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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.


Update

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.

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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

$$\frac{\delta\phi(y)}{\delta\phi(x)}=\delta(x-y)$$

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

$$\frac{\delta^2S[\phi]}{\delta\phi(x)\delta\phi(y)}=-\frac{\delta}{\delta\phi(y)}\left[(\square_x+m^2)\phi(x)\right]=-(\square_x+m^2)\delta(x-y)$$

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!

B-B-B-BONUS ROUND

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

$$\langle\mathcal{F}[\phi](x)\rangle=\int\mathcal{D}\phi\,\mathcal{F}[\phi](x)\,e^{iS[\phi]}$$

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

$$\left(\square_y+m^2\right)\left(\square_x+m^2\right)\langle\phi(x)\phi(y)\rangle=-i\left(\square_y+m^2\right)\delta(x-y)$$

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.

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  • $\begingroup$ 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 '17 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). $\endgroup$ – Bob Knighton May 10 '17 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 '17 at 19:05
  • $\begingroup$ Exactly. The integral itself only has support on the point $x=y$. $\endgroup$ – Bob Knighton May 10 '17 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 '17 at 19:13
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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:

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

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

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

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  • $\begingroup$ 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 '17 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 '17 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 '17 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 '17 at 19:33
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    $\begingroup$ 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 '17 at 20:17

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