Angular momentum raising lowering operator

  • Thread starter v_pino
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  • #1
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Homework Statement



Derive [itex] [L_\pm , L^2]=0 [/itex]



Homework Equations



[itex] L_{\pm}=L_x \pm iL_y [/itex]


The Attempt at a Solution



[itex] [L_\pm , L^2]=[L_x,L_x^2] \pm i[L_y,L_y^2]=[L_x,L_x]L_x + L_x[L_x,L_x] \pm i([L_y,L_y]L_y+L_y[L_y,L_y]) [/itex]

Is this right so far? If so, how do I proceed from this? The books and websites I'll read always just state [itex] [L_\pm , L^2]=0 [/itex] instead of proving it.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
[itex]L^{2}=L^{2}_{x}+L^{2}_{y}+L^{2}_{z}[/itex]

Then

[itex][L_{±},L^{2}]=[L_{x}±iL_{y},L^{2}_{x}+L^{2}_{y}][/itex]

which is going to give you some cross terms. Using the fact that [itex][L_{i},L_{j}]=ih\epsilon_{ijk}L_{k}[/itex], you can move some stuff around and cancel terms to arrive at the final answer. I'm not sure how you got to where you are.
 
  • #3
169
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I'm having some trouble expanding the commutators due to the +/- sign. Is it still [A,B]=AB-BA?
 
  • #4
Is it still [A,B]=AB-BA?
This is always true.

After writing the commutation relation, and expanding, the ± just becomes [itex]\mp[/itex] when multiplied by -. If you have two like terms, one of which has the sign ± and the other of which has the sign [itex]\mp[/itex], they will cancel.
 
  • #5
169
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Is the following ok?

[itex] [L_\pm,L^2]=(L_x \pm iL_y)(L_x^2+L_y^2)-(L_x^2+L_y^2)(L_x \pm iL_y) [/itex]

[itex] [L_\pm,L^2]=L_x^3+L_xL_y^2\pm iL_x^2L_y\pm iL_y^3 -L_x^3 \mp i L_x^2 L_y -L_x L_y^2 \mp iLy^3 [/itex]

= 0
 
  • #6
169
0
I tried to do the same as what I did above to prove [itex] [L_+,L_-]=2\hbar L_z [/itex] but I got zero instead:

[itex] [L_+,L_-]=(L_x +iL_y)(L_x-iL_y)-(L_x-iL_y)(L_x+iL_y)=(L_x^2+L_y^2)-(L_x^2+L_y^2)=0 [/itex]

I found the following derivation but I don't understand how the last three equalities were arrived at:

[itex] [L_+,L_-]=[L_x+iL_y,L_x-iL_y]=i[L_y,L_x]-i[L_x,L_y]=\hbar (L_z+L_z)=2\hbar L_z [/itex]
 

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