Does this commutator commute?

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If we define:

[tex]A_{j}=\omega \hat{x}_{j}+i \hat{p}_{j}[/tex]

and

[tex]A^{+}_{j}=\omega \hat{x}_{j}-i \hat{p}_{j}[/tex]

Would it be true to say:

[tex][A_k , (A^{+}_{i}+A_i)(A^{+}_{j}-A_j)]=0[/tex]

My reasoning is that, because

[tex][\hat{x}_{j}, \hat{p}_{i}]=0[/tex]

the the ordering of the contents of commutation bracket shouldn't matter (as [tex]\hat{x}_{j} \hat{p}_{i}=\hat{p}_{i}\hat{x}_{j}[/tex]), so we simply get that:

[tex][A_k , (A^{+}_{i}+A_i)(A^{+}_{j}-A_j)]=A_{k}(A^{+}_{i}+A_i)(A^{+}_{j}-A_j)-(A^{+}_{i}+A_i)(A^{+}_{j}-A_j)A_{k}= A_{k}(A^{+}_{i}+A_i)(A^{+}_{j}-A_j)-A_{k}(A^{+}_{i}+A_i)(A^{+}_{j}-A_j)=0[/tex]

This seems obvious to me, but it would make a 10 mark exam question too easy! Would be grateful if someone could confirm whether this is right or not.

Thanks.
 

Hurkyl

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What about the case where i=j?
 
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What about the case where i=j?
The question, specifically, is to find:

[tex][A_k , \hat{L}_{ij}]=[A_k, (A^{+}_{i}+A_i)(A^{+}_{j}-A_j)][/tex]

So I am not sure we need to consider the case where i=j.

Ofcourse it would get *really* messy if any of the subscripts are the same. But if i,j and k are not the same, the commutator would be zero, right?
 

Hurkyl

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The question, specifically, is to find:

[tex][A_k , \hat{L}_{ij}]=[A_k, (A^{+}_{i}+A_i)(A^{+}_{j}-A_j)][/tex]

So I am not sure we need to consider the case where i=j.
Sounds like you do. You said the question specifically asks you to find that commutator -- not to find that commutator in the special case where i,j,k are all distinct!

Ofcourse it would get *really* messy if any of the subscripts are the same.
It might not be as bad as you think.

But if i,j and k are not the same, the commutator would be zero, right?
Yes.
 
63
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Sounds like you do. You said the question specifically asks you to find that commutator -- not to find that commutator in the special case where i,j,k are all distinct!


It might not be as bad as you think.


Yes.
Well

[tex]\hat L_{ij}:=\hat{x}_{i}\hat{p}_{j} - \hat{x}_{j}\hat{p}_{i}
[/tex]

So if i=j

[tex]\hat L_{ii}:=\hat{x}_{i}\hat{p}_{i} - \hat{x}_{i}\hat{p}_{i}=0[/tex]

So the commutator:

[tex]
[A^{+}_{j},\hat L_{ij}]
[/tex]

would also be zero in this case as well

:)
 

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