# Path integrals using Fourier transformation

While going through a book named Mirror Symmetry, I came across a path integral,

$$Z(\beta) = \int\limits_{X(t+\beta) = X(t)} DX(t) \exp\left(-\int\frac{1}{2}( \dot{X}^2 + X^2)dt\right)dt$$

where $X(t)$ has a periodicity of $\beta$.

Using orthonormal eigenfunctions of the operator $\Theta = -\frac{d^2}{dt^2}+1$, the exponential term is given as $\exp(-\frac{1}{2}\sum_n \lambda_n c^2_n)$, where the $\lambda_n$ are the eigenvalues. To perform the above integral, the measure is transformed as, $$DX(t) = \prod_n \frac{dc_n}{\sqrt{2\pi}}$$ I didn't get how is it derived, especially the $\frac{1}{\sqrt{2\pi}}$ factor in front. How is the jacobian transformation is done here ?

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An absolute constant in $Z(\beta)$ does not matter and so most people/books are sloppy about that. If I were you I would not worry. Otherwise, you first have to define $DX(t)$ properly. –  Fabian Sep 27 '12 at 9:49
@Fabian : Well I don't know if there is a rigorous definition of $DX(t)$ because it appears in path integrals, accounting for all the paths that have the above mentioned periodicity. One way to think of it is in time-slice method of evaluation of feynmann path integrals is $DX(t) = lim_{n\rightarrow \infty} \prod_n dq_n$ . –  Jaswin Sep 27 '12 at 11:41
I do not think that your definition (if I understand it correctly) is good, as you will get $\infty$ for any path integral which I know how to solve. –  Fabian Sep 27 '12 at 13:01
@Fabian : As far as I know, the definition is bit sloppy, but we physicists live with it. –  Jaswin Sep 27 '12 at 15:04

The constant $\sqrt{2\pi}$ is just there for convenience. It has been arbitrarily introduced as a normalization factor without changing anything both for mathematics or physics. The advantage to have it there is that the integral $$\int_{-\infty}^\infty\frac{dc_n}{\sqrt{2\pi}}{e^{-\frac{1}{2}\lambda_nc_n^2}}=\frac{1}{\lambda_n}$$ and you will get at the end for the partition function exactly the inverse of the determinant of the operator $\Theta$.

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