1. Apr 1, 2008

### chuckschuldiner

Hi
I am a Mech Engg student trying to study how stress is defined in quantum mechanics. I am referring to a paper where the following identity is given but i am not sure how to go about proving it
The identity is
$$$\left\{ {\hat A,\left[ {\nabla _i \nabla _i ,\delta \left( {{\bf{\hat r}} - {\bf{R}}} \right)} \right]} \right\} = - \nabla _{R,i} \left\{ {\hat A,\left\{ {\nabla _i ,\delta \left( {{\bf{\hat r}} - {\bf{R}}} \right)} \right\}} \right\}$$$

Here, $$$\nabla _i = - \frac{{\hat p_i }}{{i\hbar }}$$$

Also, $$\hat{A}$$ is any operator.

My LHS is

$$$\hat A\nabla _i \nabla _i \delta \left( {{\bf{\hat r}} - {\bf{R}}} \right) - \hat A\delta \left( {{\bf{\hat r}} - {\bf{R}}} \right)\nabla _i \nabla _i + \nabla _i \nabla _i \delta \left( {{\bf{\hat r}} - {\bf{R}}} \right)\hat A - \delta \left( {{\bf{\hat r}} - {\bf{R}}} \right)\nabla _i \nabla _i \hat A$$$

My RHS is

$$$- \nabla _{R,i} \left( {\hat A\nabla _i \delta \left( {{\bf{\hat r}} - {\bf{R}}} \right) + \hat A\delta \left( {{\bf{\hat r}} - {\bf{R}}} \right)\nabla _i + \nabla _i \delta \left( {{\bf{\hat r}} - {\bf{R}}} \right)\hat A + \delta \left( {{\bf{\hat r}} - {\bf{R}}} \right)\nabla _i A} \right)$$$

For the 1st term of the RHS, since $$\hat{A}$$ and $$\nabla _i$$ are independent of R, i can write it as
$$$\begin{array}{l} - \nabla _{R,i} \hat A\nabla _i \delta \left( {{\bf{\hat r}} - {\bf{R}}} \right) = - \hat A\nabla _i \nabla _{R,i} \delta \left( {{\bf{\hat r}} - {\bf{R}}} \right) \\ But,\nabla _{R,i} \delta \left( {{\bf{\hat r}} - {\bf{R}}} \right) = - \nabla _i \delta \left( {{\bf{\hat r}} - {\bf{R}}} \right) \\ So, - \hat A\nabla _i \nabla _{R,i} \delta \left( {{\bf{\hat r}} - {\bf{R}}} \right) = \hat A\nabla _i \nabla _i \delta \left( {{\bf{\hat r}} - {\bf{R}}} \right) \\ \end{array}$$$

As you can see this matches the 1st term of the LHS

Is this approach correct?

Similarly, i can try the third term of the RHS

$$$\begin{array}{l} - \nabla _{R,i} \nabla _i \delta \left( {{\bf{\hat r}} - {\bf{R}}} \right)\hat A = - \nabla _i \nabla _{R,i} \delta \left( {{\bf{\hat r}} - {\bf{R}}} \right)\hat A \\ {\rm{ = }}\nabla _i \nabla _i \delta \left( {{\bf{\hat r}} - {\bf{R}}} \right)\hat A \\ \end{array}$$$

This matches the 3rd term of the LHS, but i am not sure if i am correct here.

To match the other two terms,i am trying to use

2nd term of the RHS

$$$\begin{array}{l} - \nabla _{R,i} \hat A\delta \left( {{\bf{\hat r}} - {\bf{R}}} \right)\nabla _i = - \hat A\nabla _{R,i} \delta \left( {{\bf{\hat r}} - {\bf{R}}} \right)\nabla _i \\ = - \hat A\nabla _{R,i} \nabla _{R,i} \\ \end{array}$$$

2nd term of the LHS

$$$- \hat A\delta \left( {{\bf{\hat r}} - {\bf{R}}} \right)\nabla _i \nabla _i = - \hat A\nabla _{R,i} \nabla _{R,i}$$$

Similarly, i can prove the equivalence of the 4th terms of the RHS and the LHS
I am not sure but i think if i am violating the commutation principles (especially to match the 2nd terms). Is this correct ?