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In a book I am reading about QFT (Quantum field theory by Mark Srednicki ,page 48), I see the following equation: $$ \int \mathcal D p\mathcal D q \exp\left[i\int_{\mathbb R} dt (p\dot q - H_0(p,q)-H_1(p,q) +fq+hp)\right] $$ $$=\exp\left[i\int_{\mathbb R} dt (H_1(i\delta/\delta h(t),i\delta/\delta f(t)))\right]$$ $$\times \int \mathcal D p\mathcal D q \exp\left[i\int_{\mathbb R} dt (p\dot q - H_0(p,q)+fq+hp)\right] \tag{6.22} $$ where

  • $H=H_0+H_1$ is the Hamiltonian of the system, and $H_1$ is a small perturbation.
  • $\mathcal D q$ denotes path integral.
  • $\delta/\delta f(t)$ is the functional derivative operator; see here for the definition.

I do not understand how this equation works. Why could we insert differential operators in the place of $p,q$. And if I take the part $\exp\left[i\int_{\mathbb R} dt (H_1(i\delta/\delta h(t),i\delta/\delta f(t)))\right]$ under the integral sign, I find that it the operator has nothing to act on.

How to understand this equation?

Qmechanic
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Ma Joad
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1 Answers1

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Let's look at an ordinary integral \begin{equation} \mathcal Z(h) = \int dx e^{-f(x) + xh}, \end{equation} with $f(x)$ such that the integral is convergent. It should be obvious that \begin{equation} \int dx \, x \, e^{-f(x)+xh} = \frac{\partial}{\partial h} \int dx \, e^{-f(x)+xh} \end{equation} or \begin{equation} \int dx \, x^n \, e^{-f(x)+xh} = \frac{\partial^n}{\partial h^n} \int dx \, e^{-f(x)+xh}. \end{equation} Now using those you can easily get \begin{eqnarray} I(h) & = & \int dx \, g(x) \, e^{-f(x)+xh} = \int dx \, \sum_{n=0}^\infty\frac{g^{(n)}(0)}{n!}x^n \, e^{-f(x)+xh}=\sum_{n=0}^\infty \frac{g^{(n)}(0)}{n!}\int dx \, x^n \, e^{-f(x)+xh} \\ & = & \sum_{n=0}^\infty \frac{g^{(n)}(0)}{n!}\frac{\partial^n}{\partial h^n}\int dx \, e^{-f(x)+xh}=g\left (\frac{\partial}{\partial h} \right) \int dx \, e^{-f(x)+xh} \end{eqnarray} In your specific case $g(x)$ is an exponent

nwolijin
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