Relativistic Invariance

The basic equation of motion of a Newtonian particle in one dimension acting under a potential energy $V(x)$ is

$$ \frac{\d^2}{\d t^2}q(t) = \frac{1}{m}\frac{\d}{\d x}V(x) $$
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 $$ x \mapsto x + ut $$ describing the (Newtonian) coordinate change between inertial frames at relative constant velocity $u$ to each other (the coordinate $t$ remains unchanged). To wit: Under such a transformation, the derivative in the left-hand-side remains untouched, and is now evaluated at $x = q(t) - ut$; the right hand side sees a replacement of $q(t) \mapsto q(t) - ut$ and the $ut$ drops out in the second derivative, so any motion satisfying Newton's equation in one (inertial) frame will satisfy it in all (inertial) frames. In general, such a transformation may introduce a $t$-dependence in $V$; this doesn't bother us though, except for raising the specter that energy might no longer be conserved. If $V$ is a constant in $x$, this doesn't happen, of course, and solutions $q(t) = vt$$ remain solutions, albeit with different $v$. For my purposes, all I actually care about is the equation of motion in the absence of forces (when $\d V / \d x = 0$).

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:

$$ x \mapsto \frac{x + ut}{\sqrt{1 - u^2/c^2}} $$
$$ t \mapsto \frac{t + ux/c^2}{\sqrt{1 - u^2/c^2}} $$

What I'm wondering about in this entry, however, is a different equation of one-dimensional motion, this one written down by Schrodinger:

$$ i\hbar \frac{\d}{\d t} \Psi(t,x) = \frac{-\hbar}{2m} \frac{\d^2}{\d x^2} \Psi(t,x) + V(x)\Psi(t,x) $$
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:$$ \Psi(t,x) = \sum a_j e^{i(w_j t + k_j x)} $$where $-\hbar w_j = E_j = \frac{\hbar^2 k^2}{2m} + V_0$. ($E_j$ is the "eigenenergy" for the $j$th summand; $w_j$ and $k_j$ are the corresponding frequency and wave number, respectively.) The universal constant $V_0$ cannot be measured; changing it corresponds to multiplying the entire wave function by some $e^{i\theta}$. I.e. the physical system has this one real degree of degeneracy.

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 $$ -\hbar(w+ku) = \frac{\hbar^2 k^2}{2m} + V_0 $$i.e.$$ V_0 = -\hbar(w+ku) - \frac{\hbar^2 k^2}{2m} = -\hbar k u $$But the transformed $V_0$ depends on $k$, so a superposition of eigenstates does not transform to a state satisfying Schrodinger's equation.

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?