Commutator with r, p_r and angular momentum

[omyojj]

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

anyway..here`s my curiosity..

From Paul Dirac`s 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...

It`s 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 want to understand the underlying physics..

commutator(Quantum Poisson`s Bracket) with L zero?

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

:)

 

[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.

 

[omyojj]

I don`t 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!

 

[meopemuk]

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...

It`s 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 want to understand the underlying physics..

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

[tex][L_x, r] = 0 [/itex]<br /> <br /> 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<br /> <br /> $$F(t) = \exp(-\frac{i}{\hbar} Ht )F\exp(\frac{i}{\hbar} Ht )$$<br /> <br /> and $F$ does not depend on time if and only if $F$ commutes with the Hamiltonian. <br /> <br /> Eugene.[/tex]