Non-Relativistic Limit of Klein-Gordon Probability Density












2














In the lecture notes accompanying an introductory course in relativistic quantum mechanics, the Klein-Gordon probability density and current are defined as:
$$
begin{eqnarray}
P & = & dfrac{ihbar}{2mc^2}left(Phi^*dfrac{partialPhi}{partial t}-Phidfrac{partialPhi^*}{partial t}right) \
vec{j} &=& dfrac{hbar}{2mi}left(Phi^*vec{nabla}Phi-Phivec{nabla}Phi^*right)
end{eqnarray}
$$
together with the statement that:




One can show that in the non-relativistic limit, the known expressions for the probability density and current are recovered.




The 'known' expressions are:
$$
begin{eqnarray}
rho &=& Psi^*Psi \
vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
end{eqnarray}
$$



When taking a 'non-relativistic limit', I am used to taking the limit $c to infty$, which does give the right result for $vec{j}$, but for the density produces $P=0$. How should one then take said limit to recover the non-relativistic equations?






quantum-mechanics wavefunction speed-of-light schroedinger-equation klein-gordon-equation







share|cite|improve this question




















  • 1




    For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein.
    – Qmechanic♦
    5 hours ago
















2














In the lecture notes accompanying an introductory course in relativistic quantum mechanics, the Klein-Gordon probability density and current are defined as:
$$
begin{eqnarray}
P & = & dfrac{ihbar}{2mc^2}left(Phi^*dfrac{partialPhi}{partial t}-Phidfrac{partialPhi^*}{partial t}right) \
vec{j} &=& dfrac{hbar}{2mi}left(Phi^*vec{nabla}Phi-Phivec{nabla}Phi^*right)
end{eqnarray}
$$
together with the statement that:




One can show that in the non-relativistic limit, the known expressions for the probability density and current are recovered.




The 'known' expressions are:
$$
begin{eqnarray}
rho &=& Psi^*Psi \
vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
end{eqnarray}
$$



When taking a 'non-relativistic limit', I am used to taking the limit $c to infty$, which does give the right result for $vec{j}$, but for the density produces $P=0$. How should one then take said limit to recover the non-relativistic equations?






quantum-mechanics wavefunction speed-of-light schroedinger-equation klein-gordon-equation







share|cite|improve this question




















  • 1




    For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein.
    – Qmechanic♦
    5 hours ago














2












2








2


1





In the lecture notes accompanying an introductory course in relativistic quantum mechanics, the Klein-Gordon probability density and current are defined as:
$$
begin{eqnarray}
P & = & dfrac{ihbar}{2mc^2}left(Phi^*dfrac{partialPhi}{partial t}-Phidfrac{partialPhi^*}{partial t}right) \
vec{j} &=& dfrac{hbar}{2mi}left(Phi^*vec{nabla}Phi-Phivec{nabla}Phi^*right)
end{eqnarray}
$$
together with the statement that:




One can show that in the non-relativistic limit, the known expressions for the probability density and current are recovered.




The 'known' expressions are:
$$
begin{eqnarray}
rho &=& Psi^*Psi \
vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
end{eqnarray}
$$



When taking a 'non-relativistic limit', I am used to taking the limit $c to infty$, which does give the right result for $vec{j}$, but for the density produces $P=0$. How should one then take said limit to recover the non-relativistic equations?






quantum-mechanics wavefunction speed-of-light schroedinger-equation klein-gordon-equation







share|cite|improve this question















In the lecture notes accompanying an introductory course in relativistic quantum mechanics, the Klein-Gordon probability density and current are defined as:
$$
begin{eqnarray}
P & = & dfrac{ihbar}{2mc^2}left(Phi^*dfrac{partialPhi}{partial t}-Phidfrac{partialPhi^*}{partial t}right) \
vec{j} &=& dfrac{hbar}{2mi}left(Phi^*vec{nabla}Phi-Phivec{nabla}Phi^*right)
end{eqnarray}
$$
together with the statement that:




One can show that in the non-relativistic limit, the known expressions for the probability density and current are recovered.




The 'known' expressions are:
$$
begin{eqnarray}
rho &=& Psi^*Psi \
vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
end{eqnarray}
$$



When taking a 'non-relativistic limit', I am used to taking the limit $c to infty$, which does give the right result for $vec{j}$, but for the density produces $P=0$. How should one then take said limit to recover the non-relativistic equations?






quantum-mechanics wavefunction speed-of-light schroedinger-equation klein-gordon-equation




quantum-mechanics wavefunction speed-of-light schroedinger-equation klein-gordon-equation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 5 hours ago









Qmechanic♦

101k121821140




101k121821140










asked 5 hours ago









Simon

755313




755313








  • 1




    For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein.
    – Qmechanic♦
    5 hours ago














  • 1




    For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein.
    – Qmechanic♦
    5 hours ago








1




1




For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein.
– Qmechanic♦
5 hours ago




For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein.
– Qmechanic♦
5 hours ago










2 Answers
2






active

oldest

votes


















2














You can substitute $Phi = e^{-mc^2t/hbar} Psi$ and then neglect the second order time derivative of $Psi$. Drop the constant $mc^2$ and you will have recovered the Schrödinger equation.






share|cite|improve this answer





























    1














    The trick is to make the approach for the relativistic Klein-Gordon wave function
    $$ Phi(vec{r},t) = Psi(vec{r},t) e^{-frac{imc^2t}{hbar}} $$
    (The physical reasoning behind this approach is:
    The fast oscillating exponential is the solution for the particle at rest.
    And compared to that, $Psi$ will give only slow variations.)



