reference frame with a hydrogen atom at it's origin. How would the probability distribution function of an electron in 1s orbit look relative to an observer inside the spacecraft?
 A: The Schroedinger equation is only invariant under galilean boosts. Under such a boost with velocity $v$ we have
$$
\psi(x,t) \mapsto \tilde \psi = e^{i(mvx- \frac 12 mv^2t)}\psi(x-vt).
$$
The extra phase factors are because seen form the moving frame the s-wave electron  has an extra momentum $mV$ and extra energy $\frac 12 mV^2$.  It's an amusing exercise with the Schroedinger equation to show that this transformed  wave function obeys
$$
i\hbar \frac{\partial \tilde\psi}{\partial t}= -\frac{\hbar^2}{2m}\nabla^2 \tilde \psi+V({\bf x}-vt)\tilde \psi
$$
if
$
\psi$ obeys
$$
i\hbar \frac{\partial \psi}{\partial t}= -\frac{\hbar^2}{2m}\nabla^2  \psi+V({\bf x}) \psi.
$$
It is surprising because a relativistic  scalar particle obeys the Klein-Gordon equation and transforms as a scalar. The extra phases come from  the fact the Schroedinger $\psi$ is related to the KG $\phi$ by
$$
\phi(x,t)=\psi(x,t)e^{-imc^2t}
$$
because  the Schroedinger equation  does not take into account contribution of the rest mass to the energy.
Under a Lorentz boost
$$
t\to t'= \frac{(t-xv/c^2)}{\sqrt{1-v^2/c^2}}.
$$
The $1/c^2$ in the Lorentz transform cancels  the $c^2$ in the extracted phase factor $\exp\{-imc^2t\}$.
If you want to use Lorentz transformations for electrons you need to replace the Schroedinger equation by the Dirac equation, but then, as @Trebor says, you need the machinary of quantum field theory to interpret the spinor wavefunction as it  includes antiparticles.     The net effect is to squash to the charge distribution as answered by @timm, but you will  see particle antiparticle pairs in the pancaked   distribution as well as the expected electron.
A: Copy/pasted from https://physics.stackexchange.com/questions/591368/galilean-invariance-of-schrödinger-equation as a reaction on @MikeStone's answer. It is too long for a comment.
The Schrödinger equation is not Galilei covariant, even in the absence of electromagnetic fields. There is a symmetry group, the Schrödinger group, that describes the symmetry of the Schrödinger equation.  One reason for the Galilei non covariance is that the rest energy of matter is not included. For example, the Schrödinger equation for the hydrogen atom does not include the proton and electron rest energy. These can be included but in a manuscript I argue that this is not sufficient. In the same manuscript  I argue that its elements are not coordinate transformations as they depend on mass. Indeed the factor $$e^{i(m\mathbf{v\cdot r}-m\mathbf v^2t/2)/\hbar}$$ depends on mass and it does not follow from application of a Galilei transformation to the Schrödinger equation. It has to be postulated as part of an element of the Schrödinger group.  For a review of the Schrödinger group see ref 2 of my manuscript: H. Brown and P. R. Holland, Am. J. Phys. 67 (1999) 204 (behind a paywall).
A: The observer in the spaceship would "see" the s-orbital of the H-atom contracted in his frame in the direction it moves.  
A: Technically, in order to incorporate special relativity with quantum theory, you need Quantum Field Theory.
The Lorentz transformation (in particular, boosts) is not unitary. This means that the wavefunction is no longer properly normalized. This is not the case with rotations, which are unitary. Trying to directly jam relativity into quantum theory results in problems such as infinite negative energy eigenstates, negative probability and so on. In QFT such problems can be resolved.
