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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.
[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.