When you are trying to write clearly about algebraic logic, you have to follow the rules of algebra and logic. Rule one: Respect the definitions. If you are going to substitute scalars for operators, then use a different symbol: \( p,\,q,\,a,\,a^\dag ,\,H \quad \rightarrow \quad \tilde{p},\,\tilde{q},\,\tilde{a},\,\tilde{a}^\dag ,\,\tilde{H}\), so that you may then say \(\tilde{p}\tilde{q} = \tilde{q}\tilde{p}\) without contradiction. A recent poster had a lot of trouble with this in a manuscript he has revised for years, to the point where he doesn't recognize that he is at the mercy of his own statements and definitions. Rule two: Respect the equals sign. Then \(\tilde{H} = \tilde{a}\tilde{a}^\dagger = \tilde{a}^\dagger\tilde{a}\) and doesn't conflict with later statements about \(H\). Rule two is where people mess up and write angry posts about V=RI versus I=V/R. I think it would have been much more transparent to introduce the relationship \(pq -qp = [p,q] = i\) and then show how \(\omega[a,a^\dag] = i\omega[p,q] = -\omega\) while \(\omega \left{a,a^\dag \right} = \omega aa^\dag + \omega a^\dag a = 2 H\). In that sense, \(\frac{H}{\omega}\) is the symmetric part of the product of \(a\) and \(a^\dag\), which we can formalize as half the value of the anticommutator.
Since I am in your debt (for making me look a fool!) let's see if I can answer. Long before the internal combustion engine was invented I studied Chemistry at college. There, one of our first exercises was to derive the Schroedinger eqn from (classical) first principles. It's not an interesting exercise (as best I recall it uses the de Broglie conjecture, Bohr's postulate and some properties of waves; I forget the details) but it turns out you get a sort of "long-hand" version like \((-i\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}+U(x))\psi\) (here I am NOT using \(\hbar =1\). Of course life was longer back then - these days life is too short, there is so much to learn, and a lot of shit has to be taken trust, or "by definition" as you put it). Now the sum in parenthesis is an operator, the Hamiltonian, and it is known that whenever an operator can be decomposed as a sum, then the terms in the summand must also be operators. So given \(H = T +U\), where one may assume this is the sum of kinetc (T) and potential (U) energy operators, I may set \(T = -i\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}\) as the kinetic energy operator. (The potential energy operator will simply be a function of the position coordinates) Now write the Newtonian form \(\text{ke} = \frac{1}{2}mv^2 = \frac{(mv)^2}{2m}\) I easily see that just as \( \text{ke} \to mv\) (Newtonial momentum), the exact same algebra gives \(-i\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \to -i\hbar \frac {\partial}{\partial x\), which justifies calling it the "momentum operator". See how my hands wave? Wow on re-reading, this sounds terribly patronizing, sorry. But I'll leave it, in the unlikely event that some others here might find it interesting
Well historically it went Bohr, then Sommerfeld as a generalization of Bohr to more interesting cases (done by tweaking the action-angle formulation of classical mechanics), then de Broglie used an analogy with light waves to explain Sommerfeld's result. Schrodinger then decided to further the light wave analogy by starting with the free particle case (no potential energy) and writing a particle's position wavefunction as a superposition of momentum waves, each momentum wave obeying the de Broglie relation. If you imagine the wavefunction has only a single momentum component, i.e. \(\psi(\vec{x})=Ae^{i(\vec{p}\cdot \vec{x}-Et)/\hbar}\), then you have the relation \(-i\hbar\vec{\nabla}\psi(\vec{x})=\vec{p}\psi(\vec{x})\), so that's more or less how you justify the momentum operator in position space using Schrodinger's approach. He found that all wavefunctions of this sort could be written as solutions of the equation \(-\frac{\hbar^2}{2m}\nabla^2\psi=i\hbar\frac{ \partial\psi}{\partial t}\). The final step is to recognize that the right hand side of this equation is the total energy, and the left hand side is the kinetic energy, which in this case happen to be identical. In the general case, you'd want to add a potential \(V(\vec{x})\) to the left hand side, and when Schrodinger applied this assumption to the hydrogen atom, he found it worked beautifully with what was already known from earlier models and from experiment. Heisenberg had his own alternative approach to the problem which (I believe) followed up on Sommerfeld's model, but didn't have any relation to de Broglie's work. I have to learn more of the details, but I know it involves a close connection with the Poisson bracket formulation of classical mechanics, and making sure that radiation intensities at different frequencies in energized hydrogen atoms matched with classical EM predictions at large scales.
I didn't find it patronising, don't worry. I've only ever studied quantum mechanics as a mathematics option, so a lot of the history and such were left behind in favour of studying the structure of it all. I do find the early approaches to quantum theory interesting so yours and CptBork's posts were cool. Maybe one day I'll have enough time to sit down and catch up on all the holes in my knowledge.
Just wanted to mention, I forgot to add the mass factor to Schrodinger's equation... And of course in the free particle nonrelativistic case, you have \(E=\frac{p^2}{2m}\), so that's where the initial equation came from, before adding in the potential energy term. It's good to know the history of these subjects because you get a deeper understanding, and then it helps when you apply the ideas to solve real problems. Sometimes when I'm writing an exam or a homework assignment and I forget a step or two, remembering the history of the whole thing helps me recall the idea of what I'm trying to do, and why I'm doing it, and then that's usually enough of a refresher.
Thanks, this is very good advice. I'm doing theoretical physics and GR this year and I took the time over the summer to learn the background to the topics covered. Not just the mathematical background, but the historical background and development and implications of the topics. It's definitely helping out so far.
