Why spacetime translations don't affect the physics of de Broglie plane waves? I'm studying Merzbacher's Quantum Mechanics. In Chapter 2 Section 1, he "derives" the expression $\psi(x, t)=Ae^{i(kx-\omega t)}$ for the de Broglie plane waves for free particles. Basically he does the following.

*

*He assumes that $\psi$ has to be of the form $\psi(x, t)=\cos(kx-\omega t) + \delta\sin(kx-\omega t)$. (He motivated this from the experimental results of the electron double-slit experiment.)


*He then states that "An arbitrary displacement of $x$ or $t$ should not alter the physical character of these waves, ..., nor should the phase constants of these waves have any physical significance." This leads him to conclude that there exists a (possibly complex) function $a$ such that for any real $\varepsilon$,
\begin{align} 
\cos(kx-\omega t+\varepsilon) +\delta\sin(kx-\omega t+\varepsilon) &= a(\varepsilon)\left[\cos(kx-\omega t)+\delta\sin(kx-\omega t)\right]\text{, and}\\
\left| a(\varepsilon)\right| &= 1.
\end{align}


*Since $a(\varepsilon)^2\neq 1$ for some $\varepsilon$, it follows that $\delta = i$ and hence $\psi(x, t) = e^{i(kx-\omega t)}$.
Question: What justifies that "An arbitrary displacement of $x$ or $t$ should not alter the physical character of these waves"? A free particle changes its position as time progresses and I would expect the physics of its associated de Broglie wave to also change with $t$, for instance.

I know that de Broglie waves for free particle are plane waves, but I want to be able to make sense of how Merzbacher seems to reason it out from more fundamental things.
 A: When you obtain a wave function, your goal is to compute the probability of finding your particle in some region of space (or the probability of having some somemtum) at a given instant of time, so the quantity that carries out the important physical information is $|\psi(x,t)|^2$. When you compute the modulus of the wave function, any aditional complex phase that you introduce does not provide any aditional physical information: $|e^{i\varepsilon}\psi(x,t)|=|\psi(x,t)|$.
In the particular case of the plane wave, the arbitrary displacement of $x$ and $t$ is refered to this arbitrary phase. You can introduce a phase like $\varepsilon=-kx_0+\omega t_0 +\varphi$ and it does not change the properties of the plane wave: $\psi(x,t)=e^{i(k(x-x_0)-\omega(t-t_0)+\varphi)}=e^{i\varepsilon}e^{i(kx-\omega t)}$, where this $\varepsilon$ is the same as yours. You can think about this independence of displacement as an independence of the intial conditions (in this particular case).
A: Not its position changes. Its coordinates. QM systems are invariant to global phase transformations. The measurable value is the phase difference betwern waves.
A: Note that Merzbacher claims that "an arbitrary displacement of $x$ or $t$ should not alter the physical character of these waves". The physical character, not the value of the wave function (or eigenfunction), which of course varies as a function of $x$ and $t$ as you have written. This means that the "form" of the eigenfunctions of the free-particle Hamiltonian $H$ must be independent of time and spacial translations, i.e. it must be a plane wave for all values of space and time.
Why so? Well, "free particle" means exactly that the particle is not subject to any time-dependent or position-dependent potential $V(x,t)$, while it is only driven by its free kinetic energy $T=\frac{p^2}{2m}$. So its "physical character" now, in 20 years or 300 kms away must be the same.
Now let's focus on what is the infamous physical character of the particle. As explained in the other answers, this is captured by the probability density derived through the eigenfunction, i.e. $\lvert \psi(x,t)\rvert^2$. Why must it be uniform over time and space? What is the physical meaning of this? Well, here the strange behavior of quantum mechanics comes into play.
First, the energy values, i.e. the eigenvalues of $H=T$, are determined by the absolute value of the momentum $p$. In other words, $[H,p]=0$. So if we know the value of momentum we have "fixed" the eigenfunction of our free particle. But according to Heisenberg's uncertainty principle, if we know the exact value of momentum we must have infinite uncertainty on the value of position. Therefore each eigenfunction must have a uniform distribution in space. The physical intuition of this might be: we know that our particle is moving with velocity $v=p/m$ (and according to de Broglie, with fixed corresponding frequency and wave number), but we cannot have any information about its position. Consequently, after some time $t$ once again we cannot know anything about the position, and therefore the spatial distribution must be uniform and independent of space translations.
What about the time-translation symmetry? This is easier. We have said there is no time-dependent potential $V(t)$, thus the Hamiltonian $H$ cannot depend on time. Therefore, the physical character of its eigenfunctions must be independent of time as well, so that $\psi(t)=e^{i\alpha(t)}\psi(0)$. The physical intution of this follows the lines of the previous paragraph.
All that said, we cannot describe in quantum mechanics a free particle whose wave function displays a physical character that changes as a function of space and time? Of course we can, for instance by representing it as a wave packet (this is the standard description in a scattering experiment):
$$
\psi(x,t)=\int_{-\infty}^\infty  a(k) e^{i(kx-\omega(k) t)}dk,
$$
i.e. a linear combination of eigenfunctions of the free-particle Hamiltonian. Nevertheless, the wave packet must respect the uncertainty principle: the more it is localized, the more it contains spread values of velocity. In other words, if it has a certain degree of localization, and therefore $\lvert \psi(x,t)\rvert^2$ depends on space and time, it cannot be an eigenfunction of $H$.
