# 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.

• This is a basic property of "functional derivatives" though usually a $\delta$ symbol is used instead of the partial derivative $\partial$. Wikipedia isn't the worst resource for this: en.wikipedia.org/wiki/… – wsc Apr 27 '14 at 14:56

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.

• @JamalS I agree with the other changes you performed editing my answer. However, why did you remove "suggestive and effective"? – Valter Moretti Apr 27 '14 at 15:14
• In my opinion, the comment was superfluous and did not contribute anything to the overall answer. – JamalS Apr 27 '14 at 15:19
• Yes, but I am also a mathematician and that notation is problematic from a pure mathematical viewpoint. However, from the physical side it is, in fact, suggestive and effective, as everybody dealing with QFT knows very well. I thought it was worth stressing it. Do not worry in any cases, it was just a detail. It is OK as it stands now. – Valter Moretti Apr 27 '14 at 15:23

The functional differentiation identity is given by$^{\dagger}$

$$\frac{\delta}{\delta J(x)}J(y)=\delta^{(d)}(x-y) \quad \implies \frac{\delta}{\delta J(x)} \int \mathrm{d}^d y \, J(y)\phi(y)=\phi(x) \qquad (1)$$

The identity is the natural generalization, as Peskin and Schroeder state, of the discrete identities,

$$\frac{\partial}{\partial x_i}x_j = \delta_{ij} \quad \implies \frac{\partial}{\partial x_i} \sum_j x_jk_j=k_i \qquad (2)$$

$\dagger$ Note this is the opposite of the OP, as the correct identity subtracts the argument of the function differentiated from the argument of the function differentiated with respect to. In addition, the identity becomes clear when one observes the function $J(y)$ is equivalent to the functional,

$$F[J(y)]= \int \mathrm{d}^d y \, \delta^{(d)}(y-x) \,J(y)$$

using the standard delta distribution (function) identity.