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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.

user575201
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    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). – jacob1729 May 21 '19 at 17:55
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    Possible duplicates: https://physics.stackexchange.com/q/17477/2451 , https://physics.stackexchange.com/q/16812/2451 , https://physics.stackexchange.com/q/220697/2451 and links therein. – Qmechanic May 21 '19 at 17:58
  • 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. – Frederic Thomas May 21 '19 at 18:01
  • This is somewhat historically correct as to how some early attempts were made at deriving the Schrodinger equation. But physically, this is not really accurate or appropriate. –  May 21 '19 at 18:21
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    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. –  May 21 '19 at 19:08
  • @BenCrowell: Please don't treat new users with this kind of hostility I don't think that is happening here (and a discussion of it belongs in Meta, not here). We can't wrap new contributors in cotton wool either. People should be free to vote how they see fit. It's a subjective process anyway. – Gert May 21 '19 at 19:38
  • @jacob1729 de Broglie's assumption was that that matter was not only satisfying $\lambda=h/p$ but also $E=hf$. Nevertheless I revised my post. – user575201 May 22 '19 at 20:11
  • @Qmechanic Can you be more specific in what way those questions are possible duplicates? – user575201 May 22 '19 at 20:14
  • @FeynmansOutforGrumpyCat I'd be very much appreciated to know what are not accurate or appropriate. – user575201 May 22 '19 at 20:16
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    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. – knzhou May 22 '19 at 20:22
  • @knzhou I agree that this is how Schrodinger came up with the equation but it isn't appropriate to treat the energy operator as the time-derivative, right? Also, the way the OP has stated it (and the way Schrodinger derived it) is based on intuitions from the photon, I think that itself is also problematic given the inherently relativistic nature of the photon. –  May 23 '19 at 03:48
  • @user575201 The MITOCW lectures on QM I course has some discussion on this kind of motivation to "deriving" the Schrodinger equation--around the fifth lecture but I might be pretty off in estimating the lecture index. If you haven't, you might want to check it out. –  May 23 '19 at 03:53
  • @FeynmansOutforGrumpyCat Regarding your response to knzhous, that is the definition of the energy operator. What you are trying to say, I believe, is that the Hamiltonian operator $\hat{H}$ should be time-independent in the Schrödinger picture. – user575201 May 23 '19 at 04:29

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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.

  • 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. – jacob1729 May 22 '19 at 21:57
  • @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. – user575201 May 22 '19 at 23:09