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Is the evolution operator well-defined mathematically?


The formal solution of the time-dependent Schrödinger equationWhat exactly does the Hamiltonian operator tell us?How to describe time evolution in relativistic QFT?Prove time-dependent hamiltonian is hermitian from unitarity of time-evolution operatorEvolution operatorWhat's the reasons to use time-ordering operator?Time-dependent quantum mechanical pictureQM: Time evolution with $H = H(t)$Solving the Schrodinger equation with a time-dependent HamiltonianTime evolution operator in QM






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$begingroup$


We know that in order to solve the time-dependent Schrodinger equation $ipartial_t psi = H(t) psi$, we need the evolution operator
$$U(t) = T expleft(-iint_0^t H(t')dt'right)$$
where $T$ is the time ordering operator and the right hand side (RHS) denotes a formal summation.
My question: Does the evolution operator exist in a mathematically rigorous sense? We may deal with imaginary time if necessary.










share|cite|improve this question











$endgroup$




















    8












    $begingroup$


    We know that in order to solve the time-dependent Schrodinger equation $ipartial_t psi = H(t) psi$, we need the evolution operator
    $$U(t) = T expleft(-iint_0^t H(t')dt'right)$$
    where $T$ is the time ordering operator and the right hand side (RHS) denotes a formal summation.
    My question: Does the evolution operator exist in a mathematically rigorous sense? We may deal with imaginary time if necessary.










    share|cite|improve this question











    $endgroup$
















      8












      8








      8


      2



      $begingroup$


      We know that in order to solve the time-dependent Schrodinger equation $ipartial_t psi = H(t) psi$, we need the evolution operator
      $$U(t) = T expleft(-iint_0^t H(t')dt'right)$$
      where $T$ is the time ordering operator and the right hand side (RHS) denotes a formal summation.
      My question: Does the evolution operator exist in a mathematically rigorous sense? We may deal with imaginary time if necessary.










      share|cite|improve this question











      $endgroup$




      We know that in order to solve the time-dependent Schrodinger equation $ipartial_t psi = H(t) psi$, we need the evolution operator
      $$U(t) = T expleft(-iint_0^t H(t')dt'right)$$
      where $T$ is the time ordering operator and the right hand side (RHS) denotes a formal summation.
      My question: Does the evolution operator exist in a mathematically rigorous sense? We may deal with imaginary time if necessary.







      hilbert-space operators mathematical-physics schroedinger-equation time-evolution






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      edited Aug 9 at 7:58









      Qmechanic

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      asked Aug 9 at 6:57









      Andrew YuanAndrew Yuan

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          If $H(t)$ is selfadjoint and bounded (thus everywhere defined), the theory is standard and quite easy to handle using the uniform operator topology. You can find all required proofs in the first or second volume of Reed and Simon's textbook on mathematical methods. If the operators $H(t)$ are unbounded, the theory is much more difficult also in view of evident problems with domains and it uses the strong operator topology. I am sure that there are classical results by Kato in his celebrated book about the general context ("Perturbation theory of linear operators"), but I suspect there are further more modern results.






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            1 Answer
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            $begingroup$

            If $H(t)$ is selfadjoint and bounded (thus everywhere defined), the theory is standard and quite easy to handle using the uniform operator topology. You can find all required proofs in the first or second volume of Reed and Simon's textbook on mathematical methods. If the operators $H(t)$ are unbounded, the theory is much more difficult also in view of evident problems with domains and it uses the strong operator topology. I am sure that there are classical results by Kato in his celebrated book about the general context ("Perturbation theory of linear operators"), but I suspect there are further more modern results.






            share|cite|improve this answer











            $endgroup$



















              10











              $begingroup$

              If $H(t)$ is selfadjoint and bounded (thus everywhere defined), the theory is standard and quite easy to handle using the uniform operator topology. You can find all required proofs in the first or second volume of Reed and Simon's textbook on mathematical methods. If the operators $H(t)$ are unbounded, the theory is much more difficult also in view of evident problems with domains and it uses the strong operator topology. I am sure that there are classical results by Kato in his celebrated book about the general context ("Perturbation theory of linear operators"), but I suspect there are further more modern results.






              share|cite|improve this answer











              $endgroup$

















                10












                10








                10





                $begingroup$

                If $H(t)$ is selfadjoint and bounded (thus everywhere defined), the theory is standard and quite easy to handle using the uniform operator topology. You can find all required proofs in the first or second volume of Reed and Simon's textbook on mathematical methods. If the operators $H(t)$ are unbounded, the theory is much more difficult also in view of evident problems with domains and it uses the strong operator topology. I am sure that there are classical results by Kato in his celebrated book about the general context ("Perturbation theory of linear operators"), but I suspect there are further more modern results.






                share|cite|improve this answer











                $endgroup$



                If $H(t)$ is selfadjoint and bounded (thus everywhere defined), the theory is standard and quite easy to handle using the uniform operator topology. You can find all required proofs in the first or second volume of Reed and Simon's textbook on mathematical methods. If the operators $H(t)$ are unbounded, the theory is much more difficult also in view of evident problems with domains and it uses the strong operator topology. I am sure that there are classical results by Kato in his celebrated book about the general context ("Perturbation theory of linear operators"), but I suspect there are further more modern results.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Aug 9 at 15:04









                John Rennie

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                286k45 gold badges584 silver badges833 bronze badges










                answered Aug 9 at 11:03









                Valter MorettiValter Moretti

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