Is [\vec{p}^2, \vec{p} \times \vec{L}] Equal to Zero?

[LagrangeEuler]

Is there some easy way to see that
$$[\vec{p}^2, \vec{p} \times \vec{L}]$$ is equal zero? I use component method and got that.

 

[andrewkirk]

What does the expression mean? In linear algebra commutators take operators as inputs, but in the above the inputs are not operators. The first appears to be a scalar (assuming $\vec p^2$ means the dot product of $\vec p$ with itself) and the second a pseudovector.

Commutators are also defined for the binary operation of any group, but if that is the intention above, it is not clear what the group or the operation is.

 

[blue_leaf77]

What does the expression mean? In linear algebra commutators take operators as inputs, but in the above the inputs are not operators.

It is supposed to be understood as element-wise statement, that is
$$
[\mathbf p^2,(\mathbf p\times \mathbf L)_i] = 0
$$
for $i=1,2,3$.

Is there some easy way to see

You can work on it using symbolic expressions but IMO it's not too simple nor too difficult either, in fact it is also component-wise proof
$$
[p^ip^i, \epsilon_{mjk} p^j L^k] = \epsilon_{mjk} \left( p^i[p^i,p^j]L^k + [p^i, p^j]p^iL^k + p^jp^i[p^i, L^k] + p^j[p^i,L^k]p^i \right) \\
= \epsilon_{mjk} \left( 0 + 0 + p^jp^ip^l (-i\hbar \epsilon_{kil}) + p^j p^l p^i (-i\hbar \epsilon_{kil}) \right) = -i\hbar \epsilon_{kil} \epsilon_{kmj} \left( p^jp^ip^l + p^j p^l p^i \right)\\
= -i\hbar (\delta^m_i\delta^j_l - \delta^j_i\delta^m_l) \left( p^jp^ip^l + p^j p^l p^i \right) = 0
$$
The first equality results from using commutator identity for $[AB,CD]$ and the last one leading to zero final answer can be easily proved by expanding the expression to the left of the last equal sign.