Proving the Identity in Quantum Mechanics: A Comparison of LHS and RHS

[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 ?

Can somebody please correct me?

 

[Jhero]

Thank you.Hello, thank you for sharing your approach to proving the identity given in the paper. Your approach seems to be on the right track, but there are a few points that could use some clarification.

Firstly, your approach to matching the 2nd term of the RHS with the 2nd term of the LHS is not quite correct. You have correctly identified that the commutator \left[\nabla_i,\delta\left( {{\bf{\hat r}} - {\bf{R}}} \right)\right] will give you a delta function times the gradient operator \nabla_{R,i}. However, in your second step, you have used the product rule incorrectly. The correct way to expand the second term of the RHS would be:

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

The first term in this expansion is the one you have correctly identified, but the second term is not equal to the 2nd term of the LHS. Similarly, the 4th term of the RHS cannot be matched with the 4th term of the LHS using this approach.

To match the 2nd and 4th terms of the RHS with the LHS, you will need to use the fact that the commutator \left[\nabla_i,\hat A\right] is equal to the gradient of the operator \hat A. This means that the 2nd term of the RHS can be expanded as:

\[
\begin{array}{l}
- \nabla _{R,i} \hat A\delta \left