where $q(t)$ is the position (measured in $x$) of the particle at time $t$, and the left hand side is evaluated at $x=q$. (I tend to write $\d$ for $\partial$ --- my TeX files always start with \def\d\partial.) This equation can be written more succinctly as $a = F/m$.
This equation is symmetric under what I will call the common relativistic (or Galileo; "common" as an antonym to "special", "relativitistic" because this is a kind of "relativity") transformation
Einstein (most likely working in close consultation with Maric) is, of course, famous for modifying Newton's equation to instead remain invariant under the special relativistic (or Lorentz) transform:
$$ t \mapsto \frac{t + ux/c^2}{\sqrt{1 - u^2/c^2}} $$
What I'm wondering about in this entry, however, is a different equation of one-dimensional motion, this one written down by Schrodinger:
where now we must reinterpret particles. No longer is $q(t)$ "the location of our particle at time $t$"; instead, $|\Psi(t,x)|^2$ is the "probability of measuring our particle as having position $x$ at time $t$". (I retain the quotes only because really such a probability is the infinitesimal $\Psi dx$, and the probability of measuring the particle to be between $x = a$ and $x = b$ is the appropriate integral. In this discussion, I'm going to ignore the problems of normalization.)
In the case when $V(x)$ is the constant $V_0$, we can easily solve this equation:
Consider one particular eigenstate $e^{i(wt+kx)}$, and transform it a la "common" relativity. We'd get an adjusted state $e^{i((w+ku)t+kx)}$. Does this satisfy Schrodinger's equation for a constant potential? Well, yes. The wave we get is red-shifted, and we now need
This is not, actually, surprising: Replacing $\Psi(t,x)$ with $\Psi(t,x+ut)$ in Schrodinger's equation affects the time derivative as well as the space derivative: Writing $\Phi(t,x) = \Psi(t,x+ut)$ gives $\d\Phi/\d t = \d\Psi/\d t + u\d\Psi/\d x$. On the other hand, the right-hand side $\d^2/\d x^2$ does not pick up such an extra term. Only in the high-energy ($k\to 0$) limit does the extra $-ku\hbar$ term vanish, but never before the $w ~ k^2$ term does.
(Incidentally, lest ye think Schrodinger's equation is actually special-relativisticly invariant: checking the relationship between $k$ and $w$ after a Lorentz transform gives a fourth-power in $k$ on the right and only a squared $k$ on the left. And the high-energy limit is equally bad.)
So what's up? Why do we tout Schrodinger's equation when it can't even give the right formulas under the patently classical common-relativistic transform? And what's the popper formula? I take it on good authority that the Dirac Equation is special-relativistically invariant (interestingly, and unlike Einstein's special relativity equations of motion, Dirac's wave function does not transform as a vector under rotations in 3-space, but as a mystical object called a "spinor"). But I'd expect there to be a common-relativistic equation of motion spit out by various more basic Quantum-Mechanics axioms. Or is it that Schrodinger actually correct and I'm just being daft?
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Of course, now that I've taken a first course in QFT, I can partly answer the question: there are very coherent special-relativistic EOMs. In fact, the most basic classical field theory --- the continuous limit of harmonically-coupled harmonic oscillators (arranged isotropically, and coupled only to adjacent oscillators) --- is special-relativist with c set by the "spring strength" and "spring mass" ratio. A classical field, for all intents and purposes, is a first-quantized point particle. Second-quantize, and you still preserve special relativity.
But the question still remains: is there a common-relativistic theory? We could take $c\to\infty$ in QFT; the discrete springy model picks up infinities, but the continuous limit doesn't have problems. We end up with a theory of classical fields, but we can't Fourier-transform, because the waves move at the speed of light, and without transforming it's hard to use the same tricks to second-quantize.
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