How does one correctly calculate the second-order variation of an action? I have started an attempt at the calculation (restricting the scalar fields for simplicity), but I'm unsure how to proceed.

Starting with the action for a free scalar field $$S[\phi]=\int\;d^{4}x\mathcal{L}=\frac{1}{2}\int\;d^{4}x\left(\partial_{\mu}\phi(x)\partial^{\mu}\phi(x)-m^{2}\phi^{2}(x)\right)$$ with Minkowski sign convention $(+,-,-,-)$. Naively, if I expand this to second-order, I get $$S[\phi+\delta\phi]=S[\phi]+\int\;d^{4}x\frac{\delta S[\phi(x)]}{\delta\phi(x)}\delta\phi(x)+\int\;d^{4}x d^{4}y\frac{\delta^{2} S[\phi(x)]}{\delta\phi(x)\delta\phi(y)}\delta\phi(x)\delta\phi(y)$$ Now, assuming that $\phi(x)$ satisfies the equations of motion (EOM), then the first-order term vanishes, however, I'm unsure how to calculate the second-order variation. So far, my attempt is $$\delta^{2}S=\int\;d^{4}x d^{4}y\frac{\delta^{2} S[\phi(x)]}{\delta\phi(x)\delta\phi(y)}\delta\phi(x)\delta\phi(y)=\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\\=\int\;d^{4}x d^{4}y\left(\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial\phi(y)}\delta\phi(x)\delta\phi(y)+2\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial(\partial_{\mu}\phi(y)}\delta\phi(x)\delta(\partial_{\mu}\phi(y))\\+\frac{\partial^{2}\mathcal{L}}{\partial(\partial_{\mu}\phi(x))\partial(\partial_{\mu}\phi(y))}\delta(\partial_{\mu}\phi(x))\delta(\partial_{\mu}\phi(y))\right)$$ However, I am unsure how to progress (integration by parts doesn't seem to work as nicely in this case), as naively it seems as though the only term that would survive is $\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial\phi(y)}$, but I've seen references stating that $\frac{\delta^{2} S[\phi(x)]}{\delta\phi(x)\delta\phi(y)}$ is of the form $\frac{\delta^{2} S[\phi(x)]}{\delta\phi(x)\delta\phi(y)}\sim\Box +m^{2}$.

Any help would be much appreciated.


I have though about it a bit more and have come up with a general formula (for a Lagrangian with up to first-order derivatives in the fields) that I hope is correct: $$\frac{\delta^{2} S[\phi(x)]}{\delta\phi(x)\delta\phi(y)}=\frac{\delta^{2}}{\delta\phi(x)\delta\phi(y)}\int\;d^{4}z\,\mathcal{L}\left(\phi(z),\partial_{\mu}\phi(z)\right)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\=\frac{\delta}{\delta\phi(x)}\int\;d^{4}z\left[\frac{\partial\mathcal{L}}{\partial\phi(z)}\delta^{4}(z-y)+\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi(z))}\partial_{\mu}\left(\delta^{4}(z-y)\right)\right]\qquad\qquad\qquad\qquad\\=\int\;d^{4}z\left[\frac{\partial^{2}\mathcal{L}}{\partial\phi(z)^{2}}\delta^{4}(z-x)\delta^{4}(z-y)+2\frac{\partial^{2}\mathcal{L}}{\partial\phi(z)\partial(\partial_{\mu}\phi(z))}\partial_{\mu}\left(\delta^{4}(z-x)\right)\delta^{4}(z-y)\\+\frac{\partial^{2}\mathcal{L}}{\partial(\partial_{\mu}\phi(z))\partial(\partial_{\nu}\phi(z))}\partial_{\mu}\left(\delta^{4}(z-x)\right)\partial_{\nu}\left(\delta^{4}(z-x)\right)\right]\\ =\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)^{2}}\delta^{4}(x-y)+2\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial(\partial_{\mu}\phi(x))}\partial_{\mu}\left(\delta^{4}(x-y)\right)\qquad\qquad\qquad\qquad\qquad\qquad\qquad-\frac{\partial^{2}\mathcal{L}}{\partial(\partial_{\mu}\phi(x))\partial(\partial_{\nu}\phi(x))}\partial_{\mu}\partial_{\nu}\left(\delta^{4}(x-y)\right)$$ which, upon further integrations by parts (neglecting boundary terms), gives $$\frac{\delta^{2} S[\phi(x)]}{\delta\phi(x)\delta\phi(y)}=\delta^{4}(x-y)\left[\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)^{2}}-2\partial_{\mu}\left(\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial(\partial_{\mu}\phi(x))}\partial_{\mu}\right)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad-\partial_{\mu}\partial_{\nu}\left(\frac{\partial^{2}\mathcal{L}}{\partial(\partial_{\mu}\phi(x))\partial(\partial_{\nu}\phi(x))}\right)\right]$$ This seems to give the expected answer in the case of a free scalar field. Indeed, $$\frac{\partial^{2}\mathcal{L}}{\partial\phi^{2}}=-m^{2}\, ,\qquad \frac{\partial^{2}\mathcal{L}}{\partial(\partial_{\mu}\phi)\partial(\partial_{\nu}\phi)}=\eta^{\mu\nu}\, , \qquad\frac{\partial^{2}\mathcal{L}}{\partial\phi\partial(\partial_{\mu}\phi)}=0$$ and hence, $$\frac{\delta^{2} S[\phi(x)]}{\delta\phi(x)\delta\phi(y)}=-\delta^{4}(x-y)\left[\Box+m^{2}\right]$$ Any feedback on whether this is correct or not would be much appreciated.


