I have to compute the determinant of a differential operator that comes from an action variation: $\frac{\delta^2S}{\delta q(\tau)\delta q(\tau)}=-M\frac{\partial^2}{\partial \tau^2}+V''(q)$.
It is known that $\delta q(0)=0$ and $\delta q(B)=0$. What I really need is the value of $D=\mid\frac{det(-M\frac{\partial^2}{\partial \tau^2}+M w_0^2)}{det'(-M\frac{\partial^2}{\partial \tau^2}+V''(q))}\mid^{1/2}$ where $Mw_0^2=V''(0)$ and det' means that the eiganvalue equal to zero is omitted to compoute the determinant.
How can D be computed?
Thank you very much.
To compute functional determinant is quite difficult in general. However in this case, one can recall that the determinant is the product of the complete set of eigenvalues.
Now, find the eigenvalues of this operator \begin{equation} -m(\partial_T^2+\omega^2)f_n=\epsilon_n f_n \end{equation} A complete set of solution to the above differential equation is given by \begin{equation} f_n(t)=\sin\left(\frac{n\pi t}{T}\right) \quad (n=1,2,3,\cdots) \end{equation} and the eigenvalues are given by \begin{equation} m \left(\frac{\pi ^2 n^2}{T^2}-\omega ^2\right) \quad n=1,2,\cdots \end{equation} Thus, the determinant is given by \begin{equation} \det\left(-m(\partial_T^2+\omega^2) \right)^{-\frac12}=\prod_{n=1}^\infty \left[m \left(\frac{\pi ^2 n^2}{T^2}-\omega ^2\right) \right]^{-\frac{1}{2}} \end{equation}
Now, you can see several problems that might arise here. There might be zero eigenvalues and this method does not work. If you look at the last expression, the infinite product is clearly divergent.
But again, in your case, you can do the similar calculation to find the denominator and it turns out the divergent quantity cancels form the denominator and numerator and you get a nice finite answer.