If it's so easy to derive the Schrodinger equation, you won't have a problem showing me, now will you?
No, none whatever, but you won't read it, so why should I bother?
Oh shittikens......
First the Schrodinger equation is an energy eigenfunction on some state vector/function $$\psi$$.We seek a Hamiltonian operator H = K + U, where K is kinetic energy and U is potential energy. One writes (in skeletal form) $$H\psi = \epsilon |\psi\rangle$$ Note that H is an operator, and that whenever an operator can be written as the sum of terms, these terms must themselves be operators
Classically momentum for an object of mass $$m$$ is defined as $$p=mv$$ where $$v$$ is velocity of our particle
Classically kinetic energy is defined as $$K=\displaystyle \frac{1}{2}mv^2$$
Without even thinking, I can combine these 2 equalities as $$\displaystyle K= \frac{(mv)^2}{2m}= \frac {p^2}{2m}$$
Now turn to the Bohr conjecture for a bound "orbiting" particle with mass $$m$$ and an orbit radius of $$r$$.
This states that the allowable angular momenta are given by $$\displaystyle mvr=n\frac{h}{2 \pi}$$. Loosely speaking, since we don't believe in "orbits" the radius is irrelevant, and therefore (from a previous post of mine) so is the integer$$n$$
Thus $$\displaystyle mv = p = i \frac{h}{2 \pi}$$ where I have included the imaginary factor as we are doing quantum mechanics (actually there is more to it than that!).
So putting this all together one gets for the kinetic energy operator $$\displaystyle K=-\frac{h^2}{8 \pi^2 m}$$ And since as a convenient shorthand we have that $$\displaystyle \frac{h}{2 \pi} \equiv \hbar$$ we arrive at $$\displaystyle K = -\frac{\hbar^2}{2m}$$
Since this is second order, I seek a "direction". I look no further than $$\displaystyle\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$$ which we recognize as the Laplacian $$\nabla^2$$
I now seek the potential energy operator U. Since this depends only on position {x, y, z} I end up with
$$\displaystyle [-\frac{\hbar^2}{2m} \nabla^2 + U(x,y,z)]\psi = \epsilon |\psi \rangle$$
