Angular momentum eigenstates

  • Thread starter jfy4
  • Start date
  • #1
649
3

Homework Statement


Let [itex]|0,0\rangle [/itex] be the simultaneous eigenstate of [itex]\mathbf{J}^2[/itex] and [itex]J_z[/itex] with eigenvalues 0 and 0. Find
[tex]
J_x|0,0\rangle \quad\quad J_y |0,0\rangle \quad\quad [J_x,J_y]|0,0\rangle
[/tex]

2. The attempt at a solution
It seemed reasonable to write [itex]J_x[/itex] and [itex]J_y[/itex] in terms of ladder operators
[tex]
J_{+}=J_x + iJ_y
[/tex]
[tex]
J_{-}=J_x -i J_y
[/tex]
and then to have them operate on the states (for the last one I would just use the x,y,z commutation relations). But I was looking at the normalization constant out front
[tex]
J_{+}|j,m\rangle = \hbar\sqrt{(j+m+1)(j-m)} |j,m+1\rangle
[/tex]
[tex]
J_{-}|j,m\rangle = \hbar \sqrt{(j-m+1)(j+m)}|j,m-1\rangle
[/tex]
but given the initial state, these seem to be zero... Please tell me I'm doing this wrong, zero is such a unsatisfactory answer....

Thanks,
 

Answers and Replies

  • #2
Matterwave
Science Advisor
Gold Member
3,965
326
Because you are in a j=0 state, you can't actually raise or lower the m eigenvalue since m ranges between -j and +j so in this case it must stay 0.

So, it would seem like indeed you get 0 for everything.
 

Related Threads on Angular momentum eigenstates

  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
8
Views
3K
Replies
1
Views
1K
Replies
2
Views
702
  • Last Post
Replies
4
Views
2K
Replies
0
Views
970
Replies
6
Views
6K
Replies
1
Views
699
Replies
2
Views
626
Replies
21
Views
356
Top