- #1

saadhusayn

- 22

- 1

- Homework Statement
- This is from Hugh Osborn's 'Advanced Quantum Field Theory' (attached) 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.

- Relevant Equations
- We need to expand the exponential of the derivative as a Taylor series.

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}{4} \frac{\partial}{\partial x_{i}} A^{-1}_{ij} \frac{\partial}{\partial x_{j}} \frac{\partial}{\partial x_{k}} A^{-1}_{kl} \frac{\partial}{\partial x_{l}} + \frac{1}{12} \frac{\partial}{\partial x_{i}} A^{-1}_{ij} \frac{\partial}{\partial x_{j}} \frac{\partial}{\partial x_{k}} A^{-1}_{kl} \frac{\partial}{\partial x_{l}} \frac{\partial}{\partial x_{m}} A^{-1}_{mn} \frac{\partial}{\partial x_{n}} + \dots\Bigg) \exp(-V(x))\biggr\vert_{\underline{x} = \underline{0}}##

Which comes to

$$ Z = 1 - \frac{1}{2} A^{-1}_{ij} V_{ij} + \frac{1}{4} A^{-1}_{ij} V_{ij} A^{-1}_{kl} V_{kl} + \dots $$

But the answer is

## Z = \Bigg(1 + \frac{1}{2} \frac{\partial}{\partial x_{i}} A^{-1}_{ij} \frac{\partial}{\partial x_{j}} + \frac{1}{4} \frac{\partial}{\partial x_{i}} A^{-1}_{ij} \frac{\partial}{\partial x_{j}} \frac{\partial}{\partial x_{k}} A^{-1}_{kl} \frac{\partial}{\partial x_{l}} + \frac{1}{12} \frac{\partial}{\partial x_{i}} A^{-1}_{ij} \frac{\partial}{\partial x_{j}} \frac{\partial}{\partial x_{k}} A^{-1}_{kl} \frac{\partial}{\partial x_{l}} \frac{\partial}{\partial x_{m}} A^{-1}_{mn} \frac{\partial}{\partial x_{n}} + \dots\Bigg) \exp(-V(x))\biggr\vert_{\underline{x} = \underline{0}}##

Which comes to

$$ Z = 1 - \frac{1}{2} A^{-1}_{ij} V_{ij} + \frac{1}{4} A^{-1}_{ij} V_{ij} A^{-1}_{kl} V_{kl} + \dots $$

But the answer is