Is the Commutator of a Cross Product a Vector Operator?

[teroenza]

Homework Statement

Given that $$\vec{V}$$ and $$\vec{W}$$ are vector operators, show that $$\vec{V}\times \vec{W}$$ is also a vector operator.

2. The attempt at a solution
The only way I know how to do this is by showing that the commutator with the angular momentum vector operator ($\vec{J}$) is zero. Namely that $[\vec{V}\times \vec{W} , \vec{J}] = 0$. I want to start the problem by expressing the commutator as I would usually do by writing [A,B] = AB - BA, but I don't know exactly which type of multiplication to use here. Intuition tells me the dot product, but I want to be sure.

 

[mfb]

Dot product is right.

Edit: Or not, see below.

 

[stevendaryl]

Homework Statement

Given that $$\vec{V}$$ and $$\vec{W}$$ are vector operators, show that $$\vec{V}\times \vec{W}$$ is also a vector operator.

2. The attempt at a solution
The only way I know how to do this is by showing that the commutator with the angular momentum vector operator ($\vec{J}$) is zero. Namely that $[\vec{V}\times \vec{W} , \vec{J}] = 0$. I want to start the problem by expressing the commutator as I would usually do by writing [A,B] = AB - BA, but I don't know exactly which type of multiplication to use here. Intuition tells me the dot product, but I want to be sure.

No, it's not the dot-product. The usual meaning of $[A, \vec{J}]$ is that the result is a composite object (a tensor) with three components:

$[A, J_x]$, $[A, J_y]$, $[A, J_z]$

If $A$ is itself a vector, then you get 9 components:

$[\vec{A}, \vec{J}] = T$

where $T_{ij} = [A_i, J_j]$ and where $i$ and $j$ are either $x$, $y$, or $z$.

As to your claim that for a vector operator, $[\vec{A}, \vec{J}] = 0$, you should try an example with the momentum operator, $\vec{p}$. Try $[p_x, J_y]$.

 

[stevendaryl]

Here's a definition of a "vector operator":
http://en.wikipedia.org/wiki/Tensor_operator#Vector_operators

$[V_a, J_b] = i \hbar \epsilon_{abc} V_c$

not zero.

 

[teroenza]

Thank you. That is a mistake above. I meant that the only way I know of to show that it is a vector operator, is to show that $[\vec{V}\times \vec{W} , \vec{J}] \neq 0$. I was able to show that this is the case. This means it is not a scalar operator, but I am not sure if this is sufficient to show that it is a vector operator. It does mean that the rotation generator operator U won't commute with V x W.

Edit: I believe now that the condition to show that it is a vector operator, is to show that: $$[(\vec{V}\times \vec{W})_{i}, \vec{J}_{j}] = i \hbar \epsilon_{ijk} (\vec{V}\times \vec{W})_{k}$$