- #106

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Your response, while insightful, seems to presuppose the use of exponential functions (like the complex exponential in the wave function) rather than addressing the possibility of other bases. This feels like a circular argument, as it defines wave functions using these exponentials without considering why this representation is preferred or whether other legitimate alternatives exist. Am I correct in this observation?vanhees71 said:The physical laws look as they look predominantly due to the underlying symmetries of the mathematical model describing it. First of all in Q(F)T one exploits the space-time symmetries of the spacetime model under consideration. In both Newtonian and special-relativistic physics the space-time translations are a symmetry. So any QT model must have the space-time translations as a symmetry, and the corresponding generators of these symmetry transformations are energy and momentum. For a single particle wave function thus you get

$$\psi(x,\alpha)=\exp(-\mathrm{i} \hat{p} \cdot x) \psi(0,\alpha),$$

where ##x=(t,\vec{x})## is the space-time four-vector (which you can also use in non-relativistic physics; I also use natural units ##c=\hbar=1## for convenience).

This suggests to work with energy eigenstates, which have the time dependence ##\propto \exp(-\mathrm{i} E t)##. That's why mode decompositions lead to harmonic time behavior naturally in theories with time-translation invariance.

The same holds for momentum-space representation, where naturally plane-wave solutions ##\propto \exp(+\mathrm{i} \vec{x} \cdot \vec{p}## come up.

For free particles you have both time-translation and space-translation invariance, and you get the above mentioned mode decomposition in terms of energy-momentum eigenvalues.

In addition, if you consider elementary particles, defined by irreducible representations of the space-time symmetry group, you have a energy-momentum relation. In SR it's simply the Casimir operator ##\hat{p} \cdot \hat{p}=\hat{M}^2##. In Newtonian physics it's a bit more complicated since there mass occurs as the non-trivial central charge of the Galilei group's Lie algebra.

I seriously think my questions still remain unanswered, with all due respect.