Canonical Operators Commutation Relation

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?

• Could you show how you got that?
– J.G.
May 27, 2020 at 22:17
• Sorry, there was a typo but I fixed it. Is it clear how did I get my result now? May 27, 2020 at 22:19
• Do you know the value of $[p,\,q]$? Do you know identities such as $[a,\,bc]=[a,\,b]c+b[a,\,c]$?
– J.G.
May 27, 2020 at 22:30
• 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. May 27, 2020 at 23:01

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$$[p^3,\,q^2]=[p^3,\,q]q+q[p^3,\,q]=3(p^2q+qp^2),\,[p,\,q^2]=2q.$$