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















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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
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oldest

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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























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      2 Answers
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      2 Answers
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      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 11 mins ago

























                  answered 1 hour ago









                  Thomas Fritsch

                  18419




                  18419






























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