Commutator with r, p_r and angular momentum

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The discussion centers on the commutation relations between angular momentum operators and position/momentum in quantum mechanics, specifically referencing Dirac's Principles of Quantum Mechanics. It highlights that the angular momentum L and its magnitude L^2 commute with the position r and radial momentum p_r, indicating their scalar nature. The participants explore the implications of these commutators, particularly in relation to the conservation of angular momentum and the concept of Casimir operators. The conversation also touches on the role of angular momentum operators as generators of rotations, emphasizing their invariance under such transformations. Understanding these relationships is crucial for grasping the underlying physics of quantum mechanics.
omyojj
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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?

:)
 
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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.
 
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!
 
omyojj said:
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

[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.
 
Last edited:
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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