How to prove a relation between the evolution operators in the Schrödinger and interaction pictures of...












0












$begingroup$


Consider a Hamiltonian $H$ of two components, a time-independent $H_0$ and a time-dependent $V(t)$,



$$H(t)=H_0+V(t),.$$



In the Schrödinger picture of quantum mechanics, the state $left|psi(t)rightrangle_{mathrm{S}}$ evolves according to:



$$mathrm{i}hbarfrac{partial}{partial t}left|psi(t)rightrangle_{mathrm{S}}=H(t)left|psi(t)rightrangle_{mathrm{S}},.$$



This equation admits a formal solution given by,
$$left|psi(t)rightrangle_{mathrm{S}}=U(t,t_0)left|psi(t_0)rightrangle_{mathrm{S}},.$$



$U(t,t_0)$ is the evolution operator corresponding to $H(t)$, defined by
$$U(t,t_0)=mathcal{T}expleft(-frac{mathrm{i}}{hbar}intlimits_{t_0}^{t}mathrm{d}t'H(t')right),,$$
where $mathcal{T}$ is the time-ordering operator.



One define the states and operators in the interaction picture as
$$left|psi(t)rightrangle_{mathrm{I}}=expleft(frac{mathrm{i}}{hbar}H_{0}tright)left|psi(t)rightrangle_{mathrm{S}},,$$
$$A_{mathrm{I}}(t)=expleft(frac{mathrm{i}}{hbar}H_{0}tright)A_{mathrm{S}}(t)expleft(-frac{mathrm{i}}{hbar}H_{0}tright),.$$



$A_{mathrm{S}}(t)$ is the operator in the Schrödinger picture, such as $V(t)$, which in general could depend on time.



One could prove that
$$mathrm{i}hbarfrac{partial}{partial t}left|psi(t)rightrangle_{mathrm{I}}=V_{mathrm{I}}(t)left|psi(t)rightrangle_{mathrm{I}},.$$



This resolves into
$$left|psi(t)rightrangle_{mathrm{I}}=U'(t,t_0)left|psi(t_0)rightrangle_{mathrm{I}},,$$



where
$$U'(t,t_0)=mathcal{T}expleft(-frac{mathrm{i}}{hbar}intlimits_{t_0}^{t}mathrm{d}t'V_{mathrm{I}}(t')right),.$$



I know that $U$ and $U'$ are related to each other by
$$U'(t,t_0)=expleft(frac{mathrm{i}}{hbar}H_{0}tright)U(t,t_0)expleft(-frac{mathrm{i}}{hbar}H_{0}t_0right),.$$



This can be proved by a differentiation procedure (and essentially this is how the interaction picture is derived from the Schrödinger one). But how can one prove this based on the expansion in Neumann's series of the evolution operators only?






mathematical-physics







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Consider a Hamiltonian $H$ of two components, a time-independent $H_0$ and a time-dependent $V(t)$,



    $$H(t)=H_0+V(t),.$$



    In the Schrödinger picture of quantum mechanics, the state $left|psi(t)rightrangle_{mathrm{S}}$ evolves according to:



    $$mathrm{i}hbarfrac{partial}{partial t}left|psi(t)rightrangle_{mathrm{S}}=H(t)left|psi(t)rightrangle_{mathrm{S}},.$$



    This equation admits a formal solution given by,
    $$left|psi(t)rightrangle_{mathrm{S}}=U(t,t_0)left|psi(t_0)rightrangle_{mathrm{S}},.$$



    $U(t,t_0)$ is the evolution operator corresponding to $H(t)$, defined by
    $$U(t,t_0)=mathcal{T}expleft(-frac{mathrm{i}}{hbar}intlimits_{t_0}^{t}mathrm{d}t'H(t')right),,$$
    where $mathcal{T}$ is the time-ordering operator.



    One define the states and operators in the interaction picture as
    $$left|psi(t)rightrangle_{mathrm{I}}=expleft(frac{mathrm{i}}{hbar}H_{0}tright)left|psi(t)rightrangle_{mathrm{S}},,$$
    $$A_{mathrm{I}}(t)=expleft(frac{mathrm{i}}{hbar}H_{0}tright)A_{mathrm{S}}(t)expleft(-frac{mathrm{i}}{hbar}H_{0}tright),.$$



    $A_{mathrm{S}}(t)$ is the operator in the Schrödinger picture, such as $V(t)$, which in general could depend on time.



