Evaluation of the commutator $[frac{partial}{partial x_j},frac{x_ix_j-r^2delta_{ij}}{r^3}]=-2x_i/r^3$.












0












$begingroup$


I am trying to evaluate a commutator of the form $[frac{partial}{partial x_j},frac{x_ix_j-r^2delta_{ij}}{r^3}]=-2x_i/r^3$. Bt acting it on a function $f(vec{r})$, I get that the commuator is equal to zero. Can you help?






group-theory quantum-mechanics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Any function $f(vec{r})$.
    $endgroup$
    – mithusengupta123
    Jan 17 at 20:49










  • $begingroup$
    Can you present your calculations? It's not easy to provide constructive feedback if we don't know your approach.
    $endgroup$
    – Luke
    Jan 17 at 21:01
















0












$begingroup$


I am trying to evaluate a commutator of the form $[frac{partial}{partial x_j},frac{x_ix_j-r^2delta_{ij}}{r^3}]=-2x_i/r^3$. Bt acting it on a function $f(vec{r})$, I get that the commuator is equal to zero. Can you help?






group-theory quantum-mechanics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Any function $f(vec{r})$.
    $endgroup$
    – mithusengupta123
    Jan 17 at 20:49










  • $begingroup$
    Can you present your calculations? It's not easy to provide constructive feedback if we don't know your approach.
    $endgroup$
    – Luke
    Jan 17 at 21:01














0












0








0





$begingroup$


I am trying to evaluate a commutator of the form $[frac{partial}{partial x_j},frac{x_ix_j-r^2delta_{ij}}{r^3}]=-2x_i/r^3$. Bt acting it on a function $f(vec{r})$, I get that the commuator is equal to zero. Can you help?






group-theory quantum-mechanics







share|cite|improve this question











$endgroup$




I am trying to evaluate a commutator of the form $[frac{partial}{partial x_j},frac{x_ix_j-r^2delta_{ij}}{r^3}]=-2x_i/r^3$. Bt acting it on a function $f(vec{r})$, I get that the commuator is equal to zero. Can you help?






group-theory quantum-mechanics




group-theory quantum-mechanics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 20:49







mithusengupta123

















asked Jan 17 at 20:41









mithusengupta123mithusengupta123

1139




1139












  • $begingroup$
    Any function $f(vec{r})$.
    $endgroup$
    – mithusengupta123
    Jan 17 at 20:49










  • $begingroup$
    Can you present your calculations? It's not easy to provide constructive feedback if we don't know your approach.
    $endgroup$
    – Luke
    Jan 17 at 21:01


















  • $begingroup$
    Any function $f(vec{r})$.
    $endgroup$
    – mithusengupta123
    Jan 17 at 20:49










  • $begingroup$
    Can you present your calculations? It's not easy to provide constructive feedback if we don't know your approach.
    $endgroup$
    – Luke
    Jan 17 at 21:01
















$begingroup$
Any function $f(vec{r})$.
$endgroup$
– mithusengupta123
Jan 17 at 20:49




$begingroup$
Any function $f(vec{r})$.
$endgroup$
– mithusengupta123
Jan 17 at 20:49












$begingroup$
Can you present your calculations? It's not easy to provide constructive feedback if we don't know your approach.
$endgroup$
– Luke
Jan 17 at 21:01




$begingroup$
Can you present your calculations? It's not easy to provide constructive feedback if we don't know your approach.
$endgroup$
– Luke
Jan 17 at 21:01










2 Answers
2






active

oldest

votes


















1












$begingroup$

Calculate first this version



begin{eqnarray}
require{cancel}
left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right]f &=&
frac{partial}{partial x_k}left(frac{x_i x_j - r^2delta_{ij}}{r^3}f right) - frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k} \
&=& left(frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3}right)f + cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}} - cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}}
end{eqnarray}



So the problem reduces to



begin{eqnarray}
left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=&frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3} \
&=& frac{partial}{partial x_k} frac{x_i x_j}{r^3} - delta_{ij}frac{partial}{partial x_k}frac{1}{r} \
&=& left(frac{delta_{ik}x_j}{r^3} + frac{x_idelta_{jk}}{r^3}- 3frac{x_ix_j x_k}{r^5} right) + delta_{ij}frac{x_k}{r^3} \
&=& frac{1}{r^3}left(delta_{ik} x_j + x_idelta_{jk} + delta_{ij} x_kright) - 3 frac{x_i x_j x_k}{r^5} tag{1}
end{eqnarray}



Now replace $kto j$ and sum over $j$



begin{eqnarray}
left[frac{partial}{partial x_j}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=& frac{1}{r^3} left(delta_{ij} x_j + x_idelta_{jj} + delta_{ij} x_jright) - 3 frac{x_i x_j x_j}{r^5} \
&=& frac{1}{r^3} left(2x_i + 3x_iright) - 3 frac{x_i r^2}{r^5} \
&=& 2frac{x_i}{r^3}
end{eqnarray}






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Consider any mathematical Expression $A$ and apply the commutator on it. Then



    $[frac{partial}{partial x_j},f]A = frac{partial}{partial x_j}(fA)-f frac{partial}{partial x_j}A$



    Note that in this calculation, the first term on the right Hand side is a composition of two operations:



    First multiplication of $f$ with A because the $f$ is directly left next to the $A$



    then applying the differential Operator,since it is left next to all other expressions.



