The two key ingredients are the Planck equation for the energy of a photon, and the de Broglie wavelength of a matter wave.
The Planck equation for the energy, \(E\) of a photon with frequency \(f\) is:
\[ E = h f \]
The de Broglie wavelength, \(\lambda\), of the matter wave asoociated with a particle with momentum \(p\) is:
\[ \lambda = \frac{h}{p} \]
It's convenient to recast this in terms of the wave number \(k = 2\pi / \lambda\) and the normalized frequency \(\omega = 2\pi f\). (Conceptually, \(k\) represents the number of wave cycles in an interval of length \(2\pi\), and \(\omega\) represents the numbers of cycles in a time interval of length \(2\pi\).) In these terms, the Planck equations and the de Broglie wavelength formulas become respectively \[ E = \hbar \omega \qquad\text{and}\qquad p = \frac{h}{\lambda} = \hbar\frac{2\pi}{\lambda} = \hbar k \] where \(\hbar = h/2\pi\) is the adjusted Planck constant, approximately \(1.05 \times 10^{-34} J\cdot s\).
The Schrödinger Equation is motivated by looking for a wave-type equation which has as a solution the simplest possible candidate for a matter wave, i.e., a plane wave of the form \[ \Psi(x,t) = e^{i(kx - \omega t)} \]
Then compute \[ \frac{\partial\Psi}{\partial t} = -i\omega \Psi = \frac{-iE}{\hbar}\Psi \] and so \[ i\hbar\frac{\partial\Psi}{\partial t} = E\Psi \]
Next we calculate \[ \frac{\partial^2\Psi}{\partial x^2} = -k^2 \Psi \] and then, using the classical (non-relativistic) relationship \[ E = \frac{p^2}{2m} \] together with the de Broglie formula, we obtain \[ \frac{-\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} = \frac{\hbar^2 k^2}{2m} \Psi = \frac{p^2}{2m} \Psi = E\Psi \]
Putting together the two formulas for \(E\Psi\) we obtain the free Schrödinger Equation: \[ i\hbar\frac{\partial\Psi}{\partial t} = E\Psi = \frac{-\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} \]
Finally, since we are interpreting \[ \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \] as the energy operator for a free particle, it makes sense that the energy operator for a particle in potential \(V(x, t)\) is \[ \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x,t) \] and so we reassemble the equations for a particle in a potential as \[ i\hbar\frac{\partial\Psi}{\partial t} = \left[ \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x,t) \right] \Psi \]
Suppose that \(V=V(x)\) does not depend on \(t\). Then the Schrödinger Equation becomes a separable PDE. Write \(\Psi(x,t) = \phi(t)\psi(x)\) and substitute: \[ i\hbar\frac{d\phi(t)}{dt}\psi(x) = - \phi(t)\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} + V(x)\phi(t)\psi(x) \] Rearraging this gives \[ i\hbar\frac{1}{\phi(t)}\frac{d\phi(t)}{dt} = \left(-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} + V(x)\psi(x)\right) \frac{1}{\psi(x)} \] The lhs doesn't depend on \(x\) and the rhs doesn't depend on \(t\) and so both must equal a common constant value, \(E\). From this \[ \frac{d\phi(t)}{dt} = \frac{-iE}{\hbar}\phi(t) \] and so \[ \phi(t) = e^{-iEt/\hbar} \] (the multiplicative constant of integration is absorbed into \(\psi\) for convenience). This leaves the time-independent Schrödinger equation: \[ -\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x) \]