Is my derivation of the Schrödinger equation valid? We are in natural units. Let's suppose we have a wavefunction $\phi(t,\vec{x}) \in C^{\infty}(\mathbb{R}^{1,3})$ (because the same derivation can be made for the impulsion operator). The operation of time translation is  given by $t \rightarrow t-t'$. Now one wants to express the time translation like this:
\begin{equation}
\phi(t-t',\vec{x})=(T_{t'}\times\phi)(t,\vec{x}).
\end{equation}
Since we have just a translation we thus have the following equality:
\begin{equation}
\frac{\partial}{\partial t'}\phi(t-t',\vec{x})=-\frac{\partial}{\partial t}\phi(t-t',\vec{x}).
\end{equation}
So, the operator of time translation can be expressed as: $T_{t'} = e^{-t'\partial_t}$. Now let's take a look at the Fourier transform of our wave-function $\phi(t,\vec{x})\equiv \phi(x^\sigma)$:
\begin{equation}
\varphi(k^{\sigma})= \int d^4x\,\phi(x^\sigma)e^{ik_\mu x^\mu}.
\end{equation}
Since this integral is invariant under the change $t \rightarrow t-t'$ in the integrand, one has the following equalities:
\begin{equation}
\varphi(k^\sigma)=\int d^4x\,\phi(t-t',\vec{x})e^{-ik_0t'}e^{k_\mu x^\mu}=\int d^4x\,e^{-t'\partial_t}e^{-ik_0t'}\phi(x^\sigma)e^{k_\mu x^\mu}.
\end{equation}
Expanding the exponential of the differential operator as polynomial in $t'$ gives:
\begin{equation}
\sum_{n=0}^\infty \frac{(-t')^n}{n!} \frac{\partial^n}{\partial t^n}\phi(x^\sigma)=\sum_{n=0}^\infty \frac{(t')^n}{n!}(ik_0)^n\phi(x^\sigma).
\end{equation}
Finally the equality being term by term, and $k_0=E$ in natural units, we have the Schrödinger equation:
\begin{equation}
i\frac{\partial}{\partial t} \phi(x^\sigma)=E\,\phi(x^\sigma).
\end{equation}
Is this derivation of the Schrödinger equation valid? I heard that there is no mathematical proof of this equation so maybe I am wrong in my derivtaion. Maybe stating that the wave function has to be infinite-differentiable is a too strong condition?
 A: $k_0=E$ (Planck's famous postulate), and linearity / the superposition principle, are the key physical assumptions in your argument that (as far as we know today) can't be derived from anything else -- they are just facts about the Universe. Given these assumptions, it is possible to give a plausibility argument that the Schrodinger equation implements Planck's postulate using a linear differential equation. This is essentially what you have done.
A: No, this is not a valid derivation for a couple of reasons:

*

*Not the least of which is that the equation you arrive at is not the Schrodinger equation. The Schrodinger equation reads $\partial_t\psi=-iH\psi$. You have obtained $\partial_t\psi=-iE\psi$ which is just not the same as the former unless the wavefunction happens to be an energy eigenfunction.

*Another main reason is that you are smuggling in $k_0=E$ without a justification. By pure Fourier analysis, $k_0$ is just the frequency mode, it knows nothing about energy. So, you have nowhere to go from $k_0$ to energy unless you postulate how the $k_0$ is associated with the energy.

More broadly speaking, you are correct in having heard that one cannot derive the Schrodinger equation from pure mathematics. It is a physical principle. However, one can derive it after having postulated something that is equivalent in its physical content to the Schrodinger equation. For example, a really neat way of seeing how the Schrodinger equation comes about from the elegant principles of symmetry is to take the postulates of Wigner's theorem (the key postulate here is the symmetry of translations in time) and arrive at the conclusion that the time evolution of the state vector must be unitary and linear. We thus write $\psi(t) = U(t)\psi(0)$ where $U(t)$ is the unitary operator. Now, we define the Hamiltonian operator such that $U(t)=\exp(-iHt)$ and we can rewrite  $\psi(t) = U(t)\psi(0)$ in terms of the Hamiltonian in the differential form as $\partial_t\psi=-iH\psi$.
