In the quantum hamiltonian, why does kinetic energy turn into an operator while potential doesn't?












12












$begingroup$


When we go from the classical many-body hamiltonian



$$H = sum_i frac{vec{p}_i^2}{2m_e} - sum_{i,I} frac{Z_I e^2 }{|vec{r}_i - vec{R}_I|} + frac{1}{2}sum_{i,j} frac{ e^2 }{|vec{r}_i - vec{r}_j|} + sum_I frac{vec{p}_I^2}{2M_I}+ frac{1}{2}sum_{I,J} frac{Z_IZ_J e^2 }{|vec{R}_I - vec{R}_J|}$$



to the quantum many-body hamiltonian



$$H = -sum_i frac{hbar^2}{2m_e}nabla_i^2 - sum_{i,I} frac{Z_I e^2 }{|vec{r}_i - vec{R}_I|} + frac{1}{2}sum_{i,j} frac{ e^2 }{|vec{r}_i - vec{r}_j|} - sum_I frac{hbar^2}{2M_I} nabla_I^2+ frac{1}{2}sum_{I,J} frac{Z_IZ_J e^2 }{|vec{R}_I - vec{R}_J|}$$



only the kinetic energy parts turn into operators. I mean the other parts are also operators but merely numbers.



Why is this the case? My guess is, it has to be with the representation we are working with, but that's as far as I go, I don't know how it affects.



Can someone give a heuristic explanation also?






quantum-mechanics classical-mechanics operators hamiltonian representation-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The potential operator is actually: $hat{V} equiv V(x)hat{I}$ with $hat{I}$ the identity operator for the 1D case here. Often you omit the $hat{I}$.
    $endgroup$
    – Dani
    Mar 11 at 14:57


















12












$begingroup$


When we go from the classical many-body hamiltonian



$$H = sum_i frac{vec{p}_i^2}{2m_e} - sum_{i,I} frac{Z_I e^2 }{|vec{r}_i - vec{R}_I|} + frac{1}{2}sum_{i,j} frac{ e^2 }{|vec{r}_i - vec{r}_j|} + sum_I frac{vec{p}_I^2}{2M_I}+ frac{1}{2}sum_{I,J} frac{Z_IZ_J e^2 }{|vec{R}_I - vec{R}_J|}$$



to the quantum many-body hamiltonian



$$H = -sum_i frac{hbar^2}{2m_e}nabla_i^2 - sum_{i,I} frac{Z_I e^2 }{|vec{r}_i - vec{R}_I|} + frac{1}{2}sum_{i,j} frac{ e^2 }{|vec{r}_i - vec{r}_j|} - sum_I frac{hbar^2}{2M_I} nabla_I^2+ frac{1}{2}sum_{I,J} frac{Z_IZ_J e^2 }{|vec{R}_I - vec{R}_J|}$$



only the kinetic energy parts turn into operators. I mean the other parts are also operators but merely numbers.



Why is this the case? My guess is, it has to be with the representation we are working with, but that's as far as I go, I don't know how it affects.



Can someone give a heuristic explanation also?






quantum-mechanics classical-mechanics operators hamiltonian representation-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The potential operator is actually: $hat{V} equiv V(x)hat{I}$ with $hat{I}$ the identity operator for the 1D case here. Often you omit the $hat{I}$.
    $endgroup$
    – Dani
    Mar 11 at 14:57
















12












12








12





$begingroup$


When we go from the classical many-body hamiltonian



$$H = sum_i frac{vec{p}_i^2}{2m_e} - sum_{i,I} frac{Z_I e^2 }{|vec{r}_i - vec{R}_I|} + frac{1}{2}sum_{i,j} frac{ e^2 }{|vec{r}_i - vec{r}_j|} + sum_I frac{vec{p}_I^2}{2M_I}+ frac{1}{2}sum_{I,J} frac{Z_IZ_J e^2 }{|vec{R}_I - vec{R}_J|}$$



to the quantum many-body hamiltonian



$$H = -sum_i frac{hbar^2}{2m_e}nabla_i^2 - sum_{i,I} frac{Z_I e^2 }{|vec{r}_i - vec{R}_I|} + frac{1}{2}sum_{i,j} frac{ e^2 }{|vec{r}_i - vec{r}_j|} - sum_I frac{hbar^2}{2M_I} nabla_I^2+ frac{1}{2}sum_{I,J} frac{Z_IZ_J e^2 }{|vec{R}_I - vec{R}_J|}$$



only the kinetic energy parts turn into operators. I mean the other parts are also operators but merely numbers.



Why is this the case? My guess is, it has to be with the representation we are working with, but that's as far as I go, I don't know how it affects.



