Commutator with r, p_r and angular momentum

1. Sep 25, 2007

omyojj

Hi, guys..This is my first time to post. and I gotta aplogize for my bad English..Im a novice..;;

anyway..heres my curiosity..

From Paul Diracs Principles of Quantum Mechanics..p.153
(section of Motion in a central field of force)

It says that

The angular momentum L of the ptl about the orgin ..and its magnitude L^2 commute with r and p_r since they are scalars...

Its not hard to verify that [L, r] = [L, p_r] = 0

(L_x=x*p_y-y*p_x, r=(x^2+y^2+z^2)^(1/2) etc.)

But I just wanna understand the underlying physics..

commutator(Quantum Poissons Bracket) with L zero?

Are they concerned with conservation of angular momentum? or else?

:)

2. Sep 25, 2007

dextercioby

L^2 is a Casimir operator for the Lie algebra of angular momentum. It just means that r and p_r are \phi independent, and \phi is the canonically conjugate variable of angular momentum.

3. Sep 26, 2007

omyojj

I dont have any idea what Casimir operator is..

But canonically conjugate variable of angular momentum phi is surely indep. of p_r, r
..

thx a lot!

4. Sep 26, 2007

meopemuk

Operators of angular momentum $(L_x, L_y, L_z)$ play the role of generators of rotations (actually, this property can be used as a definition of $\mathbf{L}$). This means that if $r$ is distance measured in the reference frame O, then

$$r' = \exp(\frac{i}{\hbar} L_x \phi )r \exp(-\frac{i}{\hbar} L_x \phi )$$.....(1)

is the distance in the reference frame O' that is rotated with respect to O by the angle $\phi$ around the x-axis. Since $r$ is a scalar, it is invariant with respect to rotations ($r'=r$). Then, by Taylor expanding exponents in the right hand side of (1) one can show that this invariance is equivalent to the vanishing commutator

$$[L_x, r] = 0 [/itex] Commutators are closely related to inertial transformations in other examples as well. For example, the Hamiltonian is the generator of time translations, i.e., for any operator $F$ its time dependence is given by [tex] F(t) = \exp(-\frac{i}{\hbar} Ht )F\exp(\frac{i}{\hbar} Ht )$$

and $F$ does not depend on time if and only if $F$ commutes with the Hamiltonian.

Eugene.

Last edited: Sep 26, 2007