Commutation for vectorial operator

Question

Let's say we have a vector operator $ \hat{\mathbf{A}} $ whose components are operators $ \hat A_1, \hat A_2, \hat A_3, ... $. How do you do the commutator between two vector operators. Is it correct to say $$ [\hat{\mathbf{A}},\hat{\mathbf{B}}]= \hat{\mathbf{A}}\cdot\hat{\mathbf{B}}-\hat{\mathbf{B}}\cdot \hat{\mathbf{A}} $$

If so, what does this dot product mean and how do I operate? Since the elements are operators they can be thought as matrix, does that mean that the components should be multiplied as so and then add up? Is there any other type of product defined for this commutator?

Answer 1

It's non-standard notation. In your notation $\mathbf{A}$ is a vector with components $A_i$. Passing to a different frame/coordinate system, the components would transform as $$A_i \mapsto R_{ij} A_j$$ where $R_{ij}$ is matrix describing the coordinate change. The most general commutator is therefore a matrix: $$[A_i,B_j] = A_i B_j - B_j A_i$$ and it transforms as you'd expect (i.e. as a rank-2 tensor). The definition you're giving is a scalar one, namely the contraction with the $\delta_{ij}$ tensor: $$\delta^{ij} [A_i,B_j] = A_i B_i - B_i A_i = \mathbf{A} \cdot \mathbf{B} - \mathbf{B} \cdot \mathbf{A}.$$ This is similar to the usual grad/div stuff. Namely suppose that you have a vector field $\mathbf{A}(\mathbf{x})$ with components $A_i(\mathbf{x})$, and $\mathbf{x}$ itself is a vector $\mathbf{x} = x_j$. Then the most general tensor ("grad") you can build is $$\nabla \mathbf{A}(\mathbf{x})$$ with components $$\nabla_i A_j(\mathbf{x}) = \frac{\partial}{\partial x^i} A_j(\mathbf{x}).$$ Of course, you can take the trace of that tensor to get the "div": $$\nabla \cdot \mathbf{A}(\mathbf{x}) = \delta^{ij} \nabla_i A_j(\mathbf{x}) = \nabla_i A_i(\mathbf{x}).$$

Answer 2

Mathematically speaking, the commutator is an operation that is generally associated with a Lie algebra, and these are defined axiomatically in the same way that vector spaces are. They are rather like vector spaces except they have a multiplication called commutation, which satisfies the Jacobi rule:

[x, [y, z]] + [y, [z, x]] + [z, [x,y]] = 0

And

[x, x] = 0

The simplest example of a Lie algebra that is physically or geometrically motivated is the usual cross product for vectors.

Now, again speaking mathematically, there is also a notion of an algebra which is essentially the notion of a vector space but with an additional operation called the product, and this satisfies the associative law that we have in ordinary arithmetic:

x(yz) = (xy)z

Here, I've denoted the product of x & y simply by placing them next to each other.

Is a Lie algebra actually an algebra in this sense? Well, no; because a Lie algebra doesn't satisfy the associative law, it has instead the Jacobi identity. However, given any algebra we can get a Lie algebra merely by defining the commutator as

[x, y] := xy - yx

It's worth ploughing through the algebra (once) to see that with this definition the Jacobi identity is satisfied, and we can see immediately that [x,x] = 0.

since the elements are operators they can be thought of as matrix, does that mean that the components should multiply as so, and then add up.

As you mention matrices, I expect you're familiar with how they operate on column vectors and, also on other matrices ie their multiplication, addition & subtraction; the expression A.B - B.A should be seen as matrix multiplication; so you matrix multiply A & B together, to get A.B, and then you matrix multiply B & A together, to get B.A; and then you subtract the difference to get A.B - B.A.