Harmonic oscillator path integral: regularizing the functional determinant

Question

From Polchinski's Vol. 1 Appendix A, we can reduce the Euclidean path integral for the 1D harmonic oscillator to computing $(\det\frac{\Delta}{2\pi})^{-1/2}$ where $$\Delta = -\partial_u^2 + \omega^2.$$

Since $f_j(u) = \sqrt{2/U} \sin(j\pi u/U)$ is an eigenbasis, we compute

$$\det \frac{\Delta}{2\pi} = \prod_{j=1}^{\infty} \frac{j^2\pi^2 + \omega^2 U^2}{2\pi U^2}.$$

Here Polchinski regularizes using Pauli-Villars; he divides by the amplitude for an oscillator with very high frequency $\Omega$ and gets

$$\det \frac{\Delta}{2\pi} \to \prod_{j=1} \frac{j^2 \pi^2 + \omega^2 U^2}{j^2 \pi^2 + \Omega^2 U^2} = \frac{\Omega \sinh \omega U}{\omega \sinh \Omega U}.$$

Now he takes the $\Omega \to \infty$ limit. Why is that the appropriate limit to take at the end?

He obtains $(\det \frac{\Delta}{2\pi})^{-1/2}\approx \left(\frac{\omega}{\sinh\omega U}\right)^{1/2} e^{\frac12(\Omega U - \ln \Omega)}$ and subtracts divergences and gets the right answer. But why is $\Omega \to \infty$ the right limit to take? When we regularize, we typically take a limit at the end to "undo" it. For instance, in QFT we take $d= 4-\epsilon$ and then take $\epsilon \to 0$ at the end.

Answer 1

You’re only interested in the dependence of the determinant on $U$ here (the numeric value for a given $U$ depends on the normalization constant in the functional measure and is therefore unphysical).

The expression for the determinant that you’ve given exhibits $U$-independence for $\omega \rightarrow \infty$ (the $j$-dependent term becomes small compared to the other summand in the numerator, and $U^2$ cancel the $U^2$ in the denominator). Hence, in the limit of infinite frequency you’re dividing by a $U$-independent constant — an operation which is equivalent to redefining the normalization constant. That is no longer true for finite frequencies. Intuitively, setting the frequency to infinity undoes the regularization like you expected.