Commutator relationship

  • #1

Homework Statement


Show that ##[\hat{L} \cdot \vec{a}, \hat{L} \cdot \vec{b}] = i \hbar \hat{L} \cdot (\vec{a} \times \vec{b})##

Homework Equations


##[\hat{L}_i, \hat{L}_j]= i \hbar \epsilon_{ijk} \hat{L}_k ##

The Attempt at a Solution


[/B]
Maybe a naive attempt, but it has been a while. I have two ways I think this can work, but both I'm not sure. they "work", but without a more trained eye, I'm not sure if they're valid.

Start with ##[\hat{L} \cdot \vec{a}, \hat{L} \cdot \vec{b}] \rightarrow [\hat{L}_ia_i, \hat{L}_jb_j] =i \hbar \epsilon_{ijk} \hat{L}_k a_ib_i \rightarrow i \hbar\hat{L}_k \epsilon_{ijk} a_i b_i = i \hbar \hat{L} \cdot (\vec{a} \times \vec{b}) ##

The other one starts similarly, but once we put it in the ##[\hat{L}_ia_i, \hat{L}_jb_j]## form, we can take out the ##a_i## and ##b_j## giving us ##a_ib_j[\hat{L}_i,\hat{L}_j]## which from here the rest follows as above, but not sure if either approach is "valid".
 

Answers and Replies

  • #2
Orodruin
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How are these different ways? They look identical to me.
 
  • #3
How are these different ways? They look identical to me.
Yes, i believe they are. Is it true that im allowed to take out the ##a_i## from the brackets? I think that is what i conclude from above.
 
  • #4
Orodruin
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Is it true that im allowed to take out the aiaia_i from the brackets?
Yes, the commutator is linear in both arguments.
 
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