Can someone give a heuristic explanation also?






quantum-mechanics classical-mechanics operators hamiltonian representation-theory







share|cite|improve this question











$endgroup$




When we go from the classical many-body hamiltonian



$$H = sum_i frac{vec{p}_i^2}{2m_e} - sum_{i,I} frac{Z_I e^2 }{|vec{r}_i - vec{R}_I|} + frac{1}{2}sum_{i,j} frac{ e^2 }{|vec{r}_i - vec{r}_j|} + sum_I frac{vec{p}_I^2}{2M_I}+ frac{1}{2}sum_{I,J} frac{Z_IZ_J e^2 }{|vec{R}_I - vec{R}_J|}$$



to the quantum many-body hamiltonian



$$H = -sum_i frac{hbar^2}{2m_e}nabla_i^2 - sum_{i,I} frac{Z_I e^2 }{|vec{r}_i - vec{R}_I|} + frac{1}{2}sum_{i,j} frac{ e^2 }{|vec{r}_i - vec{r}_j|} - sum_I frac{hbar^2}{2M_I} nabla_I^2+ frac{1}{2}sum_{I,J} frac{Z_IZ_J e^2 }{|vec{R}_I - vec{R}_J|}$$



only the kinetic energy parts turn into operators. I mean the other parts are also operators but merely numbers.



Why is this the case? My guess is, it has to be with the representation we are working with, but that's as far as I go, I don't know how it affects.



Can someone give a heuristic explanation also?






quantum-mechanics classical-mechanics operators hamiltonian representation-theory




quantum-mechanics classical-mechanics operators hamiltonian representation-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 11 at 10:10









Qmechanic♦

106k121961223




106k121961223










asked Mar 11 at 9:00









UriAcevesUriAceves

858




858








  • 2




    $begingroup$
    The potential operator is actually: $hat{V} equiv V(x)hat{I}$ with $hat{I}$ the identity operator for the 1D case here. Often you omit the $hat{I}$.
    $endgroup$
    – Dani
    Mar 11 at 14:57
















  • 2




    $begingroup$
    The potential operator is actually: $hat{V} equiv V(x)hat{I}$ with $hat{I}$ the identity operator for the 1D case here. Often you omit the $hat{I}$.
    $endgroup$
    – Dani
    Mar 11 at 14:57










2




2




$begingroup$
The potential operator is actually: $hat{V} equiv V(x)hat{I}$ with $hat{I}$ the identity operator for the 1D case here. Often you omit the $hat{I}$.
$endgroup$
– Dani
Mar 11 at 14:57






$begingroup$
The potential operator is actually: $hat{V} equiv V(x)hat{I}$ with $hat{I}$ the identity operator for the 1D case here. Often you omit the $hat{I}$.
$endgroup$
– Dani
Mar 11 at 14:57












2 Answers
2






active

oldest

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21












$begingroup$

Actually the potential is also an operator. It just so happens that, in the position representation, $hat xpsi(x)=xpsi(x)$, so that the potential energy operator $V(hat x)$ acts by multiplication: $V(hat x)psi(x)=V(x)psi(x)$.



Even though it acts “by simple multiplication”, it doesn’t commute with the momentum, a sign that it is still a legitimate operator.



In the momentum representation $hat x$ acts by differentiation so in this case the potential energy operator becomes a (usually quite complicated) differential operator since one needs to use the formal expansion of potential to convert it to a polynomial.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Pseudodifferential operators are not complicated, since they are conveniently defined by Fourier transform. And acting on $L^2$, as in quantum mechanics, they can even be defined directly by spectral calculus. No power expansion (formal or not) of the function is needed, only measurability (essentially with respect to the Lebesgue measure).
    $endgroup$
    – yuggib
    Mar 11 at 12:30



















15












$begingroup$

OP's hunch is exactly right: In QM they are all operators. However each operator may become a multiplication operator in certain representations. E.g. position operators become multiplication operators in the Schrödinger position representation, while momentum operators become multiplication operators in the momentum representation, and so forth.






share|cite|improve this answer











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    2 Answers
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    2 Answers
    2






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    21












    $begingroup$

    Actually the potential is also an operator. It just so happens that, in the position representation, $hat xpsi(x)=xpsi(x)$, so that the potential energy operator $V(hat x)$ acts by multiplication: $V(hat x)psi(x)=V(x)psi(x)$.



    Even though it acts “by simple multiplication”, it doesn’t commute with the momentum, a sign that it is still a legitimate operator.



    In the momentum representation $hat x$ acts by differentiation so in this case the potential energy operator becomes a (usually quite complicated) differential operator since one needs to use the formal expansion of potential to convert it to a polynomial.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Pseudodifferential operators are not complicated, since they are conveniently defined by Fourier transform. And acting on $L^2$, as in quantum mechanics, they can even be defined directly by spectral calculus. No power expansion (formal or not) of the function is needed, only measurability (essentially with respect to the Lebesgue measure).
      $endgroup$
      – yuggib
      Mar 11 at 12:30
















    21












    $begingroup$

    Actually the potential is also an operator. It just so happens that, in the position representation, $hat xpsi(x)=xpsi(x)$, so that the potential energy operator $V(hat x)$ acts by multiplication: $V(hat x)psi(x)=V(x)psi(x)$.



