Let say we have the classical hamiltonian of a harmonic oscillator: $$H=\frac{p_x^2+p_y^2+p_z^2}{2m}+\frac{k_1x^2+k_2y^2+k_3z^2}{2}$$ and we want to find the hamiltonian operator in quantum mechanics, we use the transformations $x\rightarrow x$, $y\rightarrow y$, $z\rightarrow z$, $p_x\rightarrow -i\hbar\frac{\partial}{\partial x}$, $p_y\rightarrow -i\hbar\frac{\partial}{\partial y}$ and $p_z\rightarrow -i\hbar\frac{\partial}{\partial z}$ to obtain: $$\hat{H}=\frac{-\hbar^2}{2m}\Delta+\frac{k_1x^2+k_2y^2+k_3z^2}{2}=\frac{-\hbar^2}{2m}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right)+\frac{k_1x^2+k_2y^2+k_3z^2}{2}$$ Now suppose I want to treat this system in sphercal coordinates, should I transform the classical hamiltonian in spherical coordinates ($p_r\rightarrow \frac{\partial}{\partial r}$, $p_{\theta}\rightarrow \frac{\partial}{\partial \theta}$, $p_{\varphi}\rightarrow \frac{\partial}{\partial \varphi}$) or should I transform the hamiltonian operator to spherical coordinates using $$\Delta=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}$$ and why?
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This 3D problem (v2) is related to this Phys.SE post. A general strategy to deal with quantization in any curved coordinate system (in any dimension) is outline there. The answer is that the appropriate quantization of the classical Hamiltonian for a free particle in spherical coordinates is $-\frac{\hbar^2}{2m}$ times the Laplacian $\nabla^2\equiv \Delta$ in spherical coordinates.