    One could prove that
    $$mathrm{i}hbarfrac{partial}{partial t}left|psi(t)rightrangle_{mathrm{I}}=V_{mathrm{I}}(t)left|psi(t)rightrangle_{mathrm{I}},.$$



    This resolves into
    $$left|psi(t)rightrangle_{mathrm{I}}=U'(t,t_0)left|psi(t_0)rightrangle_{mathrm{I}},,$$



    where
    $$U'(t,t_0)=mathcal{T}expleft(-frac{mathrm{i}}{hbar}intlimits_{t_0}^{t}mathrm{d}t'V_{mathrm{I}}(t')right),.$$



    I know that $U$ and $U'$ are related to each other by
    $$U'(t,t_0)=expleft(frac{mathrm{i}}{hbar}H_{0}tright)U(t,t_0)expleft(-frac{mathrm{i}}{hbar}H_{0}t_0right),.$$



    This can be proved by a differentiation procedure (and essentially this is how the interaction picture is derived from the Schrödinger one). But how can one prove this based on the expansion in Neumann's series of the evolution operators only?






    mathematical-physics







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      Consider a Hamiltonian $H$ of two components, a time-independent $H_0$ and a time-dependent $V(t)$,



      $$H(t)=H_0+V(t),.$$



      In the Schrödinger picture of quantum mechanics, the state $left|psi(t)rightrangle_{mathrm{S}}$ evolves according to:



      $$mathrm{i}hbarfrac{partial}{partial t}left|psi(t)rightrangle_{mathrm{S}}=H(t)left|psi(t)rightrangle_{mathrm{S}},.$$



      This equation admits a formal solution given by,
      $$left|psi(t)rightrangle_{mathrm{S}}=U(t,t_0)left|psi(t_0)rightrangle_{mathrm{S}},.$$



      $U(t,t_0)$ is the evolution operator corresponding to $H(t)$, defined by
      $$U(t,t_0)=mathcal{T}expleft(-frac{mathrm{i}}{hbar}intlimits_{t_0}^{t}mathrm{d}t'H(t')right),,$$
      where $mathcal{T}$ is the time-ordering operator.



      One define the states and operators in the interaction picture as
      $$left|psi(t)rightrangle_{mathrm{I}}=expleft(frac{mathrm{i}}{hbar}H_{0}tright)left|psi(t)rightrangle_{mathrm{S}},,$$
      $$A_{mathrm{I}}(t)=expleft(frac{mathrm{i}}{hbar}H_{0}tright)A_{mathrm{S}}(t)expleft(-frac{mathrm{i}}{hbar}H_{0}tright),.$$



      $A_{mathrm{S}}(t)$ is the operator in the Schrödinger picture, such as $V(t)$, which in general could depend on time.



      One could prove that
      $$mathrm{i}hbarfrac{partial}{partial t}left|psi(t)rightrangle_{mathrm{I}}=V_{mathrm{I}}(t)left|psi(t)rightrangle_{mathrm{I}},.$$



      This resolves into
      $$left|psi(t)rightrangle_{mathrm{I}}=U'(t,t_0)left|psi(t_0)rightrangle_{mathrm{I}},,$$



      where
      $$U'(t,t_0)=mathcal{T}expleft(-frac{mathrm{i}}{hbar}intlimits_{t_0}^{t}mathrm{d}t'V_{mathrm{I}}(t')right),.$$



      I know that $U$ and $U'$ are related to each other by
      $$U'(t,t_0)=expleft(frac{mathrm{i}}{hbar}H_{0}tright)U(t,t_0)expleft(-frac{mathrm{i}}{hbar}H_{0}t_0right),.$$



      This can be proved by a differentiation procedure (and essentially this is how the interaction picture is derived from the Schrödinger one). But how can one prove this based on the expansion in Neumann's series of the evolution operators only?






      mathematical-physics







      share|cite|improve this question









      $endgroup$




      Consider a Hamiltonian $H$ of two components, a time-independent $H_0$ and a time-dependent $V(t)$,



      $$H(t)=H_0+V(t),.$$



      In the Schrödinger picture of quantum mechanics, the state $left|psi(t)rightrangle_{mathrm{S}}$ evolves according to:



      $$mathrm{i}hbarfrac{partial}{partial t}left|psi(t)rightrangle_{mathrm{S}}=H(t)left|psi(t)rightrangle_{mathrm{S}},.$$



