## 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?