# L^2 in spherical coordinates.

1. Sep 29, 2006

### koroljov

Hi

I am trying to calculate L^2 in spherical coordinates. L^2 is the square of L, the angular momentum operator. I know L in spherical coordinates. This L in spherical coordinates has only 2 components : one in the direction of the theta unit vector and one in the direction of the phi unit vector.

I get the correct result for L^2 by substituting cartesian values for the theta and phi unit vectors in L, and then squaring and adding the components.

I do not get the correct result by simply squaring and adding the theta and phi components of L directly.Why not? Surely if this were a classical vector whose components are scalars rather than operators, I could find its norm squared in both ways, isn't it?

2. Sep 30, 2006

### masudr

Before I can answer your question, I'll need to see how you've got L in spherical co-ordinates.

3. Sep 30, 2006

### koroljov

Well

L = - i * h * (r x nabla) = - i * h * ( u_phi * d/theta - u_theta/sin(theta) * d/dphi )

where h=hbar, nabla=grad operator, u_phi and u_theta=phi and theta unit vectors, x = vector product.

Substituting cartesian values

u_theta = (cos(theta)*cos(phi), cos(theta)*sin(phi), -sin(theta))
u_phi = (-sin(phi), cos(phi), 0)

and squaring and adding the components gives the desired result for L^2:
L^2 = -h^2 * (1/sin(theta)^2 * d^2/dphi^2 + 1/sin(theta) * d/dtheta (sin(theta) * d/dtheta))

Simply squaring and adding the components of L does not seem to give this result.

4. Sep 30, 2006

### masudr

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).

5. Sep 30, 2006

### koroljov

I can't say I completely understand it though. In Quantum mechanics, as it is being taught to me, we never used classical Hamiltonian mechanics (except for the hamiltonian in the Schrödinger equation). Rather, we converted from classical mechanics to quantum mechanice by replacing the momentum with -i * h * nabla and the kinetic energy by i * h * d/dt.

6. Oct 1, 2006

### markr

I haven't done the proof, but the issue may be that the derivatives of the basis vectors are not zero.

For example d/d phi u_theta = cos theta u_phi.

7. Oct 2, 2006

### koroljov

Yes, I see, you are correct. How dumb of me to not see that. Purely out of curiosity: Is it even possible to do this in spherical coordinates directly? I mean, using the correct derivatives when "squaring" the components will give me another vector operator, but the end result (L^2) is an operator whose result is a scalar (rather than a vactor).

8. Oct 2, 2006

### masudr

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.

Last edited: Oct 2, 2006