This site documents detailed derivations and uncommon findings of different topics in classical, quantum, statistical, computational, astronomical physics. Feel free to check out the pdf compiled with latex in Kei's repository for online notes.
Relate diffraction phenomenon to quantum machanics. Practical example is Davisson Germer Experiment which confirms particle-wave duality.
In the discussion below, denote \(\left(\mathbf{E}\left(\mathbf{r},t\right), \mathbf{B}\left(\mathbf{r},t\right)\right)\) as EM waves.
Very often, we regard electromagnetic waves as ideal wave, which is however inaccurate since the energy of real wave must not diverge. The approximation, however, is enough for us to give reasonable conclusions. So in the discussion below, we will use the approximation.
Using the approximation, an ideal wave is written as
$$ \exp{\left\{i \left(\mathbf{k}\cdot\mathbf{r}-\omega t\right)\right\}}$$
From de Broglie relations,
$$ E = \hbar \omega$$,
$$ \mathbf{p} = \hbar \mathbf{k} $$
The ideal wave can be rewritten as
$$ \exp{\left\{i\mathbf{p}\cdot\mathbf{r}/\hbar - i Et/\hbar\right\}}$$
Knowing that electromagnetic waves are solutions to the wave equation s.t.
$$ \nabla^2 \mathbf{E} - \frac{1}{c^2} \partial_t^2 \mathbf{E} = 0$$
If we were to simplify the expression, we can always rewrite electromagnetic waves into the following form
\[\mathbf{E} = \vec{\mathcal{E}}\psi\]
where \(\vec{\mathcal{E}}\) is the polarization of wave, and \(\psi\) is the scalar wavefunction.
If we put the expression of ideal wave into the wave equation, it can be seen obvious that
$$ \nabla \psi = \mathbf{p} \left(\frac{i}{\hbar}\right)\psi$$.
$$ \partial_t \psi = \frac{-i}{\hbar}E\psi$$
It is thus derived informally that the two operators
$$ \hat{E} = i \hbar \partial_t, \;\;\hat{p} = \frac{\hbar}{i}\nabla \psi$$.
For light, we already know \(E=pc\). So inversely, we derive can always restore the wave-equation by considering $$ \hat{E}^2 \psi = \hat{p^2} c^2\psi$$. For non-relativistic particle with mass \(m\) in potential field \(V\left(\mathbf{r},t\right)\), the energy is $$ E = \frac{p^2}{2m} + V$$ If we treat the energy and momentum as operators, we can informally derive Schrodinger Equation. $$ i\hbar \partial_t \psi = \frac{1}{m} \left(\frac{\hbar}{i}\nabla\right)^2 \psi + V\psi$$ Disclaimer: This is NOT the original derivation. The formal derivation cannot be donw without assumptions.
Lectures on Quantum Mechanics by You-Quan Li