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I've been following the EPFL lecture notes in advanced QFT, mainly chapter 2 which involves effective potential calculations, because it's important for my thesis, and I'm looking for some clarification on the concepts presented there.

First, I'd like to know what the "second functional derivative" (like $\displaystyle \frac{\delta^2 S[\phi]}{\delta\phi(x)\delta\phi(y)}$) is, and how can I compute it (rigorously, I'm afraid the hand-wavy way some physicists do it isn't exactly clear to me) for, say, the standard quartic self-interaction Lagrangian:

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

The main issue I have is with the derivative term, I'm guessing something doesn't commute properly, and if I'm not mistaken, the second functional derivative of the other parts is:

$\displaystyle \frac{\delta^2 S_{non-derivative}[\phi]}{\delta\phi(x)\delta\phi(y)}=-m^2\delta^4(x-y)-\frac{\lambda}{2}\phi^2(y)\delta^4(x-y)$

As an aside, how to compute the functional derivative(s) of that? Do I just treat the deltas as constants, so we end up with products of deltas?

Second, I'm having trouble following the actual computation of the trace (eq. 2.39):

$tr\log\left(\Box+m^2+\frac{\lambda}{2}\phi^2\right)$

How exactly is this trace defined, as we're dealing with differential operators? The author says he will use a plane wave basis, whatever that means (I'm guessing those are the eigenfunctions of the operator under the log), and, for reasons unclear to me, treats $\phi^2$ as a constant, while from the functional derivative we see that it has some spacetime dependence other than the delta function, how does that work? Frankly, I'm quite confused, and any guidance would be appreciated, especially if it's more mathematically rigorous.

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  • $\begingroup$ To reopen this post (v1), consider to only ask one subquestion, cf e.g. this meta post. In particular, the second functional derivative and the trace should be separate questions. $\endgroup$ – Qmechanic Nov 6 '16 at 11:34
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    $\begingroup$ @Qmechanic I think it is an interesting question and perhaps good answers aren't going to be that long. Or partial answers could still be interesting enough. $\endgroup$ – TwoBs Nov 6 '16 at 12:40
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    $\begingroup$ @blueshift I usually regulate spacetime on a lattice $x\rightarrow i$, $\int dx \rightarrow \sum_i $, $\phi(x)\rightarrow \phi_i$, $\delta^2 S/\delta\phi(x)\delta\phi(y)\rightarrow \partial^2 S/\partial\phi_i \partial\phi_j$, and so on. Then, once I have gotten the desired result I map back. For example, the third derivatives of $S$ with respect to $\phi_i$, $\phi_j$ and $\phi_k$ is $\partial^3 S/\partial\phi_i \partial\phi_j \phi_k=-\lambda\phi_j \delta_{ij}\delta_{jk}$ (no summation) which maps into $-\lambda\phi(y) \delta(x-y)\delta(y-z)$. As for the trace, one can use whatever basis $\endgroup$ – TwoBs Nov 6 '16 at 12:44
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    $\begingroup$ @blueshift As a matter of fact you don't even need to know it to get the right answer. Indeed, the kinetic term is can be written as $-1/2\int dy dx \phi(y) \delta(x-y)\square \phi(x)$ which shows that it's local, i.e. $\propto \delta(x-y)$. In turn, on the lattice it will look like $1/2\sum_{i,j}\phi_i K_{ij}\phi_j$ where you don't need to know much about K_{ij} except that it's real symmetry and invertible (here it comes the $i\epsilon$ prescription, so the lattice is actually used after a Wik rotation) which is important to derive the result with the trace you are refering to... $\endgroup$ – TwoBs Nov 6 '16 at 13:40
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    $\begingroup$ In any case $\partial^2 S/\partial\phi_i \phi_j=K_{ij}$ so that mapping back to the continuum (and Wick rotating back) you get $\delta^2 S/\delta\phi(x) \delta\phi(y)=-\delta(x-y)\square$. $\endgroup$ – TwoBs Nov 6 '16 at 13:42