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.


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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.

The answer to the question

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.

  • $\begingroup$ Oh I see: the Product only applies to the differentials, and not the rest of the integrand. And each differential represents one further step along all possible paths, not an altogether different path. Thanks! $\endgroup$ – Dan Nov 3 '18 at 3:31
  • $\begingroup$ @Dan No problem $\endgroup$ – Sahand Tabatabaei Nov 3 '18 at 3:58

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