# Finding the determinant of $(\omega^2-\partial_t^2)/2D$ in path integral? [closed]

I am looking to evaluate the following path integral: $$I=\int_{\vec x(t_0)=0}^{\vec x(t)=0}\mathcal{D}x \exp\left( -\frac{1}{2} \int^t_{t_0} d\tau \; \vec x \left\{ \frac{\omega^2-\partial_t^2}{2D} \right\} \vec x\right)$$ This is a standard Gaussian integral and can therefore be evaluated to:

$$I=(2\pi)^{3/2}\left(\mathrm{Det}\left( \frac{\omega^2-\partial_t^2}{2D}\right) \right)^{-3/2}$$

My question is: How do I evaluate:

$$\mathrm{Det}\left( \frac{\omega^2-\partial_t^2}{2D}\right)$$ I am aware that I need to find the eigenfunctions and eigenvalues but subject to what conditions and in what space?

I am not sure if this is more appropriate here or on MSE? Since there doesn't even seem to be a tag on MSE for path-integrals I chose PSE.

• How can you be taking an integral if you don't know what space you're integrating over (e.g. the $\vec x(t_i) = 0$ seems to already give some conditions on the space)? What space you're working on here is something only you can know since you gave us no information except for the integral expression itself. – ACuriousMind Feb 4 '18 at 11:01
• @ACuriousMind My question is not about what space $\vec x$ is in. It is a vector in $\Bbb{R}^3$ with the boundary conditions specified. I think my question is fairly clear (if not let me know and I will edit it) that I am referring to the space of eigenfunctions used to evaluate the determinant. – Quantum spaghettification Feb 4 '18 at 11:12
• The standard formula for the Gaußian integral assumes that the integral is taken over the entire domain of definition of the matrix you end up taking the determinant of, so you need to evaluate the determinant on precisely the same space you are path-integrating over. – ACuriousMind Feb 4 '18 at 11:22
• I'm reasonably certain mathematicians refer to path integration as functional integration. – Kyle Kanos Feb 4 '18 at 12:45
• @Kyle in the definitions I have seen path integration is a sum over all possible paths between points, while functional integration is an optimization problem to find the "best" path. I would be happy to be wrong because I have taken a course in functional derivatives. – Emil Feb 4 '18 at 13:12