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Homework Help: Please help me prove this identity

  1. Apr 1, 2008 #1
    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
    [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?
     
  2. jcsd
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