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This is from Hugh Osborn's 'Advanced Quantum Field Theory' notes, Lent 2013, page 15.

I want to evaluate the expression

$$ Z = \exp\Big(\frac{1}{2} \frac{\partial}{\partial \underline{x}} . A^{-1} \frac{\partial}{\partial \underline{x}} \Big) \exp\Big(-V(x) + \underline{b}. \underline{x}\Big) \bigg\vert_{\underline{x} = \underline{0}}$$

assuming that $\underline{b} = \underline{0}$.

We use the notation

$$ V_{i_{1} i_{2} \dots i_{k}} = \frac{\partial}{\partial x_{i_{1}}} \frac{\partial}{\partial x_{i_{2}}} \dots \frac{\partial}{\partial x_{i_{k}}} V(\underline{x})\Big\vert_{\underline{x} = \underline{0}}$$

Where

$\frac{\partial}{\partial \underline{x}} \equiv \Big( \frac{\partial}{\partial x_{1}}, \dots, \frac{\partial}{\partial x_{n}}\Big)$

We also assume that $V(\underline{0}) = V_{i}(\underline{0}) = 0$.

And $A^{-1}$ is an $n \times n$ matrix. I expanded the exponential with the derivative to get:

$$ Z = \Bigg(1 + \frac{1}{2} \frac{\partial}{\partial x_{i}} A^{-1}_{ij} \frac{\partial}{\partial x_{j}} + \frac{1}{2} \frac{1}{2}\frac{\partial}{\partial x_{i}} A^{-1}_{ij} \frac{\partial}{\partial x_{j}} \frac{1}{2}\frac{\partial}{\partial x_{k}} A^{-1}_{kl} \frac{\partial}{\partial x_{l}} + \frac{1}{6}\frac{1}{2} \frac{\partial}{\partial x_{i}} A^{-1}_{ij} \frac{\partial}{\partial x_{j}}\frac{1}{2} \frac{\partial}{\partial x_{k}} A^{-1}_{kl} \frac{\partial}{\partial x_{l}}\frac{1}{2} \frac{\partial}{\partial x_{m}} A^{-1}_{mn} \frac{\partial}{\partial x_{n}} + \dots\Bigg) \exp(-V(x))\biggr\vert_{\underline{x} = \underline{0}}$$

Then this series acts on $\exp(-V(x))$, where summation notation is implied. Am I proceeding correctly?

The answer is enter image description here

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  • $\begingroup$ It is not clear if what is expected is an expansion in $V$ or in $A^{-1}$... though I would assume the former. $\endgroup$ – Adam Jul 17 '19 at 10:36
  • $\begingroup$ @Adam I don't understand. How would you take derivatives without expanding in $A^{-1}$? $\endgroup$ – saad Jul 17 '19 at 11:18
  • $\begingroup$ For instance, $e^{a\partial_x}e^{f(x)}=e^{f(x+2)}$, but that is beside the point. In any case, you didn't expand the first exponential correctly $\endgroup$ – Adam Jul 17 '19 at 11:42
  • $\begingroup$ @Adam does the expansion look correct now? $\endgroup$ – saad Jul 18 '19 at 10:35
  • $\begingroup$ Yes, this should give you the expected result $\endgroup$ – Adam Jul 18 '19 at 11:17
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The corrected expansion does indeed give you the right answer. For example,

$$ \frac{1}{2} \frac{\partial}{\partial x_{i}} A^{-1}_{ij} \frac{\partial}{\partial x_{j}} \exp(-V(x)) \big\vert_{x = 0} = \frac{1}{2} \frac{\partial}{\partial x_{i}} A^{-1}_{ij} (-\frac{\partial}{\partial x_{j}}V(x)) \exp(-V(x)) \big\vert_{x = 0} = -\frac{1}{2} A^{-1}_{ij} \Big[\frac{\partial}{\partial x_{i}} \frac{\partial}{\partial x_{j}} V(x) - \frac{\partial}{\partial x_{i}} V(x)\Big] \exp(-V(x))\Big|_{x = 0} = -\frac{1}{2}A_{ij}^{-1}V_{ij}$$

And so on and so forth for higher derivatives.

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    $\begingroup$ The second line is not correct. $\endgroup$ – Adam Jul 19 '19 at 11:22

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