2 Answers 2


A proper treatment (and how you should usually go about these things if you forget) is to remember the definition of the functional derivative. It is linear, defined to obey a chain rule, a product rule, and has the fundamental feature


Thus, in painstaking detail, we have

$$\frac{\delta S[\phi]}{\delta\phi(x)}=\frac{1}{2}\int\mathrm{d}^dy\left[\frac{\delta}{\delta\phi(x)}\left(\partial\phi(y)\cdot\partial\phi(y)\right)-m^2\frac{\delta}{\delta\phi(x)}\phi(y)^2\right]\\ =\int\mathrm{d}^dy\left[\partial_{\mu}\delta(x-y)\partial^{\mu}\phi(y)-m^2\delta(x-y)\phi(y)\right]\\ =-(\square+m^2)\phi(x)$$

Thus, we can simply differentiate again to obtain


Which is the desired result (note that $\square_x$ simply means that the derivative is only with respect to $x$ -- sometimes this matters)! Note that the delta function comes after the Klein-Gordon operator.

And that's it! No need to expand to second order or pull your hair out deciding whether you have to integrate by parts and when you can.

I hope this helps!


This type of manipulation is actually extremely useful! For instance, in the path integral formulation, we have


With this, we can use the above manipulations to find correlation functions! The key is to note that the path integral of a total functional derivative is zero. Thus, we have

$$\int\mathcal{D}\phi\,\frac{\delta^2}{\delta\phi(x)\delta\phi(y)}e^{iS[\phi]}=i\int\mathcal{D}\phi\left[\frac{\delta^2S}{\delta\phi(x)\delta\phi(y)}+i\frac{\delta S}{\delta\phi(x)}\frac{\delta S}{\delta\phi(y)}\right]e^{iS[\phi]}\\ =i\bigg\langle\frac{\delta^2S}{\delta\phi(x)\delta\phi(y)}+i\frac{\delta S}{\delta\phi(x)}\frac{\delta S}{\delta\phi(y)}\bigg\rangle=0$$

This holds for any action $S[\phi]$. In particular, in your free theory, this gives us


Eliminating $\square_y+m^2$ from each side tells you that the two point function for a free theory is the Green's function of the Klein-Gordon operator. No need for generating functionals or all that messy second quantization.