    Even though it acts “by simple multiplication”, it doesn’t commute with the momentum, a sign that it is still a legitimate operator.



    In the momentum representation $hat x$ acts by differentiation so in this case the potential energy operator becomes a (usually quite complicated) differential operator since one needs to use the formal expansion of potential to convert it to a polynomial.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Pseudodifferential operators are not complicated, since they are conveniently defined by Fourier transform. And acting on $L^2$, as in quantum mechanics, they can even be defined directly by spectral calculus. No power expansion (formal or not) of the function is needed, only measurability (essentially with respect to the Lebesgue measure).
      $endgroup$
      – yuggib
      Mar 11 at 12:30














    21












    21








    21





    $begingroup$

    Actually the potential is also an operator. It just so happens that, in the position representation, $hat xpsi(x)=xpsi(x)$, so that the potential energy operator $V(hat x)$ acts by multiplication: $V(hat x)psi(x)=V(x)psi(x)$.



    Even though it acts “by simple multiplication”, it doesn’t commute with the momentum, a sign that it is still a legitimate operator.



    In the momentum representation $hat x$ acts by differentiation so in this case the potential energy operator becomes a (usually quite complicated) differential operator since one needs to use the formal expansion of potential to convert it to a polynomial.






    share|cite|improve this answer









    $endgroup$



    Actually the potential is also an operator. It just so happens that, in the position representation, $hat xpsi(x)=xpsi(x)$, so that the potential energy operator $V(hat x)$ acts by multiplication: $V(hat x)psi(x)=V(x)psi(x)$.



    Even though it acts “by simple multiplication”, it doesn’t commute with the momentum, a sign that it is still a legitimate operator.



    In the momentum representation $hat x$ acts by differentiation so in this case the potential energy operator becomes a (usually quite complicated) differential operator since one needs to use the formal expansion of potential to convert it to a polynomial.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 11 at 10:06









    ZeroTheHeroZeroTheHero

    21k53363




    21k53363








    • 1




      $begingroup$
      Pseudodifferential operators are not complicated, since they are conveniently defined by Fourier transform. And acting on $L^2$, as in quantum mechanics, they can even be defined directly by spectral calculus. No power expansion (formal or not) of the function is needed, only measurability (essentially with respect to the Lebesgue measure).
      $endgroup$
      – yuggib
      Mar 11 at 12:30














    • 1




      $begingroup$
      Pseudodifferential operators are not complicated, since they are conveniently defined by Fourier transform. And acting on $L^2$, as in quantum mechanics, they can even be defined directly by spectral calculus. No power expansion (formal or not) of the function is needed, only measurability (essentially with respect to the Lebesgue measure).
      $endgroup$
      – yuggib
      Mar 11 at 12:30








    1




    1




    $begingroup$
    Pseudodifferential operators are not complicated, since they are conveniently defined by Fourier transform. And acting on $L^2$, as in quantum mechanics, they can even be defined directly by spectral calculus. No power expansion (formal or not) of the function is needed, only measurability (essentially with respect to the Lebesgue measure).
    $endgroup$
    – yuggib
    Mar 11 at 12:30




    $begingroup$
    Pseudodifferential operators are not complicated, since they are conveniently defined by Fourier transform. And acting on $L^2$, as in quantum mechanics, they can even be defined directly by spectral calculus. No power expansion (formal or not) of the function is needed, only measurability (essentially with respect to the Lebesgue measure).
    $endgroup$
    – yuggib
    Mar 11 at 12:30











    15












    $begingroup$

    OP's hunch is exactly right: In QM they are all operators. However each operator may become a multiplication operator in certain representations. E.g. position operators become multiplication operators in the Schrödinger position representation, while momentum operators become multiplication operators in the momentum representation, and so forth.






    share|cite|improve this answer











    $endgroup$


















      15












      $begingroup$

      OP's hunch is exactly right: In QM they are all operators. However each operator may become a multiplication operator in certain representations. E.g. position operators become multiplication operators in the Schrödinger position representation, while momentum operators become multiplication operators in the momentum representation, and so forth.






      share|cite|improve this answer











      $endgroup$
















        15












        15








        15





        $begingroup$

        OP's hunch is exactly right: In QM they are all operators. However each operator may become a multiplication operator in certain representations. E.g. position operators become multiplication operators in the Schrödinger position representation, while momentum operators become multiplication operators in the momentum representation, and so forth.






        share|cite|improve this answer











        $endgroup$



        OP's hunch is exactly right: In QM they are all operators. However each operator may become a multiplication operator in certain representations. E.g. position operators become multiplication operators in the Schrödinger position representation, while momentum operators become multiplication operators in the momentum representation, and so forth.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Mar 11 at 11:14

























        answered Mar 11 at 10:07









        Qmechanic♦Qmechanic

        106k121961223




        106k121961223






























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