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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.

user56834
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1 Answers1

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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, $$ ^{−}(iℏ∂_t+ℏ^2∇^2/2−)(^{})=0, \leadsto \\ \frac{\partial R}{\partial t} = -\frac{\hbar}{2m} \left[ R \nabla^2 S + 2 \nabla R \cdot \nabla S \right],\\ \hbar \frac{\partial S}{\partial t} = - \left[\hbar^2 \frac{\left|\nabla S\right|^2}{2m} + V - \frac{\hbar^2}{2m} \frac{\nabla^2 R}{R} \right], $$ 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.

Cosmas Zachos
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