Anguluar momentum Commutation Identity

Click For Summary
The discussion revolves around proving the commutation relation [J^2, A_i] = -2iħε_{ijk}J_jA_k - 2ħ^2A_i, starting from the given relation [A_i, J_j] = iħε_{ijk}A_k. Participants clarify the use of the identity [AB, C] = A[B, C] + [A, C]B and address sign errors in expressions related to angular momentum. There is a consensus that the problem is complex, requiring careful index manipulation, and some participants express a desire for a more elegant solution. The conversation also touches on the challenge of deriving identities from Dirac's work, indicating the problem's difficulty level. The thread highlights the intricacies of quantum mechanics and the nuances of commutation relations.
decerto
Messages
84
Reaction score
2

Homework Statement



Given that [A_i,J_j]=i\hbar\epsilon_{ijk}Ak where A_i is not invariant under rotation

Show that [J^2,Ai]=-2i\hbar\epsilon_{ijk}J_jAk-2\hbar^2A_i

Homework Equations


[AB,C]=A[B,C]+[A,C]B

[A,B]=-[B,A]

The Attempt at a Solution




[J^2,Ai]=[J_x^2,Ai]+[J_y^2,Ai]+[J_z^2,Ai]
=J_x[J_x,Ai]+[J_x,Ai]J_x+J_y[J_y,Ai]+[J_y,Ai]J_y+J_z[J_z,Ai]+[J_z,Ai]J_z
=-J_x\epsilon_{ixk}Ak-\epsilon_{ixk}AkJ_x-J_y\epsilon_{iyk}Ak-\epsilon_{iyk}AkJ_y-J_z\epsilon_{izk}Ak-\epsilon_{izk}AkJ_z

Not sure where to go from here
 
Physics news on Phys.org
You left out a factor of ##i\hbar## in getting to your last line.

Note that your final line can be written compactly as ##-\epsilon_{ijk}(J_jA_k + A_kJ_j)##
 
TSny said:
You left out a factor of ##i\hbar## in getting to your last line.

Note that your final line can be written compactly as ##-\epsilon_{ijk}(J_jA_k + A_kJ_j)##

Thanks that looks a lot easier to deal with, I guess I use that ##A_kJ_j= J_jA_k-[A_k,J_j]##?
 
decerto said:
Thanks that looks a lot easier to deal with, I guess I use that ##A_kJ_j= J_jA_k-[A_k,J_j]##?
That's the right idea, but there's a sign error in your expression ##A_kJ_j= J_jA_k-[A_k,J_j]##.
 
TSny said:
That's the right idea, but there's a sign error in your expression ##A_kJ_j= J_jA_k-[A_k,J_j]##.
I have the exact same thing written?
 
Last edited:
decerto said:
I have the exact same thing written?
Yes, I was just rewriting the same expression that you wrote. But the expression is incorrect due to sign errors.
 
TSny said:
Yes, I was just rewriting the same expression that you wrote. But the expression is incorrect due to sign errors.
Ah right sorry, I had the right expression on the page and it worked so I was confused why you were correcting me
 
TSny said:
Yes, I was just rewriting the same expression that you wrote. But the expression is incorrect due to sign errors.
To prove the full identity ##[J^2,[J^2,A]]=2\hbar^2(J^2A+AJ^2)-4\hbar^2(A\cdot J)J## can I just use a nested expression of what I just proved, as in let ##[J^2,Ai]=Ai## in my original identity
 
decerto said:
To prove the full identity ##[J^2,[J^2,A]]=2\hbar^2(J^2A+AJ^2)-4\hbar^2(A\cdot J)J## can I just use a nested expression of what I just proved, as in let ##[J^2,Ai]=Ai## in my original identity

Yes, but of course ##[J^2,Ai] \neq Ai##. I managed to get the result, but only after a couple of pages of tedious index manipulations. I suspect there is a more elegant way to get to the result, but I don't see it.
 
  • #10
TSny said:
Yes, but of course ##[J^2,Ai] \neq Ai##. I managed to get the result, but only after a couple of pages of tedious index manipulations. I suspect there is a more elegant way to get to the result, but I don't see it.
Its a bonus problem, and the last bonus problem was about 6 pages of tedious trigonometric stuff so it wouldn't surprise me if there wasn't, the identity is from dirac but I can't find his original derivation. Thanks for your help though
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
1
Views
2K
Replies
5
Views
3K
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
4K