  • $\begingroup$ Fantastic answer. I was making it way more difficult than I needed to! A couple of questions I have though: 1. One assumes that one can commute the variational derivative with the spacetime derivative. Is this always valid? 2. What is the exact justification for dropping the boundary term? I thought it was due to $\delta\phi$ vanishing on the boundary, but in this case one has a boundary term of the form $\int\;d^{d}y\partial\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi(y))}\delta^{4}(x-y)\right)$, I don't see immediately why this term vanishes? $\endgroup$
    – Will
    May 10, 2017 at 18:32
  • $\begingroup$ To answer your questions: 1.) The variational derivative will commute with the spacetime derivative since they are derivatives with respect to different things. 2.) The boundary terms vanish because distributions like $\delta(x-y)$ and its derivatives all have compact support -- that is, they and anything multiplied by them vanish at the integral boundaries (this would still be the case, even if the integrals were over a finite spacetime volume). $\endgroup$ May 10, 2017 at 18:42
  • $\begingroup$ Ok, thanks. In a heuristic argument for the $\delta$-function term can one argue that $\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi(y))}\delta^{4}(x-y)=0$ because $x^{\mu}\neq y^{\mu}$ in analogy to the identity $x\delta(x)=0$? $\endgroup$
    – Will
    May 10, 2017 at 19:05
  • $\begingroup$ Exactly. The integral itself only has support on the point $x=y$. $\endgroup$ May 10, 2017 at 19:05
  • $\begingroup$ Ok cool. Is there any way to show this explicitly? Thanks for the "Bonus Round" section by the way, it has proved very enlightening for me. $\endgroup$
    – Will
    May 10, 2017 at 19:13

In your first attempted calculation, there is an issue in the second line. Your update looks almost ok, up to the point when you say you further integrate by parts: at that point, your are no longer under an integral so you can't do that (indeed, you have already done the IBPs implicitly in lines two and three, yielding the $\partial _{\mu } (\delta ^4(z-y))$ terms). But this is not needed either, as the correct result is:

$$K(x,y) = - m^2 \delta ^4(x-y) - \eta^{\mu \nu } \partial _{\mu }\partial _{\nu } (\delta ^4(x-y))$$

which is the integral kernel of the operator $-m^2 - \square$ (as can be seen by computing $\int dx dy\; f(x) K(x,y) g(y)$ via IBP).

A different way of coming at the result (which looks cleaner to me, but that's just my own cosmetic opinion...), is to go back to the definition of $\frac{\delta ^2 S[\phi ]}{\delta [\phi (x)]\delta [\phi (y)]}$, namely:

$$S[\phi + \delta \phi ] - S[\phi ] =: \int dx\; \frac{\delta S[\phi ]}{\delta [\phi (x)]} \delta \phi (x) + \frac{1}{2} \int dx dy\; \frac{\delta ^2 S[\phi ]}{\delta [\phi (x)]\delta [\phi (y)]} \delta \phi (x) \delta \phi (y) + o(\delta \phi ^2) \tag{1}$$

Note that this equation only prescribes $\frac{\delta ^2 S[\phi ]}{\delta [\phi (x)]\delta [\phi (y)]}$ up to antisymmetric terms: it must be completed by the requirement that $\frac{\delta ^2 S[\phi ]}{\delta [\phi (x)]\delta [\phi (y)]}$ be symmetric in $x,y$ (just like $\frac{\partial ^2 f (\vec{v})}{\partial v^j \partial v^i}$ is symmetric in $i,j$ for an ordinary function of $n$ variables).

So we must calculate the second order variation of $S[\phi ]$:

$$S[\phi ] = \int d^4x\; \mathcal{L}(\phi (x),\partial \phi (x))$$

$$S[\phi +\delta \phi ] = \int d^4x\; \mathcal{L}(\phi (x)+\delta \phi (x),\partial \phi (x)+\partial \delta \phi (x))\\ = \int d^4x\; \mathcal{L}(\phi (x),\partial \phi (x)) + (\text{1st order terms = 0 on shell})\\ + \frac{1}{2} \frac{\partial ^2\mathcal{L}}{\partial \phi ^2}(\phi (x),\partial \phi (x)) \,\delta \phi (x) \,\delta \phi (x)\\ + \frac{\partial ^2\mathcal{L}}{\partial \phi \partial (\partial _{\mu }\phi )}(\phi (x),\partial \phi (x)) \,\partial _{\mu }(\delta \phi (x)) \,\delta \phi (x)\\ + \frac{1}{2} \frac{\partial ^2\mathcal{L}}{\partial (\partial _{\mu }\phi )\partial (\partial _{\nu }\phi )}(\phi (x),\partial \phi (x)) \,\partial _{\mu }(\delta \phi (x)) \,\partial _{\nu }(\delta \phi (x))$$

