# Does the vector triple-product identity hold for operators?

1. Jan 22, 2012

### thecommexokid

Does the definition of the vector triple-product hold for operators?

I know that for regular vectors, the vector triple product can be found as $\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=( \mathbf{a} \cdot\mathbf{c})\mathbf{b}-(\mathbf{a}\cdot\mathbf{b})\mathbf{c}$. Does this identity hold for vector operators as well? Specifically, does it hold for $\mathbf{\hat A}\times(\mathbf{\nabla}\times\mathbf{\hat C})$? And if not, is there any other useful identity for determining this product?

2. Jan 22, 2012

### Fredrik

Staff Emeritus
Using the Levi-Civita symbol and the convention to not write any summation sigmas (since the sum is always over the indices that appear twice), the proof for vectors in $\mathbb R^n$ is just
\begin{align}
(a\times(b\times c))_i &=\varepsilon_{ijk}a_j\varepsilon_{klm}b_l c_m =(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})a_j b_l c_m=a_j b_i c_j-a_j b_j c_i\\
&=a_j c_j b_i-a_j b_j c_i=(a\cdot c)b_i-(a\cdot b)c_i.
\end{align} Going from the first line to the second, I used that $b_i$ commutes with $a_j$. If $b_i$ doesn't commute with either $a_j$ or $c_j$, we get a different final result.

$$(A\times(\nabla\times C))_i=A_j\partial_i C_j-A_j\partial_j C_k.$$ Here you have to use the chain rule to move $\partial_i$ to the left of $A_j$, so you end up with an additional term.

3. Jan 22, 2012

### thecommexokid

Thank you for all the help so far.

I've always been pretty clumsy with working things out component-wise in general (let alone when nothing commutes), so let me do this here in public where I can be corrected on my screw-ups.

So let me make sure I understand what you've said. Your proof works fine up to here:
$${\left( {{\bf{\hat A}} \times (\nabla \times {\bf{\hat C}})} \right)_i} = {\varepsilon _{ijk}}{\hat A_j}{\varepsilon _{k\ell m}}{\partial _\ell }{\hat C_m} = ({\delta _{i\ell }}{\delta _{jm}} - {\delta _{im}}{\delta _{j\ell }}){\hat A_j}{\partial _\ell }{\hat C_m} = {\hat A_j}{\partial _i}{\hat C_j} - {\hat A_j}{\partial _j}{\hat C_i}$$
If I understand correctly what you were getting at regarding the chain rule, then at this point, we pause to notice that
$${\partial _i}({\hat A_j}{\hat C_j}) = ({\partial _i}{\hat A_j}){\hat C_j} + {\hat A_j}({\partial _i}{\hat C_j});$$
if we rearrange, we find that
$${\hat A_j}({\partial _i}{\hat C_j}) = {\partial _i}({\hat A_j}{\hat C_j}) - ({\partial _i}{\hat A_j}){\hat C_j}.$$
And if we substitute this result into our earlier equation,
$${\left( {{\bf{\hat A}} \times (\nabla \times {\bf{\hat C}})} \right)_i} = \left[{\partial _i}({\hat A_j}{\hat C_j}) - ({\partial _i}{\hat A_j}){\hat C_j}\right] - {\hat A_j}{\partial _j}{\hat C_i}.$$
Of these three terms, I recognize the first as $\partial_i(\bf{\hat A}\cdot\bf{\hat C})$, and I recognize the third as $-({\bf{\hat A}}\cdot\nabla){\hat C}_i$; but I don't recognize the second as anything. So how do I go the final yard and reassemble this thing into a vector?

4. Jan 22, 2012

### Fredrik

Staff Emeritus
You did exactly what I had in mind. I didn't think about nice ways to rewrite the final result before. Hm, I don't think there is a super nice way to do it. It can be interpreted as row i of the result of the matrix multiplication
$$\begin{pmatrix} \partial_1 A_1 & \partial_1 A_2 & \cdots\\ \partial_2 A_1 & \ddots\\ \vdots \end{pmatrix} \begin{pmatrix} C_1\\ \vdots \end{pmatrix}$$ Maybe that's as nice as it gets.

Edit: That square matrix is the transpose of the Jacobian matrix of A, so we can write the final result as $$A\times(\nabla\times C)=\big(\partial_i(A_j C_j)-(\partial_i A_j)C_j-A_j\partial_j C_i\big)e_i =\nabla(A\cdot C)-(J_A)^TC-(A\cdot\nabla)C,$$ where $(J_A)^T$ is the linear operator corresponding to the transpose of the Jacobian matrix of A.

Last edited: Jan 23, 2012
5. Jan 23, 2012

### Fredrik

Staff Emeritus
Oh, and when I said "chain rule", I actually meant "product rule". But it doesn't look like that caused any confusion.