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For the SHO, our teacher told us to scale $$p\rightarrow \sqrt{m\omega\hbar} ~p$$ $$x\rightarrow \sqrt{\frac{\hbar}{m\omega}}~x$$ And then define the following $$K_1=\frac 14 (p^2-q^2)$$ $$K_2=\frac 14 (pq+qp)$$ $$J_3=\frac{H}{2\hbar\omega}=\frac 14(p^2+q^2)$$ The first part is to show that $$Q \equiv -K_1^2-K_2^2+J_3^2$$ IS a number. My approach: $$16Q=J_3^2-K_1^2-K_2^2=(p^2+q^2)^2-(p^2-q^2)^2-(pq+qp)^2$$ $$=p^4+q^4+p^2q^2+q^2p^2-(p^4+q^4-p^2q^2-q^2p^2)-((pq)^2+(qp)^2+pqqp+qpqp)$$ $$=2p^2q^2+2q^2p^2-pqpq-qpqp-pqqp-qppq$$ At least point, I am unsure of how to simplify any further. A lot of these look like the form of anticommutators, which does not seem to provide any useful information in turning Q into a number. Any help would be appreciated!

EDIT::

This is how far I have gotten.
enter image description here

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  • $\begingroup$ As a terminological note (because I remember being confused by this), what is meant by "Q is a number" is that "Q is a number multiplied by the identity operator." $\endgroup$ Feb 11, 2014 at 2:00
  • $\begingroup$ You should post questions, not topics... This is a question and answer website, not a forum. $\endgroup$
    – Girardi
    Feb 11, 2014 at 15:59
  • $\begingroup$ Hint to the notes (v3): If you start from $p^2q^2$ (and its other orderings), and reduce via the CCR, you can only get terms $p^2q^2$, $i\hbar pq$, and $\hbar^2$ (and their other orderings). Putting $\hbar=1$, one could potentially get a term $(pq)^2$, but never get a term $i(pq)^2$ like you have in your notes. $\endgroup$
    – Qmechanic
    Feb 12, 2014 at 16:26
  • $\begingroup$ I redid it noticing that I accidentally carried that i(pq)^2 incorrectly. I still end up with an operator term that isnt a c number $\endgroup$ Feb 12, 2014 at 17:52
  • $\begingroup$ Okay. I think I solved it out. Did you happen to do it out as well and find that this results in -1? $\endgroup$ Feb 12, 2014 at 22:03

1 Answer 1

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Hint: $pq$-order$^1$ your last expression $$2(p^2q^2+q^2p^2)-(pq+qp)^2.$$ $pq$-ordering means commuting all $p$'s to the left and all the $q$'s to the right by using$^2$ the CCR formula $qp = pq +i\hbar{\bf 1}$, possibly repeatedly. (There are shorter ways, but $pq$-ordering is at least a systematic approach.) What remains will be a $c$-number. In fact, the result is $-3\hbar^2{\bf 1}$.

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$^1$ Or alternatively, $qp$-order your last expression.

$^2$ Here is an example of the $pq$-ordering procedure: $ qp^n = p^nq +i\hbar n p^{n-1}$.

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  • $\begingroup$ Not completely sure what you mean $\endgroup$ Feb 10, 2014 at 22:52
  • $\begingroup$ For example, $qp = pq+[q,p]$? $\endgroup$ Feb 10, 2014 at 22:54
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    $\begingroup$ Yes, that is an example of commuting a $p$ to the left. Everytime time you see a product $qp$ in your last formula, substitute in the right hand side of that equation for $qp$. $\endgroup$ Feb 10, 2014 at 22:55
  • $\begingroup$ So what about a case of qpqp, would that go to (pq+[q,p])(pq+[q,p]) $\endgroup$ Feb 10, 2014 at 23:03
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    $\begingroup$ Yes, and then you will have 1 c-number, two two-operator terms with all $p$'s on the left and $q$'s on the right, but then you will have a four operator term which is not $pq$ ordered. You will need to make your substition again to move the last $p$ through the $q$. $\endgroup$ Feb 10, 2014 at 23:09

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