It does seem to be possible to make sense of \psi(x,t)=\psi'(x',t') using some relativistic quantum field theory, but is that really what you're looking for?
The equation you posted looks like an attempt to find a relativistic version of the Schrödinger equation by making the substitutions
E\rightarrow i\hbar\frac d{dt}, p\rightarrow -i\hbar\frac d{dx}[/itex]<br />
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in the relativistic formula E^2=p^2c^2+m^2c^4 (with m=0) instead of in the non-relativistic formula E=p^2/2m. If you do it in the non-relativistic equation, the result is the Schrödinger eqauation. It might seem plausible that you'd get a relativistic wave equation if you start with the relativistic expression for the energy instead.<br />
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However, this doesn't work. The "psi" isn't going to be a wave function. It's a classical field, which needs to be "quantized" to a quantum field before we can do anything with it.<br />
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Here's how I would make sense of \psi(x)=\psi&#039;(x&#039;) (where x now represents all the coordinates including time), if we assume that the "wave function" represents a one-particle state in the quantum field theory of a (not necessarily massless) non-interacting scalar field:<br />
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\psi(x)=\langle 0|\phi(x)|\psi\rangle=\langle 0|U(\Lambda)^\dagger U(\Lambda)\phi(x)U(\Lambda)^\dagger U(\Lambda)|\psi\rangle=\langle 0|\phi(\Lambda x)|\psi&#039;\rangle=\psi&#039;(x&#039;)