Line element in relativistic quantum mechanics?

Question

Background

I noticed the following. Let us take a look at the massless Klien Gordon equation (in relativistic quantum mechanics) along a paticular axis  $x$:

$$  - \hbar^2\partial_t^2 \psi =  - \hbar^2 c^2 \partial_x^2 \psi$$

Writing in terms of the translation operator $\hat T$ and the unitary operator $\hat U$ and using the notation that $\delta k$ is a small shift in $k$ and cancelling factors:

$$\frac {2I - U(\delta t) - U(-\delta t)}{c^2 \delta t^2} | \psi \rangle =  \frac {2I - T(\delta x) - T(-\delta x)}{\delta x^2} | \psi \rangle $$

Taking the inner product and taking a ratio:

$$ \frac{\langle 2I - U(\delta t) - U(-\delta t) \rangle}{\langle 2I - T(\delta x) - T(-\delta x) \rangle } = \frac{c^2 \delta t^2}{\delta x^2} $$

Substracting $1$ both sides:

$$ \frac{ \langle - U(\delta t) - U(-\delta t) + T(\delta x) + T(-\delta x)  \rangle}{\langle 2I - T(\delta x) - T(-\delta x) \rangle } = \frac{c^2 \delta t^2 - \delta x^2}{\delta x^2} $$

Multplying $\delta x^2$ both sides:

$$ 0 = {c^2 \delta t^2 - \delta x^2} - {\delta x^2} \frac{ \langle - U(\delta t) - U(-\delta t) + T(\delta x) + T(-\delta x)  \rangle}{\langle 2I - T(\delta x) - T(-\delta x) \rangle } $$

This can only classically become:

$$ \delta s^2 = {c^2 \delta t^2 - \delta x^2} = 0$$

if one postulates:

$$\frac{\delta x}{\delta t} = c$$

Question

Is there a way to prove

$$\frac{\delta x}{\delta t} = c$$

 this without postulating it?

Answer 1

If you have the equation

$$(\partial_0^2-\partial_1^2)\Psi=0$$ then for any twice differentiable function $f$ a solution is $\Psi(x^0,x^1):=f(x^0-x^1)$. This is a constant shape (given by $f$) moving along the $x^1$-axis at constant speed $c$. So if I consider the same solution at a different time, $x^0+\delta x^0$, this implies $\delta x^1=\delta x^0$ for looking at the same part of the moving shape.

Introducing the operators $U$ and $T$ you write a discretisation of the partial derivatives of the KG equation. The "$=$"-sign between the l.h.s. and the r.h.s. of this discretisation implies a relation between $c\delta t$ and $\delta x$, or $\delta x^0$ and $\delta x^1$ in my above notation, which quantities are classical throughout your entire argument. As $|\Psi\rangle$ is to correspond to a solution of the KG equation, this relation is the one mentioned above, $\delta x^1=\delta x^0$.

Put somewhat differently, with the above relation between $\Psi$ and $f$ we have $$\Psi(x^0,x^1+\delta)=f(x^0-(x^1+\delta))=f((x^0-\delta)-x^1)=\Psi(x^0-\delta,x^1)$$ A shift from $x^0-\delta$ to $x^0$ is accompanied by a shift from $x^1$ to $x^1+\delta$.