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.

  • $\begingroup$ have you tried regularising the product? see Functional determinant $\endgroup$ Jan 12, 2016 at 10:24
  • $\begingroup$ I think this is normally done for propagators with some non-zero potential, where we take the ratio between this propagator and the free propagator. But how do we circumvent the infinities in the free propagator? does this mean that the functional determinant in this case is simply not able to produce any sensible results? $\endgroup$
    – M. Zeng
    Jan 12, 2016 at 12:24
  • $\begingroup$ let $\lambda_n= n\pi/\tau$; note that $\sum_{n=1}^\infty\frac{-\log\lambda_n}{\lambda_n^s}$$=(\tau/\pi)^s\log(\pi\tau)\zeta (s)+(\tau/\pi)^s\zeta'(s)\to-\frac{1}{2}\log(2\tau)$ as $s\to 0$. Taking the exponential we get $G=\frac{1}{\sqrt{2\tau}}$. This result is wrong, but because so is your expression for $\lambda_n$ (e.g, it should depend on $m$). $\endgroup$ Jan 12, 2016 at 12:46
  • $\begingroup$ anyway, the calculations from my last comment are probably wrong (i didnt do them with much care), but my point is: 1) calculate the eigenvalues of $\frac{im}{\pi\hbar}\partial_t^2$, and call them $\lambda_n$. 2) next, use the expression from the link I posted before ($\text{det}(S)=\sum \frac{\log\lambda}{\lambda^s}$), and extend this function to the complex plane. 3) Then, take the limit $s\to 0$, and finally exponentiate the result. You should get the correct result for $G$ (if you do this, post your work here and we'll check it, and it might be useful for other users in the future) $\endgroup$ Jan 12, 2016 at 12:57
  • $\begingroup$ @AccidentalFourierTransform I have added the calculation based on your suggestion, but it doesn't seem to work. plz refer to the edited version of the question. $\endgroup$
    – M. Zeng
    Jan 13, 2016 at 3:24

1 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 $$ \tag{11} Z~=~\frac{1}{\sqrt{2 \pi T}}.$$

  • $\begingroup$ it seems that the whole problem has its origin in series representation of analytic functions and the analytic continuation. Due to the functional integration technique that we employed, we have a series which diverges in the region of interest, and then we use analytic continuation to enlarge the region of convergence to eventually get some finite result that works. Is this due to the intrinsic problem of the functional integration? and is this part of the reason why people are still looking for mathematically rigorous justification of this technique? $\endgroup$
    – M. Zeng
    Jan 14, 2016 at 6:07
  • $\begingroup$ We need to use regularization to make sense of the path integral. It doesn't need to be zeta function regularization. It is possible to use other regularizations as well, such as e.g. Pauli-Villar regularization. $\endgroup$
    – Qmechanic
    Jan 14, 2016 at 20:24

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