Uncertainties & Total Angular Momentum

Click For Summary
The discussion revolves around calculating an uncertainty quantity using total angular momentum operators in the x, y, and z directions, focusing on their commutation relations. The user attempts to apply the commutator of the angular momentum operators to an eigenstate |m> of J(z), leading to the equation [J_x, J_y]|m> = iħ J_z|m> = iħ m|m>. There is uncertainty about whether this approach is correct and how to proceed further. The user acknowledges a mistake in the notation, clarifying that the correct ket is |m> instead of |n>. The thread seeks guidance on the proper methodology for addressing uncertainties in angular momentum.
Sekonda
Messages
201
Reaction score
0
Hey,

My question is on determining an 'uncertainty' quantity using total angular momentum operators in the x,y and z directions where we know the commutation relations between the x,y and z directions of the total angular momentum operators.

Uncertainty_Q.png


I'm not really sure where to go with this at all, I let the commutator given act on a state |m> i.e. an eigenfunction of J(z)

[\hat{J_{x}},\hat{J_{y}}]|m>=i\hbar\hat{J_{z}}|n>=i\hbar m|m>

So the commutator must equal i*hbar*m, right? Anyway I'm not sure if this is even a correct way to begin, but it should be pretty straightforward however I haven't really come across a question like this before.

By the way I replaced subscripts 1,2,3 with x,y and z repsectively.

Thanks for any help!
SK
 
Physics news on Phys.org
Woops, the equation should read:

Sekonda said:
[\hat{J_{x}},\hat{J_{y}}]|m>=i\hbar\hat{J_{z}}|m>=i\hbar m|m>

not 'n' ket but 'm' ket.

Sorry...

SK
 
(a) The polarisation pattern is elliptical with maximum (1,1) and minimum (-1,-1), and anticlockwise in direction. (b) I know the solution is a quarter-wave plate oriented π/4, and half-wave plate at π/16, but don't understand how to reach there. I've obtained the polarisation vector (cos π/8, isin π/8) so far. I can't find much online guidance or textbook material working through this topic, so I'd appreciate any help I can get. Also, if anyone could let me know where I can get more...

Similar threads

Replies
4
Views
2K
Replies
11
Views
2K
Replies
1
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
3
Views
2K