By the simple form of the equation (1) you wrote down, Allan really meant
$$ \frac{\partial}{\partial t} \Psi(x,t)|_{t=0} = \frac{E}{i\hbar}\Psi(x,0) $$
He just used the notation where $t=0$ is substituted from the beginning but he clearly did mean that $\Psi(x)$ is first considered as a general function of $t$, then differentiated, and then we substitute $t=0$.

This equation says that the time derivative of $\Psi(x,t)$ at $t=0$ is proportional to the same wave function. By itself, it does *not* imply that $\Psi(x,t)$ for an arbitrary later $t$ will be given by equation (2): if we only constrain the derivative at one moment $t=0$, the wave function may do whatever it wants at later (or earlier) moments $t$.

However, we may generalize (1) to any moment $t$ which is what you wrote down
$$ \frac{\partial}{\partial t} \Psi(x,t) = \frac{E}{i\hbar}\Psi(x,t) $$
and this equation *does* imply (2). If the $t$-derivative of $\Psi(x,t)$ is proportional to the same $\Psi(x,t)$, then $\Psi(x,t)$ and $\Psi(x,t')$ are proportional to each other for each $t,t'$. That implies that $\Psi(x,t)$ must factorize to $\Psi(x)f(t)$ and the function $t$ is the simple complex exponential to solve the equation with the right coefficient.

Because you are more or less writing the same things, I find it plausible that you are not missing anything.

Well, to put it shortly, it's just because $L^2$ and $L_z$ are two observables that have no $r$ dependence. Since the kinetic terms of $H$ can be written as functions of $r$ and $L^2$, and the potential depends only on $r$, it is clear that the Hamiltonian commutes with the angular momenta. As such, it makes sense to write down $Y$ in terms of these two.

But if we'd like to go a little deeper, it is a bit related to representation theory. The fact that $L^2$ and $L_z$ form a complete description of vector rotations (they represent the Lie algebra $\mathfrak{so}(3)$ of the Lie group $\mathrm{SO}(3)$) and therefore are the most natural way to express the angular dependence of a rotation-invariant space in terms of orthonormalizable functions.

After reading @ZeroTheHero 's comment, i figured it would be instructive to provide some intuition as to why angular momentum does not depend on radius. Remember the classical definition

$$\mathbf L=\mathbf r\times\mathbf p$$

Let's then give and estimate of $L$. Suppose an atom or something has some characteristic radius $R$, and the electron moves with some characteristic momentum $p$. Then we estimate

$$L\sim Rp$$
But the de Broigle relation gives us an estimate of momentum in terms of the characteristic length

$$p\sim\frac{h}{R}$$

Where $h$ is Planck's constant. As such, out estimate for angular momentum amounts to

$$L\sim h,$$

Which has no explicit dependence on characteristic lengths of the system. As such it is reasonable to say it has no $r$ dependence, even though I've not given a rigorous proof.

## Best Answer

It has to do with the Laplacian $(\nabla^2)$. When trying a separable solution

$$\psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)$$

You will get two ODE's, a radial and an angular equation. They must be equal to separation constants. Turns out the angular part is the $L^2$ operator. You can rewrite $$H \psi = E \psi$$ as $$\left( \frac{-\hbar^2}{2 m} \nabla^2 + V(r) \right) \psi = E \psi $$ But $$\nabla^2 = \left(\frac{1}{r^2}\partial_r ( r^2 \partial_r ) - \frac{1}{\sin{\theta}} \partial_{\theta} (\sin{\theta}\, \partial_{\theta}) + \frac{1}{\sin^2{\theta}} \partial_{\phi \phi} \right) = \left(\frac{1}{r^2}\partial_r ( r^2 \partial_r ) + \frac{L^2}{\hbar^2} \right) $$ Now by inserting the above relation for $\psi$ into our Schrodinger equation and finding separation constants for the ODE's you will get a second order harmonic oscillator ODE for $\Phi(\phi)$ which proves that you will have integer values of m and the associated Legendre equation that will yield $P^m_l (\cos{\theta})$. Combined these will give you the spherical harmonics which must be solutions to this angular equation, and conversely eigenfunctions of $L^2$ with eigenvalues $l(l+1)$