field theory - Is $frac{partial}{partial Phi(y)} Phi (x) = delta(x-y)$ correct?

As stated in the heading: Is $\frac{\partial}{\partial \Phi(y)} \Phi (x) = \delta(x-y)$ correct? Here denotes $\Phi(x)$ denotes a scalar field. And if yes, why? Any reference where I can read about this would be great.

Answer

It is not. The correct identity is

$$\frac{\delta}{\delta \Phi(y)} \Phi (x) = \delta(x-y)$$ where the derivative is the functional derivative. If $F : D(F)\ni \Phi \mapsto F(\Phi)\in \mathbb C$ is a function from a space of functions $D(F)$ to $\mathbb C$, the functional derivative of $F$, if it exists is the distribution $\frac{\delta F}{\delta \Phi}$ acting on smooth compact support functions $g$ such that: $$\left\langle \frac{\delta F}{\delta \Phi}, g \right\rangle := \frac{d}{d\alpha}\biggr\rvert_{\alpha=0} F(\Phi + \alpha g)\:.$$ In the considered case the functional $F$ is that associating the generic $\Phi$ with its value at the given point $x$ in its domain: $$F : \Phi \mapsto \Phi(x)\:.$$ In other words: $$F(\Phi):= \int \Phi(y) \delta(y-x) \:dy$$ hence, $$\frac{d}{d\alpha}|_{\alpha=0} F(\Phi+\alpha g) = \int \delta(y-x) g(y)\:dy$$ which can be re-written as $$\frac{\delta F}{\delta \Phi} = \delta_x$$ or, adopting the notation of physicists: $$\frac{\delta}{\delta \Phi(y)} \Phi (x) = \delta(x-y)\:.$$ Specifying better the structure of the domain $D(F)$ one can define the functional derivative as a so-called Gateaux derivative.