5
$\begingroup$

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 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?

$\endgroup$
8
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ That's horrifying. $\endgroup$ – Emilio Pisanty Feb 21 '19 at 18:17
  • 1
    $\begingroup$ Indeed a computational nighmare. But there is beauty behind the madness. $\endgroup$ – Qmechanic Feb 23 '19 at 13:28
  • $\begingroup$ The problem with that is that you need to get through the madness to get to the beauty. $\endgroup$ – Emilio Pisanty Feb 23 '19 at 13:30
  • $\begingroup$ $\uparrow$ Ha-ha. $\endgroup$ – Qmechanic Feb 23 '19 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.