# L^2 in spherical coordinates.

by koroljov
Tags: coordinates, spherical
 P: 932 L^2 in spherical coordinates. Oh I see. As far as I'm aware, the reasoning behind this is not entirely obvious. In classical Hamiltonian mechanics, the physics of a system with N degrees of freedom can be formulated in terms of 2N variables. Traditionally, these are the position and conjugate momentum in the various dimensions. However there are a class of variables called canonical variables. Any of these variables can be used used to do Hamiltonian mechanics. In going to spherical co-ordinates, you are suggesting using $r, \theta, \phi$ and the associated derivatives (for the momenta). The reason all this is important is that the prescription for going from classical to quantum mechanics is to promote the Poisson brackets to commutators, and the functions on position and momenta to functions of the associated operators. I think it boils down to which derivative operators we need for the momentum operators to be canonical variables (and I'm guessing that the extra $\sin(\theta)\mbox{'s}$ appear because of that).
 P: 932 NB. throughout, I use the following transformation $$x=r\cos{\theta}\sin{\phi}$$ $$y=r\sin{\theta}\sin{\phi}$$ $$z=r\cos{\phi}$$ Hmm. I'm sure that's not enough to explain it. The Lagrangian in spherical polars is given by: $$L=\frac{1}{2}m(\dot{r}^2+(r\dot{\theta})^2+(r\sin{\theta}\dot{\phi})^2)-V(r,\theta,\phi)$$ This gives the momenta as: $$p_r = \partial L / \partial \dot{r}} = m\dot{r}$$ $$p_\theta=\partial L / \partial \dot{\theta}} = mr^2\dot{\theta}$$ $$p_\phi = \partial L / \partial \dot{\phi}} = mr^2 \sin^2{\theta}\dot{\phi}$$ The reason this is important is because: $$[\hat{q}_i,\hat{p}_j] = i\hbar\{q_i,p_j\}=i\hbar\delta_{ij}$$ where it is understood that q, p are the variables the problem is formulated in, and $\hat{q}, \hat{p}$ are the associated position and momentum operators, and the curly brackets are Poisson brackets. What this means is that if we are to do our problem in a new set of variables, we must find what the momentum corresponds to, and then replace those with the operators $-i\hbar\partial / \partial q_i$. So: $$\hat{p}_r = -i\hbar\partial / \partial r$$ $$\hat{p}_\theta = -i\hbar\partial / \partial \theta$$ $$\hat{p}_\phi = -i\hbar\partial / \partial \phi$$ In spherical polars, the cartesian components of angular momentum are given by: $$L_x=-p_\theta \sin{\phi}\cos{\phi}\cos{\theta}-\frac{p_\phi}{\sin{\theta}}$$ $$L_y=-p_\theta \sin{\phi}\cos{\phi}\sin{\theta}-\frac{p_\phi\cos{\theta}}{\sin^2{\theta}}$$ $$L_z=p_\theta \sin^2{\theta}\end{multiline*}$$ where the $p_\theta, p_\phi$ are the canonical momenta of the $\theta, \phi$ variables. This was obtained by changing the cartesian components of L from cartesian variables (i.e. $L_x = yp_z - zp_y=my\dot{z}-mz\dot{y}$ to spherical polars). Now by doing our quantisation (i.e. replacing classical variables with their corresponding operators, whose commutators correspond to the classical Poisson bracket) $$\hat{L}_x = -i\hbar(\sin{\phi}\cos{\phi}\cos{\theta}\frac{\partial}{\partial \theta}-\frac{1}{\sin{\theta}}\frac{\partial}{\partial \phi})$$ $$\hat{L}_y=-i\hbar(\sin{\phi}\cos{\phi}\sin{\theta}\frac{\partial}{\partial \theta}-\frac{\cos{\theta}}{\sin^2{\theta}}\frac{\partial}{\partial \phi})$$ $$\hat{L}_z=-i\hbar \sin^2{\theta}\frac{\partial}{\partial \theta}$$ All that remains is to square these operators up (remembering that they apply to functions on the right hand side; this ensures that the product rule/Leibniz rule is applied accordingly) and add them up to see what $\hat{L}^2$ looks like in spherical polars. I'm not 100% if I'm on the right tracks here, but as far as I know, I haven't made any mistakes. If I had the inclination/time to square those operators and sum them, I might have found out... EDIT: lots of edits to get the $\LaTeX$ right.