The Hamiltonian of a particle moving under the influence of a central potential $V(r)$ given by $$H=\frac{\textbf{p}^2}{2m}+V(r)$$ commutes with $\textbf{L}^2\equiv L_x^2+L_y^2+L_z^2$. Without doing a long algebra, is there a way to see that the operators $\textbf{p}^2$ and $\textbf{p}^4$ commute with $\textbf{L}^2$?
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1$\begingroup$ Note that $[AB,C] = A[B,C] + [A,C] B$. So if $[p^2, L^2] = 0$, then it follows that $[p^4, L^2] = p^2 [p^2, L^2] + [p^2, L^2] p^2 = 0$. $\endgroup$– Michael SeifertCommented Jul 8, 2018 at 14:04
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$\begingroup$ You can even take a step back from the commutator rules; $[A, B] = 0$ means $AB = BA$ in an already-associative algebra. To ask if $[A^2, B]=0$ is then easily proven by $AAB = ABA = BAA.$ What remains is the $\mathbf p^2$ case which is a nontrivial thing that has to do with scalars being invariant under rotations. $\endgroup$– CR DrostCommented Jul 8, 2018 at 16:08
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$\begingroup$ Maybe the down votes are coming because the first sentence seems unrelated to the question? As such you have a single sentence question repeated both in the title and in the body of the post. $\endgroup$– hftCommented Jul 10, 2018 at 23:12
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3 Answers
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Suppose that $\mathbf{A}$ is a vector operator, i.e its components transform as a vector under rotation, $$[L_i, A_j] = i \epsilon_{ijk} A_k.$$ Using this definition gives $[\mathbf{L}, \mathbf{A}^2] = 0$, which just says that $\mathbf{A}^2$ is a rotational scalar. Then the commutator product rule implies $[\mathbf{L}^2, \mathbf{A}^2] = [\mathbf{L}^2, \mathbf{A}^4] = 0$ as desired.
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Hint: ${\bf r}$ and ${\bf p}$ are vectors (while ${\bf r}^2$ and ${\bf p}^2$ are scalars) under $SO(3)$-rotations generated by ${\bf L}$. See also this related Phys.SE post.
answered Jul 8, 2018 at 13:50
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$\begingroup$ More like the answer rather than a hint :P $\endgroup$ Commented Jul 8, 2018 at 15:25
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I assume you are happy with $[{\bf p^2},{\bf L^2}]=0$ already. Using $[AB,C]=A[B,C]+[A,C]B$ with $A=B={\bf p^2}$ and $C={\bf A^2}$ gives \begin{equation} [{\bf p^4},{\bf L^2}] ={\bf p^2}[{\bf p^2},{\bf L^2}] + [{\bf p^2},{\bf L^2}]{\bf p^2} = 0 \end{equation} and of course you can go on like that to show $[{\bf p^{2n}},{\bf L^2}]=0$ for $n=1,2,3,\dots$.
