A Operator Identity: Quantum Mechanics Explanation w/ References

[LagrangeEuler]

In Quantum mechanics, when we have momentum operator $\vec{p}$, and angular momentum operator $\vec{L}$, then
(\vec{p} \times \vec{L})\cdot \vec{p}=\vec{p}\cdot (\vec{L} \times \vec{p})
Why this relation is correct, and not
(\vec{p} \times \vec{L})\cdot \vec{p}=\vec{p}\cdot (\vec{p} \times \vec{L})
?
Could you give me some reference for this?

 

[Orodruin]

Because $\vec p$ does not commute with $\vec L$.

 

[vanhees71]

Well, let's calculate it:
$$A=(\vec{p} \times \vec{L})\cdot \vec{p}=\epsilon_{jkl} p_j L_k p_l=p_j (\vec{L} \times \vec{p})_j = \vec{p} \cdot (\vec{L} \times \vec{p}).$$
It's as simple as with usual c-number vectors, because nowhere I have to commute any operators :-)).

 

[Orodruin]

Because $\vec p$ does not commute with $\vec L$.

because nowhere I have to commute any operators :-)).

Just to follow up on this. To get to $\vec p \cdot (\vec p \times \vec L)$ you would have to commute the right $p_i$ with $\vec L$ and the commutator is non-zero.

 

[vanhees71]

Also the other order in the bracket wouldn't even be correct for c-number vectors, because the cross product is skew-symmetric, not symmetric!

 

[fresh_42]

Just a possibly stupid question. What does the dot mean? If it means a dot product, so why isn't $\vec{a} \cdot (\vec{a} \times \vec{*})=0\,?$

 

[Orodruin]

That is zero, $\vec a \cdot (* \times \vec a)$ is not necessarily zero. Note that $\vec p$ and $\vec L$ are non-commuting observables.

 

[Orodruin]

To be more specific, $\vec p \cdot (\vec p \times \vec L) = \epsilon_{ijk} p_i p_j L_k = 0$ because $p_i$ commutes with $p_j$ and $\epsilon_{ijk}$ is antisymmetric in $ij$. If you instead switch places of the second $\vec p$ and the $\vec L$, you have to commute the $\vec p$ with the $\vec L$ to get to the antisymmetric expression. This leaves a term proportional to $\vec p^2$.