Determinant of Integral operators

Question

The Fredholm determinant of $1-T$ with a trace-class integral operator $T$ with kernel $K$ is ("informally") defined by \begin{equation} \text{det}_F\big(1-T\big)=\sum_{n=0}^{\infty}\frac{1}{n!}\int_{x_1\dots x_n}\det\big(K(x_j,x_k)\big)\ , \end{equation} where $1\leq j,k\leq n$. In quantum field theory one often encounters a different notion of determinant of an integral operator via Gaussian functional integrals \begin{equation} \int\mathcal{D}\phi\ e^{-\int_{x,y}\phi(x)D(x,y)\phi(y)}\sim\text{det}_I{D}^{-\frac12}\ . \end{equation} Are $\det_F$ and $\det_I$ related? Can I choose $D=1-T$ or something similar and obtain the Fredholm determinant via a functional integral and vice versa?

Answer 1

After revisiting the problem I found that the two notions of determinant are in fact related. I would, however, be very interested to hear what someone who actually knows functional analysis has to say about this.

Definitions

For in integral operator $T:f\to Tf$, $Tf(x)=\int_yK(x,y)f(y)$ we define the Fredholm determinant $$ \text{det}_F\big(1-\lambda T\big)=\sum_{n=0}^\infty\frac{(-\lambda)^n}{n!}\int_{x_1,\dots ,x_n}\det K_n\ ,\quad K_n=K(x_j,x_k)|_{j,k\in[n]}\ . $$ Alternatively, we use a Graßmann functional integral to define the functional integral determinant $$ \text{det}_I\big(T\big)=\int\mathcal{D}\bar\psi\mathcal{D}\psi e^{-\int_{x,y}\bar\psi(x)K(x,y)\psi(y)}\ . $$

Claim $$ \text{det}_F\big(1-\lambda T\big)=\text{det}_I\big(1-\lambda T\big)\ . $$

Proof $\quad$ (Assuming $\text{det}_I\big(T_1\big)\text{det}_I\big(T_2\big)=\text{det}_I\big(T_1*T_2\big)$)

We start from the definition of the Fredholm determinant and show that it is equal to the functional determinant. First rewrite $$ \det K_n=\int\text{d}\theta_1\text{d}\xi_1\dots\text{d}\theta_n\text{d}\xi_ne^{-\theta^TK_n\xi}\ , $$ where $\theta_j$ and $\xi_j$ are Graßmann numbers. Next we note that $K_n=\int_{x,y}K(x,y)d(x)d(y)^T$, where $d(x)=\big(\delta(x-x_1),\dots,\delta(x-x_n)\big)^T$ such that we can introduce Graßmann fields $$ \chi(x)=\theta^Td(x)\ ,\quad \phi(x)=\xi^Td(x) $$ to obtain $$ \det K_n=\int\text{d}\theta_1\text{d}\xi_1\dots\text{d}\theta_n\text{d}\xi_ne^{-\int_{x,y}\chi(x)K(x,y)\phi(y)}\ . $$ We now write the exponential in terms of a Graßmann functional integral using $$ e^{-\int_{x,y}\chi(x)K(x,y)\phi(y)}=\text{det}_I\big(T\big)\int\mathcal{D}\bar\psi\mathcal{D}\psi e^{-\int_{x,y}\bar\psi(x)K^{-1}(x,y)\psi(y)+i\int_x\chi(x)\psi(x)+\bar\psi(x)\phi(x)}\ , $$ where $K^{-1}$ is defined via $\int_{z}K(x,z)K^{-1}(z,y)=\delta(x-y)$. The Graßmann fields $\chi$ and $\phi$ now appear only linear in the exponent such that we get \begin{align} \det K_n=&\,\text{det}_I\big(T\big)\int\mathcal{D}\bar\psi\mathcal{D}\psi e^{-\int_{x,y}\bar\psi(x)K^{-1}(x,y)\psi(y)}\prod_{j=1}^{n}\int\text{d}\theta_j\text{d}\xi_je^{i\theta_j\psi(x_j)+i\bar\psi(x_j)\xi_j}\nonumber\\ =&\,\text{det}_I\big(T\big)\int\mathcal{D}\bar\psi\mathcal{D}\psi e^{-\int_{x,y}\bar\psi(x)K^{-1}(x,y)\psi(y)}\prod_{j=1}^{n}\left(-\bar\psi(x_j)\psi(x_j)\right)\ . \end{align} The above can now be inserted into the definition of the Fredholm determinant to obtain \begin{align} \text{det}_F\big(1-\lambda T\big)=&\,\text{det}_I\big(T\big)\int\mathcal{D}\bar\psi\mathcal{D}\psi e^{-\int_{x,y}\bar\psi(x)K^{-1}(x,y)\psi(y)}\sum_{n=0}^{\infty}\frac{(-\lambda)^{n}}{n!}\left(\int_x-\bar\psi(x)\psi(x)\right)^n\nonumber\\ =&\,\text{det}_I\big(T\big)\int\mathcal{D}\bar\psi\mathcal{D}\psi e^{-\int_{x,y}\bar\psi(x)\big(K^{-1}(x,y)-\lambda\delta(x-y)\big)\psi(y)}\nonumber\\ =&\,\text{det}_I\big(T\big)\text{det}_I\big(T^{-1}-\lambda 1\big)\nonumber\\ =&\,\text{det}_I\big(1-\lambda T\big)\ . \end{align}