Since the definition of $\frac{\delta ^2 S[\phi ]}{\delta [\phi (x)]\delta [\phi (y)]}$ requires an integral over $x$ and $y$, we introduce it by force now (keeping carefully track of the variable on which the various derivatives act):

$$ = S[\phi ] + \int d^4x \,d^4y \;\delta ^4(x-y) \Big[\\ \frac{1}{2} \frac{\partial ^2\mathcal{L}}{\partial \phi ^2}(\phi (x),\partial \phi (x)) \,\delta \phi (x) \,\delta \phi (y)\\ + \frac{\partial ^2\mathcal{L}}{\partial \phi \partial (\partial _{\mu }\phi )}(\phi (x),\partial \phi (x)) \,\partial ^{(x)}_{\mu }(\delta \phi (x)) \,\delta \phi (y)\\ + \frac{1}{2} \frac{\partial ^2\mathcal{L}}{\partial (\partial _{\mu }\phi )\partial (\partial _{\nu }\phi )}(\phi (x),\partial \phi (x)) \,\partial ^{(x)}_{\mu }(\delta \phi (x)) \,\partial ^{(y)}_{\nu }(\delta \phi (y)) \Big]$$

Now we perform a few IBPs, in the variables $x$ and $y$. The boundary terms will be proportional to either $\delta \phi (x)$ at the $x$ boundary or $\delta \phi (y)$ at the $y$ boundary, and these are typically assumed to be zero when calculating variations. So we get:

$$ = S[\phi ] + \int d^4x \,d^4y \; \Big[\\ \frac{1}{2} \frac{\partial ^2\mathcal{L}}{\partial \phi ^2}(\phi (x),\partial \phi (x)) \,\delta ^4(x-y)\\ - \partial ^{(x)}_{\mu } \left( \frac{\partial ^2\mathcal{L}}{\partial \phi \partial (\partial _{\mu }\phi )}(\phi (x),\partial \phi (x)) \,\delta ^4(x-y) \right)\\ + \frac{1}{2} \partial ^{(x)}_{\mu }\partial ^{(y)}_{\nu } \left( \frac{\partial ^2\mathcal{L}}{\partial (\partial _{\mu }\phi )\partial (\partial _{\nu }\phi )}(\phi (x),\partial \phi (x)) \,\delta ^4(x-y) \right) \Big] \,\delta \phi (x) \,\delta \phi (y) \tag{2}$$

Note that we have used that $\partial ^{(x)}_{\mu }(\delta \phi (y)) = 0$ (that's why we had to somewhat artificially introduce a distinction between the $x$ and $y$ variables before performing the IBPs, otherwise we would be left with $\partial \delta \phi $ terms).

Now, for any function $F$, $F(x)\,\delta ^4(x-y)$ is symmetric with respect to $x \leftrightarrow y$, while the symmetric part of $\partial _{\mu }^{(x)} \left( F(x) \, \delta ^4(x-y) \right)$ is:

$$\frac{1}{2} \big[ \partial _{\mu }^{(x)} \left( F(x) \, \delta ^4(x-y) \right) + \partial _{\mu }^{(y)} \left( F(y) \, \delta ^4(y-x) \right) \big] = \\ = \frac{1}{2} \big[ \left( \partial _{\mu }^{(x)} F(x) \right) \delta ^4(x-y) + F(x) \left( \partial _{\mu }^{(x)} \delta ^4(x-y) \right) + \partial _{\mu }^{(y)} \left( F(y) \, \delta ^4(y-x) \right) \big]\\ = \frac{1}{2} \big[ \left( \partial _{\mu }^{(x)} F(x) \right) \delta ^4(x-y) - F(x) \left( \partial _{\mu }^{(y)} \delta ^4(x-y) \right) + \partial _{\mu }^{(y)} \left( F(y) \, \delta ^4(y-x) \right) \big]\\ = \frac{1}{2} \big[ \left( \partial _{\mu }^{(x)} F(x) \right) \delta ^4(x-y) - \partial _{\mu }^{(y)} \left( F(x) \, \delta ^4(x-y) \right) + \partial _{\mu }^{(y)} \left( F(x) \, \delta ^4(x-y) \right) \big]\\ = \frac{\partial _{\mu }^{(x)} F(x)}{2} \delta ^4(x-y)$$

and, similarly, the symmetric part of $\partial _{\mu }^{(x)} \partial _{\nu }^{(y)} \left( F(x) \, \delta ^4(x-y) \right)$ is:

