When solving Schrödinger's equation by separation of variables, why is the separation constant taken as the energy? For simplicity, let's take the 1D Schrödinger's equation for a single non-relativistic particle:
$$-\frac{\hbar^2}{2m} \frac{\partial^2 \Psi(x,t)}{\partial x^2} + V(x) \Psi(x,t) = i\hbar \frac{\partial\Psi(x,t)}{\partial t}$$
Applying separation of variables, $\Psi(x,t)=\psi (x)\phi (t)$, we get the time dependent solution.
$$-\frac{\hbar^2}{2m} \frac{\psi'' (x)}{\psi (x)} + V(x)= i\hbar \frac{\phi' (t)}{\phi (t)}=C$$
$$\phi (t)=Ae^{-iCt/\hbar }$$
Here, the separation constant $C$ is taken as the energy of the particle, $E$. I see that this is convenient cause the exponent must be dimensionless. However, do we have further arguments for asserting it?
 A: The reason that we take it to be the energy is that this is closely related to the classical energy, when one follows the standard rules and conventions of "quantizing" the classical quantities.
In a similar manner, you can take any classical quantity $A$, and then from the classical concept of the Hamiltonian (which is the classical energy $E_K+V$ in standard systems) you know that
$$\dot{A} = \{ A, H \} +\frac{ \partial A }{\partial t}$$
with $\{ a,b\}$ the Poisson brackets. Following quantization rules we replace the Poisson brackets with commutation relations (divided by $i\hbar$), and all quantities in operators, to get
$$\frac{d}{dt} \langle A \rangle = -\frac{i}{\hbar}\langle \left[ A, H \right] \rangle + \frac{ \partial }{\partial t}\langle A \rangle$$
and you can see how $H$ here plays the role of the energy. This is just the Heisenberg picture, which is equivalent to the Schroedinger picture, so we can identify the term as a Hamiltonian, which measures the energy.
Another (closely related) way to see this: start from the fundamental relations of QM $[x,p]=i\hbar$, from which you can derive $[x, f(p)]=i\hbar \partial_p f$ and $[f(x),p]=i\hbar \partial_x f$. Then you take Hamilton equations from classical mechanics and quantize them
$$ \frac{d}{dt}\langle x \rangle = \langle \frac{\partial H}{\partial p} \rangle = -\frac{i}{\hbar}\langle \left[x, H\right]\rangle$$
and again you got Heisenberg picture, from which you can derive Schroedinger equation, with the $H$ the energy as it comes from classical mechanics.
A: This is essentially the definition of energy in quantum mechanics -- it is $\hbar$ times the rate of change of phase. That's one of the fundamentally new ideas in quantum mechanics, so it can't be derived from anything you already know classically.
If that's not very satisfying, we can say the same thing in more steps. The energy of an energy eigenstate is defined to be the eigenvalue of the Hamiltonian. And we have
$$H \Psi(x, t) = i \hbar \frac{\partial}{\partial t} \Psi(x, t) = i \frac{\partial}{\partial t} (\psi(x) \phi(t)) = i \hbar \left(- \frac{iC}{\hbar}\right) \Psi(x, t) 
 = C \Psi(x, t).$$
So $C$ is the energy. 
Of course, this just reduces the question to "why is the time-dependent Schrodinger equation true", and the answer is, again, that in quantum mechanics energy is $\hbar$ times the rate of change in phase.
