# Scalar operators and commutators

Given a scalar operator $S$ and vector operators $V_1, V_2$, show that the commutator $[S,V_1\times V_2]= [S,V_1]\times V_2+V_1\times [S,V_2]$. I don't quite understand what a scalar operator is. But the question would be trivial if by scalar it means that, for example, $V_1S\times V_2=V_1\times SV_2$. Thanks in advance.

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Could you please share the source of the problem? This is likely to help provide useful context. –  Jonas Meyer Sep 4 '11 at 8:10
@Jonas: this is from a worksheet I have. I have actually copied the question out word for word :-). –  hil Sep 4 '11 at 8:12
hil: Thank you. Do you have any way of getting in contact with the creator of the worksheet? If there is at least one object in the problem whose definition you do not know, the first problem is to find out what that definition is. If, for example, this were an assignment for a class, then the teacher of the class would be the best source of information on what the definitions are. –  Jonas Meyer Sep 4 '11 at 8:18
I have found something in my notes. Apparently, it means that the operator is invariant under rotations. But how does this work when applied to this question? –  hil Sep 4 '11 at 8:54

Since this probably arises in the context of a physics problem, a scalar operator is an operator that transforms as a scalar under symmetry operations (i.e. changes of frame of reference). Whereas a vector operator will transform as a vector.

But I think that as far as this problem is concerned, you only need to know that a vector operator has three components, so if $V_1=(V_{1,x},V_{1,y},V_{1,z})$ and likewise for $V_2$, such that

$V_1 \times V_2 = \left(V_{1,y}V_{2,z} - V_{1,z}V_{2,y} \; ; \; V_{1,z}V_{2,x} - V_{1,x}V_{2,z} \; ; \; V_{1,x}V_{2,y} - V_{1,y}V_{2,x} \right) \;$

is the outer vector product.

On the other hand, a scalar operator just has one component, the scalar operator itself.

The rest of the exercise just amounts to checking out that the right hand side expression is indeed equal to the left hand side expression, which you can verify by simply working them out separately. I'll make an edit later if you're still struggling.

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