      This equation admits a formal solution given by,
      $$left|psi(t)rightrangle_{mathrm{S}}=U(t,t_0)left|psi(t_0)rightrangle_{mathrm{S}},.$$



      $U(t,t_0)$ is the evolution operator corresponding to $H(t)$, defined by
      $$U(t,t_0)=mathcal{T}expleft(-frac{mathrm{i}}{hbar}intlimits_{t_0}^{t}mathrm{d}t'H(t')right),,$$
      where $mathcal{T}$ is the time-ordering operator.



      One define the states and operators in the interaction picture as
      $$left|psi(t)rightrangle_{mathrm{I}}=expleft(frac{mathrm{i}}{hbar}H_{0}tright)left|psi(t)rightrangle_{mathrm{S}},,$$
      $$A_{mathrm{I}}(t)=expleft(frac{mathrm{i}}{hbar}H_{0}tright)A_{mathrm{S}}(t)expleft(-frac{mathrm{i}}{hbar}H_{0}tright),.$$



      $A_{mathrm{S}}(t)$ is the operator in the Schrödinger picture, such as $V(t)$, which in general could depend on time.



      One could prove that
      $$mathrm{i}hbarfrac{partial}{partial t}left|psi(t)rightrangle_{mathrm{I}}=V_{mathrm{I}}(t)left|psi(t)rightrangle_{mathrm{I}},.$$



      This resolves into
      $$left|psi(t)rightrangle_{mathrm{I}}=U'(t,t_0)left|psi(t_0)rightrangle_{mathrm{I}},,$$



      where
      $$U'(t,t_0)=mathcal{T}expleft(-frac{mathrm{i}}{hbar}intlimits_{t_0}^{t}mathrm{d}t'V_{mathrm{I}}(t')right),.$$



      I know that $U$ and $U'$ are related to each other by
      $$U'(t,t_0)=expleft(frac{mathrm{i}}{hbar}H_{0}tright)U(t,t_0)expleft(-frac{mathrm{i}}{hbar}H_{0}t_0right),.$$



      This can be proved by a differentiation procedure (and essentially this is how the interaction picture is derived from the Schrödinger one). But how can one prove this based on the expansion in Neumann's series of the evolution operators only?






      mathematical-physics




      mathematical-physics






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 22 at 9:22









      Willi TschauWilli Tschau

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          $begingroup$

          It is not correct that $$U'(t,t_0)=expleft(frac{mathrm{i}}{hbar}H_{0}tright)U(t,t_0)expleft(-frac{mathrm{i}}{hbar}H_{0}t_0right),.$$ You can simply see this when setting $Vequiv 0$. We then have that
          $$ U(t,t_0) = expleft[-i (t-t_0) H_0right]$$
          while $$U'(t,t_0)= 1 neqexpleft(frac{mathrm{i}}{hbar}H_{0}tright)U(t,t_0)expleft(-frac{mathrm{i}}{hbar}H_{0}t_0right),.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Respectfully disagree (I wanted to say Holy Cow!). In the case you considered, couldn't you just split $U(t,t_0)$ into $exp(-frac{mathrm{i}}{hbar}H_0t)timesexp(frac{mathrm{i}}{hbar}H_0t_0)$. $H_0$ commutes with itself!
            $endgroup$
            – Willi Tschau
            Jan 22 at 15:51













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          1 Answer
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          active

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          active

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          active

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          $begingroup$

          It is not correct that $$U'(t,t_0)=expleft(frac{mathrm{i}}{hbar}H_{0}tright)U(t,t_0)expleft(-frac{mathrm{i}}{hbar}H_{0}t_0right),.$$ You can simply see this when setting $Vequiv 0$. We then have that
          $$ U(t,t_0) = expleft[-i (t-t_0) H_0right]$$
          while $$U'(t,t_0)= 1 neqexpleft(frac{mathrm{i}}{hbar}H_{0}tright)U(t,t_0)expleft(-frac{mathrm{i}}{hbar}H_{0}t_0right),.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Respectfully disagree (I wanted to say Holy Cow!). In the case you considered, couldn't you just split $U(t,t_0)$ into $exp(-frac{mathrm{i}}{hbar}H_0t)timesexp(frac{mathrm{i}}{hbar}H_0t_0)$. $H_0$ commutes with itself!
            $endgroup$
            – Willi Tschau
            Jan 22 at 15:51


















