Free space propagator: reconciling two results

Question

In quantum mechanics, the free space propagator $G(q_f=0,q_i=0;\tau)$ can be easily calculated to be $$\sqrt{\frac{m}{2\pi i \hbar \tau}}$$ by inserting an identity operator.

However if we use functional integral, we get \begin{equation} \begin{split} G(q_f=0,q_i=0;\tau)&=\int Dq e^{-\frac{i}{\hbar}\int_0^{\tau} dt \frac{m}{2}\dot{q}^2}\\ &=\int Dr e^{-\frac{i}{\hbar}(S[q_{cl}]+S[r(t)])}\\ &=\int Dr e^{-\frac{i}{\hbar}S[r(t)]}\\ &=\int Dr e^{\frac{i}{\hbar}\int dt r(t)\partial_t^2 r(t)}\\ &=(\det[\frac{i}{\pi\hbar}\partial_t^2])^{-1/2} \end{split} \end{equation} where the classical trajectory $q_{cl}(t)=0$ due to the boundary conditions and $r(t)$ is the fluctuation. If we solve for the eigenstates and eigenvalues of $\partial_t^2$: $$\partial_t^2 r_n(t)=\lambda_nr_n(t)$$with $r_n(0)=r_0(\tau)=0$, we get $r_n(t)=\sin(n\pi t/\tau)$ and $\lambda_n=(n\pi/\tau)^2$. Therefore, we have $$\det(\partial_t^2)=\prod_{j=1}^{\infty}(n\pi/\tau)^2$$ which goes to infinity and as a result the propagator seems to go to 0.

I'm not sure where went wrong for this calculation. Any help is appreciated.

Edit: suggested by @AccidentalFourierTransform, below is the zeta function approach, which still doesn't seem to work.

for simplicity we set all the irrelevant constants to 1, and thus $\lambda_n=n^2$, then we have $$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{\lambda_n^s}=\sum_{n=1}^{\infty}\frac{1}{n^{2s}}$$ and then we need to calculate the derivative of the zeta function and then taking the limit of $s$ goes to 0 followed by exponentiation in order to obtain the determinant.

$$\zeta'(s)=\sum_{n=1}^{\infty}-\frac{ln\lambda_n}{n^{2s}}$$ I tried numerically by taking $s$ to 0 both from the real axis and the imaginary axis, but both seems to diverge, i.e the same problem remains.

Answer 1

OP's underlying question is essentially the same as this Phys.SE post, although the detailed calculation is slightly different and interesting to compare.

I) The action for a free non-relativistic point particle with mass $m=1$ reads:

$$\tag{1} S ~=~\frac{1}{2}\int_0^T\! dt~ \dot{x}(t)^2~=~ \frac{1}{2}\langle x,Ax \rangle~=~\frac{1}{2}\sum_{n\in \mathbb{N}} \lambda_n c_n^2 . $$

Here we have assumed Dirichlet boundary conditions (DBC)

$$\tag{2} x(0)~=~0~=~x(T).$$

Moreover, here

$$\tag{3} \langle f,g \rangle ~:=~ \int_0^T\! dt ~f(t)g(t) $$

is an inner product over $\mathbb{R}$.

II) In eq. (1) we have also introduced a positive operator

$$\tag{4} A~:=~-\partial_t^2 $$

with positive eigenvalues

$$\tag{5} \lambda_n~=~\left(\frac{\pi n }{T}\right)^2~>~0, \qquad n\in \mathbb{N}.$$

The determinant becomes via zeta-function regularization

$$\tag{6}\det(A)~=~\prod_{n\in\mathbb{N}} \lambda_n~=~\left(\prod_{n\in\mathbb{N}} \frac{\pi n }{T}\right)^2 ~=~2T ,$$

using e.g. eq. (7) in my Phys.SE answer here.

III) The normalized eigenfunctions are

$$\tag{7} x_n(t) ~=~ \sqrt{\frac{2}{T}} \sin \frac{\pi n }{T}t , \qquad n\in \mathbb{N}. $$

An arbitrary virtual path $t\mapsto x(t)$ that satisfies the DBC (2) is a linear combination

$$\tag{8} x ~=~ \sum_{n\in \mathbb{N}} c_n x_n, $$

where $c_n\in\mathbb{R}$ are arbitrary coefficients, which we should integrate over in the path integral.

IV) Now let us consider quantum mechanics. Let us assume $\hbar=1$ for simplicity. The path integral measure is

$$\tag{9} {\cal D}x~:=~N \prod_{n\in\mathbb{N}} \frac{\mathrm{d}c_n}{\sqrt{2\pi}} , $$

where $N$ is a normalization factor. So the Euclidean path integral is an infinite-dimensional Gaussian integral

$$\tag{10} Z~=~\int_{DBC} \!{\cal D}x ~e^{-S}~=~\frac{N}{\sqrt{\det (A)}} ~=~ \frac{N}{\sqrt{2T}}. $$

Apparently we should chose the normalization factor $N=\frac{1}{\sqrt{\pi}}$ in order to achieve the Euclidean version of OP's first formula $$ \tag{11} Z~=~\frac{1}{\sqrt{2 \pi T}}.$$