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I'm currently reading the book Quantum Field Theory and The Standard Model, and on the section on path integrals it talks about the variational partial derivative of the generating functional. It states that: $$\frac{\delta J(x)}{\delta J(y)}=\delta(x-y).\tag{1}$$ I tried to convince myself of this fact so I integrated both sides to give $$\int \frac{\delta J(x)}{\delta J(y)} \, dJ(y)=\int \delta(x-y) \, dJ(y)\tag{2}$$ $$J(x) =\int \delta(x-y) \, dJ(y)\tag{3}$$ To make my life easier I stated that $\gamma = J(y)$ which implies $y = J^{-1}(\gamma)$ $$J(x) =\int \delta(x-J^{-1}(\gamma)) \, d\gamma$$ Then we can use another substitution $\alpha = J^{-1}(\gamma)$ which is exactly what $y$ is also equal to. So $dy = \frac{dJ^{-1}(\gamma)}{d\gamma} d\gamma$ rearranged gives $ \frac{d\gamma}{dJ^{-1}(\gamma)} dy = d\gamma$. Since $\gamma = J(y)$ and $y = J^{-1}(\gamma)$ this implies$ \frac{dJ(y)}{dy} dy = d\gamma$. $$J(x) =\int \delta(x-y) \, \frac{dJ(y)}{dy} dy$$ Now we can complete the integration which gives $$J(x) = \frac{dJ(x)}{dx}$$ Which is not true for all functions $J(x)$. Where did I go wrong, and is there any convincing proofs of this identity. Thanks

Qmechanic
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Joshua Pasa
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2 Answers2

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Matt is trying to help you by skipping the functional derivative variation δ-symbol, conflating it with ordinary partial derivatives, but he ends up confusing you, instead, although his wording is crystal clear; it didn't work.

First think of a finite number of xs: $x_1,x_2,x_3,...$. You then correspondingly have a finite-dimensional vector with n independent components, $(J(x_1),J(x_2),J(x_3), ...,J(x_n))$.

Now, $$ J(x_i)=\sum_k \delta_{ki} J(x_k) \qquad \leadsto \\ {\partial J(x_i)\over \partial J(x_j)} = \delta_{ji}~. $$ Transcribing this for $n\to \infty$, you get the obvious generalization $$ J(x)=\int \!\!dz~~\delta(x-z) ~J(z) \qquad \leadsto \\ {\delta J(x)\over \delta J(y)} = \delta(x-y)~. $$

You went astray already in your second formula.

Mutatis mutandis work out $$ {\delta \int\!\!dz~ J(z) \phi(z)\over \delta J(y)} = \phi(y)~. $$

Cosmas Zachos
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To prove the identity, you should decide which definition of functional derivative to use, and that depends on which mathematical object the field is.

Usually, the field is taken to be a smooth function (a rapidly decreasing function for example), however this makes things slightly more complicated in talking about functional derivatives (one should use the so-called Gateaux derivative).

An easier example is given by functions in a normed space (e.g., in some Lebesgue space). In this case, there is the convenient definition of Fréchet derivative. So let $U$ be a normed space, and $f:U\to U$ a function (the two spaces can be taken to be different, however it is not crucial for the understanding here). The Fréchet derivative $Df(u)$ of $f$ at a point $u\in U$ is the linear operator $T$, if such a linear operator exists, from $U$ to $U$ such that for any $h\in U$, $h\to 0$, $$f(u+h)= f(u) + Th + o(h)$$ where $o(h)$ means something "that goes to zero faster than h", i.e. $$\lVert o(h)\rVert_U / \lVert h\rVert_U \to 0$$ as $h\to 0$.

So now, take the identity function $\mathrm{id}:U\to U$, acting as $$\mathrm{id}(u)=u\;.$$ What is the Fréchet derivative, at any point, of the identity function? The identity function itself of course (that is a linear operator), by the trivial identity $$u+h=\mathrm{id}(u+h)=\mathrm{id}(u) + \mathrm{id}(h)\;.$$

So, $D\mathrm{id}= \mathrm{id}$ (since it is true at any point, we forget about specifying one as in usual derivatives).

Physicists prefer to write the functional derivative by writing the integral kernel of the linear operator defined by the Fréchet derivative. So, the notation $$Df=T$$ is replaced by $$\frac{\delta f(u(x))}{\delta u(y)}=t(x-y)\; ,$$ where $t(x-y)$ is the integral kernel of the linear operator $T$ satisfying $$(Tu)(x)=\int t(x-y) u(y)\;.$$ In general, the integral kernel could be a distribution. Now, the integral kernel of the identity operator $\mathrm{id}$ is indeed $\delta(x-y)$; therefore, by the above result we have that $$\frac{\delta u(x)}{\delta u(y)}=\delta(x-y)\; .$$

yuggib
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