    Finally use product rule and lookwhich factor is in front of $A$. So the result of the commutator $R = [frac{partial}{partial x_j},f]$ is



    $[frac{partial}{partial x_j},f]A = RA$






    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077490%2fevaluation-of-the-commutator-frac-partial-partial-x-j-fracx-ix-j-r2-de%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Calculate first this version



      begin{eqnarray}
      require{cancel}
      left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right]f &=&
      frac{partial}{partial x_k}left(frac{x_i x_j - r^2delta_{ij}}{r^3}f right) - frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k} \
      &=& left(frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3}right)f + cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}} - cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}}
      end{eqnarray}



      So the problem reduces to



      begin{eqnarray}
      left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=&frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3} \
      &=& frac{partial}{partial x_k} frac{x_i x_j}{r^3} - delta_{ij}frac{partial}{partial x_k}frac{1}{r} \
      &=& left(frac{delta_{ik}x_j}{r^3} + frac{x_idelta_{jk}}{r^3}- 3frac{x_ix_j x_k}{r^5} right) + delta_{ij}frac{x_k}{r^3} \
      &=& frac{1}{r^3}left(delta_{ik} x_j + x_idelta_{jk} + delta_{ij} x_kright) - 3 frac{x_i x_j x_k}{r^5} tag{1}
      end{eqnarray}



      Now replace $kto j$ and sum over $j$



      begin{eqnarray}
      left[frac{partial}{partial x_j}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=& frac{1}{r^3} left(delta_{ij} x_j + x_idelta_{jj} + delta_{ij} x_jright) - 3 frac{x_i x_j x_j}{r^5} \
      &=& frac{1}{r^3} left(2x_i + 3x_iright) - 3 frac{x_i r^2}{r^5} \
      &=& 2frac{x_i}{r^3}
      end{eqnarray}






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Calculate first this version



        begin{eqnarray}
        require{cancel}
        left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right]f &=&
        frac{partial}{partial x_k}left(frac{x_i x_j - r^2delta_{ij}}{r^3}f right) - frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k} \
        &=& left(frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3}right)f + cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}} - cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}}
        end{eqnarray}



        So the problem reduces to



        begin{eqnarray}
        left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=&frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3} \
        &=& frac{partial}{partial x_k} frac{x_i x_j}{r^3} - delta_{ij}frac{partial}{partial x_k}frac{1}{r} \
        &=& left(frac{delta_{ik}x_j}{r^3} + frac{x_idelta_{jk}}{r^3}- 3frac{x_ix_j x_k}{r^5} right) + delta_{ij}frac{x_k}{r^3} \
        &=& frac{1}{r^3}left(delta_{ik} x_j + x_idelta_{jk} + delta_{ij} x_kright) - 3 frac{x_i x_j x_k}{r^5} tag{1}
        end{eqnarray}



        Now replace $kto j$ and sum over $j$



        begin{eqnarray}
        left[frac{partial}{partial x_j}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=& frac{1}{r^3} left(delta_{ij} x_j + x_idelta_{jj} + delta_{ij} x_jright) - 3 frac{x_i x_j x_j}{r^5} \
        &=& frac{1}{r^3} left(2x_i + 3x_iright) - 3 frac{x_i r^2}{r^5} \
        &=& 2frac{x_i}{r^3}
        end{eqnarray}






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Calculate first this version



          begin{eqnarray}
          require{cancel}
          left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right]f &=&
          frac{partial}{partial x_k}left(frac{x_i x_j - r^2delta_{ij}}{r^3}f right) - frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k} \
          &=& left(frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3}right)f + cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}} - cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}}
          end{eqnarray}



          So the problem reduces to



          begin{eqnarray}
          left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=&frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3} \
          &=& frac{partial}{partial x_k} frac{x_i x_j}{r^3} - delta_{ij}frac{partial}{partial x_k}frac{1}{r} \
          &=& left(frac{delta_{ik}x_j}{r^3} + frac{x_idelta_{jk}}{r^3}- 3frac{x_ix_j x_k}{r^5} right) + delta_{ij}frac{x_k}{r^3} \
          &=& frac{1}{r^3}left(delta_{ik} x_j + x_idelta_{jk} + delta_{ij} x_kright) - 3 frac{x_i x_j x_k}{r^5} tag{1}
          end{eqnarray}



