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
[tex](\vec{p} \times \vec{L})\cdot \vec{p}=\vec{p}\cdot (\vec{L} \times \vec{p}) [/tex]
Why this relation is correct, and not
[tex](\vec{p} \times \vec{L})\cdot \vec{p}=\vec{p}\cdot (\vec{p} \times \vec{L}) [/tex]
?
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##.