# Am I reading the Madelung equations for a 1D quantum particle right?

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{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}

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.

• Check the dimensions (missing ℏs). Your S is dimensionless where ℏ was absorbed in it. Commented Apr 29 at 16:02
• @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 $\frac {\partial S}{\partial t}=-\frac {k^2}{2m}$ Commented Apr 29 at 18:17
• You failed to check the dimensional consistency of the linked post! When one redimensionalizes, one checks the dimensions. Commented Apr 29 at 19:16
• @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. Commented Apr 29 at 19:19
• 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. Commented Apr 29 at 19:21

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, $$𝑒^{−𝑖𝑆}(iℏ∂_t+ℏ^2∇^2/2𝑚−𝑉)(𝑅𝑒^{𝑖𝑆})=0, \leadsto \\ \frac{\partial R}{\partial t} = -\frac{\hbar}{2m} \left[ R \nabla^2 S + 2 \nabla R \cdot \nabla S \right],\\ \hbar \frac{\partial S}{\partial t} = - \left[\hbar^2 \frac{\left|\nabla S\right|^2}{2m} + V - \frac{\hbar^2}{2m} \frac{\nabla^2 R}{R} \right],$$ 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.
• 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! Commented Apr 29 at 19:43