          Now replace $kto j$ and sum over $j$



          begin{eqnarray}
          left[frac{partial}{partial x_j}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=& frac{1}{r^3} left(delta_{ij} x_j + x_idelta_{jj} + delta_{ij} x_jright) - 3 frac{x_i x_j x_j}{r^5} \
          &=& frac{1}{r^3} left(2x_i + 3x_iright) - 3 frac{x_i r^2}{r^5} \
          &=& 2frac{x_i}{r^3}
          end{eqnarray}






          share|cite|improve this answer









          $endgroup$



          Calculate first this version



          begin{eqnarray}
          require{cancel}
          left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right]f &=&
          frac{partial}{partial x_k}left(frac{x_i x_j - r^2delta_{ij}}{r^3}f right) - frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k} \
          &=& left(frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3}right)f + cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}} - cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}}
          end{eqnarray}



          So the problem reduces to



          begin{eqnarray}
          left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=&frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3} \
          &=& frac{partial}{partial x_k} frac{x_i x_j}{r^3} - delta_{ij}frac{partial}{partial x_k}frac{1}{r} \
          &=& left(frac{delta_{ik}x_j}{r^3} + frac{x_idelta_{jk}}{r^3}- 3frac{x_ix_j x_k}{r^5} right) + delta_{ij}frac{x_k}{r^3} \
          &=& frac{1}{r^3}left(delta_{ik} x_j + x_idelta_{jk} + delta_{ij} x_kright) - 3 frac{x_i x_j x_k}{r^5} tag{1}
          end{eqnarray}



          Now replace $kto j$ and sum over $j$



          begin{eqnarray}
          left[frac{partial}{partial x_j}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=& frac{1}{r^3} left(delta_{ij} x_j + x_idelta_{jj} + delta_{ij} x_jright) - 3 frac{x_i x_j x_j}{r^5} \
          &=& frac{1}{r^3} left(2x_i + 3x_iright) - 3 frac{x_i r^2}{r^5} \
          &=& 2frac{x_i}{r^3}
          end{eqnarray}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 17 at 21:25









          caveraccaverac

          14.6k31130




          14.6k31130























              0












              $begingroup$

              Consider any mathematical Expression $A$ and apply the commutator on it. Then



              $[frac{partial}{partial x_j},f]A = frac{partial}{partial x_j}(fA)-f frac{partial}{partial x_j}A$



              Note that in this calculation, the first term on the right Hand side is a composition of two operations:



              First multiplication of $f$ with A because the $f$ is directly left next to the $A$



              then applying the differential Operator,since it is left next to all other expressions.



              Finally use product rule and lookwhich factor is in front of $A$. So the result of the commutator $R = [frac{partial}{partial x_j},f]$ is



              $[frac{partial}{partial x_j},f]A = RA$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Consider any mathematical Expression $A$ and apply the commutator on it. Then



                $[frac{partial}{partial x_j},f]A = frac{partial}{partial x_j}(fA)-f frac{partial}{partial x_j}A$



                Note that in this calculation, the first term on the right Hand side is a composition of two operations:



                First multiplication of $f$ with A because the $f$ is directly left next to the $A$



                then applying the differential Operator,since it is left next to all other expressions.



                Finally use product rule and lookwhich factor is in front of $A$. So the result of the commutator $R = [frac{partial}{partial x_j},f]$ is



                $[frac{partial}{partial x_j},f]A = RA$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Consider any mathematical Expression $A$ and apply the commutator on it. Then



                  $[frac{partial}{partial x_j},f]A = frac{partial}{partial x_j}(fA)-f frac{partial}{partial x_j}A$



                  Note that in this calculation, the first term on the right Hand side is a composition of two operations:



                  First multiplication of $f$ with A because the $f$ is directly left next to the $A$



                  then applying the differential Operator,since it is left next to all other expressions.



                  Finally use product rule and lookwhich factor is in front of $A$. So the result of the commutator $R = [frac{partial}{partial x_j},f]$ is



                  $[frac{partial}{partial x_j},f]A = RA$






                  share|cite|improve this answer









                  $endgroup$



                  Consider any mathematical Expression $A$ and apply the commutator on it. Then



                  $[frac{partial}{partial x_j},f]A = frac{partial}{partial x_j}(fA)-f frac{partial}{partial x_j}A$



                  Note that in this calculation, the first term on the right Hand side is a composition of two operations:



                  First multiplication of $f$ with A because the $f$ is directly left next to the $A$



                  then applying the differential Operator,since it is left next to all other expressions.



                  Finally use product rule and lookwhich factor is in front of $A$. So the result of the commutator $R = [frac{partial}{partial x_j},f]$ is



                  $[frac{partial}{partial x_j},f]A = RA$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 17 at 21:05









                  kryomaximkryomaxim

                  2,3571825




                  2,3571825






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077490%2fevaluation-of-the-commutator-frac-partial-partial-x-j-fracx-ix-j-r2-de%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

                      Image
                      28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
                      Read more

                      Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

                      Image
                      up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
                      Read more

                      A Topological Invariant for $pi_3(U(n))$

                      Image
                      3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
                      Read more