$$\frac{1}{2} \big[ \partial _{\mu }^{(x)} \partial _{\nu }^{(y)} \left( F(x) \, \delta ^4(x-y) \right) + \partial _{\mu }^{(y)} \partial _{\nu }^{(x)} \left( F(y) \, \delta ^4(y-x) \right) \big] = \\ = \frac{1}{2} \big[ \partial _{\mu }^{(x)} \left( F(x) \, \partial _{\nu }^{(y)} \delta ^4(x-y) \right) + \partial _{\mu }^{(y)} \partial _{\nu }^{(x)} \left( F(x) \, \delta ^4(x-y) \right) \big]\\ = \frac{1}{2} \big[ - \partial _{\mu }^{(x)} \left( F(x) \, \partial _{\nu }^{(x)} \delta ^4(x-y) \right) - \partial _{\nu }^{(x)} \left( F(x) \, \partial _{\mu }^{(x)} \delta ^4(x-y) \right) \big]\\ = \frac{1}{2} \big[ - \left( \partial _{\mu }^{(x)} F(x) \right) \left( \partial _{\nu }^{(x)} \delta ^4(x-y) \right) - F(x) \left( \partial _{\mu }^{(x)} \partial _{\nu }^{(x)} \delta ^4(x-y) \right) - \left( \partial _{\nu }^{(x)} F(x) \right) \left( \partial _{\mu }^{(x)} \delta ^4(x-y) \right) - F(x) \left( \partial _{\nu }^{(x)} \partial _{\mu }^{(x)} \delta ^4(x-y) \right) \big]\\ = - \frac{1}{2} \big[ \left( \partial _{\mu }^{(x)} F(x) \right) \left( \partial _{\nu }^{(x)} \delta ^4(x-y) \right) + \left( \partial _{\nu }^{(x)} F(x) \right) \left( \partial _{\mu }^{(x)} \delta ^4(x-y) \right) \big] - F(x) \left( \partial _{\mu }^{(x)} \partial _{\nu }^{(x)} \delta ^4(x-y) \right)$$

So, identifying eq. (2) with the definition of $\frac{\delta ^2 S[\phi ]}{\delta [\phi (x)]\delta [\phi (y)]}$ above (eq. (1)), we get:

$$\frac{\delta ^2 S[\phi ]}{\delta [\phi (x)]\delta [\phi (y)]} = \left( \frac{\partial ^2\mathcal{L}}{\partial \phi ^2}(\phi (x),\partial \phi (x)) - \partial ^{(x)}_{\mu } \frac{\partial ^2\mathcal{L}}{\partial \phi \partial (\partial _{\mu }\phi )}(\phi (x),\partial \phi (x)) \right) \delta ^4(x-y)\\ - \left( \partial ^{(x)}_{\mu } \frac{\partial ^2\mathcal{L}}{\partial (\partial _{\mu }\phi )\partial (\partial _{\nu }\phi )}(\phi (x),\partial \phi (x)) \right) \partial ^{(x)}_{\nu } \delta ^4(x-y)\\ - \left( \frac{\partial ^2\mathcal{L}}{\partial (\partial _{\mu }\phi )\partial (\partial _{\nu }\phi )}(\phi (x),\partial \phi (x)) \right) \partial ^{(x)}_{\mu }\partial ^{(x)}_{\nu } \delta ^4(x-y)$$

Check: We can check that we get the same result (and that we get it much faster...) with Bob Knighton's method:

$$S[\phi ] = \int d^4x\; \mathcal{L}(\phi (x),\partial \phi (x))$$

$$\frac{\delta S[\phi ]}{\delta [\phi (x)]} = \int d^4y\; \frac{\partial \mathcal{L}}{\partial \phi }(\phi (y),\partial \phi (y)) \frac{\delta \phi (y)}{\delta [\phi (x)]} + \frac{\partial \mathcal{L}}{\partial (\partial _{\mu } \phi }(\phi (y),\partial \phi (y)) \partial ^{(y)}_{\mu } \frac{\delta \phi (y)}{\delta [\phi (x)]}\\ = \frac{\partial \mathcal{L}}{\partial \phi }(\phi (x),\partial \phi (x)) - \left( \partial ^{(x)}_{\mu } \frac{\partial \mathcal{L}}{\partial (\partial _{\mu } \phi }(\phi (x),\partial \phi (x)) \right)$$

