I am going through Groenewold's theorem and in his book: On The Principles of Elementary Quantum Mechanics, page 45, eq. 4.11:

$\frac{1}{6}\left[\left(\mathbf{p}^{3}+3 c_{1} \mathbf{p}+d_{1}\right),\left(\mathbf{q}^{2}+c_{2}\right)\right]=\frac{1}{2}\left(\mathbf{p}^{2} \mathbf{q}+\mathbf{q} \mathbf{p}^{2}\right)+c_{1} \mathbf{q}$

where $\mathbf{p}$ and $\mathbf{q}$ are the canonical operators.

I cannot get the right-hand side. What I get is $\frac{1}{6}([\mathbf{p}^3,\mathbf{q}^2]+3c_1[\mathbf{p},\mathbf{q}^2])$. Help?

  • $\begingroup$ Could you show how you got that? $\endgroup$
    – J.G.
    May 27, 2020 at 22:17
  • $\begingroup$ Sorry, there was a typo but I fixed it. Is it clear how did I get my result now? $\endgroup$
    – Quantally
    May 27, 2020 at 22:19
  • $\begingroup$ Do you know the value of $[p,\,q]$? Do you know identities such as $[a,\,bc]=[a,\,b]c+b[a,\,c]$? $\endgroup$
    – J.G.
    May 27, 2020 at 22:30
  • $\begingroup$ Thank you for reminding me of this identity. I am trying to use it but I'm still not able to get the final result. $\endgroup$
    – Quantally
    May 27, 2020 at 23:01

1 Answer 1


From $[p,\,q]=1$, we can prove $[p,\,q^n]=nq^{n-1}$ by induction, using $[a,\,bc]=[a,\,b]c+b[a,\,c]$. For our purposes, we don't need to realize or prove this general result; we only need to verify the $n=2$ case, then the $n=3$ case. So$$[p^3,\,q^2]=[p^3,\,q]q+q[p^3,\,q]=3(p^2q+qp^2),\,[p,\,q^2]=2q.$$


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