1. Jun 13, 2014

### carllacan

Hi.

$\hat{\vec{\nabla}}(\hat{\vec{A}})f=(\vec{\nabla}\cdot\vec{A})f + \vec{A}\cdot(\vec{\nabla}f)$ for an arbitrary vector operator $\hat{\vec{A}}$

So if we set $\vec{A} = \vec{\nabla}$ this should be correct

$\hat{\vec{\nabla}}(\hat{\vec{\nabla}})f=(\vec{\nabla}\cdot\vec{\nabla})f + \vec{\nabla}\cdot(\vec{\nabla}f) = 2\vec{\nabla}^2f$, but apparently its not. Why?

I mean, $grad( div(f)) = div(grad(f)) = \Delta f$, right?

Where did I go wrong?

2. Jun 13, 2014

### Erland

First, it is not clear to me what an "arbitrary vector operator" is. What is its general definition?

Second, it is certainly wrong that grad(div(f))=div(grad(f)). The left side is not even defined, since div(F) is only defined for vector fields, not scalar functions.

3. Jun 14, 2014

### carllacan

By vector operator I mean an operator represented by a vector that, if applied to a scalar, works by multiplying itself for it, and if applied to a vector works by dot-multiplying itself with the vector.

To add some context this doubt comes from here: https://www.physicsforums.com/showthread.php?t=754798 When I try to develop the square in the Hamiltonian operator I have to apply $\hat{\nabla}$ to itself and to $\vec{A}$.

In the $\hat{\nabla}(\vec{A})$ case I found that according to the rules of differentiation this product is $\hat{\vec{\nabla}}(\hat{\vec{A}})f=(\vec{\nabla}\cdot\vec{A})f + \vec{A}\cdot(\vec{\nabla}f)$, and I understand why it is so.

However its not the same with $\hat{\nabla}(\hat{\nabla})$, and I don´t see why.

PD: You´re right about the grads and divs, maybe it should be $laplacian(f) = div(grad(f)) = \Delta f$?

4. Jun 14, 2014

### chogg

In geometric calculus, div(f) is perfectly well-defined for a scalar function f: it is zero everywhere. See Macdonald's excellent text for more details.