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General derivative of the exponential operator w.r.t. a parameter


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$begingroup$


I am interested in the calculation of the general $N$th derivative w.r.t. a parameter $lambda$ of a quantum mechanical exponential operator with the following structure:
begin{equation*}
frac{mathrm d^N}{mathrm d lambda^N} e^{-beta hat H(lambda)}
end{equation*}
In the case $n=1$ this reads as
begin{equation*}
sum_{n=0}^{+infty}sum_{m=0}^{n-1}frac{1}{n!}[hat H(lambda)]^{m}frac{mathrm dhat H(lambda)}{mathrm d lambda}[hat H(lambda)]^{n-m-1}
end{equation*}
Is there a compact fashion to rearrange the expression above for the general $N$th derivative?






quantum-mechanics operators mathematical-physics hamiltonian differentiation







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I am interested in the calculation of the general $N$th derivative w.r.t. a parameter $lambda$ of a quantum mechanical exponential operator with the following structure:
    begin{equation*}
    frac{mathrm d^N}{mathrm d lambda^N} e^{-beta hat H(lambda)}
    end{equation*}
    In the case $n=1$ this reads as
    begin{equation*}
    sum_{n=0}^{+infty}sum_{m=0}^{n-1}frac{1}{n!}[hat H(lambda)]^{m}frac{mathrm dhat H(lambda)}{mathrm d lambda}[hat H(lambda)]^{n-m-1}
    end{equation*}
    Is there a compact fashion to rearrange the expression above for the general $N$th derivative?






    quantum-mechanics operators mathematical-physics hamiltonian differentiation







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      0



      $begingroup$


      I am interested in the calculation of the general $N$th derivative w.r.t. a parameter $lambda$ of a quantum mechanical exponential operator with the following structure:
      begin{equation*}
      frac{mathrm d^N}{mathrm d lambda^N} e^{-beta hat H(lambda)}
      end{equation*}
      In the case $n=1$ this reads as
      begin{equation*}
      sum_{n=0}^{+infty}sum_{m=0}^{n-1}frac{1}{n!}[hat H(lambda)]^{m}frac{mathrm dhat H(lambda)}{mathrm d lambda}[hat H(lambda)]^{n-m-1}
      end{equation*}
      Is there a compact fashion to rearrange the expression above for the general $N$th derivative?






      quantum-mechanics operators mathematical-physics hamiltonian differentiation







      share|cite|improve this question











      $endgroup$




      I am interested in the calculation of the general $N$th derivative w.r.t. a parameter $lambda$ of a quantum mechanical exponential operator with the following structure:
      begin{equation*}
      frac{mathrm d^N}{mathrm d lambda^N} e^{-beta hat H(lambda)}
      end{equation*}
      In the case $n=1$ this reads as
      begin{equation*}
      sum_{n=0}^{+infty}sum_{m=0}^{n-1}frac{1}{n!}[hat H(lambda)]^{m}frac{mathrm dhat H(lambda)}{mathrm d lambda}[hat H(lambda)]^{n-m-1}
      end{equation*}
      Is there a compact fashion to rearrange the expression above for the general $N$th derivative?






      quantum-mechanics operators mathematical-physics hamiltonian differentiation




      quantum-mechanics operators mathematical-physics hamiltonian differentiation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 34 mins ago









      Qmechanic♦

      105k121881202




      105k121881202










      asked 1 hour ago









      GrazGraz

      363




      363






















          1 Answer
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          4












          $begingroup$

          Hint: Use repeatedly the identity for the 1st derivative
          $$ frac{d}{dlambda}e^{that{A}} ~=~ int_0^t!dt_1~ e^{(t-t_1)hat{A}}frac{dhat{A}}{dlambda}e^{t_1hat{A}}~=~ iint_{mathbb{R}^2_+}!dt_1~dt_2~delta(t!-!t_1!-!t_2)~ e^{t_2hat{A}}frac{dhat{A}}{dlambda}e^{t_1hat{A}} , qquad t~in~mathbb{R}_+ .tag{1} $$
          For a proof of eq. (1) see e.g. my Phys.SE answer here. Then the 2nd derivative becomes
          $$begin{align} frac{d^2}{dlambda^2}e^{that{A}}
          &~=~2iiint_{mathbb{R}^3_+}!dt_1~dt_2~dt_3~delta(t!-!t_1!-!t_2!-!t_3)~ e^{t_3hat{A}}frac{dhat{A}}{dlambda}e^{t_2hat{A}}frac{dhat{A}}{dlambda}e^{t_1hat{A}} cr
          &~+~ iint_{mathbb{R}^2_+}!dt_1~dt_2~delta(t!-!t_1!-!t_2)~ e^{t_2hat{A}}frac{d^2hat{A}}{dlambda^2}e^{t_1hat{A}} , qquad t~in~mathbb{R}_+ ,end{align}tag{2} $$
          and so forth.






