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This is my first question at Physics SE so please be kind. I am well versed in the etiquette over at Math SE, but not so much here. Anyway, I thought this question was better suited to this site because it is less a problem of understanding mathematical computations and more of a problem about understanding why two results should be the same.

MY QUESTION:

I am taking a quantum mechanics class and I have been given an assignment. It is as follows:

Let $\Psi(\mathbf{r},t)$ be the wavefunction for a single particle satisfying Schrodinger's equation, $$i\hbar \partial_t\Psi=\left(\frac{-\hbar^2}{2m}\nabla^2+V\right)\Psi$$ Write $\Psi$ in polar form, $\Psi=Re^{i\Theta}$. Show that: $$\partial_t R+\frac{\hbar}{2mR}\nabla\boldsymbol{\cdot}\left(R^2\nabla\Theta\right)=0$$ Define $S=\hbar \Theta$. Show that $S$ satisfies the Quantum Hamilton-Jacobi equation: $$\partial_t S+\frac{\Vert\nabla S\Vert^2}{2m}+V=\frac{i\hbar}{2m}\nabla^2S.$$

After some lengthy calculations taking the time derivative and Laplacian of a composite function, I ended up with a big equation $$i\hbar \partial _{t} R-\hbar R\partial _{t} \Theta =\frac{-\hbar ^{2}}{2m}\left( \nabla ^{2} R+i R\nabla ^{2} \Theta -R\Vert \nabla \Theta \Vert ^{2} +2i \nabla R\boldsymbol{\cdot } \nabla \Theta \right) +VR.$$ Taking the Imaginary part of both sides of the equation, we do end up with the first of the required equations $$\partial _{t} R+\frac{\hbar }{2mR} \nabla \boldsymbol{\cdot }\left( R^{2} \nabla \Theta \right) =0$$ However taking the real part, we do not get the equation my professor wants. We instead get (using $S=\hbar \theta$) $$\partial _{t} S+\frac{\Vert \nabla S\Vert ^{2}}{2m} +V=\frac{\hbar ^{2}}{2m}\frac{\nabla ^{2} R}{R}.$$

So... is my professor wrong? My results agree with this other Physics SE post. If my professor isnt't wrong, can anyone explain why $$\hbar \frac{\nabla^2 R}{R}=i\nabla^2S~?$$ Might be true?

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I have solved this problem myself. If we write $$\Psi=e^Z$$ Where $Z$ is now allowed to be complex, we get $$i \hbar \partial _{t} Z=\frac{-\hbar ^{2}}{2m}\left( \nabla ^{2} Z+\Vert \nabla Z\Vert ^{2}\right) +U$$ Letting $Z=\frac{iS}{\hbar}$ will yield the desired result.

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