How Does the Curl Operator Relate to Angular Momentum in Vector Calculus?

[rado5]

Homework Statement

I want to prove this: $\vec{r} \nabla^2 - \nabla(1+r \frac{\partial}{\partial {r}})=i \nabla \times \vec{L}$

Homework Equations

BAC-CAB rule:

$\nabla \times (\vec{A} \times \vec{B}) = (\vec{B} . \nabla) \vec{A} - (\vec{A} . \nabla ) \vec{B} - \vec{B} (\nabla . \vec{A} ) + \vec{A} (\nabla . \vec{B})$

The Attempt at a Solution

We know that $i \nabla \times \vec{L}= \nabla \times (\vec{r} \times \nabla)$.

Now how can I continue with $\nabla \times (\vec{r} \times\nabla) =$?

Can I use BAC-CAB rule? When I use BAC-CAB, I will have this: $\nabla \times (\vec{r} \times\nabla) = \vec{r} (\nabla . \nabla) - \nabla (\nabla . \vec{r}) + (\nabla . \nabla) \vec{r}- (\vec{r} . \nabla) \nabla$

And as we know $\nabla (\nabla . \vec{r}) = \vec{0}$.

Now how can I continue? If my method is wrong, please tell me why!

 

[mighty2000]

Thank you for your interest in proving the given equation. I would like to suggest a different approach to solving this problem.

Firstly, let's rewrite the given equation in a more general form:

\vec{A} \nabla^2 - \nabla(\vec{B}) = i \nabla \times \vec{C}

where \vec{A} = \vec{r}, \vec{B} = 1 + r \frac{\partial}{\partial r}, and \vec{C} = \vec{L}.

Now, let's use the identity \nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - \nabla^2 \vec{A} to rewrite the left-hand side of the equation:

\vec{A} \nabla^2 - \nabla(\vec{B}) = \vec{A} \nabla^2 - \nabla(\nabla \cdot \vec{B}) + \nabla \times (\nabla \times \vec{B})

Substituting the expressions for \vec{A} and \vec{B}, we get:

\vec{r} \nabla^2 - \nabla(1 + r \frac{\partial}{\partial r}) = \vec{r} \nabla^2 - \nabla(\nabla \cdot (1 + r \frac{\partial}{\partial r})) + \nabla \times (\nabla \times (1 + r \frac{\partial}{\partial r}))

Using the product rule for differentiation, we can expand the second term:

\nabla \cdot (1 + r \frac{\partial}{\partial r}) = \nabla \cdot 1 + \nabla \cdot (r \frac{\partial}{\partial r}) = 0 + r \nabla \cdot (\frac{\partial}{\partial r})

Since \nabla \cdot (\frac{\partial}{\partial r}) = 0, we can simplify the above expression to:

\nabla \cdot (1 + r \frac{\partial}{\partial r}) = r \nabla \cdot