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

*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?

## 1 comment:

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

cset 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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