          share|cite|improve this answer











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            active

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            4












            $begingroup$

            Hint: Use repeatedly the identity for the 1st derivative
            $$ frac{d}{dlambda}e^{that{A}} ~=~ int_0^t!dt_1~ e^{(t-t_1)hat{A}}frac{dhat{A}}{dlambda}e^{t_1hat{A}}~=~ iint_{mathbb{R}^2_+}!dt_1~dt_2~delta(t!-!t_1!-!t_2)~ e^{t_2hat{A}}frac{dhat{A}}{dlambda}e^{t_1hat{A}} , qquad t~in~mathbb{R}_+ .tag{1} $$
            For a proof of eq. (1) see e.g. my Phys.SE answer here. Then the 2nd derivative becomes
            $$begin{align} frac{d^2}{dlambda^2}e^{that{A}}
            &~=~2iiint_{mathbb{R}^3_+}!dt_1~dt_2~dt_3~delta(t!-!t_1!-!t_2!-!t_3)~ e^{t_3hat{A}}frac{dhat{A}}{dlambda}e^{t_2hat{A}}frac{dhat{A}}{dlambda}e^{t_1hat{A}} cr
            &~+~ iint_{mathbb{R}^2_+}!dt_1~dt_2~delta(t!-!t_1!-!t_2)~ e^{t_2hat{A}}frac{d^2hat{A}}{dlambda^2}e^{t_1hat{A}} , qquad t~in~mathbb{R}_+ ,end{align}tag{2} $$
            and so forth.






            share|cite|improve this answer











            $endgroup$


















              4












              $begingroup$

              Hint: Use repeatedly the identity for the 1st derivative
              $$ frac{d}{dlambda}e^{that{A}} ~=~ int_0^t!dt_1~ e^{(t-t_1)hat{A}}frac{dhat{A}}{dlambda}e^{t_1hat{A}}~=~ iint_{mathbb{R}^2_+}!dt_1~dt_2~delta(t!-!t_1!-!t_2)~ e^{t_2hat{A}}frac{dhat{A}}{dlambda}e^{t_1hat{A}} , qquad t~in~mathbb{R}_+ .tag{1} $$
              For a proof of eq. (1) see e.g. my Phys.SE answer here. Then the 2nd derivative becomes
              $$begin{align} frac{d^2}{dlambda^2}e^{that{A}}
              &~=~2iiint_{mathbb{R}^3_+}!dt_1~dt_2~dt_3~delta(t!-!t_1!-!t_2!-!t_3)~ e^{t_3hat{A}}frac{dhat{A}}{dlambda}e^{t_2hat{A}}frac{dhat{A}}{dlambda}e^{t_1hat{A}} cr
              &~+~ iint_{mathbb{R}^2_+}!dt_1~dt_2~delta(t!-!t_1!-!t_2)~ e^{t_2hat{A}}frac{d^2hat{A}}{dlambda^2}e^{t_1hat{A}} , qquad t~in~mathbb{R}_+ ,end{align}tag{2} $$
              and so forth.






              share|cite|improve this answer











              $endgroup$
















                4












                4








                4





                $begingroup$

                Hint: Use repeatedly the identity for the 1st derivative
                $$ frac{d}{dlambda}e^{that{A}} ~=~ int_0^t!dt_1~ e^{(t-t_1)hat{A}}frac{dhat{A}}{dlambda}e^{t_1hat{A}}~=~ iint_{mathbb{R}^2_+}!dt_1~dt_2~delta(t!-!t_1!-!t_2)~ e^{t_2hat{A}}frac{dhat{A}}{dlambda}e^{t_1hat{A}} , qquad t~in~mathbb{R}_+ .tag{1} $$
                For a proof of eq. (1) see e.g. my Phys.SE answer here. Then the 2nd derivative becomes
                $$begin{align} frac{d^2}{dlambda^2}e^{that{A}}
                &~=~2iiint_{mathbb{R}^3_+}!dt_1~dt_2~dt_3~delta(t!-!t_1!-!t_2!-!t_3)~ e^{t_3hat{A}}frac{dhat{A}}{dlambda}e^{t_2hat{A}}frac{dhat{A}}{dlambda}e^{t_1hat{A}} cr
                &~+~ iint_{mathbb{R}^2_+}!dt_1~dt_2~delta(t!-!t_1!-!t_2)~ e^{t_2hat{A}}frac{d^2hat{A}}{dlambda^2}e^{t_1hat{A}} , qquad t~in~mathbb{R}_+ ,end{align}tag{2} $$
                and so forth.






                share|cite|improve this answer











                $endgroup$



                Hint: Use repeatedly the identity for the 1st derivative
                $$ frac{d}{dlambda}e^{that{A}} ~=~ int_0^t!dt_1~ e^{(t-t_1)hat{A}}frac{dhat{A}}{dlambda}e^{t_1hat{A}}~=~ iint_{mathbb{R}^2_+}!dt_1~dt_2~delta(t!-!t_1!-!t_2)~ e^{t_2hat{A}}frac{dhat{A}}{dlambda}e^{t_1hat{A}} , qquad t~in~mathbb{R}_+ .tag{1} $$
                For a proof of eq. (1) see e.g. my Phys.SE answer here. Then the 2nd derivative becomes
                $$begin{align} frac{d^2}{dlambda^2}e^{that{A}}
                &~=~2iiint_{mathbb{R}^3_+}!dt_1~dt_2~dt_3~delta(t!-!t_1!-!t_2!-!t_3)~ e^{t_3hat{A}}frac{dhat{A}}{dlambda}e^{t_2hat{A}}frac{dhat{A}}{dlambda}e^{t_1hat{A}} cr
                &~+~ iint_{mathbb{R}^2_+}!dt_1~dt_2~delta(t!-!t_1!-!t_2)~ e^{t_2hat{A}}frac{d^2hat{A}}{dlambda^2}e^{t_1hat{A}} , qquad t~in~mathbb{R}_+ ,end{align}tag{2} $$
                and so forth.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 26 mins ago

























                answered 35 mins ago









                Qmechanic♦Qmechanic

                105k121881202




                105k121881202






























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