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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

[tex]

\[

\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\}

\]

[/tex]

Here, [tex]\[

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

\][/tex]

Also, [tex]\hat{A}[/tex] is any operator.

My LHS is

[tex]

\[

\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

\]

[/tex]

My RHS is

[tex]

\[

- \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)

\]

[/tex]

For the 1st term of the RHS, since [tex]\hat{A}[/tex] and [tex]\nabla _i[/tex] are independent of R, i can write it as

[tex]

\[

\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}

\]

[/tex]

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

[tex]

\[

\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}

\]

[/tex]

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

[tex]

\[

\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}

\]

[/tex]

2nd term of the LHS

[tex]

\[

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

\]

[/tex]

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?

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# Homework Help: Please help me prove this identity

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