          0












          $begingroup$

          It is not correct that $$U'(t,t_0)=expleft(frac{mathrm{i}}{hbar}H_{0}tright)U(t,t_0)expleft(-frac{mathrm{i}}{hbar}H_{0}t_0right),.$$ You can simply see this when setting $Vequiv 0$. We then have that
          $$ U(t,t_0) = expleft[-i (t-t_0) H_0right]$$
          while $$U'(t,t_0)= 1 neqexpleft(frac{mathrm{i}}{hbar}H_{0}tright)U(t,t_0)expleft(-frac{mathrm{i}}{hbar}H_{0}t_0right),.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Respectfully disagree (I wanted to say Holy Cow!). In the case you considered, couldn't you just split $U(t,t_0)$ into $exp(-frac{mathrm{i}}{hbar}H_0t)timesexp(frac{mathrm{i}}{hbar}H_0t_0)$. $H_0$ commutes with itself!
            $endgroup$
            – Willi Tschau
            Jan 22 at 15:51
















          0












          0








          0





          $begingroup$

          It is not correct that $$U'(t,t_0)=expleft(frac{mathrm{i}}{hbar}H_{0}tright)U(t,t_0)expleft(-frac{mathrm{i}}{hbar}H_{0}t_0right),.$$ You can simply see this when setting $Vequiv 0$. We then have that
          $$ U(t,t_0) = expleft[-i (t-t_0) H_0right]$$
          while $$U'(t,t_0)= 1 neqexpleft(frac{mathrm{i}}{hbar}H_{0}tright)U(t,t_0)expleft(-frac{mathrm{i}}{hbar}H_{0}t_0right),.$$






          share|cite|improve this answer









          $endgroup$



          It is not correct that $$U'(t,t_0)=expleft(frac{mathrm{i}}{hbar}H_{0}tright)U(t,t_0)expleft(-frac{mathrm{i}}{hbar}H_{0}t_0right),.$$ You can simply see this when setting $Vequiv 0$. We then have that
          $$ U(t,t_0) = expleft[-i (t-t_0) H_0right]$$
          while $$U'(t,t_0)= 1 neqexpleft(frac{mathrm{i}}{hbar}H_{0}tright)U(t,t_0)expleft(-frac{mathrm{i}}{hbar}H_{0}t_0right),.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 22 at 10:59









          FabianFabian

          19.8k3774




          19.8k3774












          • $begingroup$
            Respectfully disagree (I wanted to say Holy Cow!). In the case you considered, couldn't you just split $U(t,t_0)$ into $exp(-frac{mathrm{i}}{hbar}H_0t)timesexp(frac{mathrm{i}}{hbar}H_0t_0)$. $H_0$ commutes with itself!
            $endgroup$
            – Willi Tschau
            Jan 22 at 15:51




















          • $begingroup$
            Respectfully disagree (I wanted to say Holy Cow!). In the case you considered, couldn't you just split $U(t,t_0)$ into $exp(-frac{mathrm{i}}{hbar}H_0t)timesexp(frac{mathrm{i}}{hbar}H_0t_0)$. $H_0$ commutes with itself!
            $endgroup$
            – Willi Tschau
            Jan 22 at 15:51


















          $begingroup$
          Respectfully disagree (I wanted to say Holy Cow!). In the case you considered, couldn't you just split $U(t,t_0)$ into $exp(-frac{mathrm{i}}{hbar}H_0t)timesexp(frac{mathrm{i}}{hbar}H_0t_0)$. $H_0$ commutes with itself!
          $endgroup$
          – Willi Tschau
          Jan 22 at 15:51






          $begingroup$
          Respectfully disagree (I wanted to say Holy Cow!). In the case you considered, couldn't you just split $U(t,t_0)$ into $exp(-frac{mathrm{i}}{hbar}H_0t)timesexp(frac{mathrm{i}}{hbar}H_0t_0)$. $H_0$ commutes with itself!
          $endgroup$
          – Willi Tschau
          Jan 22 at 15:51




















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