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Short question: What is the generalization of the BAC-CAB rule for operators?

Longer question and context: please read below

I was reading Schiff's book on Quantum Mechanics (3rd Edition) and on page 236, he has defined a generalized Runge-Lunz vector for a central force as

[tex]\vec{M} = \frac{1}{2\mu}(\vec{p} \times \vec{L} - \vec{L} \times \vec{p}) - \frac{k}{r}\vec{r}[/tex]

Here, [itex]\mu[/itex] is the reduced mass, [itex]\vec{p}[/itex] is the momentum operator, [itex]\vec{L}[/itex] is the angular momentum operator and [itex]\vec{r}[/itex] is the position operator. [itex]k[/itex] is a scalar constant.

I was trying to prove the following identities, which are also listed on the same page:

[tex][\vec{M}, H] = 0[/tex]

[tex]\vec{L} \bullet \vec{M} = \vec{M} \bullet \vec{L} = 0[/tex]

[tex]\vec{M}^2 = \frac{2H}{\mu}(\vec{L}^2 + \hbar^2) + k^2[/tex]

In trying to prove the first one, i.e. [itex][\vec{M}, H] = 0[/itex], I come across terms of the form:

[tex](\vec{p} \times \vec{L})\vec{p}^2[/itex]

Now, the term in brackets can be written as

[tex]\vec{p} \times (\vec{r} \times \vec{p})[/tex]

If these were normal vectors, this would be easy:

[tex]\vec{A} \times (\vec{B} \times \vec{C}) = \vec{B}(\vec{A} \bullet \vec{C}) - \vec{C}(\vec{A} \bullet \vec{B})[/tex]

In the case of vectors, due to commutativity of the dot product we could write

[tex]\vec{A} \times (\vec{B} \times \vec{C}) = \vec{B}(\vec{C} \bullet \vec{A}) - \vec{C}(\vec{B} \bullet \vec{A})[/tex]

but this isn't valid if A, B, C are operators. It seems that the only way to find [itex]\vec{p} \times \vec{L}[/itex] is to express them both in cartesian coordinates, take the cross product the "usual" way and simplify everything.

I have two questions:

1. What is the generalization of the vector triple product to triple products of vectors operators?

2. Is there a more efficient way of computing terms like [itex]\vec{p} \times \vec{L}[/itex]?

Thanks in advance.

-Vivek.

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# BAC - CAB for operators?

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