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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 d\hat 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?

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Hint: Use repeatedly the identity for the 1st derivative $$ \frac{d}{d\lambda}e^{t\hat{A}} ~=~ \int_0^t\!dt_1~ e^{(t-t_1)\hat{A}}\frac{d\hat{A}}{d\lambda}e^{t_1\hat{A}}~=~ \iint_{\mathbb{R}^2_+}\!dt_1~dt_2~\delta(t\!-\!t_1\!-\!t_2)~ e^{t_2\hat{A}}\frac{d\hat{A}}{d\lambda}e^{t_1\hat{A}} , \qquad t~\in~\mathbb{R}_+ .\tag{1} $$ For a proof of eq. (1) see e.g. my Phys.SE answer here. Then the 2nd derivative becomes $$\begin{align} \frac{1}{2!}\frac{d^2}{d\lambda^2}e^{t\hat{A}} &~=~\iiint_{\mathbb{R}^3_+}\!dt_1~dt_2~dt_3~\delta(t\!-\!t_1\!-\!t_2\!-\!t_3)~ e^{t_3\hat{A}}\frac{d\hat{A}}{d\lambda}e^{t_2\hat{A}}\frac{d\hat{A}}{d\lambda}e^{t_1\hat{A}} \cr &~+~ \iint_{\mathbb{R}^2_+}\!dt_1~dt_2~\delta(t\!-\!t_1\!-\!t_2)~ e^{t_2\hat{A}}\frac{1}{2!}\frac{d^2\hat{A}}{d\lambda^2}e^{t_1\hat{A}} , \qquad t~\in~\mathbb{R}_+ ,\end{align}\tag{2} $$ the 3rd derivative becomes $$\begin{align} \frac{1}{3!}\frac{d^3}{d\lambda^3}e^{t\hat{A}} &~=~\iiiint_{\mathbb{R}^4_+}\!dt_1~dt_2~dt_3~dt_4~\delta(t\!-\!t_1\!-\!t_2\!-\!t_3\!-\!t_4)~ e^{t_4\hat{A}}\frac{d\hat{A}}{d\lambda}e^{t_3\hat{A}}\frac{d\hat{A}}{d\lambda}e^{t_2\hat{A}}\frac{d\hat{A}}{d\lambda}e^{t_1\hat{A}} \cr &~+~\iiint_{\mathbb{R}^3_+}\!dt_1~dt_2~dt_3~\delta(t\!-\!t_1\!-\!t_2\!-\!t_3)~ e^{t_3\hat{A}}\frac{d\hat{A}}{d\lambda}e^{t_2\hat{A}}\frac{1}{2!}\frac{d^2\hat{A}}{d\lambda^2}e^{t_1\hat{A}} \cr &~+~\iiint_{\mathbb{R}^3_+}\!dt_1~dt_2~dt_3~\delta(t\!-\!t_1\!-\!t_2\!-\!t_3)~ e^{t_3\hat{A}}\frac{1}{2!}\frac{d^2\hat{A}}{d\lambda^2}e^{t_2\hat{A}}\frac{d\hat{A}}{d\lambda}e^{t_1\hat{A}} \cr &~+~ \iint_{\mathbb{R}^2_+}\!dt_1~dt_2~\delta(t\!-\!t_1\!-\!t_2)~ e^{t_2\hat{A}}\frac{1}{3!}\frac{d^3\hat{A}}{d\lambda^3}e^{t_1\hat{A}} , \qquad t~\in~\mathbb{R}_+ ,\end{align}\tag{3} $$ and so forth. The $N$th derivative $$\frac{1}{N!}\frac{d^N}{d\lambda^N}e^{t\hat{A}}~=~\sum\stackrel{t_{n+1}}{\rule{1cm}{.5mm}}\fbox{$k_n$}\stackrel{t_n}{\rule{1cm}{.5mm}}\cdots \stackrel{t_3}{\rule{1cm}{.5mm}}\fbox{$k_2$}\stackrel{t_2}{\rule{1cm}{.5mm}}\fbox{$k_1$}\stackrel{t_1}{\rule{1cm}{.5mm}}\tag{N}$$ is a sum of possible Feynman diagrams with Schwinger propagators $$\stackrel{t_i}{\rule{1cm}{.5mm}}~=~e^{t_i\hat{A}}, \qquad t_i~\in~\mathbb{R}_+,\tag{P}$$ and 2-vertices $$\fbox{$k_i$}~=~\frac{1}{k_i!}\frac{d^{k_i}\hat{A}}{d\lambda^{k_i}}, \qquad k_i~\in~\mathbb{N}.\tag{V}$$ Each diagram in the sum has weight 1.

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  • $\begingroup$ That's horrifying. $\endgroup$ Feb 21, 2019 at 18:17
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    $\begingroup$ Indeed a computational nighmare. But there is beauty behind the madness. $\endgroup$
    – Qmechanic
    Feb 23, 2019 at 13:28
  • $\begingroup$ The problem with that is that you need to get through the madness to get to the beauty. $\endgroup$ Feb 23, 2019 at 13:30
  • $\begingroup$ $\uparrow$ Ha-ha. $\endgroup$
    – Qmechanic
    Feb 23, 2019 at 14:20

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