Non-Relativistic Limit of Klein-Gordon Probability Density












3












$begingroup$


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











$endgroup$








  • 1




    $begingroup$
    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.
    $endgroup$
    – Qmechanic♦
    Dec 23 '18 at 21:39
















3












$begingroup$


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











$endgroup$








  • 1




    $begingroup$
    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.
    $endgroup$
    – Qmechanic♦
    Dec 23 '18 at 21:39














3












3








3


1



$begingroup$


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











$endgroup$




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




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edited Dec 23 '18 at 21:37









Qmechanic♦

103k121861182




103k121861182










asked Dec 23 '18 at 21:21









SimonSimon

771313




771313








  • 1




    $begingroup$
    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.
    $endgroup$
    – Qmechanic♦
    Dec 23 '18 at 21:39














  • 1




    $begingroup$
    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.
    $endgroup$
    – Qmechanic♦
    Dec 23 '18 at 21:39








1




1




$begingroup$
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.
$endgroup$
– Qmechanic♦
Dec 23 '18 at 21:39




$begingroup$
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.
$endgroup$
– Qmechanic♦
Dec 23 '18 at 21:39










2 Answers
2






active

oldest

votes


















3












$begingroup$

The trick is to make the approach for the relativistic Klein-Gordon wave function
$$ Phi(vec{r},t) = Psi(vec{r},t) e^{-imc^2t/hbar} $$
(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
$$
begin{eqnarray}
frac{partialPhi}{partial t} &=& left( frac{partialPsi}{partial t} - frac{imc^2}{hbar}Psi right) e^{-imc^2t/hbar} \
vec{nabla}Phi &=& vec{nabla}Psi e^{-imc^2t/hbar}
end{eqnarray}
$$



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











$endgroup$





















    2












    $begingroup$

    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









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      The trick is to make the approach for the relativistic Klein-Gordon wave function
      $$ Phi(vec{r},t) = Psi(vec{r},t) e^{-imc^2t/hbar} $$
      (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
      $$
      begin{eqnarray}
      frac{partialPhi}{partial t} &=& left( frac{partialPsi}{partial t} - frac{imc^2}{hbar}Psi right) e^{-imc^2t/hbar} \
      vec{nabla}Phi &=& vec{nabla}Psi e^{-imc^2t/hbar}
      end{eqnarray}
      $$



      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











      $endgroup$


















        3












        $begingroup$

        The trick is to make the approach for the relativistic Klein-Gordon wave function
        $$ Phi(vec{r},t) = Psi(vec{r},t) e^{-imc^2t/hbar} $$
        (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
        $$
        begin{eqnarray}
        frac{partialPhi}{partial t} &=& left( frac{partialPsi}{partial t} - frac{imc^2}{hbar}Psi right) e^{-imc^2t/hbar} \
        vec{nabla}Phi &=& vec{nabla}Psi e^{-imc^2t/hbar}
        end{eqnarray}
        $$



        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











        $endgroup$
















          3












          3








          3





          $begingroup$

          The trick is to make the approach for the relativistic Klein-Gordon wave function
          $$ Phi(vec{r},t) = Psi(vec{r},t) e^{-imc^2t/hbar} $$
          (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
          $$
          begin{eqnarray}
          frac{partialPhi}{partial t} &=& left( frac{partialPsi}{partial t} - frac{imc^2}{hbar}Psi right) e^{-imc^2t/hbar} \
          vec{nabla}Phi &=& vec{nabla}Psi e^{-imc^2t/hbar}
          end{eqnarray}
          $$



          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











          $endgroup$



          The trick is to make the approach for the relativistic Klein-Gordon wave function
          $$ Phi(vec{r},t) = Psi(vec{r},t) e^{-imc^2t/hbar} $$
          (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
          $$
          begin{eqnarray}
          frac{partialPhi}{partial t} &=& left( frac{partialPsi}{partial t} - frac{imc^2}{hbar}Psi right) e^{-imc^2t/hbar} \
          vec{nabla}Phi &=& vec{nabla}Psi e^{-imc^2t/hbar}
          end{eqnarray}
          $$



          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 Jan 7 at 23:38

























          answered Dec 24 '18 at 0:54









          Thomas FritschThomas Fritsch

          34929




          34929























              2












              $begingroup$

              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









              $endgroup$


















                2












                $begingroup$

                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









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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









                  $endgroup$



                  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 Dec 24 '18 at 0:32









                  my2ctsmy2cts

                  4,9422618




                  4,9422618






























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