    From that you find its derivatives
    $$
    begin{eqnarray}
    frac{partialPhi}{partial t} &=& left( frac{partialPsi}{partial t} - frac{imc^2}{hbar}Psi right) e^{-frac{imc^2t}{hbar}} \
    vec{nabla}Phi &=& vec{nabla}Psi e^{-frac{imc^2t}{hbar}}
    end{eqnarray}
    $$



    Plug these into the definitions of the Klein-Gordon probability density and current ($P$ and $vec{j}$) and you get
    $$
    begin{eqnarray}
    P &=& Psi^* Psi + frac{ihbar}{2mc^2} left( Psi^*frac{partialPsi}{partial t} - Psi frac{partialPsi^*}{partial t} right) \
    vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
    end{eqnarray}
    $$



    Now you can do the limit $c to infty$ and get the 'known' expressions of the non-relativistic Schrödinger probability density and current.






    share|cite|improve this answer























      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: "151"
      };
      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: false,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: null,
      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%2fphysics.stackexchange.com%2fquestions%2f450076%2fnon-relativistic-limit-of-klein-gordon-probability-density%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









      2














      You can substitute $Phi = e^{-mc^2t/hbar} Psi$ and then neglect the second order time derivative of $Psi$. Drop the constant $mc^2$ and you will have recovered the Schrödinger equation.






      share|cite|improve this answer


























        2














        You can substitute $Phi = e^{-mc^2t/hbar} Psi$ and then neglect the second order time derivative of $Psi$. Drop the constant $mc^2$ and you will have recovered the Schrödinger equation.






        share|cite|improve this answer
























          2












          2








          2






          You can substitute $Phi = e^{-mc^2t/hbar} Psi$ and then neglect the second order time derivative of $Psi$. Drop the constant $mc^2$ and you will have recovered the Schrödinger equation.






          share|cite|improve this answer












          You can substitute $Phi = e^{-mc^2t/hbar} Psi$ and then neglect the second order time derivative of $Psi$. Drop the constant $mc^2$ and you will have recovered the Schrödinger equation.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          my2cts

          4,4482617




          4,4482617























              1














              The trick is to make the approach for the relativistic Klein-Gordon wave function
              $$ Phi(vec{r},t) = Psi(vec{r},t) e^{-frac{imc^2t}{hbar}} $$
              (The physical reasoning behind this approach is:
              The fast oscillating exponential is the solution for the particle at rest.
              And compared to that, $Psi$ will give only slow variations.)



              From that you find its derivatives
              $$
              begin{eqnarray}
              frac{partialPhi}{partial t} &=& left( frac{partialPsi}{partial t} - frac{imc^2}{hbar}Psi right) e^{-frac{imc^2t}{hbar}} \
              vec{nabla}Phi &=& vec{nabla}Psi e^{-frac{imc^2t}{hbar}}
              end{eqnarray}
              $$



              Plug these into the definitions of the Klein-Gordon probability density and current ($P$ and $vec{j}$) and you get
              $$
              begin{eqnarray}
              P &=& Psi^* Psi + frac{ihbar}{2mc^2} left( Psi^*frac{partialPsi}{partial t} - Psi frac{partialPsi^*}{partial t} right) \
              vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
              end{eqnarray}
              $$



              Now you can do the limit $c to infty$ and get the 'known' expressions of the non-relativistic Schrödinger probability density and current.






              share|cite|improve this answer




























                1














                The trick is to make the approach for the relativistic Klein-Gordon wave function
                $$ Phi(vec{r},t) = Psi(vec{r},t) e^{-frac{imc^2t}{hbar}} $$
                (The physical reasoning behind this approach is:
                The fast oscillating exponential is the solution for the particle at rest.
                And compared to that, $Psi$ will give only slow variations.)



                From that you find its derivatives
                $$
                begin{eqnarray}
                frac{partialPhi}{partial t} &=& left( frac{partialPsi}{partial t} - frac{imc^2}{hbar}Psi right) e^{-frac{imc^2t}{hbar}} \
                vec{nabla}Phi &=& vec{nabla}Psi e^{-frac{imc^2t}{hbar}}
                end{eqnarray}
                $$



                Plug these into the definitions of the Klein-Gordon probability density and current ($P$ and $vec{j}$) and you get
                $$
                begin{eqnarray}
                P &=& Psi^* Psi + frac{ihbar}{2mc^2} left( Psi^*frac{partialPsi}{partial t} - Psi frac{partialPsi^*}{partial t} right) \
                vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
                end{eqnarray}
                $$



                Now you can do the limit $c to infty$ and get the 'known' expressions of the non-relativistic Schrödinger probability density and current.






                share|cite|improve this answer


























                  1












                  1








                  1






                  The trick is to make the approach for the relativistic Klein-Gordon wave function
                  $$ Phi(vec{r},t) = Psi(vec{r},t) e^{-frac{imc^2t}{hbar}} $$
                  (The physical reasoning behind this approach is:
                  The fast oscillating exponential is the solution for the particle at rest.
                  And compared to that, $Psi$ will give only slow variations.)



