# Functional Derivative of action

Consider the action of free Klein-Gordon theory

$S[\phi]=\frac{1}{2}\displaystyle\int d^4y(\partial_\mu\phi(y)\partial^\mu\phi(y)-m^2\phi^2(y))$

Integrating by parts in the first term gives me

$S[\phi]=-\frac{1}{2}\displaystyle\int d^4y(\phi(y)\partial_\mu\partial^\mu\phi(y)+m^2\phi^2(y))$

Now I take the functional derivative $\frac{\delta S[\phi]}{\delta\phi(x)}$ to get

$\frac{\delta S[\phi]}{\delta\phi(x)}=-\frac{1}{2}(\partial_\mu\partial^\mu\phi(x)+2m^2\phi(x))$

Setting this equal to zero should give me the Klein-Gordon equation, but the first term is off by a factor of 2. What's wrong with what I've done?

I know there are other ways to do this. I want to know what's wrong with this way.

• Comment to the question (v2): Note that $\partial\phi$ appears twice in the kinetic term $\frac{1}{2}(\partial\phi)^2$. – Qmechanic Jan 13 '16 at 13:58
• $\frac{\delta}{\delta\phi}\int(\phi \partial^2\phi)\neq \partial^2\phi$, ie, the same @Qmechanic says... – AccidentalFourierTransform Jan 13 '16 at 14:01
• @Qmechanic Sure, but I still don't see how that changes anything. $\partial_\mu\phi\partial^\mu\phi=\partial_\mu(\phi\partial^\mu\phi)-\phi \partial_\mu \partial^\mu \phi$. Then drop the boundary term. – Okazaki Jan 13 '16 at 14:15
• @ryanp16 your doubt is actually pretty common afaik. Jut try to explicitly write the definition of funcional derivative (whatever definition you like; eg, $\phi(x)\to\phi(x)+\alpha \eta(x)$ for arbitrary $\eta(x)$ and infinitesimal $\alpha$), and see what you get. Hint: $\frac{\delta}{\delta\phi}\partial^2\phi\neq0$ – AccidentalFourierTransform Jan 13 '16 at 16:46
• @ryanp16 yeah, you almost got it. Functional and partial derivatives commute because they depend on different spacetime variables: the partial derivative is w.r.t. $y$ and the functional derivative is w.r.t $\phi(x)$ ($x$ vs $y$). So what you actually want to calculate is $\frac{\delta}{\delta\phi(x)}\int\mathrm dy\ \phi(y)\left(\frac{\partial}{\partial y}\right)^2 \phi(y)$: note the spacetime dependencies. – AccidentalFourierTransform Jan 13 '16 at 18:33

Let me give a pedagogical answer. You're confused about the functional derivative

$$\frac{\delta}{\delta \phi(y)}\int dx\, \phi(x) \partial^2 \phi(x).$$

We can compute such derivatives by perturbing a functional:

$$F[\phi + \epsilon \chi] = F[\phi] + \epsilon \int dx \, \frac{\delta F[\phi]}{\delta \phi(x)}\chi(x) + O(\epsilon^2).$$

Now let $F[\phi] = \int dx\, \phi(x) \partial^2 \phi(x)$. Then

$$F[\phi + \epsilon \chi] - F[\phi] = \epsilon \int dx \left[ \chi(x) \partial^2 \phi(x) + \phi(x) \partial^2 \chi(x) \right] + O(\epsilon^2).$$ But this is not of the correct form, due to the $\partial^2 \chi(x)$ term. However, you can always integrate by parts (exercise): $$\int dx \, \phi(x) \partial^2 \chi(x) = \int dx \, \chi(x) \partial^2 \phi(x) \, + \, \text{boundary terms}.$$ By assumption, the boundary terms do not contribute. Bringing everything together, we can rewrite the formula above as

$$F[\phi + \epsilon \chi] - F[\phi]= 2 \times \epsilon \int dx \, \chi(x) \partial^2 \phi(x) + O(\epsilon^2).$$ At this point, we conclude that

$$\frac{\delta F[\phi]}{\delta \phi(x)} = 2 \partial^2 \phi(x).$$

• +1 It is surprising that even many "pedagogical" QFT texts do not explicitly explain this. – Arturo don Juan Oct 8 '18 at 15:57

Consider $$A = \frac{\delta}{\delta \phi(x)} \int d^{4}y \phi(y) \partial^{2} \phi(y)$$

Using the Leibniz property,

$$A = \int d^4 y \Big[\frac{\delta \phi(y)}{\delta \phi(x)} \partial^{2} \phi(y) + \phi(y) \frac{\delta}{\delta \phi(x)} \partial^{2} \phi(y)\Big]$$

Using the fact that partial and functional derivatives commute, we have:

$$A = \int d^4 y \Big[\delta^{(4)} (y-x) \partial^{2} \phi(y) + \phi(y)\partial^{2} \Big( \frac{\delta}{\delta \phi(x)} \phi(y) \Big)\Big]$$

Thus,

$$A = \partial^{2} \phi(x) + \int d^{4}y \phi(y)\partial^{2} \Big( \delta^{(4)}(x-y) \Big)$$

In the second term, the second derivative is on the wrong function. It should be on $$\phi(y)$$.

Integrate by parts and use the fact that both $$\phi$$ and $$\delta^{(4)}$$ disappear at the boundaries to show that

$$A = 2 \partial^{2} \phi(x)$$