This is a question spawning from a comment made to my previous question. There I was asking about taking some functional derivative in the effective action of the non-linear sigma model. The comment made it clear to me that I have not enough familiarity with functional determinants, functional traces and such objects and I want to clarify some issues. I do not look forward to having a mathematically rigorous treatment, but I do want something conceptually complete. Let's set the stage. After integrating the $\sigma$ fields in the generating functional in the $O(N)$ non-linear sigma model we are left with $$ Z[J^a]=\int[d\alpha]\exp(\frac{-N}{2} tr\log(-\partial^2-i\alpha)-\frac{i}{2}\frac{N}{t_0}\int d^dx\alpha+\frac{1}{2}\int d^dx\,d^dyJ^a(x)\Delta(x-y)J^a(y)) $$ I should point out that $d=2+\epsilon$ since I am using dimensional regularization, although for the purposes of this question $d$ could be thought to be 2. I am not interested in most of the stuff above. I want to just focus on $$ \int[d\alpha]\exp(\frac{-N}{2} tr\log(-\partial^2-i\alpha)) $$ And completely understand how this comes from the following functional integral $$ \int[d\sigma']\exp\bigg(\int d^dx\bigg[-\frac{1}{2}\sigma'^aL\sigma'^a\bigg]\bigg) $$ where I have defined $L\equiv(-\partial^2-i\alpha)$. In the exponent above we have a sum over the $N$ sigma fields so the functional integral is nothing but a $N$ times the same functional integral $$ =\bigg(\int[d\sigma'^1]\exp\bigg(\int d^dx\bigg[-\frac{1}{2}\sigma'^1L\sigma'^1\bigg]\bigg)\bigg)^N $$ So far I am happy. Now, let's consider the following integral where $a$ is a postive constant $$ \int_{-\infty}^{\infty}dx\exp\bigg(-\frac{1}{2}xax\bigg)=\frac{(2\pi)^{1/2}}{a^{1/2}} $$ This is of course a Gaussian integral that is very easy. This integral can be generalized to $\mathbb{R}^n$ where we would have $$ \int_{\mathbb{R}^n}d^nx\exp\bigg(-\frac{1}{2}x_iA_{ij}x_j\bigg)=\frac{(2\pi)^{n/2}}{(\det A)^{1/2}} $$ and $A$ is a symmetric positive matrix and as usual repeated indices are summed over. The functional integral I am interested in is supposed to be the generalization of the above case, where we consider the continuously infinite case. $$ =\int[d\phi]\exp\bigg(\int d^dx\bigg[-\frac{1}{2}\phi(x) L(x)\phi(x)\bigg]\bigg) $$ Nonetheless I perceive some differences. First, in the $\mathbb{R}^n$ case we had summation over two indexes, $i$ and $j$. In the continuously infinite case the role of $x_i$ is played by $\phi(x)$ and that of $A_{ij}$ by $L(x)$. Here I see a difference. In the continuously infinite case we only have one index, namely $x$. I have thought that perhaps we can introduce one by doing $$ =\int[d\phi]\exp\bigg(\int d^dx\,d^dy\bigg[-\frac{1}{2}\phi(x)\delta(x-y) L(x)\phi(y)\bigg]\bigg) $$ and in this case we have summation over two indexes and the role of $A_{ij}$ is played by $\delta(x-y)L(x)$. Nonetheless, it doesn't look satisfactory to me because the "index" $y$ only enters in the generalized matrix through the Dirac delta, and one could say that in the finite dimensional case $i$ and $j$ played similar roles. What am I missing?

  • $\begingroup$ The operator $L$ should formally carry two 'indices' as it is a linear map from loosely speaking the vector space of twice differentiable functions to the vector space of continuous functions. Start with $N$ point and put periodic boundary condition. Pick a set of basis for the space of functions $R^N$. You can write the discrete version of $L$ as a matrix, which manifests the index structure. $\endgroup$ – user110373 Jun 6 '18 at 15:34
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    $\begingroup$ Abstractly, both $\sum x_iA_{ij}x_j$ and $\int d^dx\phi(x) L(x)\phi(x)$ can be thought of $(v,Av)$ for some inner product $()$ and some operator $()$.The sum of integration corresponds to $()$ and therefore one should not integrate twice unless $A$ is non-local in which case it comes with an integral in its definition. $\endgroup$ – user110373 Jun 6 '18 at 15:38

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