# Path Integral Notation [closed]

In my Statistical Field Theory lectures, I was told that

$$Z=\int \mathcal{D}\phi\ e^{-F[\phi]}=\int\prod_{k<\Lambda}d\phi_k\ e^{-F[\phi_k]}$$

I want to clarify that I understand the mathematical notation. Is it correct that: \begin{align} \int\prod_{k<\Lambda}d\phi_k\ e^{-F[\phi_k]} &\equiv \int_{\phi_1}\int_{\phi_2}\cdots\int_{\phi_\Lambda} \Big[d\phi_1\ d\phi_2\ \cdots d\phi_\Lambda\ \big(e^{-F[\phi_1]}e^{-F[\phi_2]}\cdots e^{-F[\phi_\Lambda]}\big) \Big] \\ &= \int_{\phi_1}\Big[d\phi_1\ e^{-F[\phi_1]} \Big] \int_{\phi_2}\Big[d\phi_2\ e^{-F[\phi_2]} \Big] \cdots\int_{\phi_\Lambda}\Big[d\phi_\Lambda\ e^{-F[\phi_\Lambda]} \Big]\\ &= \prod_{k<\Lambda}\Bigg[\int_{\phi_k}\Big[d\phi_k\ e^{-F[\phi_k]} \Big]\Bigg] \end{align}

I feel like this is incorrect, but I don't really understand what's going on enough to say for sure.

## closed as off-topic by Aaron Stevens, Kyle Kanos, Jon Custer, sammy gerbil, ZeroTheHeroNov 7 '18 at 2:58

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Introduction

For a completely general functional $$F[\phi]$$ your final equation is not correct. Let's pick a paticular discretization of the $$x$$ axis $$\mathcal D \equiv \{x_1,x_2,...,x_N\}$$ where $$x_{i+1}-x_i \equiv \Delta x=\frac{x_N-x_1}{N}$$ is the distance between consecutive points. Obviously, the continuum limit corresponds to the limit $$\Delta x \rightarrow 0$$, or equivalently $$N \rightarrow + \infty$$, which we're going to take at the end of our manipulations. Under this discretization scheme $$\mathcal D$$, any function $$\phi(x)$$ simply corresponds to a piece-wise constant function $$\tilde \phi_{\mathcal D}$$ with value $$\phi(x_i)$$ at $$x_i$$, which again approaches the actual function $$\phi$$ as $$\Delta x \rightarrow 0$$. Since this is a piece-wise constant function, the set of its values at all points of $$\mathcal D$$, i.e. $$(\phi({x_1}),...,\phi({x_N}))$$, uniquely specifies it.

This implies that the outcome of any functional $$F$$ acting on this discretized function is also uniquely given by the set $$(\phi({x_1}),...,\phi({x_N}))$$. In other words, I can write the outcome of this operation in terms of some multivariate function $$f$$ as $$F[\tilde\phi_{\mathcal D}] \equiv f(\phi(x_1),...,\phi(x_N))$$. To keep my notation clean, I'll use the shorthand $$\phi_i := \phi(x_i)$$ from here on out.

Using this discretization concept, the functional integral in question can be defined as: $$\int \mathcal D\phi \ e^{-F[\phi]}=\lim_{N \rightarrow +\infty} \int_{\mathbb R^N} \prod_{i=1}^N d\phi_i \ e^{-f(\phi_1,...,\phi_N)} \qquad (*)$$
Now if you could write $$f(\phi_1,...,\phi_N)$$ as the sum of some single variable functions in the form $$f(\phi_1,...,\phi_N) \equiv \sum_{i=1}^N f_i( \phi_i)$$, You would have $$e^{-f(\phi_1,...\phi_N)} = \prod _{i=1}^Ne^{-f_i(\phi_i)}$$, leading to your last equation:
$$\int \mathcal D \phi \ e^{-F[\phi]} = \lim_{N \rightarrow + \infty} \prod_{i=1}^N \int_{\mathbb R} d\phi_i \ e^{-f_i(\phi_i)}$$
However, for a general functional $$F$$, and its corresponding discretized multivariable function $$f$$, the $$e^{-f(\phi_1,...,\phi_N)}$$ cannot be necessarily written as a product of exponentials of single variables. So in general: $$e^{-f(\phi_1,...\phi_N)} \neq \prod _{i=1}^Ne^{-f_i(\phi_i)}$$ Meaning that your final equation wouldn't work anymore.