Commutators and Their Properties in Quantum Mechanics

[vertices]

If we define:

A_{j}=\omega \hat{x}_{j}+i \hat{p}_{j}

and

A^{+}_{j}=\omega \hat{x}_{j}-i \hat{p}_{j}

Would it be true to say:

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

My reasoning is that, because

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

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

[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

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]

What about the case where i=j?

 

[vertices]

What about the case where i=j?

The question, specifically, is to find:

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

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]

The question, specifically, is to find:

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

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.

 

[vertices]

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

\hat L_{ij}:=\hat{x}_{i}\hat{p}_{j} - \hat{x}_{j}\hat{p}_{i}<br />

So if i=j

\hat L_{ii}:=\hat{x}_{i}\hat{p}_{i} - \hat{x}_{i}\hat{p}_{i}=0

So the commutator:

<br /> [A^{+}_{j},\hat L_{ij}]<br />

would also be zero in this case as well

:)