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To get a better intuition of the schroedinger equation I am trying to work with it in the Madelung equation form instead. If I am reading them correctly, these are the Madelung equations for a 1D particle (interpret $\psi=r\cdot e^{iS}$): \begin{equation} \begin{aligned} \frac{\partial r}{\partial t}&=-\frac{1}{2m}\left(r\frac{\partial^2 S}{\partial x^2}+2\frac{\partial r}{\partial x}\cdot\frac {\partial S}{\partial x}\right)\\ \frac{\partial S}{\partial t}&=-\frac {1 }{2m}\left(\left|\frac{\partial S}{\partial x}\right|^2+V-\hbar^2\frac{\partial^2 r}{\partial x^2}\right) \end{aligned} \end{equation}

Here's why it confuses me: Define $k=\frac {\partial S}{\partial x}$. Then for a standing wave (stationary wave for a free particle), we have $\frac {\partial S}{\partial t}=-\frac{k^2}{2m}$.

This would suggest a wave function of $\Psi(x,t)=e^{i(kx-\frac{k^2}{2m}t)}$. But instead the formula for a standing wave given in Griffith's Intro to QM is: $\Psi(x,t)=e^{i(kx-\frac{\hbar k^2}{2m}t)}$. What explains the missing $\hbar$?

Note: I got the Madelung's equations from this SE question.

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  • $\begingroup$ Check the dimensions (missing ℏs). Your S is dimensionless where ℏ was absorbed in it. $\endgroup$ Commented Apr 29 at 16:02
  • $\begingroup$ @CosmasZachos, I don't understand. $S$ and $r$ are DEFINED by $\Psi=r\cdot e^{iS}$ (both by me and by the linked post), so $S=kx-\frac {hk^2}{2m}t$, which implies $\frac {\partial S}{\partial t}=-\frac {\hbar k^2}{2m}$, which doesn't correspond to the Madelung equations, because those imply $\frac {\partial S}{\partial t}=-\frac {k^2}{2m}$ $\endgroup$
    – user56834
    Commented Apr 29 at 18:17
  • $\begingroup$ You failed to check the dimensional consistency of the linked post! When one redimensionalizes, one checks the dimensions. $\endgroup$ Commented Apr 29 at 19:16
  • $\begingroup$ @CosmasZachos, I am not sure what "redimensionalization" is being done there? Don't they define S and r exactly the same as I do? (except they write $r=\sqrt \rho$). Sorry, I am new to this stuff. $\endgroup$
    – user56834
    Commented Apr 29 at 19:19
  • $\begingroup$ Check my answer and WP. The price you pay for a simple dimensionless S which allows you to set ℏ=1, (non-dimensionalization), is that when you reintroduce it in the answers, as you erroneously did, you need complete dimensional consistency. Most physicists do this as they breathe, but you and your answer didn't. $\endgroup$ Commented Apr 29 at 19:21

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Did you verify your TDSE, dimensionally? Your S, here, atypically, is an action over ℏ, hence dimensionless, $\psi\equiv R e^{iS}$; it produces the well-known continuity and quantum Hamilton-Jacobi equations (in a small normalization difference from WP, as commented), through plugging in, $$ 𝑒^{βˆ’π‘–π‘†}(iβ„βˆ‚_t+ℏ^2βˆ‡^2/2π‘šβˆ’π‘‰)(𝑅𝑒^{𝑖𝑆})=0, \leadsto \\ \frac{\partial R}{\partial t} = -\frac{\hbar}{2m} \left[ R \nabla^2 S + 2 \nabla R \cdot \nabla S \right],\\ \hbar \frac{\partial S}{\partial t} = - \left[\hbar^2 \frac{\left|\nabla S\right|^2}{2m} + V - \frac{\hbar^2}{2m} \frac{\nabla^2 R}{R} \right], $$ the last term in the second equation being the celebrated quantum potential, with its requisite ℏ. Check these are dimensionally consistent, now. $\hbar\nabla S$ has dimensions of momentum.

Repeat your plane wave consistency check for this. You had simply dropped ℏs.

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  • $\begingroup$ Ok! I took the equations from the link in my post (I'm not sure how to derive them myself). It seems that post made a mistake then. $\endgroup$
    – user56834
    Commented Apr 29 at 19:16
  • $\begingroup$ Sigh... it is a trivial and easily reversible mistake, and I have been hinting at it since the beginning... Isn't the WP article evidently superior? $\endgroup$ Commented Apr 29 at 19:22
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    $\begingroup$ To be clear, I haven't dropped any $\hbar$ msyelf, because I just copied the Madelung equations, without knowing how they were derived. I will derive them myself now that I know how, thanks! $\endgroup$
    – user56834
    Commented Apr 29 at 19:43

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