# BAC - CAB for operators?

1. Jan 4, 2009

### maverick280857

Hi

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

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

$$\vec{M} = \frac{1}{2\mu}(\vec{p} \times \vec{L} - \vec{L} \times \vec{p}) - \frac{k}{r}\vec{r}$$

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

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

$$[\vec{M}, H] = 0$$

$$\vec{L} \bullet \vec{M} = \vec{M} \bullet \vec{L} = 0$$

$$\vec{M}^2 = \frac{2H}{\mu}(\vec{L}^2 + \hbar^2) + k^2$$

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

$$(\vec{p} \times \vec{L})\vec{p}^2[/itex] Now, the term in brackets can be written as [tex]\vec{p} \times (\vec{r} \times \vec{p})$$

If these were normal vectors, this would be easy:

$$\vec{A} \times (\vec{B} \times \vec{C}) = \vec{B}(\vec{A} \bullet \vec{C}) - \vec{C}(\vec{A} \bullet \vec{B})$$

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

$$\vec{A} \times (\vec{B} \times \vec{C}) = \vec{B}(\vec{C} \bullet \vec{A}) - \vec{C}(\vec{B} \bullet \vec{A})$$

but this isn't valid if A, B, C are operators. It seems that the only way to find $\vec{p} \times \vec{L}$ 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 $\vec{p} \times \vec{L}$?

-Vivek.

2. Jan 4, 2009

### tim_lou

the BAC-CAB rule is most easily demonstrated (and generalized to noncommutative operators) using the levi-civta epsilons symbol shown here:
http://en.wikipedia.org/wiki/Levi-Civita_symbol

then any cross-product can be written as (under Einstein summation convention):
$$(a\times b)_k = \epsilon_{ijk}a_i b_j$$
so, the BAC-CAB rule becomes:
$$(a\times (b\times c))_k = \epsilon_{ijk}a_i(b\times c)_j$$
$$=\epsilon_{ijk}a_i \epsilon_{lmj}b_l c_m$$
If you think for a bit, you'll notice thata
$$\epsilon_{ijk}\epsilon_{lmj}=\delta_{kl}\delta_{im}-\delta_{km}\delta_{li}$$
proceed using this, you'll get:
$$(a\times (b\times c))_k=a_i b_k c_i - a_ib_ic_k$$
Since the operators don't generally commute, this is the best you can do.

(You can easily evaluate all kinds of commutation relations too if you master this notation)

This notation is also very useful in proving identities involving variations of stokes' integrals and is closely related to differential forms.

3. Jan 4, 2009

### maverick280857

Got it.

SOLVED question 2.

But question 1 (generalization of BAC-CAB rule) is still a question pending thought. Any ideas?

4. Jan 4, 2009

### maverick280857

Thanks tim_lou, I remembered the expression for the cross product in terms of the Levi Civita tensor just after clicking submit on my first post. I've worked it out. But it seems from your post that a generalization of BAC-CAB, with perhaps a few more terms thrown in, doesn't exist. But thanks anyway.

5. Jan 5, 2009

### clem

You can use the vector notation without indices. bac-cab tells you where the dot goes.
You keep the original order for the operators, but put the dot in the right place.