How to motivate Schrödinger's Equation?

Question

Schrödinger's equation is supposed to be a differential equation for the wave function of a particle. As I currently understand, De Broglie's hypothesis is a hypothesis that for particles there should be some wave function $\Psi$ with wavelength $\lambda$ such that

$$p=\dfrac{h}{\lambda}$$

Where $p$ is the particle's momentum and $h$ is Planck's constant. As I understand, this hypothesis is based on the following: for light, we know there is a wave function, but we also know that it has particle behavior and we know that $p = h/\lambda$ holds, in that case we want the converse for particles.

The hypothesis, however, doesn't tell which differential equation this wave function satisfies. In that case, Schrödinger's equation is the answer to that, saying that the differential equation is

$$i\hbar \dfrac{\partial \Psi}{\partial t}=\left(-\dfrac{\hbar}{2m}\nabla^2+V\right)\Psi$$

Now, I simply can't understand why it would be reasonable to consider that equation. I mean, if we just accept it and deduce results, then we get a feeling that it should be right. But how could one motivate it? I mean, what reasoning lead Schrödinger to that equation in specific?

Is it possible, without knowing further quantum mechanics that follows from the equation, to motivate that the wave function of a particle should satisfy that equation? I've heard that this equation can't be derived. But I believe there is some reasoning behind it.

Answer 1

An explanation for the motivation of Schrödinger's that I have heard is similar and is an extension of your own. Recall energy conservation, where the total energy $E$ is the sum of kinetic $K$ and potential energy $V$. $$K+V=E$$ We then multiply both sides by $\Psi$ and assume it has the wave form $\Psi \sim e^{i(kx-\omega t)}$. Where $k$ is the wave vector, $x$ is a distance, $\omega$ is an angular frequency, and $t$ is time. $$K\Psi+V\Psi=E\Psi$$ Recall that $K=\frac{p^2}{2m}$ and similar to what you said $p=\hbar k$. Also recall that the total energy of a wave is related to the angular frequency via $E=\hbar \omega$. We finally note that $k$ and $\omega$ can be brought out from the functional form of $\Psi$ via spatial and time derivatives respectively. $$ \partial_x \Psi = ik\Psi = i \frac{p}{\hbar} \Psi\\ \partial_t \Psi =-i \omega \Psi $$ Combining it all yields Schrödinger's equation. $$ K\Psi+V\Psi=E\Psi \\ \frac{p^2}{2m} \Psi + V \Psi = \hbar \omega \Psi \\ \frac{1}{2m}(i \hbar \partial_x)^2\Psi+V\Psi=\hbar (i\partial_t)\Psi \\ \frac{-\hbar^2}{2m}\partial^2_x \Psi + V(x)\Psi=i\hbar \partial_t \Psi $$

Answer 2

"Where did we get that (equation) from? Nowhere. It is not possible to derive it from anything you know. It came out of the mind of Schrödinger. ” —Richard Feynman

Now, if you want to see how Schrödinger got to this equation, I would suggest to read his own paper on the subject. In my opinion, and I am only an amateur in science history, the cleanest hint of how he got there is in this paper: http://www.kip.uni-heidelberg.de/matterwaveoptics/teaching/archive/ws07-08/Schroedinger%20-%20Quantisierung%20als%20Eigenwertproblem.pdf (E. Schrödinger Quantisierung als Eigenwertproblem, Erste Mitteilung: Ann. Phys. 79, 361 (1926))

Apologies, but it's in German... More importantly, there is no hand waving here. He basically gives you the solution for the hydrogen problem. At this point it doesn't matter how he guessed it. What matters is that the solution is structurally correct and it predicts the known spectrum with high precision. That's no different from Newton "guessing", that gravity follows a 1/r potential (even though it was probably not Newton who had this hunch first).

Answer 3

A nice motivation is this one: Hamilton had discovered that geometrical optics and classical mechanics were possibly completely equivalent. The optical-mechanical analogy is $$ n(x) = \sqrt{1 - \frac{V(x)}{E}} $$

Where $n(x)$ is the refraction index, and $x$ is a point of $\mathbb{R}^3$. Optical mechanics works fine, as the wavelength is smaller than spatial dimensions of the enviroment in consideration. Since they are equivalent, classical mechanics also works fine as the "wavelength" is smaller than spatial dimensions in consideration. How to fix when this is not the case? The solution to optics is to consider light is a wave, not a particle. Now you don't have geometrical optics, but electromagnetic optics, or "wave" optics.

So, our solution is consider that every particle is a "wave". Since an harmonic wave is the most simple example of a wave, we should start from it: $\Psi(x, t) = \phi(x)e^{-i\omega t}$. So, monocromatic wave equation of number $k$: $\left(\nabla^2 + k^2\right)\phi = 0$. This means: $$ \nabla^2\Psi = -k^2\Psi - \frac{p^2}{\hbar^2}\frac{2m}{2m}\Psi = \frac{2m}{\hbar^2}K\Psi \quad\Longrightarrow\quad K\Psi = \frac{\hbar^2}{2m}\nabla^2\Psi $$

Since $E = K+V$, this means: $$ E\Psi = -\frac{\hbar^2}{2m}\nabla^2\Psi + V(x)\Psi $$

But then: $$ \frac{\partial\Psi}{\partial t} = -i\omega\phi(x)e^{-i\omega t} = -i\omega\Psi(x, t) = \frac{-iE}{\hbar}\Psi \quad\Longrightarrow\quad i\hbar\frac{\partial\Psi}{\partial t} = E\Psi $$

Therefore, we then have Schrodinger equations, by substituting $E\Psi$: $$ i\hbar\frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2\Psi + V(x)\Psi $$