                  From that you find its derivatives
                  $$
                  begin{eqnarray}
                  frac{partialPhi}{partial t} &=& left( frac{partialPsi}{partial t} - frac{imc^2}{hbar}Psi right) e^{-frac{imc^2t}{hbar}} \
                  vec{nabla}Phi &=& vec{nabla}Psi e^{-frac{imc^2t}{hbar}}
                  end{eqnarray}
                  $$



                  Plug these into the definitions of the Klein-Gordon probability density and current ($P$ and $vec{j}$) and you get
                  $$
                  begin{eqnarray}
                  P &=& Psi^* Psi + frac{ihbar}{2mc^2} left( Psi^*frac{partialPsi}{partial t} - Psi frac{partialPsi^*}{partial t} right) \
                  vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
                  end{eqnarray}
                  $$



                  Now you can do the limit $c to infty$ and get the 'known' expressions of the non-relativistic Schrödinger probability density and current.






                  share|cite|improve this answer














                  The trick is to make the approach for the relativistic Klein-Gordon wave function
                  $$ Phi(vec{r},t) = Psi(vec{r},t) e^{-frac{imc^2t}{hbar}} $$
                  (The physical reasoning behind this approach is:
                  The fast oscillating exponential is the solution for the particle at rest.
                  And compared to that, $Psi$ will give only slow variations.)



                  From that you find its derivatives
                  $$
                  begin{eqnarray}
                  frac{partialPhi}{partial t} &=& left( frac{partialPsi}{partial t} - frac{imc^2}{hbar}Psi right) e^{-frac{imc^2t}{hbar}} \
                  vec{nabla}Phi &=& vec{nabla}Psi e^{-frac{imc^2t}{hbar}}
                  end{eqnarray}
                  $$



                  Plug these into the definitions of the Klein-Gordon probability density and current ($P$ and $vec{j}$) and you get
                  $$
                  begin{eqnarray}
                  P &=& Psi^* Psi + frac{ihbar}{2mc^2} left( Psi^*frac{partialPsi}{partial t} - Psi frac{partialPsi^*}{partial t} right) \
                  vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
                  end{eqnarray}
                  $$



                  Now you can do the limit $c to infty$ and get the 'known' expressions of the non-relativistic Schrödinger probability density and current.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 21 mins ago

























                  answered 2 hours ago









                  Thomas Fritsch

                  18419




                  18419






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Physics 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.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • 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.


                      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%2fphysics.stackexchange.com%2fquestions%2f450076%2fnon-relativistic-limit-of-klein-gordon-probability-density%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

                      404 Error Contact Form 7 ajax form submitting

                      Image
                      0 The contact form does not work, and affects loading after one try. The chrome console shows the error message: POST http://example.com/wp-json/contact-form-7/v1/contact-forms/50/feedback 404 (Not Found) send @ jquery.js?ver=1.12.4:4 ajax @ jquery.js?ver=1.12.4:4 wpcf7.submit @ scripts.js?ver=5.0.3:346 (anonymous) @ scripts.js?ver=5.0.3:53 dispatch @ jquery.js?ver=1.12.4:3 r.handle @ jquery.js?ver=1.12.4:3 How to solve this issue? thanks in advance javascript wordpress contact-form-7 share | improve this question edited Jul 17 '18 at 5:39 Kiran Shahi 4,745 3 16 43 ...
                      Read more

                      How to know if a Active Directory user can login interactively

                      Image
                      1 I would like to know if and how is it possible to know if a AD user can login interactively (on a server) in a domain. I need to know if I can find it out using a LDAP search. Thanks. windows active-directory login user-permissions user-profile share | improve this question edited 1 hour ago yagmoth555 ♦ 11.6k 3 17 42 asked 1 hour ago Luigi Luigi 6 1 New contributor ...
                      Read more

                      Refactoring coordinates for Minecraft Pi buildings written in Python

                      Image
                      0 $begingroup$ I'm fairly new to coding and I wrote this script in Python for a house with a roof in Minecraft Pi. I've found different ways to write the coordinates to make the code base shorter, but I'm wondering if it is possible for me to refactor it any further so I don't have to list each coordinate line by line. import mcpi.minecraft as minecraft import mcpi.block as block mc = minecraft.Minecraft.create() p = mc.player.getTilePos() mc.setBlocks(p.x+1, p.y, p.z+1, #Main structure starting coordinates (DO NOT ALTER) p.x+19, p.y+4, p.z+6, #Structure size/coordinates (Alters cube size) block.STONE_BRICK) mc.setBlocks(p.x+2, p.y+1, p.z+2, #Hollowing out main structure interior p.x+18, p.y+3, p.z+5, block.AIR) mc.setBlocks(p.x, p.y...
                      Read more