$$\frac{\delta ^2 S[\phi ]}{\delta [\phi (x)]\delta [\phi (y)]} = \frac{\delta }{\delta [\phi (y)]} \big[ \frac{\partial \mathcal{L}}{\partial \phi }(\phi (x),\partial \phi (x)) - \left( \partial ^{(x)}_{\mu } \frac{\partial \mathcal{L}}{\partial (\partial _{\mu } \phi )}(\phi (x),\partial \phi (x)) \right) \big]\\ = \frac{\delta }{\delta [\phi (y)]} \big[ \frac{\partial \mathcal{L}}{\partial \phi }(\phi (x),\partial \phi (x)) \big] - \partial ^{(x)}_{\mu } \left( \frac{\delta }{\delta [\phi (y)]} \big[ \frac{\partial \mathcal{L}}{\partial (\partial _{\mu } \phi )}(\phi (x),\partial \phi (x)) \big] \right)\\ = \frac{\partial ^2 \mathcal{L}}{\partial \phi ^2}(\phi (x),\partial \phi (x)) \, \delta ^4(x-y) + \frac{\partial ^2 \mathcal{L}}{\partial \phi \partial (\partial _{\mu } \phi )}(\phi (x),\partial \phi (x)) \left( \partial ^{(x)}_{\mu } \delta ^4(x-y) \right) - \partial ^{(x)}_{\mu } \left( \frac{\partial ^2 \mathcal{L}}{\partial \phi \partial (\partial _{\mu } \phi )}(\phi (x),\partial \phi (x)) \, \delta ^4(x-y) + \frac{\partial ^2 \mathcal{L}}{\partial (\partial _{\mu } \phi ) \partial (\partial _{\nu } \phi )}(\phi (x),\partial \phi (x)) \left( \partial ^{(x)}_{\nu } \delta ^4(x-y) \right) \right)\\ = \left( \frac{\partial ^2 \mathcal{L}}{\partial \phi ^2}(\phi (x),\partial \phi (x)) - \partial ^{(x)}_{\mu } \frac{\partial ^2 \mathcal{L}}{\partial \phi \partial (\partial _{\mu } \phi )}(\phi (x),\partial \phi (x)) \right) \delta ^4(x-y) - \left( \partial ^{(x)}_{\mu } \frac{\partial ^2 \mathcal{L}}{\partial (\partial _{\mu } \phi ) \partial (\partial _{\nu } \phi )}(\phi (x),\partial \phi (x)) \right) \left( \partial ^{(x)}_{\nu } \delta ^4(x-y) \right) - \frac{\partial ^2 \mathcal{L}}{\partial (\partial _{\mu } \phi ) \partial (\partial _{\nu } \phi )}(\phi (x),\partial \phi (x)) \left( \partial ^{(x)}_{\mu } \partial ^{(x)}_{\nu } \delta ^4(x-y) \right)$$

  • $\begingroup$ Ah ok, thanks for your comments so far. I look forward to your updates concerning $\frac{\delta^{2} S[\phi(x)]}{\delta\phi(x)\delta\phi(y)}$. Additionally, I'm not completely sure why one can neglect the boundary terms? If you could comment on this that would be great. $\endgroup$
    – Will
    May 10, 2017 at 17:58
  • $\begingroup$ Thanks for the additional details. So for the boundary terms does one assume that $x\neq y$, such that the $\delta$-function vanishes? What is the justification for this? $\endgroup$
    – Will
    May 10, 2017 at 19:19
  • $\begingroup$ I don't really need that: as far as the IBP in, say, $x$ is concerned, the functions of $y$ are completely spectators. For $\int dx dy\, F(x,y) (\partial^{(x)} \delta \phi(x)) \delta \phi(y)$ the boundary term is $\delta \phi(x) \int dy\, F(x,y) \delta \phi(y)$. $\endgroup$
    – Luzanne
    May 10, 2017 at 19:27
  • $\begingroup$ So the boundary terms vanish by virtue of being proportional to $\delta\phi(x)$ which is evaluated on the boundary and so is zero? (Sorry I couldn't accept your answer as well by the way, both answers are excellent) $\endgroup$
    – Will
    May 10, 2017 at 19:33
  • 2
    $\begingroup$ So is $\frac{\delta S}{\delta\phi(x)}$ defined implicitly through $\delta S=\int\,d^{4}x\frac{\delta S}{\delta\phi(x)}\delta\phi(x)$? $\endgroup$
    – Will
    May 10, 2017 at 20:17

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