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By 1924 it was well observed that matter (as well as light) has wave-particle duality (later named quantum), and the wavelength-momentum-energy relation of quanta $$\lambda=\frac{h}{p}\;\;\longleftrightarrow\;\;p=\hbar k\;\;\longleftrightarrow\;\;E=\hbar\omega$$ had been hypothesized (and experimentally validated) by de Broglie. Let us model a quantum as a periodic function $\Psi$ of spacial cooridnates and time, which can be expressed in the form of Fourier series $$\Psi=\sum_{n\in\mathbb{Z}}A_n\psi_n$$ where $$\psi_n=e^{i\left(\mathbf{k}_n\cdot\mathbf{x}-\omega_n t\right)}.$$ We want to get the information about the momentum and energy of such quantum, for now only its basis $\psi_n$. Consider the followings $$\begin{align*} \nabla^2\psi_n&=-k^2\psi_n& &\longleftrightarrow& -\hbar^2\nabla^2\psi_n&=p_n^2\psi_n\\ \frac{\partial}{\partial t}\psi_n&=-i\omega_n\psi_n& &\longleftrightarrow& i\hbar\frac{\partial}{\partial t}\psi_n&=E_n\psi_n. \end{align*}$$ which can be interpreted as eigenvalue problems with the operators $$\hat{p^2_n}:=-\hbar^2\nabla^2\quad\text{and}\quad\hat{E_n}:=i\hbar\frac{\partial}{\partial t}.$$ The total energy of the quantum is given $$\sum_{n\in\mathbb{Z}}E_n=\frac{1}{2m}\sum_{n\in\mathbb{Z}}p_n^2+V$$ and by superposition principle we can write down the following equation $$\hat{E_n}\psi_n=\frac{1}{2m}\hat{p_n^2}\psi_n+\hat{V}\psi_n$$ or equally $$i\hbar\frac{\partial}{\partial t}\psi_n=-\frac{\hbar^2}{2m}\nabla^2\psi_n+\hat{V}\psi_n.$$

I have never learned or seen this derivation. Other derivations were most of the time too advanced to me to follow (the math) or the equation itself was taken for granted in the first place. My question is that if this derivation makes sense and I'm doing right.

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    $\begingroup$ Where are you getting $E=\hbar \omega$ from? What is $\omega$ the frequency of? I think this is where you've loaded all the assumptions (which of course, you have to make, since the Schrodinger equation can't be be derived). $\endgroup$ – jacob1729 May 21 at 17:55
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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/17477/2451 , physics.stackexchange.com/q/16812/2451 , physics.stackexchange.com/q/220697/2451 and links therein. $\endgroup$ – Qmechanic May 21 at 17:58
  • $\begingroup$ Apart from relations like $p=\hbar k$ and $E=\hbar \omega$ there is one key assumption which enters in this "derivation", that is that particles behaves like waves. This fact can only be found by experiments. $\endgroup$ – Frederic Thomas May 21 at 18:01
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    $\begingroup$ Why in the world did someone downvote this? If you think the derivation is wrong, then that's an answer, not a reason to downvote. Please don't treat new users with this kind of hostility. $\endgroup$ – Ben Crowell May 21 at 19:08
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    $\begingroup$ It's perfectly fine, and probably very similar to how Schrodinger first came up with it! Of course, the physical content of the things you've assumed, explicitly and implicitly, is basically equivalent to what the Schrodinger equation says. $\endgroup$ – knzhou May 22 at 20:22
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Sure, this is valid, subject to some hidden assumptions. The issue would be that those assumptions haven't been explicitly stated, and it isn't trivial to figure out the complete list of hidden assumptions. Some hidden assumptions:

  1. That $E=\hbar \omega$. This is plausible on dimensional grounds and because of the relativistic analogy energy:momentum::time:space, but it gets cloudy when you remember that this is a derivation of the nonrelativistic Schrodinger equation. Einstein had already published this relation as an empirical one based on stuff like the photoelectric effect, but it wasn't obvious at the time how that fit into the structure of physics.

  2. That wavefunctions live in the field of complex numbers, not, e.g., in the reals or the quaternions. This is basically needed if you want to get conservation of probability, but that would not be obvious if you hadn't already tried, e.g., making a real Schrodinger equation and seeing conservation of probability fail.

  3. That the relevant degree of freedom is position, as opposed to something like spin.

  4. That position is an observable. This is false in quantum field theory, and even in nonrelativistic field theory there is the issue that we don't have eigenstates of position unless we allow things like Dirac deltas.

  5. That it's OK for phase and normalization to be unobservable.

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  • $\begingroup$ I think this answer would be even better if you added something to the effect of knzhou's comment that these assumptions are essentially equivalent to the usual Schrodinger equation. As such, the derivation is perhaps more of a 'motivation'. +1 all the same. $\endgroup$ – jacob1729 May 22 at 21:57
  • $\begingroup$ @jacob1729 Your argument that "assumptions are essentially equivalent to equations" can be applied to every scientific equation, because an equation is basically underlying assumptions (or hypotheses, or postulates, tomato tomahto) written in the language of mathematics. Coming up to an equation from known assumptions that governs broader or new object than assumptions do is enough to be called a derivation. In this case the de Broglie relations would be the assumption and wave function is what the derived equation governs. All this, I think, is only a matter of wording afterall. $\endgroup$ – user575201 May 22 at 23:09

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